So, consider the $n$-dimensional vector space $V$, on a field $\mathbb{F}$, with a chosen basis $\{e_i\}$ and its dual space $V^*$ with the chosen basis $\{e^j\}$ with the property $e^j(e_i)=\delta^j_i$.

Now we can define the vector space $$\underbrace{V\otimes...\otimes V}_{k\text{-copies}}\otimes\underbrace{V^*\otimes...\otimes V^*}_{l\text{-copies}}\equiv \otimes^kV\otimes^lV^*$$ with dimension $n^{k+l}$ and a basis $\{e_{a_1}\otimes...\otimes e_{a_k}\otimes e^{b_1}\otimes...\otimes e^{b_l}\}_{a_i=1,...,n}^{b_j=1,...,n}$.The elements of this vector space are all the multilinear functions
$$ T:\underbrace{V^*\times...\times V^*}_{k\text{-copies}}\times\underbrace{V\times...\times V}_{l\text{-copies}}\to \mathbb{F}, $$ called $(k,l)$-tensors $T\in\otimes^kV\otimes^lV^*$. Then, one can identify the subspace of $\otimes^kV\otimes^lV^*$ that contains all the symmetric $(k,l)$-tensors and the subspace of $\otimes^kV\otimes^lV^*$ that contains all the alternating $(k,l)$-tensors. It is known that the $(2,0)$- and $(0,2)$-tensors can be decomposed into a symmetric and an alternating part, a fact that doesn't hold for the other cases.

I can intuitively see that dimensionally-wise. For example, the subspace of $\otimes^kV\otimes^lV^*$ that contains all the symmetric $(k,l)$-tensors has a dimension of $\binom{n+k-1}{k}\binom{n+l-1}{l}$, while the one that contains the alternating tensors has a dimension of $\binom{n}{k}\binom{n}{l}$. Thus, the sum of these dimensions equals $n^{k+l}$ only for $(k=0 , l=2)$ and $(k=2 , l=0)$.

The above observation seems indicative of the fact that only $(2,0)$- and $(0,2)$-tensors can be decomposed into a symmetric and an alternating part but it is not a proof.

Is my thinking correct, regarding this dimensionality criterion? Can someone direct me to a proof for this Theorem of $(2,0)$- and $(0,2)$-tensor decomposition?

  • $\begingroup$ Your reasoning that there can't be a direct sum decomposition into symmetric and alternating tensors when $(k,l)$ is not $(2,0)$ or $(0,2)$ is correct. All that remains is a straightforward proof that you indeed have such a decomposition for $(2,0)$- and $(0,2)$-tensors. $\endgroup$
    – Christoph
    Sep 16 '18 at 12:31
  • $\begingroup$ @Christoph Thank for the response! $\endgroup$
    – Ozz
    Sep 16 '18 at 12:33

Let $\mathbb F$ be a field of characteristic different from $2$. On $V\otimes V$ you have an automorphism $T$ defined by $T(v\otimes w)=w\otimes v$ on simple tensors, since $w\otimes v$ is a bilinear expression in $(v,w)$. Note that for each $x\in V\otimes V$ you have $$ x = \frac{1}{2} \bigg(x + T(x)\bigg) + \frac{1}{2} \bigg(x - T(x)\bigg), $$ where the first summand is a symmetric tensor and the second is alternating.

Hence $V\otimes V$ is the sum of the subspaces of symmetric and alternating tensors. It is direct since the only $(2,0)$-tensor that is both symmetric and alternating is the $0$ tensor (since $\operatorname{char}(\mathbb F)\neq 2$).

The same argument applies to $V^*\otimes V^*$.

Note that the statement is false in characteristic $2$ where $v\otimes v$ is both symmetric and alternating for all $v\in V$.


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