I was trying to show that $V^{\otimes 2}=S^2(V)\oplus \Lambda^2V$ and this is what I came up with. Obviously $S^2(V)\oplus \Lambda^2V\subset V^{\otimes 2}$. Now take $a\in V^{\otimes 2}$.Let $\dim V=n$ and let the basis of $V^{\otimes 2}$ be $\{v_i\otimes v_j\}_{1\leq i,j \leq n}$. Then we can write $a=\sum\frac{1}{2}a_{ij}((v_i\otimes v_j)+(v_j\otimes v_i))+\sum\frac{1}{2}a_{ij}((v_i\otimes v_j)-(v_j\otimes v_i))$ and the first one is in $S^2(V)$ and the other in $\Lambda^2V$. Also, we see that intersection of $S^2(V)$ and $\Lambda^2V$ is just $\{0\}$.

I hope what I did was right. My question is: Can we say that $V^{\otimes k}=S^k (V)\oplus \Lambda^k V$ for say $k=3$ or higher? My intuition says no because I can't write any element in $V^{\otimes k}$ into two pieces like I did before.

  • $\begingroup$ For higher $k$, you have to consider tensors with some symmetric pairs, some antisymmetric pairs. Young tableaux enumerate you all the ways you can do this, so those will give the terms on the right hand side of your isomorphism. $\endgroup$ – ziggurism Dec 10 '15 at 6:02
  • $\begingroup$ @ziggurism, sorry I don't know anything about Young tableaux. $\endgroup$ – user96343 Dec 10 '15 at 6:17
  • $\begingroup$ No need to apologize... $\endgroup$ – ziggurism Dec 10 '15 at 6:52
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    $\begingroup$ I might add that a nice way to see that your isomorphism does not work is that $V^{\otimes k}$ is $n^k$ dimensional, while $S^k(V)\oplus \Lambda^k(V)$ is $n(n-1)/2+n+n(n-1)/2=n^2$ dimensional. For $k\neq 2$ those are obviously distinct. $\endgroup$ – ziggurism Dec 10 '15 at 6:56
  • $\begingroup$ @ziggurism, thank you! I think that works. I'd be still happy to see the argument using Young tableaux. $\endgroup$ – user96343 Dec 10 '15 at 6:59

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