1

Let G be a finite group and $\chi$ its character.

If degree (or dimension? I am not sure about terminology) of $\chi$ is $> 1$, is it always true that $\langle\chi,\chi\rangle \ge 1$?

My intuition is that this quantity is the sum of squares of dimensions of sub-representations of $\chi$.

So as long as we don't know if $\chi$ is irreducible, it should at least contain the trivial representation so $\langle\chi, \chi\rangle \ge 1$.

Does this make sense?

Thank you for your help.

Conjecture
  • 3,389
  • 2
    Are you sure you don't mean $\langle \chi,\chi\rangle$ ? – Captain Lama Apr 26 '19 at 19:03
  • Right @Captain Lama will fix it thanks – Conjecture Apr 26 '19 at 19:54
  • 1
    This being said, be careful, you seem to think that a reducible representation always contains a copy of the trivial representation, which is not true. – Captain Lama Apr 26 '19 at 20:23
  • When you say that $\chi$ contains the trivial representation, you're incorrect. However, let $\chi$ be the character of some irreducible representation $(\rho, V)$ of some finite group $G$ over a field $\mathsf{k}$. Then $\langle \chi, \chi \rangle$ is defined to be the dimension of the $\mathsf{k}$ vector space $\operatorname{End}{\mathsf{k}\left[G\right]}(V)$, where $\operatorname{End}{\mathsf{k}\left[G\right]}(V)$ denotes the linear endomorphisms of $V$ commuting with $\rho$ (called intertwining maps). Well the identity map is one such map, and so $\langle \chi, \chi \rangle \geq 1$ – Adam Higgins Apr 26 '19 at 22:59

1 Answers1

3

Let $G$ be a finite group, with $\rho : G \to \operatorname{GL}(V)$ some representation of $G$ for $V$ some finite dimensional vector space over some field $\mathsf{k}$ of some characteristic not dividing $\left| G \right|$ (feel free to assume $\mathsf{k} = \mathbb{C}, \mathbb{R}, \mathbb{Q}$). Then Maschke's Theorem states that $\rho$ is completely reducible. What does this mean? Well, if $U \leq V$ is a subrepresentation of $V$, then there exists some other subrepresentation $W \leq V$ satisfying $V = U \oplus W$. Since $V$ is finite dimensional, we can repeat this step with $U,W$ until we are left with a decomposition of the form $V = \bigoplus_{i=1}^{k} V_i$ for some subrepresentations $V_i \leq V$ that have no proper subrepresentations, i.e are irreducible. Then

$$ \operatorname{Hom}_{\mathsf{k}\left[G\right]}(V,V) \cong \bigoplus_{i,j} \operatorname{Hom}_{\mathsf{k}\left[G\right]}(V_i,V_j). $$

Then let $\chi$ denote the character of $\rho$, and $\chi_k$ denote the character of the restriction of $\rho$ to $V_k$. Then

$$ \langle \chi, \chi \rangle = \dim_{\mathsf{k}} \operatorname{Hom}_{\mathsf{k}\left[G\right]}(V,V) = \sum_{i,j} \dim_{\mathsf{k}} \operatorname{Hom}_{\mathsf{k}\left[G\right]}(V_i,V_j) = \sum_{i,j} \langle \chi_i, \chi_j \rangle. $$

Now we will write $V \cong \bigoplus_{i=1}^{\ell} n_i V_i$, for some integers $n_i$, and $V_i \not \cong V_j$ when $i \neq j$, and assume now that $\mathsf{k}$ is algebraically closed. Then by Schur's Lemma

$$ \dim_{\mathsf{k}} \operatorname{Hom}_{\mathsf{k}\left[G\right]}(V_i,V_j) = \delta_{i,j}, $$

where $\delta_{i,j}$ denotes the kronecker delta function. Thus

$$ \langle \chi, \chi \rangle = \sum_{i,j} n_i n_j \delta_{i,j} = \sum_{i,j} n_i^{2}. $$

Thus $\langle \chi, \chi \rangle$ is a sum of squared integers, and so $\langle \chi, \chi \rangle \geq 1$, with equality if and only if $\rho$ was irreducible to being with.

