If $K\cong K(X)$ then must $K$ be a field of rational functions in infinitely many variables?

If $k$ is any field, then the field $K=k(X_0,X_1,\dots)$ of rational functions in infinitely many variables satisfies $K(X)\cong K$ (by mapping $X$ to $X_0$ and $X_n$ to $X_{n+1}$).  My question is, does the converse hold?  That is:

Suppose $K$ is a field such that $K\cong K(X)$.  Must there exist a field $k$ such that $K\cong k(X_0,X_1,\dots)$?

(If such a $k$ exists, then we can in fact take $k=K$, by splitting the variables into two infinite sets.)
Note that if we were talking about polynomial rings instead of fields of rational functions, the answer would be no: there exists an integral domain $R$ such that $R\cong R[X]$ but $R\not\cong S[X_0,X_1,\dots]$ for any ring $S$.  A counterexample is given at A ring isomorphic to its finite polynomial rings but not to its infinite one.  However, for that example the field of fractions of $R$ actually is a field of rational functions in infinitely many variables, so it does not give a counterexample to this question.
In any case, I suspect the answer is no and it may be possible to find a counterexample using some idea similar to the example there (some sort of "all but finitely many..." construction), but don't have any concrete idea of how to make it work.
Some other (unanswered) questions that may be related: Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? (also on MO), The field of fractions of the rational group algebra of a torsion free abelian group.  (In particular, an answer to the former question would give a field $F$ such that $F\cong F(X,Y)$ but $F$ is not isomorphic to a field of rational functions in infinitely many variables, which is very close to this question.)

@JeremyRickard Nevermind, those are the same thing. And I see now why my idéa doesn't work.

@JeremyRickard I thought that $K(X)$ denoted $\text{Frac} (K[X])$, not field extension.

@Kilian I don't see why $K$ would be isomorphic to $K(X)$. $\mathbb{R}((X_0,X_1,\dots))$ is not generated as a field extension of $\mathbb{R}((X_1,X_2,\dots))$ by $X_0$.

Just an idea, haven't fully thought it through, but if we let $K=\mathbb{R}((X_0,X_1...))$, i.e. the field of formal Laurent series in infinitely many variables over $\mathbb{R}$, it seems like we would have $K\cong K(X)$ but I don't see how it would be isomorphic to $k(X_0,X_1,...)$ for any field $k$. Possibly a counterexample?

It’s an example of something. Either it’s a counterexample for your question, or the rational group algebras of the nonisomorphic groups $\mathbb{Z}^\omega$ and $\mathbb{Z}^\omega\oplus\mathbb{Z}^{(\omega)}$ have isomorphic fields of fractions, answering my question.

@JeremyRickard: That's a neat idea.  I don't actually recall thinking through it before (I think the constructions I was trying were all countable).

Do you know that the field of fractions of the rational group algebra of the Baer-Specker group is not a counterexample? Given the links in the question, I’m guessing you’ve thought about it.

Sure.  Given that your related question hasn't gotten an answer there I would be surprised if this one got an answer there quickly, though.  Feel free to crosspost it if you want.

Might it be worth asking this interesting question on MathOverflow?


Constructing an infinite chain of subfields of 'hyper' algebraic numbers?

This has now been cross posted to MO.
Let $F$ be a subset of $\mathbb{R}$ and let $S_F$ denote the set of values which satisfy some generalized polynomial whose exponents and coefficients are drawn from $F$.
That is, we let $S_F$ denote
$$\bigg \{x \in \mathbb{R}: 0=\sum_{i=1}^n{a_i x^{e_i}}: e_i \in F \text{ distinct}, a_i\in F \text{ non-zero}, n\in \mathbb{N}  \bigg  \}$$
Then $S_{\mathbb{\mathbb{Q}}}$ is the set of algebraic real numbers and we start to see the beginnings of a chain:
$
\mathbb{Q}
\subsetneq S_\mathbb{Q} \subsetneq
S_{S_\mathbb{Q}} $
Main Question

Does this chain continue forever? That is, we let $A_0= \mathbb{Q}$ and let $A_{n+1}=S_{A_{n}}$. Is it the case that $A_n \subsetneq A_{n+1}$ for all $n\in\mathbb{N}$?

Other curiosities:
Is $A_i$ always a field? Perhaps, the argument is analogous to this. Or maybe this is just the case in a more general setting: Is it the case that $F \subset \mathbb{R}$, a field implies that $S_F$ is a field?
Is it possible to see that $e\notin \cup A_i$? Perhaps this is just a tweaking of LW Theorem.

@JackM, $x=0$ satisfies this and is algebraic. I am not sure I understand your meaning...  But if you take $x=y^6$ then we can replace your equation with $y^3+y^2=0$ and we can always manage to do this type of change when the exponents are rational.

Why is $S_\mathbb Q$ equal to the set of algebraic real numbers? For instance, is any solution to $\sqrt x+\sqrt[3]x=0$ algebraic?

I am not so sure but maybe this piece by B Zilbner speaks to the question.

Actually... $S_F$ is clearly closed under multiplicative inverses because $-e_i$ is in a field $F$ whenever $e_i$ is in $F$

It seems like we might be able to achieve multiplicative inverses using a technique analogous to the one presented here. I haven't worked out all the details but it looks like it could be a good lead

How do we prove that $4^\sqrt3+1\in\cup A_n$? This seems like an argument for defining $S_F$ to include closure under field operations.

I don’t think we can rule out $e\in A_3$. My impression is that we know almost nothing about the transcendence of numbers like $3^\sqrt6-4^\sqrt3-1$.

@JyrkiLahtonen, for this purpose I think we can define $(-|x|)^\alpha$ as $-(|x|^\alpha)$.

Just a question: is $n$ (defined inside the definition of $S_F$ as a natural number) also arbitrary for any member of $S_F$?

(continued) We should answer $e^{i \pi \sqrt{2}}$ but all of sudden it's start look like the question is better raised in the complex numbers. But maybe there is a different polynomial that satisfies $-x$?

@JyrkiLahtonen.  Just to be clear about the concern raised: $ x^{\sqrt2}-2=0 \implies x= \sqrt{2}^{1/\sqrt{2}}$ What (generalized) polynomial is satisfied by $-x$? It could be that $ (-x)^{\sqrt2}-2=0$ is not a satisfactory answer. No. I would think this is indeed a satisfactory answer. The first polynomial has a non negative domain and the second has a non positive domain. What trouble do we get into if we take this naive approach? This route at least wins us closure under negation. Oh. The trouble we get into is when someone asks what the coefficient of the $x^{\sqrt 2}$ term is...

Mason, I had some simple worries in my mind. Like is $S_F$ necessarily closed under negation?

I'm still somewhat concerned about the definition of $x^\alpha$ when $x<0$ and $\alpha$ is not an integer. But, there is a cool question  in here (possibly a very difficult one).

$2^{\sqrt2}$ is transcendental, but it satisfies $x^{\sqrt2}-4=0$, so it's in $A_2$. I'd conjecture that $2^{2^{\sqrt2}}$ is in $A_3$ and not $A_2$, that $2^{2^{2^{\sqrt2}}}$ is in $A_4$ and not $A_3$, etcetera.

How do you propose to define $x^\alpha$ if $\alpha \notin \mathbb{Q}$?

Every set in the chain is countable, so you certainly don't hit $\mathbb{R}$ at any point.


An iterative logarithmic transformation of a power series

Consider the following iterative process. We start with the function having all $1$'s in its Taylor series expansion:
$$f_0(x)=\frac1{1-x}=1+x+x^2+x^3+x^4+O\left(x^5\right).\tag1$$
Then, at each step we apply the following transformation:
$$f_{n+1}(x)=x^{-1}\log\left(\frac{f_n(x)}{f_n(0)}\right).\tag2$$
A few initial iterations give us:
$$
\begin{array}{l}
f_1(x)=1+\frac{x}{2}+\frac{x^2}{3}+\frac{x^3}{4}+\frac{x^4}{5}+O\left(x^5\right), \\
f_2(x)=\frac{1}{2}+\frac{5 x}{24}+\frac{x^2}{8}+\frac{251 x^3}{2880}+\frac{19 x^4}{288}+O\left(x^5\right), \\
f_3(x)=\frac{5}{12}+\frac{47 x}{288}+\frac{2443 x^2}{25920}+\frac{5303 x^3}{82944}+\frac{412589x^4}{8709120}+O\left(x^5\right),
\end{array}\tag3
$$
and their initial terms form the following sequence:
$$1,\,\frac{1}{2},\,\frac{5}{12},\,\frac{47}{120},\,\frac{12917}{33840},\,\frac{329458703}{874222560},\,\dots,\tag4$$
whose denominators grow pretty quickly, but which appear to slowly converge to a value $$\lambda\stackrel{\color{#aaaaaa}?}\approx0.3678\dots\tag5$$
If we assume that the process with the iterative step $(2)$ converges to a fixed point, we can see that it must have a form:
$$f_\omega(x)=-x^{-1}\,W(-c\,x)=c+c^2 x+\frac{3\, c^3\, x^2}{2}+\frac{8\, c^4\, x^3}{3}+\frac{125\, c^5\, x^4}{24}+O\left(x^5\right),\tag6$$
where $W(\cdot)$ is the Lambert 𝑊-function, and $c$ is a coefficient that is not uniquely determined but depends on the choice of the initial function $f_0(x)$. In our case, $c=\lambda$.

Questions: Does this process actually converge to a fixed point? If yes, then what is a closed-form expression (or another useful description) for $\lambda$?

Update: An explicit recurrent formula for the coefficients (the parenthesized superscript $m$ in $a_n^{\small(m)}$ is just the second index of the coefficient; the sum $\sum_{\ell=1}^m$ is assumed to be $0$ when $m=0$):
$$f_n(x)=\sum_{m=0}^\infty a_{n\vphantom{+0}}^{\small(m)}x^m,$$
where
$$a_{0\vphantom{+0}}^{\small(m)}=1,\quad a_{n\vphantom{+0}}^{\small(m)}=\frac1{\,a_{n-1}^{\small(0)}\,}\left(a_{n-1}^{\small(m+1)}-\frac1{m+1} \sum_{\ell=1}^m\ell\;a_{n\vphantom{+0}}^{\small(\ell-1)}\,a_{n-1}^{\small(m-\ell+1)}\right).$$
The sequence of coefficients $(4)$ is $\big\{a_{n\vphantom{+0}}^{\small(0)}\big\}$.

He also gives the first terms of the next power series in the sequence. Then, he writes the first ever recorded insights on the formula for iteration of power series $f \circ \ldots \circ f(x)$, which then insipired Schröder in his 1872 paper, in which he first considered, what now call, the "Schröder equation". This is one of the first paper on iteration theory, which unfortunately doesn't seem to apply to your case.

Interestingly enough, Caylay briefly considered this problem in a note called "On Some Numerical Expansions" from the Quarterly Journal of Pure and Applied Mathematics, vol. III. (1860), pp. 366-369, but he didn't go much farther than you. What stroke me is that he got the same first few terms as you, but he was probably wrong for the last term "$\log\left(-\frac2x\log\left(\frac1x\log(1+x)\right)\right)=-\frac5{12}x+\frac{47}{288}x^2+\frac{2443}{25920}x^3-\frac{5303}{82944}x^4+\frac{19631}{580608}x^5+\&c$".

@VladimirReshetnikov It seems that, if $f_n(0)=a/b$ (in lowest terms), then the denominator of $f_{n+1}(0)$ is a small multiple ($2$ or $6$) of $ab$. Let $b_n$ be the denominator of $f_n(0)$. If $f_n(0)$ converges to a limit $\lambda$, then this would imply that $b_{n+1}$ is multiplicatively not far off from $\lambda b_n^2$, which gives double-exponential growth.

Another very interesting question is how fast do denominators grow (or would grow if we ignored occasional cancellations of some of their factors with numerators). Empirically, they appear to grow (very roughly) doubly exponentially: $2^{2^{n-0.97942\dots}}$.

Originally I found this sequence when I tried to construct the continued exponent expansion of the Fibonacci numbers: pbs.twimg.com/media/FHAQfurVQAANsLx?format=png&name=large, but later realized I can get it simply from $1/(1-x)$.

@mathworker21 Usually, OEIS keeps numerators and denominators as distinct sequences. This entry mixes them into one interleaved. Probably this is why I was not able to find the coefficients in OEIS after I computed a few of them.

@mathworker21 Thanks! It would be nice to see a rigorous proof. I haven't been able to prove it myself so far.

I concur with Claude and will go further saying this is a very nice problem in every respect.

I think $\lambda =1/e=0.367879441\ldots$. A rough argument is as follows. It seems that each $f_n(x)$ is singular at $x=1$. Since the limit function is $-x^{-1}W(-\lambda x)$ and $W(z)$ is singular at $z=-1/e$, $\lambda$ should be $1/e$ in order for $f_\omega(x)$ to be singular at $x=1$.

The formula is correct. We assume that $\sum_{\ell=1}^0(\dots)=0$, because there are no indices satisfying $1\le\ell\le0$, so the sum is empty .

is there a typo in your explicit formula? We are after $a_n^{(0)}$ and when I tried plug $m = 0,$ the sum starts at $l=1$ and ends at $l = m.$

Whatever the answer could be, this is a nice problem.

