# How to show that $\sum_{\phi \in \hat{G}} d_{\phi}^2 \leq \vert G \vert$ for a finite group $G$?

I am trying to prove that $\vert \hat{G}\vert \leq \sum_{\phi \in \hat{G}} d_{\phi}^2 \leq \vert G \vert$ for a finite group $G$ where $\hat{G}=\{\text{inequivalent irreducible unitary representation } \phi \text{ of } G\}$. I am able to prove that $\vert \hat{G}\vert \leq \vert G \vert$. By considering the map $\alpha : \hat{G} \longrightarrow L^{2}(G)$ given by $\alpha(\phi)=\text{Trace}(\phi)$ which is well defined as trace is well defined (since similar matrices have same trace) and it is also injective by the orhogonality of the character. Let $A_{\phi}=\{\phi_{ij} : 1 \leq i,j\leq d_{\phi}\}$, and $\vert A_{\phi}\vert = d_{\phi}^2$. Now consider the map $\beta: \hat{G} \longrightarrow \cup_{\phi \in \hat{G}} A_{\phi}$ defined by $\beta(\phi)=\phi_{11}$, where $\cup_{\phi \in \hat{G}} A_{\phi}$ has finitely many elements. $\beta$ is injective by schur's orhogonality relations, but I am not able to show why $\beta$ is well defined? Is there any other map which will work? It might be a small thing but I am unable to get it. Any help in this regard would be appreciated!

• What is $d_\phi$? – Sam Jaques Aug 16 '16 at 20:09
• $d_{\phi}$ is the degree of the irreducible unitary representation $\phi : G \longrightarrow GL(V)$ which is defined to be the dimension of the complex vector space $V$. – Shubham Namdeo Aug 17 '16 at 4:03
• Okay, I'm pretty sure $\sum_{\phi\in\hat{G}}d_\phi^2=\vert G\vert$, from which the inequality is straightforward. Have you seen a proof of that? Were you looking for a more direct proof? – Sam Jaques Aug 18 '16 at 13:55
• Actually, I am trying to prove the fact which you stated, and I need the well-definedness of the map $\beta$, to conclude the result. All the other remaining details I have verified. – Shubham Namdeo Aug 19 '16 at 14:50

## 1 Answer

The proof I've seen (in "Erdos-Ko-Rado Theorems: Algebraic Approaches" by Godsil and Meagher) is to look at the left-regular representation $\Lambda:G\rightarrow GL(L(G))$, where $L(G)$ is the set of all complex functions on $G$. The action is, for $\phi\in L(G)$ and $g,x\in G$, that $$(\Lambda(g)\phi)(x)=\phi(g^{-1}x)$$ In other words, $\Lambda(\phi)=\phi'$, where $\phi'$ is a new function such that $\phi'(x)=\phi(g^{-1}x)$ (you can check that it's a representation).

Then we take functions $\delta_g\in L(G)$, defined as $\delta_g(g)=1$, $\delta_g(x)=0$ if $x\neq g$. These form a basis for $L(G)$, and we note that $(\Lambda(a)\delta_g)(x)=\delta_g(a^{-1}x)=\delta_{ag}$. Then if we define $\lambda(g)=Tr(\Lambda(g))$: $$\lambda(a)=\sum_{g\in G}\langle \delta_g,\Lambda(a)\delta_g\rangle=\sum_{g\in G}\langle \delta_g,\delta_{ag}\rangle$$ And the terms in the sum are only non-zero if $ag=g$, so it's only non-zero if $a=e$. Thus we have that $\lambda(e)=\vert G\vert$, and $\lambda(x)=0$ for $x\neq e$. Using this, we can use the fact that the irreducible characters are an orthonormal basis of all characters and say that: $$\lambda=\sum_{\phi\in\hat{G}}m_\phi \phi$$ And then to find the coefficients, use the inner product: $$m_\phi=\langle \lambda,\phi\rangle=\frac{1}{\vert G\vert}\sum_{g\in G}\phi(g^{-1})\lambda(g)=\frac{\phi(e)}{\vert G\vert}$$ (where all but one of the terms in the sum disappeared since $\lambda(g)=0$ if $g\neq e$). This means that $\lambda=\frac{1}{\vert G\vert}\sum_{\phi\in\hat{G}}\phi(e)\phi$. Then since $\lambda(e)=1$: $$1=\lambda(e)=\frac{1}{\vert G\vert}\sum_{\phi\in\hat{G}}\phi(e)\phi(e)$$ And $\phi(e)=d_\phi$, so that gives that $\sum_{\phi\in\hat{G}}d_\phi^2=\vert G\vert$.

