# Decomposition of $\mathbb{C}[G]$ / Orthogonality relations

I am currently working on Jean-Pierre Serre's "Linear representations" german translation in chapter 6 and I do not understand the last part of the proof of the following theorem.

a) Let $\rho_i : G \longrightarrow GL(V_i)$ for $i = 1,...,n$ be distinct irreducible representations of the group G. The map $\rho: \mathbb{C}[G] \longrightarrow \prod_{i}^{n} End(V_i)$, defined by the $\rho_i$, is surjective.

Proof:

Suppose the Image of $\rho$ is a proper subspace of $\prod_{i}^{n} End(V_i)$. Then there exists a non zero linear form $\phi : \prod_{i}^{n} End(V_i) \longrightarrow \mathbb{C}$ that vanishes on the Image of $\rho$, i.e. $\phi (\sum_{i}^{n} \rho_i(s)) = 0$ for all $s \in G$.

This will give us a relation of the form $\sum_{k} \lambda_k m_k(s) = 0$ for all $s \in G$. Here, the $m_k(s)$ are the coefficients in the Matrix of the representation $\rho_i(s)$.

So far the proof makes sense to me, but I do not understand the following part at all. We want to show that all $\lambda_k$ are zero, because then we have found our contradiction to the non zeroness of $\phi$. Serre only writes:

Due to the orthogonality relations of the coefficients we can conclude that $\lambda_k = 0$ for all $k$.

The orthogonality relations from chapter 2.2 in the english version are meant here, but I just cannot see how I have to apply them to get that all $\lambda_k = 0$.

I have found another thread to the same theorem, but the proof is different at the end and I hope someone can help me with my problem.

Help would be highly appreciated.

• I don't know what orthogonality relations you are talking about, but if you take the trace on that equality, you get a relation between characters of distinct (I'm assuming up to isomorphism) irreducible representations, which are known to be linearly independent (precisely because they are orthogonal) – Max May 30 '18 at 21:22
• Exercise: If $v_1,\cdots,v_k$ are orthogonal vectors in an inner product space, then they are linearly independent (that is, $\lambda_1v_1+\cdots+\lambda_kv_k=0$ implies $\lambda_1,\cdots,\lambda_k$ are all $0$). – anon May 30 '18 at 22:56
• I see your point Max, but I don't know how I am supposed to apply the trace on $\sum_{k} \lambda_k m_k(s)$, as $\lambda_k$ and the $m_k(s)$ are only complexe numbers ? – Gilligans May 31 '18 at 13:12

Pick an $i \in \{1, \dots, n\}$. We multiply both sides of the equation $\sum_{k} \lambda_k m_k(s)= 0$ by $\bar m_i(s)$, and we take a sum over all $s \in G$, giving $$\sum_k \left( \lambda_k \sum_{s \in G} m_k(s) \bar m_i(s) \right)= 0$$ But by orthogonality, $$\sum_{s \in G} m_k(s) \bar m_i(s) = \begin{cases} |G| & {\rm if \ } k = i \\ 0 &{\rm if \ } k \neq i \end{cases}$$ So $$\sum_k \left(\lambda_k \sum_{s \in G} m_k(s) \bar m_i(s) \right) = \lambda_i |G|.$$ Thus we have shown that $\lambda_i|G| = 0$, i.e. $\lambda_i = 0$. Since the choice of $i$ was arbitrary, this completes the proof.
• what exactly are the $\bar{m}_i(s)$ ? And for the orthogonality: do you assume that each representation is given as a unitary matrix ? I don't really know why $m_k(s) \bar{m}_i(s) = 1~ or~ 0$ – Gilligans May 31 '18 at 13:09
• @H.Sch $\bar m_i(s)$ is the complex conjugate of $m_i(x)$. And it's not true that $m_k(x) \bar m_i(s) = 1$ or $0$ for each $s \in G$. However, it is true that $\sum_{s \in G} m_k(s) \bar m_i(s) = |G|$ or $0$ (and this is precisely the statement of orthogonality). – Kenny Wong May 31 '18 at 13:36