By the mean value theorem, taking $x\to 0,$ you get $$f_{n+1}(0)=\frac{f_n’(0)}{f_n(0)}.$$ Not sure if that helps in any way.


Smallest Subset of $\mathbb{R}_{>0}$ Closed under Typical Operations

Let $S$ denote the smallest subset of $\mathbb{R}_{>0}$ which includes $1$, and is closed under addition, multiplication, reciprocation, and the function $x,y \in \mathbb{R}_{>0} \mapsto x^y.$
Questions:
1. Do either $S$ or its elements have an accepted name?
2. Where can I learn more about the set $S$ and it elements? (Reference Request)
3. Do there exist positive, real algebraic numbers which are not in $S$?
4. Are either $e$ or $\pi$ elements of $S$?
What I Already Know:
By the Gelfond-Schneider Theorem, $S$ includes some transcendental numbers, like $2^{\sqrt{2}}$.

Could we build this set by transfinite recursion and then study its properties (like for the Borel Hierarchy)?

@EricWofsey, if S is closed under subtraction it would include $2^\sqrt{2}-1$, and that seems unlikely.

S is a subset of the positive "elementary-logarithmic" or "closed-form" numbers, as defined by Chow.  (timothychow.net/closedform.pdf)  So his paper proves, assuming Schanuel's conjecture, that the positive roots of $2x^5-10x+5$ are not in S.

@YoTengoUnLCD, in any partially ordered set $P$, every subset $A$ of $P$ has at most one minimum element (but may have many "minimal" elements). So let $P$ denote the poset $(\mathcal{P}(\mathbb{R}_{>0}),\subseteq).$ Let $A$ denote those elements of $P$ that are closed under the relevant operations. Then $A$ has at most one least element. Regarding existence of a minimum element, it can be constructed as $\bigcap A$.

Just a question: what tells you that this characterization determines $S$? As inclusion is not a total order...

I might add another question: is $S$ closed under subtraction (when the difference is positive)?

@Jihad, I suppose that any incomputable would not belong to $S,$ and there exist incomputable numbers that can be defined explicitly. Although my knowledge of computability theory is sufficiently poor that I am not 100% certain of this statement. More interestingly perhaps, I have hunch that $e$ and $\pi$ do not belong to $S$, and that there might exist a fifth-degree polynomial (or higher) over $\mathbb{R}$ with a solution that does not belong to $S$.

Do you know any number that does not belong to this set?

@Jihad, it certainly is. That doesn't tell us much, though, since the set of algebraic numbers is also countable.

Seems like this set should be countable.


Determining the kernel of a Vandermonde-like matrix

The kernel of a Vandermonde matrix can be determined using this formula.
The following type of matrix has a similar structure, and should also have a one-dimensional kernel.
$$V=
\begin{bmatrix}
1 & 1 & 1 & \ldots & 1 \\
x_1 & x_2 & x_3 & \ldots & x_n \\
x_1^2 & x_2^2 & x_3^2 & \ldots & x_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_1^{m-1} & x_2^{m-1} & x_3^{m-1} & \ldots & x_n^{m-1}\\
y_1 & y_2 & y_3 & \ldots & y_n \\
y_1x_1 & y_2x_2 & y_3x_3 & \ldots & y_nx_n \\
y_1x_1^2 & y_2x_2^2 & y_3x_3^2 & \ldots & y_nx_n^2 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y_1x_1^{m-1} & y_2x_2^{m-1} & y_3x_3^{m-1} & \ldots & y_nx_n^{m-1}\\
y_1^2x_1 & y_2^2x_2 & y_3^2x_3 & \ldots & y_n^2x_n\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y_1^{m-1}x_1^{m-1} & y_2^{m-1}x_2^{m-1} & y_3^{m-1}x_3^{m-1} & \ldots & y_n^{m-1}x_n^{m-1}\\
\end{bmatrix} \in \mathbb{R}^{(n-1)\times n}$$
where $n = m^2+1$ and $(x_i, y_i) \neq (x_j, y_j)$ for $i \neq j$; i.e. there are $m$ groups of $m$ rows with all possible combinations of powers $y^ax^b$ and one more column than rows.

Does a similar analytical form exist for it? Or, would additional constraints be required, like $x_i^ay_i^b \neq x_i^cy_i^d$ for $i \neq j$?

(Crossposted to MO)

Wouldn't there be an additional row of $y_1^2 \cdots y_n^2$ before the row $y_1^2x_1 \cdots y_n^2x_n$?

Assuming you're in the complex field, what you're asking for is exactly which are the two-variable polynomials in $\mathbb{C}[x,y]$ with degree $m-1$ in both $x$ and $y$, such that they are zero on a given set of points $\{ (x_1,y_1), (x_2,y_2) , ... (x_n, y_n)\}$. This is just Lagrange interpolation.

Questions: have you tried using Lagrange interpolation, as in the regular proof of the kernel of the van der monde matrix? Also, have you tried getting people on mathematica.stackexchange to experiment with different $m$?

There are $m$ groups of $m$ rows (All possible combinations of powers $x^iy^j$); the matrix should have one more column than rows to give a 1-dimensional kernel, therefore $n=m^2+1$.

Shouldn't $n=2m+1$ rather than $n=m^2+1$?


Is the power of $2$ in the Euclidean norm related to the fact that the algebraic closure of the reals is $2$-dimensional?

Consider any local field $K$, endowed with its topological field structure. We define the function $| \cdot | : K \to \mathbb{R}_{\ge 0}$ as
$$|x| = \frac{\mu(xS)}{\mu(S)},$$
where $\mu$ is any Haar measure (which exists because a local field is additively a locally compact group), and $S$ is any measurable set of nonzero measure (see e.g. Definition 12.28 here). This definition can be shown to be independent of $\mu$ and $S$.
Now consider a vector space $V \simeq K^n$ over $K$, and equip it with the $p$-length function $$\|(x_1,x_2,\ldots,x_n)\|_p = (|x_1|^p+|x_2|^p+\ldots+|x_n|^p)^{1/p}$$ for some $p \in [1,\infty]$ (where as usual we define $\| \cdot \|_\infty$ as the limit of $\| \cdot \|_p$ as $p\to \infty$). There is a unique critical value $p=p^*(K)$ for which this length function becomes especially symmetric, in that its associated group of isometries (i.e., the subgroup of $GL(V)$ that preserves $\| \cdot \|_{p^*(K)}$) acts transitively on the corresponding unit sphere. The amazing fact is that this critical value $p^*(K)$ always coincides with the degree $[\bar{K}:K]$ of the algebraic closure of $K$!
Here is a proof sketch:
From the classification of local fields, it is known${}^{(\dagger)}$ that $K$ is either $\mathbb{R}$, $\mathbb{C}$ or a non-Archimedean local field (that is, a finite extension of either $\mathbb{Q}_P$ or $\mathbb{F}_P((t))$ for some prime $P$).

If $K=\mathbb{R}$, $|\cdot|$ is the usual absolute value, and the critical length function turns out to be $\| \cdot \|_2$ (the usual Euclidean length), invariant under $O(n)$. On the other hand, $[\bar{\mathbb{R}}:\mathbb{R}]=[\mathbb{C}:\mathbb{R}]=2$.

If $K=\mathbb{C}$, we have $|x|=\bar{x} x$, and the corresponding critical length function is $\| \cdot \|_1$, invariant under $U(n)$.${}^{(\ddagger)}$ On the other hand, $[\bar{\mathbb{C}}:\mathbb{C}]=[\mathbb{C}:\mathbb{C}]=1$. Note that neither $|\cdot|$ nor $\|\cdot\|$ satisfy the triangle inequality in this particular case (this is why I avoided using the terms absolute value and norm throughout the question).

If $K$ is non-Archimedean, $|\cdot|$ is a suitably normalized $\chi$-adic absolute value for some $\chi$, and the corresponding critical length function is the supremum norm $\| \cdot \|_\infty$, which can be shown to be invariant under $GL(\mathcal{O}_K^n)$, where $\mathcal{O}_K$ is the ring of integers of $K$. On the other hand, it is known that the algebraic closure of $K$ has infinite degree, i.e. $[\bar{K}:K]=\infty$.

Thus in all cases we have $p^*(K) = [\bar{K}:K]$.
A problem with this proof is that it gives no indication as to why the relationship holds: after all, the values could just happen to be the same by complete coincidence. For that reason, I am wondering if there exists an alternative proof of this fact that is "$K$-agnostic", i.e. a proof that does not use at any point the Artin-Schreier theorem, nor the classification of local fields, nor any properties from a specific such field (such as the existence of a symmetric/Hermitian inner product) other than the degree of an algebraic closure and properties derived from it.
Such a proof would allow one to meaningfully speak of a relationship between those two quantities, and to state things like "in a field with a $3$-dimensional algebraic closure, the natural analogue of the Pythagorean theorem would be $|a|^3+|b|^3=|c|^3$" without appealing to the principle of explosion. I know it is hard in general to formally determine whether a proof does or does not use a fact, hence why I'm labeling this with the soft-question tag, but hopefully the kind of proof I want is clear, at least informally.

In summary, my question is:
Given a local field $K$, with $| \cdot |$ defined as above, is there any $K$-agnostic proof that the group of isometries of $\|\cdot\|_p$ (defined in terms of $| \cdot |$ as above) acts transitively on unit spheres if and only if $p = [\bar{K}:K]$?
It seems reasonable to try proving first that $p^*(K) = 1$ iff $K$ is algebraically closed. That is where I am currently stuck. The thing that makes linear algebra over an algebraically closed field "special" is that every linear transformation has an eigenvalue, but I am not sure how to apply that fact to show that the critical length function is $\|\cdot\|_1$.
UPDATE: As suggested by Torsten Schoeneberg in the comments, perhaps it will be easier to prove the relationship $p^*(K) = [L:K]\: p^*(L)$ for a finite field extension $L/K$. Consider such an extension with $[L:K]=n$, and suppose we know $L$ has a critical $p$-length function for some $p$. Now, $L \simeq K^n$ is additively a vector space over $K$, which is equipped with the product topology. This means that, if we identify $k\in K $ with its image under the inclusion $K \subseteq L$, we have
$$|k|_L = \frac{\mu_L(k S^n)}{\mu_L(S^n)} = \left(\frac{\mu_K(k S)}{\mu_K(S)}\right)^n = |k|_K^n.$$
On the other hand, note that the subgroup of $GL_1(L) = L^\times$ preserving $|\cdot|_L$, i.e. the group of invertible elements of norm $1$, acts transitively on the associated unit sphere (which coincides with the group itself) in a tautological way, since any element $x$ of norm $1$ can be sent to any other element $y$ of norm $1$ through multiplication by $x^{-1}y$. Since $GL_1(L) \subseteq GL_n(K)$, this proves that the length function on $K^n$ defined through the identification $L \simeq K^n$ as $\|\cdot\| = |\cdot|_L^{1/n}$ is a critical length function, that moreover satisfies the homogeneity property $\|kv\|=|k|_L \|v\|$ (which follows by the above relationship between the Haar measures and multiplicativity of $|\cdot|$). In a similar way, from the inclusion $GL_m(L) \subseteq GL_{mn}(K)$ we can prove that any critical length function $||\cdot||$ on $L^m$ induces a corresponding critical length function on $K^{mn}$.
The only ingredient left would be to prove is that there always exists some basis $\{1=a_0, a_1, \ldots, a_{n-1}\}$ of $L$ such that the above defined length functions are of the required form, that is, $|\sum_{i=0}^{n-1} k_i a_i|_L = \sum_{i=0}^{n-1} |k_i|_K^n$. Once we have that, we can show using isotropy subgroups that the length function induced from $L$ is critical in any dimension, not just multiples of $n$, so that $p^*(K) = [L:K]\: p^*(L)$. However, I don't know how to prove the existence of such a basis in a $K$-agnostic way.

EDIT: A possible way to restate the problem is by noticing that for any $(v,w)$ in the direct sum $V \oplus W$, we have $\|(v,w)\|^p_p = \|v\|^p_p + \|w\|^p_p$, i.e. the $p$th power of $\|\cdot\|_p$ acts additively on "independent" vectors (for finite $p$, of course).
Given a length function on vector spaces $V$, we can define the associated Gaussian function as an integrable function $g : V \to \mathbb{R}_{\ge 0}$ depending only on length (i.e. factorizing as the composition $f \circ \|\cdot\|_p : V \to \mathbb{R}_{\ge 0} \to \mathbb{R}_{\ge 0}$ for some $f$), and satisfying the independence property
$$g( (v,w) ) = g(v) g(w).$$
With the above choice of maximally symmetric length functions, using the aforementioned additivity property immediately leads to the usual Gaussian function $\exp(-k \|x\|_2^2)$ for $\mathbb{R}$ and to $\exp(-k' \|x\|_1)$ for $\mathbb{C}$ (where the constants $k, k' >0$ depend on the normalization convention). For $p=\infty$, this reasoning can't be used directly, but if we interpret it as a limit like before, we do recover the standard Gaussian function for non-Archimedean fields
$$\lim_{p\to \infty} \exp(-k'' \|x\|_p^p) = \begin{cases} 1 & \|x\|_p \le 1 \\ 0 & \text{otherwise} \end{cases} = \mathbf{1}_\mathcal{O_K^n}(x),$$
which is the indicator function of $\mathcal{O}_K^n$ in $V=K^n$. Focusing on the $1$-dimensional case $V=K$, these Gaussian functions coincide with the standard ones as used e.g. in Tate's thesis.
So if I'm not mistaken, the problem can be restated as proving that $[\bar{K} : K] = p$ if and only if the standard Gaussian function associated to $K$ is $\exp(-k \|x\|_p^p)$ (with a limit intended if $p=\infty$). A possible advantage of this restatement is that standard Gaussian functions have different characterizations that do not previously assume any length function (for example, they are their own Fourier transform, or satisfy analogues of the central limit theorem, see e.g. here).

