Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G=GL(2,p)$ be the group of linear transformations. It acts on the set $X=F_p^2$. Then $C^X$ is a linear representation of $G$. What is the decomposition of this representation into irreducible representations?

Note: Let $d_i$ denote the number of isomorphic irreducible representations. Then $\sum d_i^2$ is equal to the number of orbits of $G$ when $G$ acts on $X\times X$. The number of orbits of $G$ in $X\times X$ is equal to $p+1$.

share|improve this question
I don't know the answer, but the character table for $G$ can be found in various texts, for instance the third edition of Lang's Algebra. The character of the permutation representation on $X$ is easy to compute. Using the orthogonality relations one gets the decomposition of the permutation character into irreducibles. With a little more work, one can get the decomposition on the level of actual representations. –  Robin Chapman Jul 5 '10 at 6:39
Acctually I want to know how $C^X$ decomposed and not just what irreducible represintations it contain, i.e. I want to know not only irreducible sub-repesintation in $C^X$ but also homomorphism from it to C^X. –  Klim Efremenko Jul 5 '10 at 8:22
To find the representations all you need is the characters. Each character $\chi$ has a corresponding central idempotent $e_\chi$ in the group algebra. Hitting your representation with $e_\chi$ will give the sum of all irreps with character $\chi$ inside it. –  Robin Chapman Jul 5 '10 at 8:53
What I've got so far: there are two obvious copies of the trivial representation: one coming from the fact that G fixes the origin, and the other coming from the fact that G fixes the sum over all points besides the origin. There is also a subrepresentation spanned by sums over all the points contained in each line through the origin; the action of G on this subrepresentation factors through PSL(2, p), and since the action of PSL(2, p) on the projective line is doubly transitive this subrepresentation is irreducible. –  Qiaochu Yuan Jul 5 '10 at 9:44
Another nice source for the computation of the character table is in Etingof's lectures on representation theory: www-math.mit.edu/~etingof/cltrunc.pdf –  bavajee Jul 5 '10 at 10:38

2 Answers 2

up vote 3 down vote accepted

Note: This was written up concurrently with David's answer, but wasn't proofread and didn't get past the captcha stage due to technical problems.

It is more common to consider the representation of $G=GL(2,\mathbb{F}_p)$ on $\mathbb{C}^Y$, where $Y=X\setminus 0$. This is a direct sum over all multiplicative characters $\chi$ of $\mathbb{F}_p$ of the induced representations of $G$ in the homogeneous functions of homogeneity degree $\chi:$

$$\mathbb{C}^Y=\bigoplus_{\chi\in \mathbb{F}_p^\ast}(\mathbb{C}^Y)_{\chi}, \quad (\mathbb{C}^X)_{\chi}=\{f:Y\to \mathbb{C}: f(\lambda a)=\chi(\lambda)f(a)\ \text{for all}\ \lambda\in \mathbb{F}_p \},$$

and $\mathbb{C}^X=\mathbb{C}\oplus\mathbb{C}^Y,$ where the first summand corresponds to the functions supported at $0\in\mathbb{F}_p^2.$

Each induced representation has dimension $p+1.$ For the trivial character, the induced representation contains a trivial subrepresentation of $GL(2,\mathbb{F}_p)$ (constants), with irreducible $p$-dimensional quotient (Steinberg representation). All other induced representations are irreducible and pairwise non-isomorphic (they all have different central characters). The situation is a bit more complicated for restrictions to $H=SL(2,\mathbb{F}_p)$ (denoted the same by abuse of notation): for any $\chi,\ (\mathbb{C}^Y)_{\chi}\simeq (\mathbb{C}^Y)_{\chi^{-1}}$ and if $p$ is odd and $\psi$ is a character of order $2$ then the representation $(\mathbb{C}^Y)_{\psi}$ is a direct sum of two non-isomorphic $(p+1)/2$-dimensional representations, all other representations remain irreducible and pairwise non-isomorphic. These facts can be verified ad hoc by the "sum of squares" calculation or, more systematically, by using the Mackey theory for finite groups. This accounts for close to the half of irreducible representations of $H$; to get the rest, one needs to induce from the non-split torus $\simeq \mathbb{F}_{p^2}^\ast/\mathbb{F}_{p}^\ast. $

share|improve this answer
Phew! Sudden power loss just a second after posting. –  Victor Protsak Jul 8 '10 at 17:30
Thanks, This answers my question about GL(2,p). Does the same works for GL(n,p)? Where can I read about Mackey theory for finite groups? –  Klim Efremenko Jul 9 '10 at 6:05
Yes, this works for $GL(n,p),$ but the representations you'll get will be more special (i.e. they will not look like "typical" representations of this group). I think Kirillov's "Elements of representation theory" describes Mackey theory, but this may be worth a separate MO question: I remember scratching my head trying to come up with a good reference. For this specific question, David's answer really walks you through the procedure without mentioning the big theorems that justify it in general. –  Victor Protsak Jul 9 '10 at 7:36