Running with the method you were trying, I don't think $\beta$ is well-defined. For any irreducible representation, there is an equivalent one where I just swap any two vectors. By the orthogonality, we know that for any representation $\phi$, if $i\neq j$, then $\phi_{ii}\neq\phi_{jj}$. Then I can construct an equivalent representation $\phi'=M^{-1}\phi M$, where $M$ swaps $v_1$ and $v_j$, then $\beta(\phi)\neq\beta(\phi')$. But what stops you from simply choosing a representative of each equivalence class of irreducible representations and constructing a set $\hat{G}'$ of these? Then you can define $\beta':\hat{G}'\rightarrow \cup A_\phi$ in the same way and avoid problems of equivalence.

• Firstly, thanks for the answer and for your valuable time Sam Jaques. In my problem, by $\hat{G}$, I meant equivalence classes of irreducible representations of $G$ which I forgot to mention about. Sorry! For that. The proof you which gave using regular representation and characters I knew it, but I was trying to prove the injectivity of $\beta$ without using characters and I wanted to show that $\vert \hat{G}\vert \leq \sum_{\phi \in \hat{G}}^{s} d_{\phi}^{2} \leq \vert G \vert$. – Shubham Namdeo Aug 21 '16 at 3:54
• I was also discussing the same problem with my friend Senthil Rani and she sujjested me the following answer yesterday. $\beta$ is well defined : Let $\{e_1, \ldots, e_n\}$ be the standard basis for $\mathbb{C}^n$ and $\langle, \rangle$ be the standard inner product on $\mathbb{C}^n$. We will show that, if $\phi \sim \psi$ (i.e. $\exists$ $P \in GL_{n}(\mathbb{C})$ such that $\phi = P^{-1} \psi P$) then $\phi_{11}=\psi_{11}$. On contrary, assume that $\phi_{11} \neq \psi_{11} \Rightarrow$ $\exists$ $g \in G$ such that – Shubham Namdeo Aug 21 '16 at 4:04
• \begin{alignat*}{2} \quad && \phi_{11}(g) & \neq \psi_{11}(11)\\ \Rightarrow \quad && \langle \phi_{g}e_{1},e_{1} \rangle & \neq \langle \psi_{g}e_{1},e_{1} \rangle \\ \Rightarrow \quad && \langle \phi_{g}e_{1},e_{1} \rangle & \neq \langle (P^{-1}\phi_{g}P) e_{1},e_{1} \rangle \\ \Rightarrow \quad && \langle [\phi_{g}-(P^{-1}\phi_{g}P)]e_{1} ,e_{1} \rangle & \neq 0 \\ \Rightarrow \quad && \phi_{g} - P^{-1}\phi_{g}P & \neq 0\\ \Rightarrow \quad && \phi_{g} & \neq P^{-1}\phi_{g} P\end{alignat*} – Shubham Namdeo Aug 21 '16 at 4:08
• Why is it a contradiction that $\phi_g\neq P^{-1}\phi_g P$? – Sam Jaques Aug 22 '16 at 16:41
• $\beta'$ uses representatives from each equivalence class, instead of the class itself (I called the set of representatives $\hat{G}'$). It's defined explicitly for each element of $\hat{G}'$, so it can't be ill-defined. There's a small issue of whether we can choose such representatives, but since $\hat{G}'$ is finite and every equivalence class is non-empty, that's no problem. – Sam Jaques Aug 22 '16 at 21:13