$(\dagger)$ Note that by the Artin-Schreier theorem, the only degrees $[\bar{K}:K]$ that could possibly occur for any field $K$ are $1, 2, \infty$ (this, in turn, seems to be related to the fact that a nonzero integer can only have multiplicative order $1, 2$ or $\infty$, but that is a topic for another question).
$(\ddagger)$ As a fun fact, this can be shown to imply that the hypervolume of the $n$-dimensional complex unit ball $|x_1|+|x_2|+\ldots+|x_n| = 1$ , i.e. the ordinary $2n$-dimensional unit ball, equals $V(B_{2n})=V(B_{2})^n /n!=\pi^n /n!$; compare with the hypervolume $V(X_{n})=V(X_{1})^n /n!=2^n /n!$ of the $n$-dimensional cross-polytope $|x_1|+|x_2|+\ldots+|x_n| = 1$, where $| \cdot |$ now denotes the real absolute value.

The question has been posted on MathOverflow.

I also suggest asking on MO, as Nolord suggested. You can repost it there and link the questions to each other (for example, with a brief heading saying "Cross-posted to/from MO/MSE.").

@user76284 Edited, thanks for the suggestion!

Minor suggestion: You can typeset double vertical bars as \|. This shows up as $\|x\|$ rather than $||x||$ (notice the lack of excess space between the bars).

I guess in the last paragraph should be $|x_1|+|x_2|+\ldots+|x_n|  {\color{red}\le}1 $ instead of $|x_1|+|x_2|+\ldots+|x_n| = 1$.

@TorstenSchoeneberg That seems like a good idea, thank you! I've updated the question with some of my thoughts about it.

Well, maybe there is a "field-agnostic"/"Artin-Schreier-less" proof that if $L \vert K$ is a finite dimensional field extension, $p^*(K) =[L:K] \cdot p^*(L)$. (Or show it at least for some, e.g. cyclotomic, extensions. But as you say it actually holds even for skew field extensions.) Then we would be done in the non-archimedean case once we show further that the possible $p^*$'s are bounded away from zero.

@TorstenSchoeneberg As I mention in a footnote, the fact that the multiplicative order of $n\in\mathbb{Z}$ (and thus the order of this generator) can only be $1, 2, \infty$ seems to be the "source" of many (if not all?) appearances of the trichotomy in the theory.

@TorstenSchoeneberg Yes, that trichotomy is realized as $\mathrm{trivial}, \mathbb{Z}_2, \mathbb{Z}$ in the Brauer groups of local fields. There are many things to say about this which I want to save for my next question, but here is something I can't resist mentioning: for all local $K$, the topological generator of the Galois group of $K^{ur}$ acts as $\omega \mapsto \omega^n$ on roots of unity for some $n$. In the nonarchimedean case this is Frobenius, and $n$ is a prime power; for $\mathbb{R}$ it is complex conjugation with $n=-1$, and for $\mathbb{C}$ it is the identity ($n=1$).

@TorstenSchoeneberg Good questions! For the quaternions one can mimic the construction of $|\cdot |$, with the slightly weird result that the critical length function invariant under $USp(n)$ looks like $||(x_1,\ldots,x_n)||_{1/2} = (\sqrt{|x_1|}+\ldots+\sqrt{|x_n|})^2$. I haven't checked it but I think the same should happen for the octonions with $||\cdot ||_{1/4}$ invariant under several exceptional groups, with the usual caveat that octonionic vector spaces are only well-behaved for $n\le 3$.

By the way, what happens over the quaternions $\mathbb H$? --- Superficially related, the trichotomy $1,2,\infty$ also matches the cardinality of the Brauer group $Br(k)$ for the respective local fields $k$, right?

I have a feeling that at least in the two archimedean cases, this is actually related to the fact that what we have decided on as codomain for measures, $\mathbb R_{\ge 0}$, is made up of the squares in $(\mathbb R, \cdot)$, or the norms from $\mathbb C \vert \mathbb R$. --- Anyway, maybe one can get insight by checking what happens in a non-local case like some cubic or higher extension of $\mathbb Q$, or $\mathbb Q(i)$, with restricting the possible Haar measures (they still exist, just not unique) to those.


Can a 4D spacecraft, with just a single rigid thruster, achieve any rotational velocity?

It seems preposterous at first glance. I just want to be sure. Even in 3D the behaviour of rotating objects can be surprising (see the Dzhanibekov effect); in 4D it could be more surprising.
A 2D or 3D spacecraft (with no reaction wheels or gimbaling etc.) needs at least two thrusters to control its spin. See my answer on space.SE.

For any two $4\times4$ matrices $X$ and $Y$, define the commutator $[X,Y]=XY-YX$, the anticommutator $\{X,Y\}=XY+YX$, and the Frobenius inner product $\langle X,Y\rangle=\operatorname{tr}(X^\top Y)$, where $\operatorname{tr}$ is the trace and $\top$ is the transpose.
Let $M$ be a symmetric positive-definite $4\times4$ matrix, and $T=\mathbf f\,\mathbf r^\top-\mathbf r\,\mathbf f^\top$ an antisymmetric $4\times4$ matrix. ($M$ describes the distribution of mass in the spacecraft, $\mathbf r$ is a vector locating the thruster relative to the centre of mass, $\mathbf f$ is the force produced by the thruster, and $T$ is the torque produced by the thruster. More details here. Everything is described in the rotating reference frame.)
The angular velocity $\Omega(t)$, an antisymmetric $4\times4$ matrix, changes with time $t$ according to Euler's equation
$$\{M,\Omega'(t)\}+[\Omega(t),\{M,\Omega(t)\}]=f(t)\,T$$
where $f(t)\geq0$ is a function (continuous, piecewise-constant, or just integrable) describing when and how strongly the thruster is used.
Question: Can $M$ and $T$ be chosen such that, for any two antisymmetric $4\times4$ matrices $\Omega_0$ and $\Omega_1$, there exist $t_1>0$ and $f$ such that the solution $\Omega$ to Euler's equation with initial value $\Omega(0)=\Omega_0$ has final value $\Omega(t_1)=\Omega_1$?
To simplify things, we may take $f$ to be a combination of Dirac deltas instead of an ordinary function. Then the angular velocity satisfies $\{M,\Omega'(t)\}+[\Omega(t),\{M,\Omega(t)\}]=0$ except at a finite set of times when the angular momentum $L(t)=\{M,\Omega(t)\}$ changes by a positive multiple of $T$, or equivalently when $\Omega(t)$ changes by a positive multiple of $A=\{M,\}^{-1}(T)$, where $A$ is the angular acceleration produced by the thruster.

Here is Euler's equation in terms of the components $\omega_{ij}$ of the angular velocity. Assume that $M$ is diagonal, with components $m_i>0$.
$$(m_1+m_2)\,\omega_{12}'(t)+(m_2-m_1)\big(\omega_{13}(t)\,\omega_{32}(t)+\omega_{14}(t)\,\omega_{42}(t)\big)=f(t)\,\tau_{12}$$
(And permute the indices to get 6 equations like this.) Equivalently:
$$\omega_{12}'(t)=\frac{m_1-m_2}{m_1+m_2}\big(\omega_{13}(t)\,\omega_{32}(t)+\omega_{14}(t)\,\omega_{42}(t)\big)+f(t)\,\alpha_{12}$$
If $m_1=m_2$, then $\omega_{12}'$ has constant sign; $\omega_{12}$ is either always non-increasing, or always non-decreasing. So, if the difference between initial and final values of $\omega_{12}$ has the wrong sign compared to $\tau_{12}$, then there is no solution. Thus, we must take $m_1\neq m_2$, and similarly $m_i\neq m_j$ for $i\neq j$.

The Frobenius inner product of two antisymmetric matrices has every term duplicated: $\sum_{i,j}x_{ij}y_{ij}=2\sum_{i<j}x_{ij}y_{ij}$ (since $x_{ji}=-x_{ij}$ and $y_{ji}=-y_{ij}$). So it's natural to take half of this.
The angular momentum's squared magnitude is $\tfrac12\langle L(t),L(t)\rangle$; the derivative of this is $\langle L(t),f(t)T\rangle$. So the angular momentum has constant magnitude whenever $f(t)=0$. (The angular momentum itself would be constant in an inertial reference frame, but here we're using a rotating reference frame.)
The rotational energy is $\tfrac14\langle\Omega(t),L(t)\rangle$; the derivative of this is $\tfrac12\langle\Omega(t),f(t)T\rangle$. So the energy is constant whenever $f(t)=0$.
All of that applies in any dimension. But in 4D there's another constant, the angular momentum bivector's exterior square, or equivalently its inner product with its Hodge dual.
Here are those constants in component form. Squared magnitude of angular momentum:
$$(m_1+m_2)^2\omega_{12}^2+(m_1+m_3)^2\omega_{13}^2+(m_2+m_3)^2\omega_{23}^2\\+(m_1+m_4)^2\omega_{14}^2+(m_2+m_4)^2\omega_{24}^2+(m_3+m_4)^2\omega_{34}^2$$
Doubled energy:
$$(m_1+m_2)\omega_{12}^2+(m_1+m_3)\omega_{13}^2+(m_2+m_3)\omega_{23}^2\\+(m_1+m_4)\omega_{14}^2+(m_2+m_4)\omega_{24}^2+(m_3+m_4)\omega_{34}^2$$
Halved exterior square of angular momentum:
$$(m_1+m_2)(m_3+m_4)\omega_{12}\omega_{34}-(m_1+m_3)(m_2+m_4)\omega_{13}\omega_{24}+(m_2+m_3)(m_1+m_4)\omega_{23}\omega_{14}$$
(Its derivative is the exterior product of angular momentum and torque: $(m_1+m_2)\omega_{12}\,f\,\tau_{34}+\cdots$ .)
These three constants define two ellipsoids and another quadratic surface (with signature $+^3-^3$) in the 6D space of antisymmetric matrices. When $f=0$, $\Omega$ must remain on the intersection of these three quadratic surfaces.
In fact there are more than three, as I recently found! And these aren't specific to 4D. The general form is $Q_i=\sum_{j\neq i}\frac{m_i+m_j}{m_i-m_j}\omega_{ij}^2$; check that $Q_i'=0$, and also note $\sum_iQ_i=0$. Then define $P_i=\sum_{j\leq i}Q_j$; these are positive-semidefinite, on the assumption (WLOG) that $m_i$ are sorted, $m_1>m_2>m_3>m_4>0$. The doubled energy is $\sum_i(m_i-m_{i+1})P_i$, and the momentum's squared magnitude is $\sum_i(m_i^2-m_{i+1}^2)P_i$. But what we should focus on are the quadratic forms $N_i=P_{i-1}-P_i+P_{i+1}$ which are not positive-semidefinite but have a single negative term:
$$N_1=\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2+\frac{m_2+m_4}{m_2-m_4}\omega_{24}^2-\frac{m_1+m_2}{m_1-m_2}\omega_{12}^2$$
$$N_2=\frac{m_1+m_2}{m_1-m_2}\omega_{12}^2+\frac{m_1+m_4}{m_1-m_4}\omega_{14}^2+\frac{m_3+m_4}{m_3-m_4}\omega_{34}^2-\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2$$
$$N_3=\frac{m_1+m_3}{m_1-m_3}\omega_{13}^2+\frac{m_2+m_3}{m_2-m_3}\omega_{23}^2-\frac{m_3+m_4}{m_3-m_4}\omega_{34}^2$$
These are all constant. The first expression (if it's equated to $0$) defines a cone, or a pair of opposite cones, around the $\omega_{12}$ axis in 6D. The second expression defines a pair of opposite cones around the $\omega_{23}$ axis; and the third, around the $\omega_{34}$ axis. If the applied acceleration $A$ is inside one of these cones, and the initial angular velocity $\Omega_0$ is in the same cone, then $\Omega$ stays in that cone, and there's no solution with the final angular velocity $\Omega_1$ outside of the cone. Thus, we must take $A$ to be outside of all six of these cones. That is, all three expressions (with $\alpha_{ij}$ in place of $\omega_{ij}$) must be positive.