A few comments. When you say "Let $G$ be a finite group and $\chi$ its character", what you mean is "a character". A finite group will have many different characters, infinitely many in fact. What you wish to say is "Let $G$ be a finite group, with $\chi$ the character of some representation $\rho$ of $G$. Also, as others have pointed out, reducible representations need not contain a copy of the trivial representation. A contrived example would be the character of the representation

$$ \rho : C_2 = \langle s \mid s^{2} \rangle \to \operatorname{GL}(\mathbb{C}^{2}) \ : \ s \mapsto \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}. $$

Then $\rho$ is the direct sum of two copies of the sign representations of $C_2$, so is reducible and does not contain any copies of the trivial representation.

Adam Higgins
  • 2,271
  • Thank you! For the counter example, does it contain a double copy of the sign representation because $\chi_{\rho}(s)= -2=2 \chi_{sign}(s)$ and $\chi_{\rho}(e)= 2 =\chi_{sign}(e)$? – Conjecture Apr 27 '19 at 09:39
  • 1
    You're welcome, but no that's not why it contains two copies of the sign representation (though it could be a clue). The reason it contains two copies of the sign representation (besides the fact I constructed it so it did) it that $\langle \chi_\rho, \chi_{\operatorname{sign}} \rangle = 2$, and $\rho_{\operatorname{sign}}$ is irreducible (it's one-dimensional). You can also see that $\rho$ is exactly $\rho_{\operatorname{sign}} \oplus \rho_{\operatorname{sign}}$ since $\langle \chi_\rho, \chi_\rho \rangle = 4$. – Adam Higgins Apr 27 '19 at 09:43
  • Your appeal to Schur's lemma requires the field to be algebraically closed. – Joshua P. Swanson Aug 29 '19 at 04:47
  • @JSwanson You'll notice that I stated "assume now that $\mathsf{k}$ is algebraically closed. I did so to simplify the proof a little. But one could use this result: https://math.stackexchange.com/questions/2566754/is-it-true-that-langle-psi-omega-rangle-is-the-dimension-of-homv-w instead. – Adam Higgins Aug 29 '19 at 07:50
  • Oh, so you did! I think I missed it because it was buried in the middle. The fact that $\dim \mathrm{Hom}_G(U, V) = \langle \chi_U, \chi_V\rangle$ is the thing that's doing the work in your argument anyway, but it's being hidden by an appeal to Schur's lemma. Schur's lemma is a bit of a red herring since the $\langle\chi, \chi\rangle \geq 1$ result follows immediately from the above equality, regardless of being algebraically closed. It also shows that $\langle \chi, \chi\rangle$ is a linear combination of squares rather than a literal sum of squares in general. – Joshua P. Swanson Aug 29 '19 at 14:33
  • @JSwanson I accept that Schur’s lemma is hiding some of the detail here. My choice of using it was because Schur’s lemma is far easier to prove than the dimension-character pairing formula. – Adam Higgins Aug 29 '19 at 14:37
  • The proof of the version of Schur's lemma you're using involves the dimension-character formula, so Schur's lemma is strictly "harder" (though barely). The proof has two steps: (1) apply the dimension-character formula; (2) show that $\dim \mathrm{End}_G(U) = 1$ if $U$ is irreducible and the field is algebraically closed. The additional difficulty from step (2) is minimal: $\mathrm{End}_G(U)$ is a finite-dimensional division ring over an algebraically closed field, hence it must be the base field. Proving the dimension-character formula is not hard either, but it is harder than (2). – Joshua P. Swanson Aug 29 '19 at 16:13
  • Maybe I’m going wrong here but the proof I have in mind of the form of Scur’s lemma I’m using does not involve the dimension-character formula. It only uses that endomorphisms of finite dimensional vector spaces over algebraically closed fields always have at least one eigenvector. – Adam Higgins Aug 29 '19 at 16:16
  • At some point you have to relate $\dim \mathrm{End}G(U)$ and $\langle \chi_U, \chi_U\rangle$, the latter by definition being $\frac{1}{|G|} \sum{g \in G} \chi_U(g^{-1}) \chi_U(g)$. The thing you're thinking of is essentially an explicit version of an argument for (2). – Joshua P. Swanson Aug 29 '19 at 16:18