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If $G$ is a finite group, then it is well known that there are finitely many inequivalent irreducible representations of $G$ over $\mathbb{C}$; moreover the sum of squares of dimensions of the representations is equal to $|G|$. Also, the dimension of representation divides $|G|$.

If we consider representations over $\mathbb{Q}$, are the dimensions of irreducible representations related to $|G|$ in such a nice way?

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3 Answers 3

The best you can get if you take the character theory done over $\mathbb C$ and attempt to redo it over an arbitrary field $\Bbbk$ of characteristic $0$ (the usual character theory does not work in positive characteristic because $p$-dimensional representations could end up with the $0$ character), is that the group ring $\Bbbk G$ as a left module over itself is isomorphic to $\bigoplus_V \frac{\dim V}{\dim Hom_\Bbbk(V,V)^G} V$ where $V$ ranges over all irreducible representations and $Hom_\Bbbk(V,V)^G$ is the space of $G$-linear endomorphisms of $V$. This gives you $|G|=\sum_V\frac{(\dim V)^2}{\dim Hom_\Bbbk(V,V)^G}$.

All of the above follows immediately from noticing that the projection formula $\dim V^G=\frac1{|G|}\sum_{g\in G}\chi_V(g)$ still holds (because that formula says nothing more than that $\frac1{|G|}\chi_\Bbbk^*=\sum_{g\in G}g$ is an idempotent with $1$-dimensional image in the ring $\Bbbk G$), that the characters are additive and multiplicative on direct sums and tensor products, and that if $V$ is a representation with character $\chi_V$, then the dual $V^*$ is a representation with character $\chi_{V^*}(g)=\chi_V(g^{-1})$. Then the 'product' $(\alpha,\beta)=\frac1{|G|}\sum_{g\in G}\alpha(g)\beta^{-1}(g)$ is linear in the first variable, so for irreducible representations $V$ and $W$ we get $(\chi_V,\chi_W)=\frac1{|G|}\sum_{g\in G}\chi_V(g)\chi_W(g^{-1})=\frac1{|G|}\sum_{g\in G}\chi_{W^*\otimes V}(g)=\dim(W^*\otimes_{\Bbbk} V)^G=\dim Hom_\Bbbk(V,W)^G$. Since we still have that a $G$-linear map between irreducibles $V$ and $W$ either has trivial kernel or trivial image (as they are subrepresentations), it follows that if $V\neq W$, then $(\chi_V,\chi_W)=0$. Then linearity in the first variable tells us that for an arbitrary representation $R$ we have a unique decomposition $\chi_R=\sum_V\frac1{\dim Hom_\Bbbk(V,W)^G}(\chi_R,\chi_V)\chi_V$, which implies that characters of irreducible representations are linearly independent. This establishes that the decomposition of a representation into irreducibles corresponds to the represenation of the character as the sum of irreducible characters. Also, because characters are class functions, and the space of class function is generatd by the linearly independent projections of conjugacy classes, we get that there are at most as many irreducible representations as there are conjugacy classes.

Then the usual observation that for the regular representation $R$ of the group ring $\Bbbk G$ as a left module over itself we have $\chi_R(e)=\dim\Bbbk G=|G|$ and $\chi_R(g)=0$ otherwise (look at the traces of the matrices with basisi $G$ for $\Bbbk G$ over $\Bbbk$) implies that $(\chi_R,\chi_V)=\frac1{|G|}|G|\chi_V^*(e)=\dim V$ as usual and hence $\chi_R=\sum_V\frac{\dim V}{\dim Hom_\Bbbk(V,W)^G}\chi_V$ which gives us the direct sum decomposition of $\Bbbk G$ that I claimed in the beginning.

Note that when $\Bbbk$ is algebraically closed, then surely $\dim_\Bbbk(V,V)^G=1$ for all irreducible representations $V$ since any endomorphism of $V$ must have an eigenvalue and the eigenspace for that eigenvalue for $G$-linear endomorphism must be all of $V$, i.e. the $G$-linear endomorphism must be scalar multiplication, which gives you $\Bbbk G=\bigoplus_V(\frac{\dim V}1)V$ and hence $|G|=\sum_V(\dim V)^2$.

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I think you mean $\dim Hom_\Bbbk(V,V)^G$ in the denominator. Regards. –  awllower Jun 23 '13 at 10:37
    
A lot of the character theory used here I feel is unnecessary. One can more directly compute $\dim\hom_{k[G]}(k[G],W)$ for irreducibles $W$ in two different ways, and then equating the results straight away gives us the multiplicities of irreducibles in the group algebra. I posted this computation in my own answer. –  blue Jul 6 at 7:59

No. That's not even true over $\mathbb{R}$. For example the irreducible real representations of $C_3$ are the trivial one and a two-dimensional one, the sum of the conjugate 1-dimensional non-trivial complex representations. This two-dimensional representation is also defined over $\mathbb{Q}$.

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One doesn't need character theory calculations to decompose the group algebra into irreducibles.

Suppose $G$ is a finite group and $k$ a field in which $|G|$ is invertible. Then Maschke's theorem applies so all finite-dimensional representations are expressible as a direct sum of irreducibles. Write

$$k[G]\cong \bigoplus_{V\in\widehat{G}} V^{\oplus m(V)}$$

for some unknown multiplicities $m(V)$, as $V$ ranges over irreducibles in $\widehat{G}$. Fix an irreducible $W$, and consider an intertwiner $k[G]\to W$. Such a map is determined by where $1$ is sent, and then conversely $1$ may be sent to any element of $W$. Hence $\dim\hom_{k[G]}(k[G],W)=\dim W$.

On the other hand, using distributivity of $\hom$,

$$\hom_{k[G]}\left(\bigoplus_{V\in\widehat{G}} V^{\oplus m(V)},W\right)\cong\bigoplus_{V\in\widehat{G}}\hom_{k[G]}(V,W)^{\oplus m(V)}\cong {\rm End}_{k[G]}(W)^{\oplus m(W)}$$

(since $\hom_{k[G]}(V,W)=0$ if $V\not\cong W$) which has dimension $m(W)\dim{\rm End}_{k[G]}(W)$.

Equating $\dim W=m(W)\dim{\rm End}_{k[G]}(W)$ gives us the multiplicities $m(W)$. Hence

$$k[G]\cong\bigoplus_{V\in\widehat{G}} V^{\oplus (\dim V)/(\dim{\rm End}_{\large k[G]}(V))}\implies |G|=\sum_{V\in\widehat{G}}\frac{(\dim V)^2}{\dim{\rm End}_{k[G]}(V)}.$$


If $k$ is algebraically closed, $\dim V$ divides $|G|$ for each $V\in\widehat{G}$. This fails in general. However there is a weaker version which still holds: $\dim V$ divides $|G|\varphi(\exp G)$ for each $V$. Again, if $k$ is algebraically closed the number of irreducible representations equals the number of conjugacy classes (there is no generic, canonical bijection - instead they are "dual"). This fails if $k$ isn't algebraically closed, or if $|G|$ isn't invertible. And again a weaker version holds: the number of irreducible representations equals the number of $K$-conjugacy classes of $K$-regular elements.

These facts and more should be discussed at the GroupProps Wiki.

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This is very nice. –  Vladimir Sotirov Jul 11 at 6:06

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