Notice that Euler's equation has a time symmetry: Given a solution $\Omega(t)$, we can construct another solution $\tilde\Omega(t)=-\Omega(t_1-t)$, so that $\{M,\tilde\Omega'(t)\}+[\tilde\Omega(t),\{M,\tilde\Omega(t)\}]=f(t_1-t)\,T$, and the initial and final values are swapped (and negated): $\tilde\Omega(0)=-\Omega_1$ and $\tilde\Omega(t_1)=-\Omega_0$. Therefore, any angular velocity can be reached from any other, if and only if any angular velocity can be reached from $0$. (We can reverse a path from $0$ to $-\Omega_0$ to get a path from $\Omega_0$ to $0$, and concatenate that with a path from $0$ to $\Omega_1$, to get a path from $\Omega_0$ to $\Omega_1$.)
Also, Euler's equation has a scale symmetry: Given a solution $\Omega(t)$, for any $k>0$ we can construct another solution $\tilde\Omega(t)=k\,\Omega(kt)$, so that $\{M,\tilde\Omega'(t)\}+[\tilde\Omega(t),\{M,\tilde\Omega(t)\}]=k^2f(kt)\,T$, and the final value is $\tilde\Omega(t_1/k)=k\,\Omega(t_1)=k\,\Omega_1$. Therefore, any angular velocity can be reached from $0$, if and only if any angular velocity in a small neighbourhood of $0$ can be reached from $0$.
(It looks like these two symmetries are related by $k=-1$, but we should keep a distinction between positive time and negative time.)

This has the form of a quadratic differential equation, $\mathbf x'(t)=\mathbf x(t)\odot\mathbf x(t)$ where $\odot$ is some bilinear function. (Specifically, for antisymmetric matrices $X$ and $Y$, define $X\odot Y=-\{M,\}^{-1}([X,\{M,Y\}])$, so Euler's equation is $\Omega'=\Omega\odot\Omega$, as long as $f=0$. Alternatively, define $X\odot Y=-[\{M,\}^{-1}(X),Y]$, so Euler's equation is $L'=L\odot L$.)
(I ought to move this power series stuff to a different page, since it's applicable far beyond this particular Question, and anyway it's bloating this page.)
Fix a norm on the space, and find some constant $\lVert\odot\rVert>0$ such that $\lVert\mathbf x\odot\mathbf y\rVert\leq\lVert\odot\rVert\lVert\mathbf x\rVert\lVert\mathbf y\rVert$ for all $\mathbf x,\mathbf y$.
Given initial value $\mathbf x(0)=\mathbf a$, the equation $\mathbf x'=\mathbf x\odot\mathbf x$ has the power series solution
$$\mathbf x(t)=\sum_{n=0}^\infty t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}$$
where the coefficient of $t^n$ is the average of all possible ways of evaluating $\mathbf a^{n+1}$ using the product $\odot$. For example, the coefficient of $t^3$ is $\tfrac16$ of
$$\sum^6\mathbf a^4=((\mathbf a\mathbf a)\mathbf a)\mathbf a+(\mathbf a(\mathbf a\mathbf a))\mathbf a+2(\mathbf a\mathbf a)(\mathbf a\mathbf a)+\mathbf a((\mathbf a\mathbf a)\mathbf a)+\mathbf a(\mathbf a(\mathbf a\mathbf a))$$
$$\newcommand{\aaaa}[3]{\mathbf a\underset{#1}\odot\mathbf a\underset{#2}\odot\mathbf a\underset{#3}\odot\mathbf a} =\aaaa{1}{2}{3}+\aaaa{2}{1}{3}\begin{matrix}{}+\aaaa{1}{3}{2} \\ {}+\aaaa{2}{3}{1}\end{matrix}+\aaaa{3}{1}{2}+\aaaa{3}{2}{1}.$$
The terms in the series can be obtained recursively:
$$\mathbf u_n=\frac{t^n}{n!}\sum^{n!}\mathbf a^{n+1}$$
$$=\frac tn\sum_{k=0}^{n-1}\mathbf u_{n-1-k}\odot\mathbf u_k.$$
We also have $\lVert\mathbf u_n\rVert\leq|t|^n\lVert\odot\rVert^n\lVert\mathbf a\rVert^{n+1}$, which ensures absolute convergence, for small $t$:
$$\sum_{n=0}^\infty\left\lVert t^n\frac{\sum^{n!}\mathbf a^{n+1}}{\sum^{n!}1}\right\rVert\leq\sum_{n=0}^\infty|t|^n\lVert\odot\rVert^n\lVert\mathbf a\rVert^{n+1}=\frac{\lVert\mathbf a\rVert}{1-|t|\lVert\odot\rVert\lVert\mathbf a\rVert},$$
$$|t|<\frac{1}{\lVert\odot\rVert\lVert\mathbf a\rVert}.$$
And at certain times (when the thruster is used) an impulse may be applied to $\mathbf x$, with a fixed direction $\mathbf b$ but an arbitrary magnitude $c>0$, thus: $\mathbf x(t+)=\mathbf x(t-)+c\mathbf b$. These impulses, and the time intervals between them, are many variables that we can control. If the number of variables is at least $6$ (or the dimension of the space $\mathbf x$ is in), then there is hope of surrounding $0$ in an open set, and thus reaching everywhere in the space (according to the previous section).
$$\mathbf x(0+)=0+c_0\mathbf b$$
$$\mathbf x(t_1-)=\sum_{n=0}^\infty\frac{t_1^n}{n!}\sum^{n!}\mathbf x(0+)^{n+1}$$
$$\mathbf x(t_1+)=\mathbf x(t_1-)+c_1\mathbf b$$
$$\mathbf x(t_1+t_2-)=\sum_{n=0}^\infty\frac{t_2^n}{n!}\sum^{n!}\mathbf x(t_1+)^{n+1}$$
$$\mathbf x(t_1+t_2+)=\mathbf x(t_1+t_2-)+c_2\mathbf b$$
$$\mathbf x(t_1+t_2+t_3-)=\sum_{n=0}^\infty\frac{t_3^n}{n!}\sum^{n!}\mathbf x(t_1+t_2+)^{n+1}$$
$$c_0,t_1,c_1,t_2,c_2,t_3\geq0$$
$$\mathbf x(t_1+t_2+t_3-)\approx0\quad?$$

Cross-posted to MO: mathoverflow.net/questions/448773/…

Oh, I didn't explain what $M$ is (though the linked physics.SE post did explain it). It's the matrix of second moments of mass: $M=\int\mathbf r\,\mathbf r^T\,\rho\,dV$, where $\mathbf r$ is position relative to the centre of mass, $\rho$ is density, and $V$ is volume. But that's not necessary for understanding this Question.

Given that $M$ is diagonal (which is no loss of generality, since any symmetric matrix has an orthogonal eigenbasis), the inertia tensor is also diagonal, with components $m_1+m_2$, $m_1+m_3$, $m_2+m_3$, $m_1+m_4$, $m_2+m_4$, $m_3+m_4$.

And of course I'm considering bivectors as antisymmetric matrices.

The inertia tensor is the function $\Omega\mapsto L=\{M,\Omega\}=M\Omega+\Omega M$.

What's the relationship of $M$ with the inertia tensor? My understanding was that the inertia tensor is a linear transformation on bivectors, which in this case are 6-dimensional, so it would be a symmetric 6x6 matrix. Is that not correct?


Semirings induced by symmetric monoidal categories with finite coproducts

A symmetric monoidal category with finite coproducts is by definition a symmetric monoidal category $(\mathcal{C},\otimes,1,\dotsc)$ such that the underlying category $\mathcal{C}$ has finite coproducts (this includes an initial object $0$) with the property that $\otimes$ preserves finite coproducts in each variable. In other words, we have a distributive law
$$X \otimes (\bigoplus_{i \in I} Y_i) \cong \bigoplus_{i \in I} (X \otimes Y_i).$$
This is also known as a finitary distributive symmetric monoidal category. Notice that to each symmetric monoidal category with finite coproducts we may associate a commutative semiring: The elements are the isomorphism classes of objects in $\mathcal{C}$. The addition is $[X] + [Y] := [X \oplus Y]$, the multiplication is $[X] \cdot [Y] := [X \otimes Y]$. The zero element is $[0]$, the multiplicative unit is $[1]$. To avoid set-theoretic issues, let us assume that $\mathcal{C}$ is essentially small. As a consequence, we may think of symmetric monoidal categories with finite coproducts as a possible categorification of semirings.
A familar example is the symmetric monoidal category of vector bundles on a space $X$, which produces the semiring of isomorphism classes of vector bundles on $X$. If $X$ is a point, we see that the semiring associated to $\mathsf{FinSet}$ is $\mathbb{N}$.
Note, however, that not every commutative semiring arises from the construction above. Basically this is because we have agreed that the addition is not induced by some symmetric monoidal structure which distributes over $\otimes$, but rather from the coproduct. The coproduct satisfies $X \oplus Y \cong 0 \Rightarrow X \cong Y \cong 0$. It follows that in the associated semiring only $0$ has an additive inverse.
Question. How can we classify those commutative semirings which are induced by essentially small symmetric monoidal categories with finite coproducts?
Every commutative monoid with the property $a+b=0 \Rightarrow a=b=0$ arises from an essentially small category with finite coproducts, see SE/834869. But it is not clear if we can use this here.

@Arducode This does not define a functor for $\otimes$. Or how do you define the tensor product of morphisms? Also notice that it is required that the tensor product is distributive - this is not guaranteed by arbitrary isomorphisms, they have to be the canonical ones. Therefore I do not think that the skeleton will help us.

If you have a commutative semiring $(R, +, \cdot)$ that satisfies $a+b = 0 \Rightarrow a=b=0$, can't you just take an essentially small category $C$ realizing $(R,+)$, take a skeleton $\mathrm{Sk}(C)$ of it and define on $\mathrm{Sk}(C)$ the symmetric monoidal structure induced by $(R,\cdot)$? That is, $[x] \otimes [y] \colon= [x \cdot y]$ (where $[x]$ denotes the object in $\mathrm{Sk}(C)$ that is associated to the object $x \in R$)


Can you obtain $\pi$ using elements of $\mathbb{N}$, and finite number of basic arithmetic operations + exponentiation?

Is it possible to obtain $\pi$ from finite amount of operations
$\{+,-,\cdot,\div,\wedge\}$ on $\mathbb{N}$ (or $\mathbb{Q}$, the answer will still be the same)? Note that the set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental).
Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $\mathbb{N}$) ?

As mentioned above, it is probably false, but a disproof is difficult. A hope could be to describe a subfield of $\Bbb{R}$ containing all elements generated by the method you describe, but not containing $\pi$. How such a field would look is difficult to imagine though.

The bonus question is the same question except allowing the use of $\log$, since all the operations are invertible, and their inverse is included in the allowed operations - except for $^$, which needs $\log$.

If you only use +,-,x,/, you can't get $\pi$ because it will only yield a rational number, so you have to use the exponential operator sometime.

It can be obtained using calculus, check Wikipedia.

Could Machin-like formulas be of any use?

Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special.

Of course not, but I can't prove it.  There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way.  Unless a real has some reason to be expressible it almost certainly can't be.

(this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well)


Does there exist a function $f$, holomorphic on the whole complex plane $\mathbb{C}$, such that $f\left(\mathbb{Z}\right)=\{0,1,2,3\}$ and
$\forall z\in\mathbb{C}\ (f(z)\in\{0,1,2,3\}\Rightarrow z\in\mathbb{Z})$?
If yes, is it possible to have an explicit construction?
Note that, for example, $h(z)=\frac{1}{6} \left(9-8 \cos \left(\frac{\pi z}{3}\right)-\cos (\pi z)\right)$ is not a valid solution, since, in particular, certain roots of the equation $h(z)=0$ are not integers, but complex numbers.
Also note that this question is answered in positive regarding the function $g$ such that $g\left(\mathbb{Z}\right)=\{0,1,2\}$ and $\forall z\in\mathbb{C}\ (g(z)\in\{0,1,2\}\Rightarrow z\in\mathbb{Z})$. In this case, an example of such function is $g(z)=1-\cos \left(\frac{\pi z}{2}\right)$.

Crossposted to MathOverflow: A holomorphic function sending integers (and only integers) to $\{0,1,2,3\}$

I suggest you post this to MathOverflow, adding a link to this MSE post to let readers know you've crossposted.

I'd ask for a Laurent polynomial solution in a separate post (ie. $g\in \Bbb{C}[q,q^{-1}]$ such that $g^{-1}(\{0,1,2,3\}) = \mu_n$ for some $n$)

@user1950 - one has to be careful as the cited theorem applies only for multiplicity one values, so in other words if $f^{-1}(\{0,1,2,3\})=\mathbb Z$ and $f'(n) \ne 0$ for any $n \in \mathbb Z$; probably not hard to generalize if the multiplicities are uniformly bounded but unclear what happens if multiplicities at integers can be unbounded

So from @reuns remark it follows that the sought function, if it exists, must be a Laurent polynomial of $e^{\frac{2\pi i z}{n}}$. Looks like the problem is within reach of an expert in Galois theory.

