3
$\begingroup$

The document I am reading about tensors claims the following:

In practice it will turn out to be very useful to also introduce this convention [raised and lowered indices] for matrices. Without further argumentation (this will be given later) we note that the matrix $A$ can be written as: $$A:\quad{A^\mu}_{\nu}.\tag{2.19}$$ The transposed version of this matrix is $$A^T:\quad{A_\nu}^\mu.\tag{2.20}$$

However, I cannot find this “further argumentation.” When I have a rank-$2$ tensor, how do I know which indices to raise, if any, and which indices to lower, if any? Does the raising of one index and the lowering of the other indicate that the object is a matrix specifically?

As I understand it, one raised and one lowered index is supposed to evoke that the object has mixed variant and contravariant transformational properties. I can not be for sure, however, because the paper mentions every mixture of index positions.

Edit: If you could explain how, when looking at a $3\times3$ rank-$2$ tensor / array, one knows where to locate the indices, that would also be very helpful. For example, when dealing with a contravariant/column vector, I immediately know the index is raised (at least before manipulation).

$\endgroup$
3
$\begingroup$

Hmm this is a good question! I mean to make things a bit easier...I have always heard that the best way to denote the different placement of the indices with dots...so instead of just having $A^\mu_{\phantom{...}\nu}$ or $A^{\phantom{...}\nu}_{\mu}$ its always goot to put a dot above the index which has been altered. So for example if you have the matrix $A^\mu_{\phantom{..}\nu}$ and lower the first index, then the result is $A^{\bullet}_{\mu\nu}$ and if you then raised the second, the result is $A^{\bullet\phantom{.}\nu}_{\mu\bullet}$...thats just how I learned though! So up to you whatever feels good. The only issue that could happen if you don't is that for some tensors, the position of the indices is VERY important so for example things could get confusing if you dont put dots with tensors such as $A^{\mu\phantom{..}\nu\phantom{..}\sigma}_{\phantom{..}\lambda\phantom{..}\zeta}$ (just as an example). So I will be doing this

Now for example I suppose the best way to prove this for tensors is just to show you the linear algebra 'sense' of it all. We know that a transpose will essentially 'reflect' all the components of the matrix. This makes it easy for a Rank 2-tensor of a single order ($A^{\mu\nu}$ or $A_{\mu\nu}$)...for example: $$(A^{\mu\nu})^T=A^{\nu\mu}$$ but when it is mixed things are trickier. The only way I can think of explaining it is that you know that the metric tensor and its inverse raises and lowers indices and you also know that: $$(AB)^T=B^TA^T$$ The cool thing about dealing with matrix components is that you dont need to worry about the placement of the components (in the sense of which matrix is multiplied first) so consider $A^\mu_{\phantom{..}\nu}=g^{\mu\sigma}A^\bullet_{\sigma\nu}$ and then transopose each individually (since the metric tensor is symmetric it is its own transpose)...youll find that you'll get the book's result! Keep at it though! Thats the document I used myself as my introduction to tensor calculus

$\endgroup$
  • $\begingroup$ I think I pretty much understand what you’re saying. For matrices, there isn’t much difference between ${A_\mu}_\nu}$, ${A_\mu}^\nu$, ${A^\mu}_\nu$, and ${A^\mu}^\nu$, right? $\endgroup$ – gen-z ready to perish Nov 18 '17 at 2:05
  • 1
    $\begingroup$ Pretty much! I mean except for the placement of the indices after you transpose them. So for clarity: $(A_{\mu\nu})^T=A_{\nu\mu}$ , $(A_{\mu}^{\phantom{..}\nu})^T=A^{\nu\bullet}_{\bullet\mu}$ , $(A^{\mu}_{\phantom{..}\nu})^T=A_{\nu\bullet}^{\bullet\mu}$ , $(A^{\mu\nu})^T=A^{\nu\mu}$ $\endgroup$ – Keith Afas Nov 18 '17 at 8:13

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.