User jon cohen - MathOverflow

Character table entries and sums of roots of unity

2013-03-06T09:53:38Z

It is well-known that the entries of the character table of a finite group are sums of roots of unity.

Question: Is the converse true? Explicitly, given $z\in \mathbb{Z}[\mu_\infty]$, can I find a finite group $G$ and irreducible character $\chi$ with $z=\chi(g)$ for some $g\in G$?

It's certainly true if I dropped the irreducibility; in this case one can take $G$ to be abelian, and just hit each summand of $z$ one at a time and take a direct sum. Also noteworthy is that if one has a single $G$ one can compose with a surjection to $G$, so it will be true for infinitely-many groups if it is for one.

A related question (possibly less trivial): If it's YES, is there a natural "minimal" subclass of finite groups that suffice for this purpose?

If it's "NO," are the obstructions completely understood?

This is partially meant as a (hopefully) easier relative to a previous question: http://mathoverflow.net/questions/90085/a-realization-problem-for-character-tables. I apologize in advance if it is trivial (one way or the other), as I fear it may be.

Can sine be made into a homomorphism?

2013-02-26T22:58:34Z

Consider the usual sine function $\mathbb{R}\rightarrow \mathbb{R}$. Is there some (single) group structure we can put on $\mathbb{R}$ with respect to which sine becomes a homomorphism?

I suspect the answer is either no for a trivial reason, or yes by a simple set-theoretic argument (probably providing a great many such group structures of no interest).

This latter seems plausible if I can "replace" the reals (and the sine function) by some arbitrary equal-sized set (and sufficiently similar function). Indeed, if I only asked that we have a pair of group structures $*_1$ and $*_2$ so that sine is a homomorphism from one to the other, then such an argument does go through (if I'm not mistaken).

So I'm putting this question forward mostly in case there's a pleasant surprise. Depending on the answer, one could of course ask for further restrictions (abelian, continuous (probably impossible), torsion, torsion-free etc), but for now I'll leave as is.

How many ways can a number be Fortunate?

2012-09-05T02:49:53Z

An integer $m$ is Fortunate if it can be written as $q-P$, where $P$ is a primorial and $q$ is the smallest prime greater than $P+1$. It is conjectured that Fortunate numbers are always prime.

It is easy to see that there are only finitely many possible primorials $P$ for which a given $m$ can be decomposed in the above manner (this is because $m$ must be greater than the largest prime dividing $P$).

QUESTION: Is the number of such representations of an integer $m$ uniformly bounded above?

Irreducibility of Coxeter Graphs as a Function of Generating Sets

2013-01-17T23:29:51Z

Given a Coxeter system $(W, S)$, we can form its Coxeter graph, and say that the system is irreducible if the graph is connected. Now, irreducibility is not solely a function of $W$; it depends also on $S$. E.g., for $W=D_{12}$ we could have $|S|=2$ or $3$, giving the systems $G_2$ or $A_1 \times A_2$.

Question: For which (other) irreducible $(W,S)$ is there an $S'$ with $(W,S')$ reducible? When this phenomena occurs, does it always change $|S|$ as above? In particular, is it obvious one way or the other for $W=S_n$?

I think this should at least be answerable for finite Coxeter groups since these are completely classified, but I'd also be happy to hear examples in the infinite case (e.g. affine Weyl groups or hyperbolic Coxeter groups). The systems for which it cannot be done would seem to be in some sense "more irreducible," or at least the irreducibility is more purely algebraic than geometric.

Can we promote to a Lie Group Isomorphism?

2012-11-26T22:30:39Z

We regard an isomorphism of Lie groups to mean a group isomorphism which is simultaneously a diffeomorphism of the underlying smooth manifold. I'm wondering about how much rigidity is imposed by this definition.

Question: If we have maps $f, g: G\rightarrow H$, where $G, H$ are Lie groups, $f$ is an abstract group isomorphism, and $g$ is a diffeomorphism, must $G$ and $H$ be isomorphic as Lie groups?

I think there should be a (possibly easy) counterexample, but neither I nor the professors I've asked could immediately find one.

EDIT: assume $G$ and $H$ are connected.

Cyclic Subgroups of the Symmetric Group

2012-11-11T22:13:30Z

If we write a partition $n=k_1+...+k_r$, then we can create a $(k_1,...,k_r)$-cycle in $S_n$ with order equal to the least common multiple of the $k_i$'s. It is clear that every cyclic subgroup will arise this way, by considering the cycle type of a generator. So this appears to give a classification of which cyclic subgroups can occur in $S_n$, namely $Z/mZ \hookrightarrow S_n$ iff $m$ is the lcm of numbers whose sum is $\leq n$. I'd like a (computationally at least) cleaner criterion. So...