It follows from Theorem 1 of G.G. Gundersen and C.C. Yang's "On the Preimage Sets of Entire Functions" that the sought function must be periodic. (projecteuclid.org/journalArticle/…)

one way to go is to interpolate the values $0,1,2,3$ evenly at integers (first for example the values may be symmetrized and the function made odd say on the integers as that gives better interpolation convergence) and get a function $f(z)=\frac{\sin \pi z}{\pi}(f(0)/z +\sum_{n \ne 0} (-1)^n f(n)/(z-n)+c)$ of order $1$ where $f(n)$ are precisely the choices you make and see if some clever choice of the interpolation values, the function obtained can be show to satisfy the property that only integers are sent there

why does it need to be periodic? the above applies only to periodic functions

@Conrad: So any valid solution to the main question must be of the form $\sum_{n=a}^{b} a_n q^n$, where $-\infty<a,b<+\infty$ and $q=e^{\frac{2\pi i z}{k}}$?

the problem with @joricki function is that is of infinite order so any $f(z)=a$ should have tons of solutions (many more than integers); for example $f(z)=0$ means $\frac{\pi z}3+\alpha\sin \frac{2\pi z}3=(2k+1)\pi$ and for a fixed large $k$ that has only a few integral solutions since $z/3$ needs to be around $2k+1$ as the other term is bounded on the integers, but (by Picard) it must have infinitely many solutions for all but at most one $k$; a function with the given property kind of have to be of finite order $1$ so the roots of $f(z)=0,1,2,3$ are the right number in any disc of radius $R$

@joriki $f$ is entire $n$-periodic iff $g(q) = f(\frac{n\log(q)}{2i\pi })$ is analytic on $\Bbb{C}^*$. Due to great Picard's theorem $g(q)$ must be in the Laurent polynomials ring $\Bbb{C}[q,q^{-1}]$.

@reuns: I'm afraid I don't quite follow. What are $q$ and $\mathbb C[q,q^{-1}]$? (And in case this is an argument that my proposed function takes on the desired values at non-integers, why doesn't it apply to the $4$-periodic function in the question?)

@joriki Your function is $6$-periodic. If there is a $n$-periodic solution $f(z)=g(e^{2i\pi z/n})$ then $g$ is analytic on $\Bbb{C}^*$ so it must be in $\Bbb{C}[q,q^{-1}]$ as otherwise $g(q)$ or $g(q^{-1})$ has an essential singularity at $0$ and we can apply great Picard's theorem

@user1950: I'm sorry, I was missing a factor of $2$, it should be $\frac32\left(1+\cos\left(\frac{\pi z}3+\alpha\sin \frac{2\pi z}3\right)\right)$ (with the same $\alpha$).

@joriki - can you recheck the formula? With $z=2$ I get $f(2)=3/2 (1 + \sin[π/6 - \arccos[1/3]])$, which is not an integer.

A candidate would be $\frac32\left(1+\cos\left(\frac{\pi z}3+\alpha\sin \frac{\pi z}3\right)\right)$ with $\alpha=\left(\arccos\left(\frac13\right)-\frac\pi3\right)/\sin\left(\frac\pi3\right)\approx0.21219$, which takes the right values on the integers, but I suspect that it also takes them elsewhere, and I don't know how to easily check for that. Wolfram|Alpha doesn't return any other solutions.

@Conrad - well, first I've noticed the fact that this is easily solvable if the discrete set is composed of all the $n$-th roots of unity. And then started to check if generalizations are possible.

@Dmitry - the function is entire so is analytic on the plane; the condition just talks about specific values of the function at specific points - it is very easy to construct a function like that st it sends the integers surjectively to the given set, the problem is to ensure that no other numbers go there and that is harder; for the smaller set $0,1,2$ a simple solution is actually presented

No, such a function does not exist. A holomorphic function on the complex plane is infinitely differentiable, and thus must have derivatives of all orders. However, since the set of integers is countable, a function that is only defined on the integers cannot be infinitely differentiable. Therefore it cannot be holomorphic on the whole complex plane.


A conjecture about Catalan sequence

For $n=2$, consider the free abelian group generated by polynomials corresponding to $\frac{(2 n)!}{2^n n!}=3$ partitions of $2n=4$ vertices into pairs (chords):
$$
(x_1-x_3)(x_2-x_4),(x_1-x_2)(x_3-x_4),(x_1-x_4)(x_2-x_3)
$$
Euler's Identity: $
(x_1-x_3)(x_2-x_4)=(x_1-x_2)(x_3-x_4)+(x_1-x_4)(x_2-x_3)$.

So$$\{(x_1-x_2)(x_3-x_4),(x_1-x_4)(x_2-x_3)\}\tag1\label2$$
is a basis.
So the rank of the free abelian group is $2$.

For $n=3$, consider the free abelian group generated by polynomials corresponding to $\frac{(2 n)!}{2^n n!}=15$ partitions of $2n=6$ vertices into pairs (chords):
\begin{array}r
\{(x_1-x_2) (x_3-x_4) (x_5-x_6),\\(x_1-x_2) (x_3-x_5) (x_4-x_6),\\(x_1-x_2) (x_4-x_5) (x_3-x_6),\\(x_1-x_3) (x_2-x_4) (x_5-x_6),\\(x_1-x_3) (x_2-x_5) (x_4-x_6),\\(x_1-x_3) (x_4-x_5) (x_2-x_6),\\(x_2-x_3) (x_1-x_4) (x_5-x_6),\\(x_1-x_4) (x_2-x_5) (x_3-x_6),\\(x_1-x_4) (x_3-x_5) (x_2-x_6),\\(x_2-x_3) (x_1-x_5) (x_4-x_6),\\(x_2-x_4) (x_1-x_5) (x_3-x_6),\\(x_3-x_4) (x_1-x_5) (x_2-x_6),\\(x_2-x_3) (x_4-x_5) (x_1-x_6),\\(x_2-x_4) (x_3-x_5) (x_1-x_6),\\(x_3-x_4) (x_2-x_5) (x_1-x_6)\}
\end{array}
Using Euler's Identity we can decompose the product of intersecting chords into the sum of products of the non-intersecting chords. For example $(x_1-x_4)(x_2-x_5)(x_3-x_6)$ as in Fuhrmann's Theorem,

There are 5 partitions of 6 vertices into non-intersecting chords
$$\tag2\label1\begin{array}l\{(x_1-x_6)(x_2-x_5)(x_3-x_4),\\ (x_1-x_6)(x_2-x_3)(x_4-x_5),\\ (x_1-x_4)(x_2-x_3)(x_5-x_6),\\(x_1-x_2)(x_3-x_6)(x_4-x_5),\\(x_1-x_2)(x_3-x_4)(x_5-x_6)\}\end{array}$$
Under lexicographic order, the leading terms of \eqref{1} are distinct: $\{x_1x_2x_3,x_1x_2x_4,x_1x_2x_5,x_1x_3x_4,x_1x_3x_5\}$. Suppose $c_1(x_1-x_6)(x_2-x_5)(x_3-x_4)+\dots+c_5(x_1-x_2)(x_3-x_4)(x_5-x_6)=0$ for some $c_1,\dots,c_5\in\mathbb{Z}$.
Comparing coefficients of term $x_1x_2x_3$ to get $c_1=0$, comparing coefficients of term $x_1x_2x_4$ to get $c_2=0$, comparing coefficients of term $x_1x_2x_5$ to get $c_3=0$, comparing coefficients of term $x_1x_3x_4$ to get $c_4=0$, comparing coefficients of term $x_1x_3x_5$ to get $c_5=0$, so $c_1=\dots=c_5=0$, \eqref{1} is linearly independent.
So the rank of the free abelian group is 5.

Here is a proof that the number of non-intersecting chords is Catalan number $\frac1{n+1}\binom{2n}n$.
Each polynomial in \eqref{1} can be written as a $2×n$ "non-intersecting tableaux"
$$
\Biggl\{\begin{array}{|c|c|}\hline1&2&3\\\hline6&5&4\\\hline\end{array},
\begin{array}{|c|c|}\hline1&2&4\\\hline6&3&5\\\hline\end{array},
\begin{array}{|c|c|}\hline1&2&5\\\hline4&3&6\\\hline\end{array},
\begin{array}{|c|c|}\hline1&3&4\\\hline2&6&5\\\hline\end{array},
\begin{array}{|c|c|}\hline1&3&5\\\hline2&4&6\\\hline\end{array}\Biggr\}
$$
To show that a "non-intersecting tableux" is determined by its first row, we give a procedure to fill in the second row.
For example $\begin{array}{|c|c|}\hline1&2&5&7\\\hline?&?&?&?\\\hline\end{array}$ to fill $3,4,6,8$ in the second row such that there are no intersecting chords, we must put $3$ under the largest number less than $3:\begin{array}{|c|c|}\hline1&2&5&7\\\hline?&3&?&?\\\hline\end{array}$ then we must put $4$ under the largest remaining number less than $4:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&?&?\\\hline\end{array}$, then we must put $6$ under the largest remaining number less than $6:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&6&?\\\hline\end{array}$, then we must put $8$ under the largest remaining number less than $8:\begin{array}{|c|c|}\hline1&2&5&7\\\hline4&3&6&8\\\hline\end{array}$
For all $n$ we have shown the first rows of $\frac1{n+1}\binom{2n}n$ number of "non-intersecting tableaux" are distinct.
Under lexicographic order, the leading terms of \eqref{1} correspond to the first row of tableaux, so the leading terms are distinct, so the set of polynomials corresponding to "non-intersecting tableux" are linearly independent for all $n$.
So we have proved:

For all $n$, the rank of the above free abelian group in $2n$ variables is $\frac1{n+1}\binom{2n}n$.

For $n=4$, similarly decompose the product of intersecting chords into the sum of products of the non-intersecting chords. For example $(x_1-x_5)(x_2-x_6)(x_3-x_7)(x_4-x_8)$

As you can see, there will be two identical pairs in the end, so after collecting terms there will be two terms with coefficient $2$
\begin{align}  (x_1 - x_5)(x_2 - x_6)(x_3 - x_7)(x_4 - x_8)\\
=   (x_1 - x_4)(x_2 - x_3)(x_5 - x_8)(x_6 - x_7)\\
  + \color{red}{2}(x_1 - x_8)(x_2 - x_3)(x_4 - x_5)(x_6 - x_7)\\
  + (x_1 - x_2)(x_3 - x_8)(x_4 - x_7)(x_5 - x_6)\\
  + (x_1 - x_2)(x_3 - x_8)(x_4 - x_5)(x_6 - x_7)\\
  + \color{red}{2}(x_1 - x_2)(x_3 - x_4)(x_5 - x_6)(x_7 - x_8)\\
  + (x_1 - x_2)(x_3 - x_4)(x_5 - x_8)(x_6 - x_7)\\
  + (x_1 - x_4)(x_2 - x_3)(x_5 - x_6)(x_7 - x_8)\\
  + (x_1 - x_8)(x_2 - x_3)(x_4 - x_7)(x_5 - x_6)\\
  + (x_1 - x_2)(x_3 - x_6)(x_4 - x_5)(x_7 - x_8)\\
  + (x_1 - x_8)(x_2 - x_7)(x_3 - x_4)(x_5 - x_6)\\
  + (x_1 - x_8)(x_2 - x_7)(x_3 - x_6)(x_4 - x_5)\\
  + (x_1 - x_8)(x_2 - x_5)(x_3 - x_4)(x_6 - x_7)\\
  + (x_1 - x_6)(x_2 - x_3)(x_4 - x_5)(x_7 - x_8)\\
  + (x_1 - x_6)(x_2 - x_5)(x_3 - x_4)(x_7 - x_8)\end{align}

Conjecture. Express $(x_1-x_{n+1})(x_2-x_{n+2})\dots(x_n-x_{2n})$ as a linear combination of polynomials corresponding to the non-intersecting chords, then the term $(x_1-x_2)(x_3-x_4)\cdots(x_{2n-1}-x_{2n})$ and the term $(x_{2n}-x_1)(x_2-x_3)\cdots(x_{2n-2}-x_{2n-1})$ have largest coefficient in absolute value.

Mathematica code to compute the largest coefficients in absolute value
Needs["Combinatorica`"];
coeff[n_]:=Max[Abs[PolynomialReduce[Product[Indexed[x,i]-Indexed[x,n+i],{i,n}],Table[Module[{t1=t[[1]],t2=t[[2]],poly=1,i=1},Do[i=1;While[Length[t1]>i&&t1[[i+1]]<t2[[1]],i++];poly*=Indexed[x,t1[[i]]]-Indexed[x,t2[[1]]];t1=Delete[t1,i];t2=Rest[t2],n];poly],{t,Tableaux[Table[n,2]]}],Table[Indexed[x,i],{i,2n}]][[1]]]]

the terms $n=2,\dots,9$ in the sequence agree with A099960:
\begin{array}{c|c}
n&2&3&4&5&6&7&8&9\\\hline
\text{largest abs coefficient}&1&1&2&3&8&17&56&155
\end{array}

@DietrichBurde The first example you gave only starts failing at the $x^{10}$ term. $132$ and $429$, as well as the two terms that follow $1430$ and $4862$, are Catalan numbers. The second example evaluates to $3$ when $k=3$. Though I do get your point that it's best not to be too confident about integer sequences when you only know the first few terms.

