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Let $\mathfrak{g}$ be a complex Lie algebra of type $A_{n-1}$. Consider representation of $\mathfrak{g}$ on direct sum of complex vector spaces which is given by the highest weight $\omega_1\oplus\omega_1$ where $\omega_1$ is the first fundamental representation. So, it is standard representation on vector but twice. Denote by $V$ the direct sum of vector spaces. I am interesting in second symmetric power of this representation. First, I am studying second tensor power and its decomposition: $\otimes^2(\omega_1\oplus\omega_1)=\oplus_4(\omega_1\otimes\omega_1)$, where $\omega_1\otimes\omega_1$ decomposes to $2\omega_1\oplus\omega_2$. In the representation $2\omega_1\oplus\omega_2$ is symmetric part naturally $2\omega_1$. So, I claim $\odot^2(\omega_1\otimes\omega_1)=\oplus_42\omega_1$. But if I compare dimensions of the last equation it does not fit (difference -1). By dimensionality issues following equation holds, $\odot^2(\omega_1\otimes\omega_1)=(\oplus_32\omega_1)\oplus\omega_2$. It is very strange for me. Can it be right?

user323461
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1 Answers1

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For any $v,w \in \omega_1$ we have that $(v,0)\otimes(0,w) + (w,0)\otimes(0,v) \in S^2(\omega_1 \oplus \omega_1)$ in particular this gives us a copy of $\omega_1 \otimes \omega_1$ in $S^2(\omega_1 \oplus \omega_1)$ which you miss in your initial count.

I think it is easier if you break symmetry and think of the two copies of $\omega_1$ as being different. In general you should have: $S^n(A\oplus B) = S^n(A) \oplus (S^{n-1}(A)\otimes B) \oplus (S^{n-2}(A)\otimes S^2(B)) \oplus \dots \oplus S^n(B)$.

Nate
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  • Thank you. Nice rule, I have never seen it before. So, decomposition $\odot^2(\omega_1\otimes\omega_1)=(\oplus_32\omega_1)\oplus\omega_2$ holds. Am I right? Could you please give me a reference for that rule or a name? – user323461 Mar 16 '16 at 19:34
  • Yea that looks right to me. I'm not sure that rule really has a name, I've just referred to it as the symmetric power direct sum formula. There is also an (equivalent) exterior power version here which I think might be more well known. – Nate Mar 16 '16 at 20:52
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    It follows from some general nonsense that we should have a formula like this: the full symmetric algebra is left adjoint to the forgetful functor from commutative $\mathbb{g}$-algebras to $\mathbb{g}$-modules, hence commutes with colimits. Colimits in the module category is direct sum and in the algebra category is tensor, so $S(A \oplus B) = S(A) \otimes S(B)$. Taking the $n$-th degree part gives the above formula. – C.D. Nov 03 '22 at 03:04