Suppose one starts with an (infinite) multiset of positive integers $\mathcal{A} = \{a_i\}_{i\geq 0}$ such that:

$1=a_0\leq a_1\leq a_2\leq\ldots$

Can one always find a (necessarily infinite) group $G$ such that the set of degrees of its finite dimensional irreducible complex representations is exactly $\mathcal{A}$?

Clearly the answer to the above question is no for arbitrary $\mathcal{A}$; for example if $a_1 = N>1$ (i.e. the trivial representation is the only 1-dimensional representation of the potential group), then a simple argument involving the decomposition of the tensor square of this $N$-dimensional representation shows that $N\leq a_2\leq \frac{N^2+N}{2}$. Using more general forms of such arguments involving tensor powers of decompositions of $GL_n(\mathbb{C})$-representations, one can find other restrictions ruling out certain potential degree sequences.

$\bf{Question:}$ Given $\mathcal{A}$ such that there are no obstructions arising from tensor power considerations, is it always possible to find/construct a group $G$ whose irreducible degree sequence is exactly $\mathcal{A}$?

I assume that in general this is still a hard question, so I am also interested in partial results. I would also be interested in negative results along the lines "Even though this sequence $\mathcal{A}$ has no obstructions, it still cannot be the degree sequence of any group for some deeper reason."

  • $\begingroup$ I have a feeling that if you allow your group to have many quotients, there are not so many obstructions, so it is my feeling that this question is more interesting if you require the group to be simple... $\endgroup$ – Igor Rivin Sep 12 '11 at 19:56
  • $\begingroup$ @Igor: Having many quotients is one way to help achieve a given $\mathcal{A}$, but at the same time it introduces obstructions of the form $\it{every}$ irrep dimension of a quotient must appear in $\mathcal{A}$. On the other hand, requiring the group to be simple also can introduce obstructions (most notably that a simple group cannot have a 2-dimensional irrep), so it seems that the answer in general will be between these two extremes. $\endgroup$ – ARupinski Sep 12 '11 at 21:56

There are lots of results like this, but they're pretty piecemeal. For example, the number of 1-dimensional representations divides the size of the group. There are some results about how large the largest dimensions can be (see Durfee & Jensen (J. Alg, 2011) for the most recent results). There are also some results about the small end of the distribution (e.g. if you have a 2-dimensional representation then you can use the classification of subgroups of SU(2) to get info).

In general if you write down a sequence that passes all the obvious tests then no one has any idea whether a group exists which realizes that.


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