The expansion of $(1-8x+21x^2-20x^3+5x^4)/(1-9x+28x^2-35x^3+15x^4-x^5)$ also produces $1,1,2,5,14,42$ as Catalan, but then $132,429$, which is not Catalan. Also $binomial(fibonacci(k) + 1, 2)$ produces $1,1,2,5,14,42,132$ and so on.


How do I find the common invariant subspaces of a span of matrices?

Let $G_1, \ldots, G_n$ be a set of $m\times m$ linearly-independent complex matrices.
Let $\mathcal{G} = \operatorname{span}\left\{ G_1, \ldots , G_n\right\}$ be the vector space that spans the set of $G$ matrices, and let $\mathcal{S}$ be any linear subspace of $\mathcal{G}$.
Finally, let $\mathcal{I}$ be a linear subspace of $\mathbb{C}^m$.
Is there a way to determine all pairs $\left( \mathcal{S}, \mathcal{I}\right)$ such that for all $S \in \mathcal{S}$, the subspace $\mathcal{I}$ is an invariant subspace of $S$? In other words, is there a way to determine the linear subspaces of $\mathcal{G}$ whose elements share a (non-trivial) invariant subspace?
This problem is vaguely related to this one posted in 2013.
I know that if we arbitrarily pick an $\mathcal{I}$, we can determine the largest subspace $\mathcal{S}$ for which $\mathcal{I}$ is invariant. The problem is that there are infinitely many possible $\mathcal{I}$ to choose from. Likewise, if we arbitrarily choose an $\mathcal{S}$, we can deduce $\mathcal{I}$, but there are infinitely many $\mathcal{S}$ to choose from.
I haven't made any progress trying to determine both subspaces simultaneously. Any assistance would be greatly appreciated. Even some suggestions on where I might look to find a solution would be much appreciated.

How about computing products $G_1G_2\cdots G_k$, $\forall k \in \{1,N\}$ and multiply a random vector with that. Then put all vectors with some large enough norm into a new matrix and then iterate by the new matrix with the sequence of matrices. Then iterate the procedure. This (I hope) will give something similar to the "power method" for finding largest eigenvalue, eigenvector pair. Oh yes, probably some re-normalization should be done also.

@DavidSpeyer - Many thanks for the heads up. I'll look into it.

More at math.stackexchange.com/questions/380730/… and math.stackexchange.com/questions/184960/…


Minimal time gossip problem

The gossip problem (telephone problem) is an information dissemination problem
where each of $n$ nodes of a communication network has a unique piece of information
that must be transmitted to all the other nodes using two-way communications
(telephone calls) between the pairs of nodes. Upon a call between the given two
nodes, they exchange the whole information known to them at that moment.
The minimum number of calls needed to guarantee the spread of whole information is $2n-4$.
http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/gossips.pdf
Assuming that the calls between non-overlapping pairs of nodes can take place simultaneously, the minimum amount of time $T(n)$ required to complete gossiping is
$\lceil \log_2 n \rceil$ for even $n$ and $\lceil \log_2 n \rceil + 1$ for odd $n$.
http://onlinelibrary.wiley.com/doi/10.1002/net.3230180406/abstract
As it is written in the paper by R. Labahn, "Kernels of minimum size gossip schemes",
Discrete Mathematics 143 (1995) 99-139:
Theorem 5.3 Any gossip scheme on $n \geq 4$ vertices with $2n-4$ calls has at least $2\lceil \log_2 n \rceil - 3$ rounds.
My question is following. Is there a result on the problem to find the minimum number of calls of a gossip scheme with time (rounds) $T(n)$.

Yes, this paper is about fault-tolerance of the gossip schemes. I have some ideas, and I want simply find any results about minimum number of calls of a gossip scheme with minimal time $T(n)$ to develop the theory. Presumably it is an open question.

It was this one:arxiv.org/abs/1304.5633 but I didn't read the second half of the title. Are you looking for a reference to add to your paper?

@BrianRushton, thank you for the comment. Which paper you mean ?

It looks like your own paper has such results.


The limit of $(\sin(n!)+1)^{1/n}$ as n approaches infinity

Calculate the limit
$$
\lim_{n\rightarrow\infty}(\sin(n!)+1)^{1/n}
$$
or prove that the limit does not exist.
This appeared as a problem in my mathematical analysis test, and the answer was that the limit exists and it was $1$. But later the teacher found a mistake in his proof and eventually removed the problem from the test. But I'm just curious. Does this problem have a certain answer?
The biggest question for me, is that I can't show that there does not exist any $n_0$ so that $\sin(n_0!)$ is close to $-1$ enough so that the original term might not converge. Any help would be appreciated!

One way to show that this sequence diverges would be to construct a subsequence which diverges.

AFAIK, $1$ will be an accumulation point (limit of a subsequence). But it is much harder to say that $0$ is not another one...

Factorial is not defined on not-integers. If the gamma function is used to extend the factorial function, the function oscillates between 0 and 1.

Related math.stackexchange.com/questions/677719/… and mathoverflow.net/questions/45665/…

I was being sloppy with my language, I meant one thing and wrote another @AdamRubinson . I was thinking: "... if $\sin(n!)$ is bounded away from $-1$, then..." but although: equidistributed --> not bounded away from $-1$, the converse fails! You are quite right, it's unjustified at present

@FShrike Please can you justify your claim? It seems wrong, or at best, unjustified to me. For example, it's not clear to me why we cannot have: $\ (\sin(n!))_n\ $ not equi-distributed and $\liminf_{n\rightarrow\infty}(\sin(n!)+1)^{1/n}<1.$

@FShrike: The  problem of the density or lacunarity of the limit points of the sequence $\sin n!$ is probably as hard as that of $\cos(n!)$.

@MartinR we know $(\sin(n))_n$ is equidistributed, but do we know $(\sin(n!))_n$ is equidistributed? If it isn't, then we know for sure that the limit is $1$

Here is a similar question about $\cos(n!)$. The answer (as I understand it) is “we don't know.”


Wave equation: predicting geometric dispersion with group theory

Context
The wave equation
$$
\partial_{tt}\psi=v^2\nabla^2 \psi
$$
describes waves that travel with frequency-independent speed $v$, ie. the waves are dispersionless. The character of solutions is different in odd vs even number of spatial dimensions, $n$. A point source in odd-$n$ creates a disturbance that propagates on the light cone and vanishes elsewhere: if the point source is a flash of light, an observer sees darkness, then a flash, then darkness. When $n$ is even, a disturbed media never returns to rest: the observer sees darkness, then brightness that lingers for all $t$. This phenomena is known as geometric dispersion.
Question
Is it possible to show that geometric dispersion is predicted by the wave equation, using group theory? For a point source at the origin, we would be searching for spherically symmetric solutions, and the rotation group $SO(n)$ has a different structure depending on whether $n$ is odd or even. In particular, I am interested in doing this without actually solving the wave equation. Unfortunately, I don't know enough group theory to know if this is even possible.
What I know
I can 'show' geometric dispersion by solving the wave equation with an initial condition, or computing the Green's function for the wave equation and noting that it is either supported only on the light cone (odd $n$), or everywhere within the light cone (even $n$).
I know some group theory 'for physicists'.
Related
This unanswered question is similar. I think my question is more specific: I'm asking about a way to predict (rather than explain) geometric dispersion using group theory.
Update: (thanks to comments of Alp Uzman and GiuseppeNegro)
It appears to be possible using group theoretic machinery, described in the book Nonabelian harmonic analysis by Howe and Tan. The relevant section is 4.3.1. So an equivalent question becomes: can someone explain the result from Howe and Tan in a more accessible way? The book is beyond my level of group theory at the moment.

@GiuseppeNegro I had tried studying the Howe-Tan account of the Huygens' principle about five years ago and at the time I too could not handle the representation theory involved (I'm afraid I still can't). I don't really have any useful insights, as such I can't claim any credits, though thank you for acknowledging my comment elsewhere.

I wonder if we can @AlpUzman to notify them. It's them, not me, who gave that useful reference

@GiuseppeNegro Thank you, that reference seems to answer precisely my question- I do not understand it yet either

Following the comments in the linked question, I arrived at Theorem 4.3.1 in the book "Nonabelian harmonic analysis" of Howe and Tan (thank you Alp Uzman). That theorem contains exactly the answer to this question. Unfortunately, it uses heavy machinery from representation theory and I cannot fully understand it.

I would love to see an answer to this. Group theory is one of my weakest areas though, so I won't be able to produce one....


Binomial coefficients modulo a prime

Consider an odd prime $p\equiv1 \pmod {16}$ and set $M=\frac{p-1}{2}$ for notational convenience. Then is there even a single prime $p$ of the above form for which the following congruence holds? $$\binom{M}{M/2}\binom{M}{M/4}\equiv \pm \binom{M}{3M/8}\binom{M}{7M/8}\pmod p ?$$
Here the symbol $\binom{n}{r}$ is a binomial coefficient and is defined as $\frac{n!}{r!(n-r)!}$
I am trying to prove something significantly more abstract but have managed to reduce it down to proving that the above statement can never hold. I have written some code to check this for primes of the above form less than $1000$ and found no counter examples. My attempts to find results on binomial coefficients $\pmod p$ have mostly involved Lucas' theorem which does not seem very applicable here. I would appreciate a solution, reference or possible strategies I might try.

Added later: The more general problem.
Let $$f_a(x)=x(x^4-1)(x^4+ax^2+1)$$ be a family of polynomials in $
\mathbb F_p[t]$ in the parameter $a\in \overline {\mathbb F}_p$.
Let $$\sum c_{\alpha}x^\alpha := f_a(x)^{\frac{p-1}{2}} $$ and note that the $c_\alpha$ are polynomials in $a$. Now make the $4
\times 4$ matrix whose $(i,j)$th entry is $c_{pi-j}$ for $1\leq i,j \leq 4$. Then I want to show that this matrix is non-invertible for values of $a$ other than $a=2,-2$. In other words, I want to show that the determinant $d(a)$ has irreducible factors other than $(a-2)$ and $(a+2)$.
What I have managed to prove is that $d(a)$ is a polynomial of degree $\frac{3(p-1)}{2}$ and also that $(a-2)$ and $(a+2)$ occur with the same multiplicity in the factorization of $d(a)$ in $\mathbb F_p[a]$. So for a contradiction, suppose $d(a)=\alpha(a^2-4)^{\frac{3(p-1)}{4}}$ for some $\alpha\in \mathbb F_p$. Then, the ratio of the leading term and the constant term must be 1. The leading term is given by the square of the right hand side in the top most equation and the constant term is given by the square of the left hand side. The other coefficients are very difficult to extract and it is lucky that I even managed to get the leading and constant terms. So I am hoping that if by some combinatorial argument I can rule out this congruence, then I will have proved that the determinant has to have an irreducible factor other than $(a-2)$ and $(a+2)$.

@JyrkiLahtonen it's been a while but yes the presence of that last factor is critical. For some more information, this is about superspecial hyperelliptic curves in the family $y^2 =x(x^4-1)(x^4+ax^2+1)$. I was trying to prove that for all primes $p = 1,7 \mod 8$, and $p>7$, there is a superspecial curve in this family.

You can find a bunch of congruences of similar shape in Binomial coefficients and Jacobi sums, by Hudson and Williams.

I still wonder what the original problem is? The exponent in $f_a(x)^{(p-1)/2}$ somehow suggests a character sum involving a quadratic character, but I may be off :-)

Just to check that I have the right determinant. When $p=17$, the determinant of that 4x4 matrix is $$8(a+2)^8(a-2)^8(a^2-5)^4,$$ and the presence of that last factor is crucial for your purpose?

@mathworker21 Per your request, I have added some additional information.

@RKD can you say the significantly more abstract thing you want to prove? that might actually help to prove the thing you asked as your question

(Added for completeness of comment: I wrote some code to check up to $10^7$ - didn't find anything)

(cont.) My point being that strong computational evidence would require that one examine way more than the primes up to $10000$; I got Mathematica to go up to $244737$ and found nothing, but computing the binomial coefficients seems to be rather expensive (...though maybe less so if I could tell Mathematica to do this computation mod $p$, though I still don't know of any methods better than the usual formula, which uses $O(p)$ multiplication mod $p$ - which is really bad if you want a long search)

A heuristic note: if we let $P'_n$ be the $n^{th}$ prime congruent to $1$ mod $16$ and if we chose two numbers $(a_n,b_n)$ at random mod each $P'_n$, the probability that no solution to $a_n=\pm b_n$ exists in the first $N$ terms is bounded below by $$\prod_{n=1}^{N}(1-2/P_n).$$ This product is zero in the limit, suggesting a counterexample exists unless a pattern does - but of the primes up to $10^8$, this product is still $0.619$ - so even if we checked that far on a random sequence, we would still be less likely to find a $a_n=\pm b_n$ pair than not to.

I have no clue. They seem to be completely at random.