$\bullet$ What are the (orders of) maximal cyclic groups occurring in $S_n$ (under containment)? In particular, what is the largest-order cyclic subgroup in $S_n$?

In light of the above, the latter question amounts to asking for a formula for the maximum value (over all partitions) of the function from partitions of $n$ to $\mathbb{N}$ which outputs the lcm of the terms in the partition. The former question asks for local maximum values (under divisibility) of this function. Statements about the partitions giving rise to these values would also be of interest.

In a slightly different direction, I also wondered about the following:

$\bullet$ How many different (up to iso) cyclic subgroups are in $S_n$? This is deducible from an answer to Q1 in terms of the number-of-divisors function, but I'd be interested in arguments that go a different route.

Recovering a group from its number of symmetric embeddings?

2012-10-16T22:03:41Z

Fix a finite group $G$, and let $f_G:\mathbb{N}\rightarrow\mathbb{N}$ be defined by setting $f_G(n)$ to be the largest $m$ such that $S_n$ contains $m$ disjoint (pairwise-intersecting at 1) copies of $G$. In other words, $f_G(n)$ is the optimal number of $G$'s that can be packed into $S_n$ without unnecessary overlap. Clearly $f_G$ starts with a string of zeros, monotonically increases and is unbounded. It would seem spectacularly ambitious to actually compute this function for $G$ remotely complicated.

Q1 - Does $f_G$ determine $G$?

In case the affirmative does hold generally, I'd also be curious to know if a stronger result is true:

Q2 - Does there necessarily exist an $n$ (depending on $G$) such that $f_G(n)\neq f_{H}(n)$ for all $H$ distinct from $G$? In other words, can we be assured that a finite amount of computation would be sufficient to distinguish $G$ from all other finite groups?

An affirmative answer to Q2 would mean that $G$ is classified by a pair of (hopefully computable) integers.

Both answers are affirmative for groups of (very) small order since they're the only ones that even fit in the first few symmetric groups, but I don't see any hope to prove/disprove them in general.

Distribution of fibers of Euler-Phi

2012-10-02T09:25:44Z

Motivation: It's an elementary exercise to show that the number of solutions to the equation $\phi(x)=n$ is finite for any $n$, where $\phi$ is the Euler-phi function. Of course, counting the number of solutions is another matter.

Question: For a given $n$, can we describe the number of solutions to $|\phi^{-1}(n)|=m$? In other words, I want to know about how often the number of solutions to the equation $\phi(x)=n$ is $m$, for a given $m$, as $n$ varies. Since it wouldn't surprise me if this is way too hard, can we at least say something similar about finiteness?

2 Possible Generalizations of Cayley's Theorem?

2012-09-26T09:45:56Z

I'm wondering about the following 2 generalizations of Cayley's Theorem (every group embeds in a symmetric group). If these are known to be true/false, references would be appreciated.

1) (Weak Version) Given any finite collection of (not necessarily distinct) finite groups, can we embed them simultaneously in a (finite) group so that they have pairwise-disjoint intersection (i.e., intersect only at the identity)?

2) (Strong Version) Given an arbitrary set of arbitrary groups, can we embed them simultaneously in some group so that they have pairwise-disjoint intersection? It seems like this version may run into set-theoretic difficulties; if so an explanation/reference for those would also be welcome.

Note of course that if such a group exists for a given collection, then we can embed it in a symmetric group by Cayley's Theorem (so these are indeed generalizations).

Edit: To make it more interesting / rule out the obvious answer pointed out in the comments, can we arrange the embeddings so that the groups' normalizers are pairwise disjoint?

Why doesn't this group have a name?

2012-09-20T01:37:18Z

$$\{A\in GL_n(\mathbb{C}) : |det(A)|=1\}$$ This seems to me to be a perfectly natural group to study; it is easy to define and contains $U(n), SL_n$, and all the torsion. Is there any good reason why this group isn't among the usual classical groups that are so well-understood and thoroughly discussed? I understand that most of those are studied/defined by looking at groups preserving particular inner products, but it still surprises me that I've never heard of any interesting results/properties of this group. The only guess I currently have is it's not compact.

I could ask the same question with $\mathbb{R}$ but then "morally" the group is just two copies of $SL_n(\mathbb{R})$ so I understand why it's less interesting.

A perfectly acceptable answer is that I'm totally misinformed and this group is perfectly understood, classical, named, etc., in which case any reference would be appreciated.

When is a manifold a tangent bundle?