There are $2$ copies of the trivial representation, and $p-1$ other representations. The two trivials are spanned by the characteristic function of $(0,0)$, and by the function which is $1$ on all the nonzero vectors and $0$ on $(0,0)$. The other representations are indexed by the characters of $\mathbb{F}_p^*$. Namely, for any character $\chi$, consider those functions $f:\mathbb{F}_p^2 \to \mathbb{C}$ such that $f(g \cdot v)=\chi(g) f(v)$ for $g \in \mathbb{F}_p^*$. This is a subrepresentation. It is irreducible except when $\chi$ is trivial, in which case the two trivials split off as summands.

How to do this computation: Let $V$ be a representation of $G$ over $\mathbb{C}$, and let $V:=\bigoplus W_{\chi} \otimes V_{\chi}$ be the decomposition into isotypic components, where $V_{\chi}$ are representatives for the isomorphism classes of $G$-irreps. Then $$\mathrm{Hom}_G(V,V) = \bigoplus W_{\chi} \otimes W_{\chi}. \quad (*)$$ This formula is particularly useful when $V$ is a permutation representation $\mathbb{C}^X$, as then $\mathrm{Hom}_{\mathbb{C}}(V, V)$ is $\mathbb{C}^{X \times X}$ and the subalgebra of $G$-equivariant Homs is spanned by the $G$ orbits in $X \times X$. So the left hand side of $(*)$ has dimension equal to the number of $G$ orbits in $X \times X$. (In particular, we obtain the corollary mentioned by Qiaochu in a now-deleted answer: if the action of $G$ on $X \times X$ is doubly transitive, then $V$ is a trivial representation and one other irrep.)

In this case, there are $(p-1)+3+1$ orbits in $X \times X$. For every $g \in \mathbb{F}_p^*$, the set $\{ (v,w) : v=gw,\ v,w \neq 0 \}$ is an orbit. Also, we have three orbits by imposing that $v$, $w$ or both, be zero. Finally, there is the large orbit where $v$ and $w$ are linearly independent.

If we have already guessed the decomposition into irreducibles, as above, then we can confirm it by checking that the dimensions of both sides of $(*)$ match. In this case, that says $(p-1)+3+1=(p-1)+2^2$, which is true.

If we didn't already know the answer, we could deduce it from $(*)$. Let $R$ be the ring in $(*)$. We can describe $R$ as the vector space of $G$-invariant functions on $X \times X$, with the convolution product $(f*g)(u,w)=\sum_{v \in X} f(u,v) g(v,w)$. We must find the decomposition of $R$ into simple summands, and decompose $V$ as an $R$-module. The $R$-submodules of $V$ are the isotypic components for the $G$-action.

In this case, let $\Omega_g$ be the orbit $\{ (v,w) : v=gw,\ v,w \neq 0 \}$. It is easy to check that the $\Omega_g$ span a subalgebra isomorphic to the group algebra $\mathbb{C}[\mathbb{F}_p^*]$. This, of course, is isomorphic to $\bigoplus_{\chi} \mathbb{C}$, where the sum is over characters of $\mathbb{F}_p^*$. The idempotents for this decomposition project onto the corresponding $\chi$ representations above. The other semisimple summand of $R$ is a little harder to guess. Let $a_{00}$, $a_{10}$, $a_{01}$ and $a_{11}$ be the characteristic functions of the subsets of $X \times X$ where the two components are zero or nonzero according to whether the index is $0$ or $1$. Then the $a_{ij}$ span an algebra isomorphic to $\mathrm{Mat}_{2\times 2}(\mathbb{C})$, corresponding to the trivial summands.

You might enjoy reading the notes from the 2007 Quantum Gravity Seminar, which (despite its title) mostly focused on representation theory of finite groups. This is where I learned how to systematically use the ring $R$ to study permutation representations.

share|improve this answer
I am sorry I did not follow: "The idempotents for this decomposition project onto the corresponding representations above" Can you, please, explain it? I do understand how the ring R looks like, but I do not understand how to deduce from it irreducible represintations of $C^X$. –  Klim Efremenko Jul 7 '10 at 12:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.