Do you have any clue what LHS/RHS (modulo $p$) might be equal to?


$f\colon I\rightarrow G$ and Gromov $\delta$-hyperbolicity

Please recall that $\left|\int_0^1 f(t)\,dt -w\right|\leq \int_0^1|f(t)-w|\,dt$. In general, let $(X,d)$ be a metric space. Given a function $f:I\to X$ let $m_f\in X$ be such that $d(m_f,w)\leq \int_0^1 d(f(t),w)\,dt$ for every $w\in X$. If such $m_f\in X$ exists then we call $f$ as $D$-integrable with $D$-integral $m_f$. ($I=[0,1]$.)
Question: Given a finitely generated group $G$ (with a usual word distance), does there exist a condition on the set of $D$-integrable functions $f: I\to G$ (which uses $D$-integrability) for the Gromov $\delta$-hyperbolicity of $G$? I.e, what is the characterization of hyperbolicity in terms of $D$-integrability?
(It seems to me that the real question here should be: how can one extract group theoretic intel from $D$-integrability?)


Lowest dimensional faithful representation of a finite group

How does one compute the lowest dimensional faithful representation of a finite group?
This question originated in the context of given a finite group $G$: trying to find the lowest dimensional shape whose rotational/reflection symmetries form $G$. (Formally stated as finding the lowest dimensional faithful representation of $G$ into the orthogonal group $O(n)$.)
Now just listing out examples is hard, because even in just $\mathbb{R}^2$ outside of the symmetries of simple polygons we see things like $\mathbb{Z}_2 \times \mathbb{Z}_2 $ crop up, yet it appears that $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ doesn't have any obvious two dimensional representations. Grouppropswiki doesn't even have a single representation for it at all.
So I thought it was wise before I tackle the orthogonal group question I should know how to just, in general, find a low dimensional faithful representation. Some trivialities are that the dimension for a group $G$ will be less than or equal to the smallest $j$ such that $G \subset S_j$ since each symmetric group can be realized as the symmetries of a $j$-dimensional simplex space due to Burnside.
But this doesn't say much because even something like $\mathbb{D}_{\text{Graham's Number}}$ can be realized in $\mathbb{R}^2$ (and we still are dealing with isometries here, what about represenatations that transcend that!?)

For future visitors: some research developments since this question was asked mathoverflow.net/questions/351938/… as well it is interesting to ask even whether there are small irreducible representations, as discussed here: mathoverflow.net/questions/400864/… Finally, note that $j$ in the question statement is called the minimal degree in group theory; computing is still open in many cases.

Is there a list of these smallest dimensions somewhere for small groups? @DerekHolt

It's a difficult problem in general over any field. For an abelian group with invariant factor decomposition $C_{k_1} \times \cdots C_{k_r} \times C_2^s$ with each $k_i > 2$, I think the smallest dimension of a real faithful representation is $2r+s$. It is certainly $s$ for $C_2^s$.

I'm focusing on real representations. Does that make matters worse?

Are you talking about complex representations, or representations over an arbitrary field? For complex representations, the smallest dimension of a faithful representation of a finite abelian group with minimal generating set of size $d$ is $d$. But $C_2 \times C_2 \times C_2$ has a faithful $2$-dimensional representation over the finite field of order $8$. In general, this is a difficult problem - even finding the smallest degree faithful permutation representation can be difficult.


Dividing the whole into a minimal amount of parts to equally distribute it between different groups.

Suppose we have a finite amount of numbers $x_1, x_2, ..., x_n$ ($x_i\in\mathbb{N}$) and an object that should be divided into parts in such a way that it can be without further dividing distributed in $x_i$ equal piles for any $1\leq i\leq n$.
The question is what is the minimal amount of parts for some $x_1, x_2, ..., x_n$ that would allow us to do that?
Example: we have a cake and expect some friends, but we don't know for sure how many of them are coming. But somehow we know, that the total amount of people will be either 5 or 6 $(x_1 = 5, x_2 = 6)$. So we divide a cake beforehand into $5\times 6 = 30$ parts, and so in each scenario we can just give the guest his share of the cake.
Optimization. Of course, 30 parts is too many. In the example above we could just divide the cake into 5 parts, then ignore it and divide the cake (like it was whole) into 6 parts. First we do 4 cuts, second we do 5 cuts; they don't coincide because 5 and 6 are coprimes. Total 9 cuts gives us 10 pieces. So instead of cutting the cake in 30 pieces, we can only cut it in 10. And that is the minimum here.
Simple case. Let $x_1$ and $x_2$ be two coprime numbers. Then, using the same argumentation as above with the case $x_1=5, x_2=6$, we can note that the minimal amount of parts is $(x_1-1)+(x_2-1)+1$ which is equal to $x_1+x_2-1$.
Generalization. First I thought that for $n$ coprime numbers the solution is simple and beautiful:
$$
\sum_{i=1}^n (x_i-1) + 1
$$
But this formula is wrong for $n>2$ as shows the following counterexample.
Counterexample. For $x_1=3, x_2 = 4, x_3 = 5$ the formula gives the answer 10. But that's not the minimum, as we can divide at least in nine portions:
$$
\frac{1}{60} + \frac{2}{60} + \frac{4}{60} + \frac{5}{60} + \frac{7}{60} + \frac{8}{60} + \frac{10}{60} + \frac{11}{60} + \frac{12}{60} = 1
$$
so we can split it equally in 3 pieces:
$$
\left(\frac{1}{60} + \frac{2}{60} +  \frac{7}{60} + \frac{10}{60}\right) + \left(\frac{4}{60} + \frac{5}{60} +\frac{11}{60}\right) + \left(\frac{8}{60} + \frac{12}{60}\right) = 1
$$
in 4 pieces:
$$
\left(\frac{1}{60} + \frac{2}{60} +  \frac{12}{60}
\right) + \left(\frac{4}{60} +\frac{11}{60}\right) + \left(\frac{5}{60} + \frac{10}{60}\right) + \left(\frac{7}{60} + \frac{8}{60}\right) = 1
$$
in 5 pieces:
$$
\frac{12}{60}+\left(\frac{1}{60} +  \frac{11}{60}
\right) + \left(\frac{2}{60} +\frac{10}{60}\right) + \left(\frac{4}{60} + \frac{8}{60}\right) + \left(\frac{5}{60} + \frac{7}{60}\right) = 1
$$
To sum up, the posed question has an evident answer in the case $n=1,2$, but in the case $n=3$ (and higher) I didn't discover any pattern.
[Edit] Related Topics:
Minimum Cake Cutting for a Party
https://puzzling.stackexchange.com/questions/19870/nine-gangsters-and-a-gold-bar
https://mathoverflow.net/questions/214477/minimal-possible-cardinality-of-a-a-1-a-k-distributable-multiset

Note that the same question has been asked at math.stackexchange.com/questions/1383406/…

OK, so for $n$ coprime numbers, we can always do it with $1+\sum(x_i-1)$ pieces, and the question is, under what circumstances can we do it with fewer, and how many fewer. So I wonder whether the $3,4,5$ construction generalizes to, say, $3,3k+1,3k+2$, or to $2k-1,2k,2k+1$, or some other infinite family of examples.

@rywit, but if we shift a little the second series of cuts like this (dividing with shift), we will use one additional cut but at the same time we will be able to assemble one-third parts and no more cuts will be needed.

@rywit, about the counterexample. Dividing into 3, 4 and 5 part using the formula in the question (let's call it the naive formula), can be graphically presented like this: naive dividing. When we divide the object into 5 parts and then into 6, it can be pictured like this (numbers are sizes of pieses in one-sixtieth portions): first two series of cuts. We cannot assemble any one-third parts using this pieses.

@GerryMyerson, at least that small (you can make it as big as you want by dividing into $kx_1x_2$ parts with any positive integer $k$). In order to divide into $x_1$ parts we do $x_1 - 1$ cuts. To divide into $x_2$ parts we do $x_2-1$ cuts. Number of pieces is the number of cuts plus 1, so we have $(x_1 - 1)+(x_2-1)+1 = x_1 +x_2-1$. Geometrical point of view shows that this is minimum, but that is more intuitive approach than a strict proof; that's why I wrote "seems to be".

@rywit, I think we're trying to minimize the number of pieces, rather than the number of physical cuts.

When you write, in the "simple case", that the answer "seems to be" $x_1+x_2-1$, do you mean you can prove it's at least that big? or do you mean you can prove it's at least that small? or what?

Also, in your example with 5 and 6 people you say you make 4 cuts and then separately 5 cuts for a total of 9. But you could align the first cut in each set. So you could make 4 cuts (for 5 people) and then 4 more cuts (reusing one of the existing cuts) for 6 people. Thus a total of only 8 cuts is required in this case. Am I right?

How did you discover this counterexample?


Circles of radius $1, 2, 3, ..., n$ all touch a middle circle. How to make the middle circle as small as possible?

Non-overlapping circles of radius $1, 2, 3, ..., n$ are all externally tangent to a middle circle.

How should we arrange the surrounding circles, in order to minimize the middle circle's radius $R$?

Take $n=10$ for example.

On the left, going anticlockwise the radii are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, and $R\approx 10.77$.
On the right, going anticlockwise the radii are $10, 2, 9, 4, 7, 6, 5, 1, 8, 3$ and $R\approx 9.98$.
How should the circles be arranged to minimize $R$?
Here is a desmos graph where you can try different arrangements for the case $n=10$.
My attempt
For general $n$, call the radii of the surrounding circles going anti-clockwise, $r_1, r_2, r_3, ..., r_n$.
Draw line segments from the centre of the middle circle to the centre of each surrounding circle. So we have $n$ angles at the centre of the middle circle. Each of these angles can be expressed in terms of $R, r_k, r_{k+1}$, using the law of cosines. The sum of the $n$ angles is $2π$. So we have:
$$\sum\limits_{k=1}^n \arccos{\left(\frac{(R+r_k)^2+(R+r_{k+1})^2-(r_k+r_{k+1})^2}{2(R+r_k)(R+r_{k+1})}\right)}=2\pi \text{ (where }r_{n+1}=r_1)$$
This simplifies to:
$$\sum\limits_{k=1}^n \arccos{\left(1-\frac{2r_k r_{k+1}}{(R+r_k)(R+r_{k+1})}\right)}=2\pi$$
We want to assign each $r$ a unique value among $1, 2, 3, ..., n$ so that $R$ is minimized. But how? Is there a general pattern?
Conjectured answer
I think the following general procedure will make the middle circle as small as possible. Take $n=10$ for example. We ignore the $1$ at first; it will be placed last. First put down the numbers $10, 9, 8, 7, 6$ in pyramid-fashion, from top to down and from left to right, like the red numbers below. Then put down the numbers $2, 3, 4, 5$ in pyramid-fashion also, but between the previous rows, like the blue numbers below.

This gives the order of radii going around the circle: $7, 4, 9, 2, 10, 3, 8, 5, 6$. Then put the circle with radius $1$ anywhere you like, as long as it fits without disturbing other circles.
For larger values of $n$, there may be multiple small circles that can fit between other circles without disturbing them. Put those last.
Intuitively, this procedure minimizes the tendency of the large surrounding circles to take up space around the middle circle. You can see this in the diagrams above with $n=10$, looking specifically at the two largest surrounding circles in each case.
Anyway, it's just a conjecture for now. Notice that this conjecture implies that, for $n=10$, the arrangement I showed above on the right, yields the minimum radius of the middle circle. I don't think the middle circle can be any smaller.

I wrote a program that loops over all permutations of 2...10 (excluding the circle of radius 1), and finds an $R$ that solves the equation that you wrote (the sum of the arc-cosines). It confirmed that the optimal permutation is the one that you conjectured: 2, 9, 4, 7, 6, 5, 8, 3, 10, where the radius of the inner circle is ~9.979907. I had to exclude the circle of radius $1$ from the sum of course.

It seems the right drawing the small circle may be removed without impacting the result.

@VarunVejalla Good point. I took this point into account in my desmos graph, but forgot to mention this in my question. We should ignore the smallest few circles, because they will each fit between some two other circles that touch each other. How many circles should be ignored, depends on $n$, with an upper limit of $n/4$ (based on Descartes' Circle Theorem).

The problem with the current sum of $\arccos$ is that you can have gaps between circles. For example, in the arrangement on the right, there would have to be some wiggle room for the circle of radius $1$ between the circles of radius $5$ and $8$ (i.e. it can't be tangent to both of those and the middle circle), and numerically solving for $R$ using the sum of $\arccos$ yields $\approx 9.90$, different from the $\approx 9.98$ that it would actually be.

@Dan Yes, absolutely. Actually, I think there's no point trying that. For small numbers (say, $n=7,8,9,10$) one might solve the equation numerically for all the possible combinations of $r_k$. Perhaps that gives a clue on how the $r_k$ must be arranged for general $n$.

@Dog_69 That approach is very difficult due to the $\sum$, I think.

First thing tthat comes to my mind is to take the derivative with respect to $R$ and see the condition for it to vanish.


