Homework Help: Quick question regarding matrix/index notation

1. Jun 13, 2015

Anchovy

Attached is a screenshot of a text I'm trying to follow. However, the author does something that I don't quite understand in line (3.5). They equate the following:
$$A^{\mu}_{\hspace{2 mm}\alpha} x^{\alpha} (A^{-1})^{\beta}_{\hspace{2 mm}\mu} y_{\beta} = (A^{-1})^{\beta}_{\hspace{2 mm}\mu} A^{\mu}_{\hspace{2 mm}\alpha} x^{\alpha} y_{\beta}$$
So they've taken the $(A^{-1})^{\beta}_{\hspace{2 mm}\mu}$ that was initially operating on $y_{\beta}$ and moved it backwards so that it now operates on $A^{\mu}_{\hspace{2 mm}\alpha} x^{\alpha} y_{\beta}$. My issue is that I don't know why this is allowed, it would not occur to me to do that. This 'trick', if one could call it that, also appears to be repeated in (3.7). Can someone explain what's going on here / why this stuff is allowed?

Another thing that keeps bothering me is regarding dummy indices, specifically, when to introduce a new one. In this example (line (3.5)) we see
$$s' = x'^{\mu}y'_{\mu} = A^{\mu}_{\hspace{2 mm}\alpha} x^{\alpha} (A^{-1})^{\beta}_{\hspace{2 mm}\mu} y_{\beta}$$
So for the coordinate transformation $x'^{\mu} = A^{\mu}_{\hspace{2 mm}\alpha} x^{\alpha}$ they've introduced one new dummy index, namely $\alpha$, which is fair enough, but then for $y'_{\mu} = (A^{-1})^{\beta}_{\hspace{2 mm}\mu} y_{\beta}$ they've introduced the index $\beta$. Why not just use $\alpha$ here?

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2. Jun 13, 2015

Dick

For the first question, you have the product of components of tensors. Components of tensors are just numbers. Numbers are commutative. You can order them any way you like. For the second, remember the Einstein summation convention. If an index occurs twice, it is assumed to be summed over. Your expression has two separate sums. Hence two separate repeated dummy indices. If they were all the same, that would imply only one sum.

3. Jun 13, 2015

Anchovy

Ok thanks Dick. A further question I have is, the author has done the following:
$$(A^{-1})^{\beta}_{\hspace{2 mm}\mu} A^{\mu}_{\hspace{2 mm}\alpha} = \delta^{\beta}_{\hspace{2 mm}\alpha}$$
Now if we were talking about a straightforward product of a matrix, say M, and its inverse, we'd have $M^{-1}M = \mathbb{1}$. That's simple enough. However, I can see that we're again dealing with $A$ and it's inverse $A^{-1}$, but now we're dealing not with whole matrices, only with just components/numbers like you say, as well as fiddly little indices --> the fact that $(A^{-1})^{\beta}_{\hspace{2 mm}\mu} A^{\mu}_{\hspace{2 mm}\alpha}$ results in a simple Kronecker delta does not immediately leap off the page at me. How come this is true?

4. Jun 13, 2015

Dick

Because you know that product is the $\alpha$, $\beta$ component of the identity matrix. The identity matrix is ones along the diagonal (when $\alpha=\beta$) and zero off the diagonal. Isn't that a Kronecker delta?

5. Jun 13, 2015

Anchovy

How do I know this if I don't know explicitly what A looks like?

6. Jun 13, 2015

Dick

You don't need to know what $A$ looks like. You only need to know what the identity matrix looks like. The product of a matrix and its inverse is always the identity. Doesn't matter what the matrix is. You DO have the case $A^{-1}A=I$, that arrangement of indices denotes a matrix product. Look up the definition of matrix product.

7. Jun 14, 2015

Fredrik

Staff Emeritus
Don't forget that the definition of matrix multiplication is $(AB)^i_j=A^i_k B^k_j$. (This is how it's written when we use the summation convention and the convention to write the row index upstairs and the column index downstairs).

Last edited: Jun 14, 2015
8. Jun 18, 2015

Anchovy

thanks guys I'm doing OK now.