Index Gymnastics: Matrix Representations & Rank-2 Tensor Components

[SiennaTheGr8]

I'm trying to get the hang of index gymnastics, but I think I'm confused about the relationship between rank-2 tensor components and their matrix representations.

So in Hartle's book Gravity, there's Example 20.7 on p. 428. We're given the following metric:

$g_{AB} = \begin{bmatrix} F & 1 \\ 1 & 0 \end{bmatrix} \qquad \qquad g^{AB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix}$

And we're given the following components of a tensor:

$t_{AB} = \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix}$

Then we're asked to calculate ${t^A}_B$ and $t^{AB}$ (etc.).

Okay:

${t^A}_B = g^{AC} t_{CB} = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} G & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}$

That's correct.

Now, for $t^{AB}$ I was thinking:

$t^{AB} = g^{AC} {t_C}^B$,

which is correct, but I didn't have ${t_A}^B$ yet, so I had to approach it another way. I thought maybe:

$t^{AB} = g^{CB} {t^A}_C = \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix}$.

But that's actually the wrong order for the matrix multiplication (though if I'm not mistaken, $g^{CB} {t^A}_C = {t^A}_C g^{CB}$ -- is that right?). Instead, the correct answer is:

$t^{AB} = \begin{bmatrix} -1 & 0 \\ G+F & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & -F \end{bmatrix} \Big( \stackrel{?}{=} {t^A}_C g^{CB} \Big)$.

So I suppose what I'm asking is, first, whether it's indeed true that $g^{CB} {t^A}_C = {t^A}_C g^{CB}$, and then second, what should have tipped me off that I needed to "reverse" the order of the matrices representing $g^{CB}$ and ${t^A}_C$ when calculating $t^{AB}$?

Hope that's clear enough.

 

[Orodruin]

So I suppose what I'm asking is, first, whether it's indeed true that $g^{CB} {t^A}_C = {t^A}_C g^{CB}$, and then second, what should have tipped me off that I needed to "reverse" the order of the matrices representing $g^{CB}$ and ${t^A}_C$ when calculating $t^{AB}$?

Yes, that is true. What should have tipped you off is that the index that is contracted with $g^{CB}$ is the second index of $t$, not the first. In the matrix representation, the first index is the row index and the second the column index. Note that the metric is symmetric so it does not matter whether you multiply by it or its transpose. What matters is where you put it when you deal with the matrix representation.