A curious identity on $q$-binomial coefficients

Let's first recall some notations:

The $q$-Pochhammer symbol is defined as
$$(x)_n = (x;q)_n := \prod_{0\leq l\leq n-1}(1-q^l x).$$

The $q$-binomial coefficient (also known as the Gaussian binomial coefficient) is defined as $$\binom{n}{k}_q := \frac{(q)_{n}}{(q)_{n-k}(q)_{k}}.$$

I found the following curious identity on $q$-binomial coefficients, and I would like to ask for some ideas on how to prove it.
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} (q)_n$$
I am familiar with the $q$-binomial theorem and $q$-Vandermonde identity, which I think might be useful. If there are some other well-known identities on $q$-binomial coefficients that might be helpful in figuring this out, I would like to know them too. Thank you!

Some thoughts:

The classical version of this identity is simply $$\sum_{0\leq j\leq 2n}(-1)^{j}(2n+1-j)\binom{2n}{j} = \sum_{0\leq j\leq 2n}(-1)^{j}(j+1)\binom{2n}{j} = \delta_{n,0},$$
which can be proved simply by $$\sum_{0\leq j\leq 2n}(-1)^{j}\binom{2n}{j} + \sum_{0\leq j\leq 2n}(-1)^{j}j\binom{2n}{j} = (1-x)^{2n}+\frac{d}{dx}(1-x)^{2n} \bigg\vert_{x=1} = \delta_{n,0}.$$
There is a natural $q$-analogue of the above identity, obtained by replacing $(1-x)^{2n}$ by $(x)_{2n}$ and the derivative by the $q$-derivative, but unfortunately this is not the desired identity.

There seems to be other closely related identities:
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n-1)+\frac{j(j+1)}{2}}\binom{2n}{j}_q \overset{?}{=} \delta_{n,0},$$
$$\sum_{0\leq j\leq k\leq 2n}(-1)^{j}q^{k(k-j-n+1)+\frac{j(j-1)}{2}}\binom{2n}{j}_q \overset{?}{=} q^{n}(q)_{2n}.$$
Proving these other identities (or a family of similar identities) might be helpful if we were to try to use induction on $n$.

So we begin with a combinatorial interpretation : Consider all $n$-tilings using $k$ green tiles and $n-k$ red tiles, such that every green tile's weight is $q^r$ where $r$ is the number of red tiles preceding this green tile. The weight of a tiling is the product of weights of all green tiles, and the sum over all such tilings of these weights equals $\binom{n}{k}_q$. We get that the LHS is a sum over all tilings of length $2n$ and $k \geq $ the number of green tiles, of sth. involving $k$ and the green tiles. Using $(-1)^j$ and some work, cancellations can be figured out that eliminate $k$.

I managed to obtain a partial explanation using combinatorics. I don't have a full answer to fall back upon ,but I eliminated the role of $k$ and figured out some further cancellations using the ideas. I don't want to post a partial answer (I'd rather it be complete and worthy of evaluation), but if someone's interested I'll put the details up in the next comment(s).

It might be useful to post the same question on Mathoverflow. It is definitely worth it.

@Henry That is nice. I will have a look today at it. The quadratic form on the exponent is  unusual. Nice identity tho!

@Phicar Okay, I have turned the desired identity into a single-sum identity and was able to verify it using qZeil. This is nice, and thanks for the suggestion. Still, it would be nice to have a non-algorithmic proof of the identity.

@SarveshRavichandranIyer Unfortunately I am not aware of any useful reference either.

@Phicar Can qZeil handle double summations? I thought it was mostly for single summations.

Can you provide some source for these identities? I had a general look and unfortunately there's very little on double-sum identities that I found lying around for the $q$-binomial.

Perhaps $q-$Zeilbergs gets you a nice recursion? risc.jku.at/sw/qzeil


What is $\int_0^1 \left(\tfrac{\pi}2\,_2F_1\big(\tfrac13,\tfrac23,1,\,k^2\big)\right)^3 dk$?

As in this post, define the ff:
$$K_2(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac12,\tfrac12,1,\,k^2\right)}$$
$$K_3(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac13,\tfrac23,1,\,k^2\right)}$$
$$K_4(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac14,\tfrac34,1,\,k^2\right)}$$
$$K_6(k)={\tfrac{\pi}{2}\,_2F_1\left(\tfrac16,\tfrac56,1,\,k^2\right)}$$
We find that,
$$\int_0^1 K_2(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;1,\tfrac32;1\right)}=2G$$
$$\int_0^1 K_3(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac13,\tfrac23;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}2\, \ln2$$
$$\int_0^1 K_4(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac14,\tfrac34;1,\tfrac32;1\right)}=2\ln(1+\sqrt2)$$
$$\int_0^1 K_6(k)\, dk = {\tfrac{\pi}{2}\,_3F_2\left(\tfrac12,\tfrac16,\tfrac56;1,\tfrac32;1\right)}=\tfrac{3\sqrt3}4\, \ln(2+\sqrt{3})$$
where $G$ is Catalan's constant. However, this post gives a third power,
$$\int_0^1\big(K_2(k)\big)^3 dk=\frac35 \bigg(\tfrac{\pi}2\,_2F_1\big(\tfrac12,\tfrac12,1,\tfrac12\big)\bigg)^4=\frac35 \big(K(k_1)\big)^4 = \frac{3\,\Gamma (\frac14)^8}{1280\,\pi^2} \approx 7.0902$$
where $K(k_d)$ is an elliptic integral singular value while YuriyS in his comment below gives,
$$\int_0^1\big(K_3(k)\big)^3 dk \approx 6.53686311168760876289835638374 $$

Questions:

What are the closed-forms of:
$$\int_0^1\big(K_n(k)\big)^m dk=\,?$$
for power $m=2$ or $m=3$?
Or at least their numerical evaluation up to 20 digits?

P.S. My old version of Mathematica can't evaluate it with sufficient precision, nor does WolframAlpha. (Enough digits may make it amenable to the Inverse Symbolic Calculator.)

At one point, I convinced myself that dlmf.nist.gov/15.5 (at least a couple of them) worked for fractional derivatives.  If so, then the order might be reduced To 1F1(a,b;k^2), which might be easier to evaluate (and invert).  If you're still interested, I might take a stab at that.  No warranties implied :)

A related case with a closed form in this question: math.stackexchange.com/q/3325678/269624

@YuriyS: I revised the post to give a clearer version of the problem.

Mathematica gives $I_3=6.53686311168760876289835638374$ with WorkingPrecision->30 $$ $$ It has no problem evaluating the integral, only takes a few seconds

@YuriyS: The closed-form is not the titular integral $I_3$. Note the closed-form involves $\frac12,\tfrac12$, while the titular integral $I_3$ involves $\frac13,\frac23$. P.S. If you can provide a numerical evaluation of $I_3$ for 20 digits or more, that would be nice.

Could you clarify: does your 2nd question ask for numerical evaluation of the titular integral? Or the general integral from the 1st question? Above you have already provided the closed form for the titular integral, so I'm confused

Please use \left and \right rather than \bigg, especially in titles.


Is there a group $G$ for which $\mathrm{Aut}(G) \simeq (\mathbb{R},+)$?

I know the classic theorem that $(\mathbb{Q},+)$ cannot be expressed as an automorphism group, i.e. there is no group $G$ such that $\mathrm{Aut}(G)\simeq (\mathbb{Q},+)$.

Theorem A. If $L$ is a locally cyclic group with no element of order $2$, then $L$ cannot be expressed as an automorphism group.

But how about $(\mathbb{R},+)$?
I think the answer might be no, and the proof proceeds by showing that if $\mathrm{Aut}(G) \simeq \mathbb{R}$, then some subgroup or quotient $H$ of $G$ will satisfy $\mathrm{Aut}(G) \simeq \mathbb{Q}$, which contradicts Theorem A.
So how can we construct this $H$? Assuming the axiom of choice, we can say $\mathbb{R} = \mathbb{Q}\oplus B$ for some additive subgroup $B$ of $\mathbb{R}$ (just by picking a $\mathbb{Q}$-basis for $\mathbb{R}$). Now I'm tempted to do some Galois-type thing, where you use a "fixed" subgroup $$H=\mathrm{Fix}(B)=\{g\in G : b(g) = g \text{ for all } b\in B \},$$ and then try to say something about $\mathrm{Aut}(H)$ or $\mathrm{Aut}(G/H)$ (where, for the latter, maybe we take the normal closure of $H$). But I can't complete this line of reasoning. It seems key that $\mathrm{Aut}(G)$ splits as a direct sum --- that seems special.
Am I just totally off-base here? Is there some obvious group $G$ whose automorphism group is $\mathbb{R}$?
Related questions
If I can prove that $\mathbb{R}$ is not an automorphism group, then the same would hold for the isomorphic group $\mathbb{R}_{>0}$ of positive reals under multiplication. But how about $\mathbb{R}^\times \simeq \mathbb{R}_{>0}\times C_2\simeq \mathbb{R}\times C_2$ --- can this be achieved as an automorphism group? If $\mathrm{Aut}(G)$ is a direct product of abelian groups, what can be said about $G$?
A related observation is that if $\mathrm{Aut}(G)$ is abelian, then $G$ is nilpotent of rank $\leq 2$. So there is a sense in which $G$ is "almost" abelian.
I know $\mathbb{Q}^\times$ is the automorphism group of $(\mathbb{Q},+)$, and that $\mathbb{R}^\times$ constitutes the continuous automorphisms of $(\mathbb{R},+)$ ... Then there's multiplicative/additive groups of other rings/fields ...
I can't even resolve this question in the case of $\mathrm{Aut}(G) \simeq \mathbb{Z}\times \mathbb{Z}$. For this one I can show that $G'$ would necessarily by cyclic, but that's about it.
I'm very curious about which types of groups can be achieved as automorphism groups.
A quick observation
If $\mathrm{Aut}(G) \simeq \mathbb{R}$, then $G$ must be infinitely generated (or else its automorphism group is countable) and nilpotent of rank $\leq 2$ (this follows for any group whose automorphism group is abelian).

Thanks, I did already know that! Same idea to prove that $\mathbb{Q}$ is not an automorphism group.

Partial answer: $G$ cannot be abelian. If $G$ is abelian, then $x\mapsto x^{-1}$ is an automorphism of $G$ of order $\leq 2$. $\mathbb{R}$ has no nonidentity elements of finite order, so $x=x^{-1}$ for all $x\in G$. This implies that $G$ is the underlying abelian group of an $\mathbb{F}^2$-vector space $V$, and we have $\operatorname{Aut}(G)=\operatorname{Aut}_{\mathbb{F}_2}(V)$. We must have $\dim V\geq 2$, but this means $\operatorname{Aut}(G)$ contains $\operatorname{GL}_2(\mathbb{F}_2)$ as a subgroup. This would make  $\operatorname{Aut}(G)$ nonabelian, contradiction!

>it's consistent with ZF that every endomorphism of R is measurable  --- Wow.

@Ehsaan: re: your related questions, it's consistent with ZF that every endomorphism of $\mathbb{R}$ is measurable, hence continuous; if this is true then $\text{Aut}(\mathbb{R}) \cong \mathbb{R}^{\times} \cong \mathbb{R} \times C_2$. (With the axiom of choice $\text{Aut}(\mathbb{R})$ is much larger and in particular nonabelian.)

@DietrichBurde thank you --- that's related for sure. I wonder which if there are groups $A$ for which it is true, but difficult, to prove that $A$ cannot be achieved as an automorphism group.

Yes, it would have to be some special property of $\mathbb{R}$ that comes up here. Subgroup or quotient would be OK here --- just as long as we can use the fact that $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ to reduce to the case of $\mathbb{Q}$ (which is a known exercise).

It's certainly not the case 'generically' that if $H\subset Aut(G)$ then there's some subgroup $G'$ of G with $Aut(G')\equiv H$, even if $Aut(G)=H\times K$; consider the automorphism group of $C_p$ for $p$ prime. I don't see anything in your reasoning offhand that uses any properties of $\mathbb{R}$ that don't mirror to this case.


If $r\in\mathbb{Q}\setminus\mathbb{Z}$ is it possible that $r^{r^{r^r}}\in \mathbb{Q}$?

It's straightforward to prove that $r^r\notin\mathbb{Q}$, which furthermore allows us to use the Gelfond-Schneider theorem to prove that $r^{r^r}\notin\mathbb{Q}$.

Is it true that $r^{r^{r^r}}\notin\mathbb{Q}$?

It seems like it ought to be true, though my guess would be that this is an open problem. If not, does anyone know a proof? And if so, do there exist any discussions of it in the literature?

en.wikipedia.org/wiki/… - the techniques for the proof by infinite descent here can be generalized in a straightforward way

@DarkMalthorp how do u prove $r^r \not \in \mathbb{Q}$?

Yes, indeed it is a stronger result :) Of course, I fully expect that $r^{r^{r^r}}$ is also transcendental.

@darkmalthorp For the record, Gelfond-Schneider says that $r^{r^r}\notin\overline{\Bbb Q}$, not just $\notin\Bbb Q$ (though I presume you meant that).

I would also guess that this is an open problem. It is even unknown whether the power-tower is an integer for $\ r=\pi\ $ (which would however be a miracle)