2012-08-29T16:50:54Z

Given a (smooth) manifold $M$, are there any sufficient, intrinsic properties that would tell you there exists a (smooth) manifold $N$ such that $M$ is diffeomorphic to $TN?$ There are some obvious necessary conditions (like $M$ needs to be even-dimensional), but I'm more interested in whether any sufficient conditions are known.

j-invariant fixed point?

2012-04-12T12:50:38Z

If we view the j-invariant of a lattice as a map from the upper-half plane to the complexes by $\tau\mapsto j([1,\tau])$, then it is surjective, holomorphic, and has quite a number of other wonderful properties (see the third part of Cox's Primes of the Form for a great introductory reference).

My question is: does $j$ have any fixed points? If so, do we know what any/all of them are?

I'm in particular curious what goes into the proof. Specifically, whether the answer is immediate from some complex analysis, or whether you need to have a good handle on $j$ itself (or both!). A professor I asked suggested thinking about $j(\tau)-\tau$ on the compactification of the fundamental domain of $SL(2,\mathbb{Z})$, but we weren't able to clean it up.

Number of Generators of [G,G] when G is free?

2012-08-18T22:35:27Z

If $G$ is a free group on $n>1$ generators, then $[G,G]$ is also free, being a subgroup of a free group. Is there a formula for the number of generators of this free group in terms of n?

Answer by Jon Cohen for Proofs without words

2012-08-16T22:01:48Z

The pathspace of any topological space is contractible.

Pf (as given in my homotopy theory class): slurp spaghetti.

Solubility of the quintic?

2012-04-09T08:22:21Z

Over the p-adics, every Galois group is solvable. Does this imply that the quintic (and higher-order polynomials for that matter) can be solved by radicals over $\mathbb{Q}_p$?

EDIT: The original place I learned that the p-adic galois groups were solvable was in Milne's Algebraic Number Theory text (Chapter 7, Cor 7.59).

As was pointed out the comments, I should clarify that I meant to ask 2 questions. Namely, whether the general quintic can be solved by radicals in this context (still no) and whether any given one can be (which I now believe is yes).

Proof of a Simple Converse in Algebraic Number Theory

2012-04-30T22:49:04Z

If $L/K$ is a Galois extension, then any prime $\mathfrak{p}$ of $K$ splits into a product ${\mathfrak P}_1^e\cdots {\mathfrak P}_g^e$ of primes in $L$, and the exponents on the primes are equal since the Galois group acts transitively on the primes dividing $\mathfrak{p}$.

Question: Is the converse true? Namely, if $L/K$ is an extension such that every prime of $K$ has a single associated exponent in this manner, then is $L/K$ necessarily Galois?

I believe the answer is yes but I was wondering in particular if there was an easy way to see this.

"Stably" finite-fiber maps of the integers

2012-04-12T00:37:49Z

Suppose $\varphi:\mathbb{N}\rightarrow\mathbb{N}$ is a finite-to-one map. We can then define a function $\varphi_1:a\mapsto |\varphi^{-1}(a)|$. If this function is finite-to-one, we can similarly define $\varphi_2=(\varphi_1)_1: a \mapsto |\varphi_1^{-1}(a)|$ and so on.

For example, the identity function on $\mathbb{N}$ would have its first induced function be the constant function 1, and thus would not have a second induced function.

Question: Is there a function $\varphi$ such that $\varphi_n$ exists for all $n$? If so, is there any obvious classification (i.e. necessary and sufficient conditions)?

EDIT: If the answer is yes, then for (possibly) an added challenge, can we make all the $\varphi_n$ surjective?

A Realization Problem for Character Tables

2012-03-02T23:16:01Z

Given an $n\times n$ table of complex numbers, are there known sufficient conditions for the table to be the character table of a finite group? Representation theory gives plenty of necessary conditions, but I can't imagine they'd be enough in general.

Comment by Jon Cohen

2012-11-11T22:30:16Z

Yes, up to abstract isomorphism.

Comment by Jon Cohen

2012-10-18T15:02:10Z

@Derek: couldn't it still be the case that given $n$ there are groups whose sequences agree with $G$'s up $n$?

Comment by Jon Cohen

2012-10-17T02:50:09Z

@Andreas: you're right. I don't know what I was thinking.

Comment by Jon Cohen

2012-09-27T12:49:50Z

@Francois: Thanks for the suggestion; I'm surprised that the infinite case is easier.

Comment by Jon Cohen

2012-09-06T15:36:48Z

Thanks for the correction; I fixed the question so that it doesn't contradict Euclid.

Comment by Jon Cohen

2012-04-12T04:22:20Z

Thanks! Exactly what I was hoping for.