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Is it known how many faithful linear representations a finite group G has on a complex vector space of given dimension? What if G is abelian?

I would even be interested in this special case: the smallest dimension of a faithful representation of an abelian group over C is the number of factors in the group's invariant factor decomposition, how many faithful representations are there of that dimension? It is not just the sum of φ(mi) (where mi are the orders of the "invariant factors") (e.g. Z/2 × Z/2 has 3 faithful two-dimensional representations).

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It's "known" in the sense that if you give me G and n I can look at the character table of G and then work the answer to your question out. And more generally I can write a computer program in magma which will give you an answer. Is this what you mean? I bet it's not. In fact I've never really understood what questions like this "mean" on some level. I am guessing that the comment above is not the answer you want, but it's not clear to me what distinguishes an answer which is in some sense correct but not "the right one" from "the right one". Can you elaborate? –  Kevin Buzzard Nov 5 '09 at 18:55
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As for the second question, I think Z/2 x Z/2 is a counterexample to your guess... –  Kevin Buzzard Nov 5 '09 at 18:59
    
That's not quite the answer I'm looking for (though I'm happy to know it now). My question is related to Jordan's answer to my question about the notion of conductor in a non-abelian number field. I was thinking one could look at the Artin conductors of faithful representations of the Galois group over Q and take the minimal one. If one wanted to count some invariant of number fields in the conductor aspect by varying over such faithful representations one would need to know the multiplicity of overcounting occurring. –  Rob Harron Nov 5 '09 at 21:53
    
Answering this question may not be necessary to deal with the problem I'm talking about, but it certainly seemed like an interesting question whose answer could be helpful. And indeed Z/2 x Z/2 presents a counterexample. Apparently, I should've tried a basic example before putting that guess up there. Thanks. –  Rob Harron Nov 5 '09 at 21:54
    
I think the question has a life of itself now. –  Ilya Nikokoshev Nov 5 '09 at 21:57
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2 Answers

I think that for Abelian groups, there is an explicit formula, but it is not pretty, and probably of limited practicial value in all but the smallest cases. Let the minimal non-trivial subgroups of $G$ be $M_1,M_2,\ldots, M_s.$ In this case, these are just the subgroups of prime order of $G$ (if trying to generalize this for a general finite group, we would look at the minimal normal subgroups). Let $r$ be the dimension we are interested in. We are looking for all characters of the form $\chi = \lambda_1 + \lambda_2 + \ldots + \lambda_r$ with each $\lambda_i$ linear such that $\chi$ is faithful. If $\chi$ is not faithful, there must a minimal non-trivial subgroup $M$ which is in the kernel of $\lambda_i$ for each $i$. So we need to pick the $\lambda_i$ so that for each $j$, there is at least one $\lambda_i$ whose kernel does not contain $M_j$.

For any choice of $j$, there $|G|- [G:M_j]$ linear characters of $G$ which do not contain $M_j$ in their kernels. Let $I = \{1,2,\ldots,s\}$, and for each subset $J$ of $I$ (including the empty set), let $M_J = \langle M_j : j \in J \rangle$.` Notice that $\lambda_i$ contains $M_J$ in its kernel if and only if it contains $M_j$ in its kernel for each $j \in J$. For each choice of of $J$, then there are $[G:M_J]$ choices for $\lambda_i$ which contain $M_J$ in their kernels. It is almost easier just to write the formula down than explain it.

Can't do the latex right, but there are $\sum_{J \subseteq I} (-1)^{|J|}$ ($[G:M_J]$ choose $r$) choices of $r$-tuple of linear characters such that the kernel of the sum is trivial. We start with all possible $r$-tuples, corresponding to $J = \emptyset$, then for each $j$, we subtract all $r$ tuples where the kernel of each $\lambda$ occurring contains some $M_j$. Then we have to put back $r$ tuples where the kernel of each $\lambda$ contains $M_{J}$ for some subset $J$ of cardinality $2$, etc.

An explicit formula for a general group could be done, but it would get rather messy.

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I haven't heard this question before, but I would approach it in a following way: you know some basic relations between irreducible representations of a groups — e.g. their # is the # of conjugacy classes — and let's say we've written the kernels of representations and their dimensions. The generating function (*) that counts all representations is $((1-x^{r\_1})(1-x^{r\_2})\dots (1-x^{r\_k}))^{-1}$ and you're asked to subtract the representations that have nontrivial kernel. This combinatorial problem is partly tractable for a given group, though I don't see a nice closed formula.

The situation simplifies for Abelian groups, and the answer you provide is in the right direction. A semisimple representation of Abelian group is a sum of characters. (I misunderstood what you asked here a bit, but bear with me.) A single character could already be faithful: for example, for $\mathbb Z\_2 \times \mathbb Z\_3 = \mathbb Z\_6$ the irreducible character that hits all 6-roots of unity is a faithful representation. And there are $\varphi(6) = \varphi(3)\cdot\varphi(2) = 2$ of those. Another example: $\mathbb Z\_2\times \mathbb Z\_2$. The same formula doesn't works here as $1\cdot 1 = 1$ but the corresponding representation isn't faithful. For two-dimensional case, you positively get two characters, but there is no reason to expect that there will be a splitting into a character of exactly first summand and exactly second summand. You should, however, be able to calculate it by the method below.

So the formula $\prod\varphi(m\_i)$ will correctly describe the answer only for a case of group having only one invariant factor.

Here's how to make a complete computation for an $n$-dimensional space and the case of $\mathbb Z\_6$. You need to take all representations of it which are neither representations of $\mathbb Z\_2$ nor of $\mathbb Z\_3$. Fortunately, this is easy to write using inclusion-exclusion principle: $(1 -x)^{-5} - (1-x)^{-1} - (1-x)^{-2} + 1$ and you can simplify this or just get the first three terms:
$1+5x+20x^2 + 5x^2 - 1 - x - x^2 - 1 - 2x - 3x^2 + 1 = 0 + 2x + 21x^2 + \dots$ The 2 here stands for 2 "simple" characters of $\mathbb Z\_6$, while 21, if I made no mistake, is the number of 2-dimensional representations: 19 of them are of the type "character as above $\oplus$ anything" and 2 of the type "character of $\mathbb Z\_2$ $\oplus$ character of $\mathbb Z\_3$ ".

Now I think using the same principle you can actually write a generating function for your sequence with an inclusion-exclusion principle for any group G. Since a representation is faithful iff it doesn't have a kernel, you can enumerate all of your normal subgroups and then find your answer as $F\_G(x) - F\_{G/H\_1}(x) - F\_{G/H\_2}(x) - \dots + F\_{G/H\_1\cup H\_2}(x) + \dots $. Here the notation $F\_G(x)$ is a generating function counting all representations and $H\_1, H\_2, \dots$ are all maximal normal subgroups.

Hope that helps.


Update: I understand now you're interested in a closed formula for a specific case. For example, the computation by the formula above for $\mathbb Z\_2\times \mathbb Z\_2$ gives $(3\cdot 4)/2 -1 -1 -1 = 3$. The normal subgroups of an Abelian group are not hard to write, so it would be plausible if this could be made into a good formula.


(*) Let me know if you're not familiar with generating functions.

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Yeah, I meant \sum &phi;(m<sub>i</sub>), but that's also wrong for Z/2 x Z/2, as I've now amended the question to say. Thanks for your answer, I'll look into it. –  Rob Harron Nov 5 '09 at 22:09
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