0
$\begingroup$

I was working on a text that made use of the assumption that, for some rotation matrix $a_{ij}$, and some tensor $U$, the tensor under rotation is represented by $U'_{\alpha}=a_{\alpha i}U_i$. It made use of this (and the outer product) to prove that: $$ T'_{\alpha\beta\gamma\delta}=a_{\alpha i}a_{\beta j}a_{\gamma k}a_{\delta l}T_{ijkl} $$

Where $T_{ijkl}=U_{ij}\otimes V_{kl}$.

Is there a simple proof to show that any tensor can be decomposed into a product of $n$-tensors? (as it seems that the proof given above heavily relies on this to be the case in order to work)

Thank you.

$\endgroup$

1 Answer 1

1
$\begingroup$

No, you need the tensor product $\otimes$ to construct the most general tensors. The wedge product $\wedge$ produces skew-symmetric objects.


Well, the space of tensors on a vector space V is constructed by taking tensor products of the form $V \otimes V \otimes \ldots V \otimes V^* \otimes V^* \ldots V^*$ where $V$ is the vector space and $V^*$ is its dual space. So an (m,n) type tensor can always be written as $$ \mathbf{T} = T^{ijk\ldots}_{pqr\ldots} \hat{e}_{i} \otimes \hat{e}_{j} \ldots \otimes \hat{f}^{p} \otimes \hat{f}^{q} \ldots $$ in terms of the basis vectors and co-vectors. So, if you insist on strict decomposition of a tensor, that's not possible always. You have to allow for linear combinations of the elements, as shown in the formula above. And that's just for decomposition into vectors / co-vectors.

A tensor of the form you've used here (T) is a very special case. A general (0,4) tensor cannot be written as the product of two (0,2) tensors.

$\endgroup$
4
  • $\begingroup$ Fixed, thank you; for some reason I just immediately thought outer-product. $\endgroup$ Aug 3, 2014 at 6:53
  • $\begingroup$ @GuillermoAngeris I've just updated my answer. $\endgroup$ Aug 3, 2014 at 7:16
  • $\begingroup$ @GuillermoAngeris I just edited my reply to something more accurate. $\endgroup$ Aug 3, 2014 at 7:23
  • $\begingroup$ okay, that's what I had imagined originally, but the book does not state. Thank you. $\endgroup$ Aug 3, 2014 at 18:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.