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I am learning some representation theory now. In the book I am reading (Group theory and physics by Sternberg), the author started with $GL(N, \mathbb{C})$, and then by using the relation, $SU(N) \subset SL(N) \subset GL(N)$, he obtained the representations of $SU(N)$.

I am wondering why he did not mention $SO(N)$.

After some thought, I realized that the Schur-Weyl duality should not hold for $SO(N)$. For example, consider $SO(3)$, and the tensor product space $V\otimes V $, with the dimension of $V$ being 3. The tensor product space $V\otimes V $ decomposes under $S_2$ into the symmetric subspace $S^2 V$ and the antisymmetric subspace $A^2 V $. The latter is 3-dimensional, and should be invariant and irreducible under $SO(3)$. However, the former, i.e., $S^2 V $ is not irreducible for $SO(3)$. It splits into a one-dimensional subspace spanned by $e_1\otimes e_1 + e_2\otimes e_2 +e_3 \otimes e_3$, and a 5-dimensional subspace.

In this example, we see that the Schur-Weyl duality fails for $SO(3)$.

But I cannot see why it holds for $GL(N)$ but fails for $SO(N)$. What is wrong with $SO$?

S. Kohn
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1 Answers1

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Exactly what you said always happens: $S^2(V)$ is not irreducible as it contains a one-dimensional invariant subspace.

If $G$ is a group and $V$ is a representation then a non-zero map from $S^2(V) \to \mathbb{C}$ is the same thing as a $G$-invariant symmetric bilinear form on $V$. $SO(n)$ preserves a symmetric bilinear form on $V$ (by definition) so therefore $S^2(V)$ has an invariant, $GL_n$ does not preserve any such form so there is no such invariant.

The orthogonal groups do satisfy a version of Schur-Weyl duality though, however you need to replace the symmetric groups with what's called a Brauer algebra.

Nate
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