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I am a college sophomore self-studying the beginnings of representation theory using Serre's Linear Representations, and I am wondering if I am proving the following identities correctly.

Let $\chi, \chi'$ be the characters of two representations $\rho, \rho'$ of $G$ into $\mathrm{GL}(V)$. Let $\chi_\sigma$ be the character of the symmetric square, $\mathrm{Sym}^2(V)$, of $V$, and let $\chi_\alpha$ be the character of the $\mathrm{Alt}^2$. Prove the formulas:

\begin{align*} (\chi + \chi')^2_\sigma = \chi_\sigma^2 + \chi_\sigma'^2 + \chi \chi' \\ (\chi + \chi')^2_\alpha = \chi_\alpha^2 + \chi_\alpha'^2 + \chi \chi' \\ \end{align*}

My Work

Let $s \in G$. Fix a basis of eigenvectors for each representation: $(e_i), (e_i')$ respectively. We have then that $\rho_se_i = \lambda_ie_i$ with $\lambda_i \in \Bbb C$. This implies $$ \chi(s) = \sum \lambda_i \qquad \qquad \chi(s^2) = \sum \lambda_i^2 $$ Additionally we have the identities \begin{align*} \chi_{\sigma}^2(s) &= {1\over 2}(\chi(s)^2 + \chi(s^2)) \\ \chi_\alpha^2(s) &= {1\over 2}(\chi(s)^2 - \chi(s^2)) \end{align*}

The work follows (for the symmetric square of the representation, the alternating square is proved similarly): \begin{align*} (\chi + \chi')_\sigma^2(s) &= {1\over 2}[(\chi + \chi')^2(s) + (\chi + \chi')(s^2)]\\ &= {1\over 2}[(\chi^2 + 2\chi\chi' + \chi'^2)(s) + \chi(s^2) + \chi'(s^2)]\\ &= {1\over 2}[\chi(s)^2 + \chi(s^2)] + {1\over 2}[\chi'(s)^2 + \chi'(s^2)] + \chi\chi'(s) \\ &= \chi_\sigma^2 + \chi_\sigma'^2 + \chi \chi' \end{align*}

My Question

Am I going about this correctly? Nowhere in this did I refer to the trace of the representation. Should that be incorporated to fix this proof?

Thanks for your time.


Would I incoporate the trace immediately after writing down the expanded form of the symmetric square of the sum of $\chi$ and $\chi'$?

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This is correct -- you need the trace in proving the two identities that you quoted. – darij grinberg Feb 20 '13 at 1:53
For an alternative proof, you could have used the canonical isomorphisms $\mathrm{Sym}^2\left(V\oplus W\right) \cong \left(\mathrm{Sym}^2 V\right) \oplus \left(V\otimes W\right) \oplus \left(\mathrm{Sym}^2 W\right)$ and $\wedge^2\left(V\oplus W\right) \cong \left(\wedge^2 V\right) \oplus \left(V\otimes W\right) \oplus \left(\wedge^2 W\right)$. – darij grinberg Feb 20 '13 at 1:54
@darijgrinberg where would the trace first be necessary? Am I able to expand the square of the characters from $(\chi + \chi')^2$ to $(\chi^2 + 2\chi \chi' + \chi'^2)$ or is this where I will need the trace? – Moderat Feb 20 '13 at 2:08
@BenjaLim thanks very much. I now see. Could you post your response as an answer? – Moderat Feb 20 '13 at 3:30
@JJR I have posted an answer. – user38268 Feb 20 '13 at 3:49
up vote 1 down vote accepted

Your answer is correct and I will try to clean up some things. Let $ρ,ρ′:G \to \text{GL}(V)$ be two representations where $G$ is a finite group and $V$ is a finite dimensional complex vector space. Let $s \in G$ be given. Firstly we note that because $\rho_s \in \text{GL}(V)$ has finite order it is diagonalisable by the Jordan Normal form. Consequently $V$ has a basis of eigenvectors $e_i$ of $ρ_s$. I should say in your answer above you have used the definition of the trace when you say that $\chi(ρ_s^2) = \sum \lambda_i^2$.

Think about the diagonal matrix $\rho_s$; what happens when you multiply two diagonal matrices together and take the trace?

Also here are a few remarks about the proof in Serre of Proposition 3 of chapter 2. Firstly, usually the second symmetric and exterior powers are constructed as quotients of $V \otimes V$; however they are isomorphic to the subspace with basis $e_i\otimes e_j + e_j \otimes e_i$ (for the symmetric power) and $e_i \otimes e_j - e_j \otimes e_i$ for the exterior power.

Now Serre used the following fact in the proof of the proposition, which I append here if you're not familiar with. Let $\rho_s : V \to V$ be an endomorphism of $V$. Then there is a unique linear operator $\rho_s \otimes \rho_s : V \otimes V \to V \otimes V$ such that $\rho_s \otimes \rho_s(v \otimes w) = \rho_s(v) \otimes \rho_s(w)$. If you have a basis for $V$ (which will give you a basis for $V \otimes V$) and write down the matrix for $\rho_s \otimes \rho_s$, that matrix is given precisely by the Kronecker Product.

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