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Let $V$ be a real finite-dimensional vector space, let $\alpha$ be a symmetric $k-$tensor on $V$, and let $\beta$ be a symmetric $l-$tensor on $V$. Define the symmetric product of $\alpha$ and $\beta$ to be the $\left(k+l\right)$-tensor $\alpha\beta$ defined by $\left(\alpha\beta\right)\left(v_{1},\ldots,v_{k+l}\right)=\frac{1}{\left(k+l\right)!}\sum_{\sigma\in S_{k+l}}\alpha\left(v_{\sigma(1)},\ldots,v_{\sigma(k)}\right)\beta\left(v_{\sigma(k+1)},\ldots,v_{\sigma(k+l)}\right)$ for all $v_{1},\ldots,v_{k+l}\in V$. Prove that $\alpha\beta=\frac{1}{2}\left(\alpha\otimes\beta+\beta\otimes\alpha\right)$, where $\otimes$ is the tensor product of multilinear maps.

I can't seem to find a way to show that they are equal. Any help would be appreciated.

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  • $\begingroup$ Maybe you can try proving this for basis tensors? Also, make sure your definition of $\alpha\beta$ that has the reciprocal of factorial is consistent, since some other authors define it without the factorial coefficient. $\endgroup$ – Vim Jun 1 '17 at 12:46
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This is false. For example let $\alpha$ be the dot product for some basis:

$$\alpha(e_i,e_j)=\delta_{ij}$$

and let $\beta=\alpha$. These are certainly symmetric. Then $\frac 12 (\alpha\otimes\beta+\beta\otimes\alpha)=\alpha\otimes\alpha$.

But for $i\neq j$

$$(\alpha\otimes\alpha)(e_i,e_i,e_j,e_j)=\alpha(e_i,e_i)\alpha(e_j,e_j)=1$$

while

$$(\alpha\otimes\alpha)(e_i,e_j,e_i,e_j)=\alpha(e_i,e_j)\alpha(e_i,e_j)=0$$

so $\alpha\otimes\alpha$ is not symmetric.

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  • $\begingroup$ It seems that this is true for 1-tensors in general, not an arbitrary k-tensors. $\endgroup$ – Eigenfield Jun 3 '17 at 2:33

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