# Need definition of symmetric and antisymmetric tensor representations of a Lie algebra

I couldn't find a definitive answer online.

Suppose we have a representation of a Lie algebra $(\pi,V)$. Consider the symmetric and antisymmetric vector subspaces of the $k$-th tensor product of $V$. I know that to extend the representation to the entire tensor product you operate using the operators

$$(\pi(g) \otimes I \otimes...\otimes I) + (I\otimes \pi(g) \otimes ...\otimes I) + ...$$

When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible?

• It is not clear why this should be a physics question - it is, currently, pure math. Mar 23, 2015 at 22:46
• i can't really argue why, other than there does exist a mathematics tag and i've seen similar math-centric questions answered here.
– SWV
Mar 23, 2015 at 22:49
• This resembles second quantization functor $d\Gamma(\pi(\cdot))$ (restricted to the $k$-th particle subspace); in that case yes, the two subspaces (symm. and antisymm.) are left invariant by the action of the ($d\Gamma$) operators and therefore the representation on the whole space is reducible. Mar 24, 2015 at 9:13

Also yes, although this doesn't quite constitute a proof until you prove that both the anti-symmetric and symmetric subspaces have positive dimension and don't intersect. In fact, for large enough $k$, $\Lambda^k(V)$ vanishes the dimension of the symmetric subspace shows that it isn't the whole of $V^k$, and so it's still true that $V^k$ decomposes. Note that $V^k$ isn't the direct sum of $\Lambda^k(V)$ and $S^k(V)$; there are tensors that are neither symmetric or antisymmetric or even a sum of symmetric and antisymmetric!