From Einstein Summation to Matrix Notation: Why?

[epovo]

TL;DR Summary

How does the change from Einstein summation convention to matrix multiplication work?

I know that if $\eta_{\alpha'\beta'}=\Lambda^\mu_{\alpha'} \Lambda^\nu_{\beta'} \eta_{\alpha\beta}$
then the matrix equation is
$$ (\eta) = (\Lambda)^T\eta\Lambda $$
I have painstakingly verified that this is indeed true, but I am not sure why, and what the rules are (e.g. the $(\eta)$ is in the middle but why?- My text book uses this without proof.
Where can I find a discussion of these topics?

 

[epovo]

Sorry that should have been $\eta^{\alpha\beta}$ at the end

 

[pasmith]

Recall the definiition of matrix multiplication: $$(AB)_{ij} = \sum_k A_{ik}B_{kj}.$$ Thus you need the index contracted over to be the second index of the first factor and the first index of the second factor. Swapping the indices is the same as taking the transpose, since $(A^T)_{ij} = A_{ji}$.

 

[epovo]

That's what I missed! Thank you very much

 

[PeroK]

I know that if $\eta_{\alpha'\beta'}=\Lambda^\mu_{\alpha'} \Lambda^\nu_{\beta'} \eta_{\alpha\beta}$

This is not correct. You should have repeated (summation) indices on the RHS. It should be:
$$\eta_{\alpha'\beta'}=\Lambda^\mu_{\alpha'} \Lambda^\nu_{\beta'} \eta_{\mu\nu}$$Although, in fact, there is no need to have primes on the new indices. You could equally have:
$$\eta_{\alpha\beta}=\Lambda^\mu_{\alpha} \Lambda^\nu_{\beta} \eta_{\mu\nu}$$

 

[epovo]

This is not correct. You should have repeated (summation) indices on the RHS. It should be:
$$\eta_{\alpha'\beta'}=\Lambda^\mu_{\alpha'} \Lambda^\nu_{\beta'} \eta_{\mu\nu}$$Although, in fact, there is no need to have primes on the new indices. You could equally have:
$$\eta_{\alpha\beta}=\Lambda^\mu_{\alpha} \Lambda^\nu_{\beta} \eta_{\mu\nu}$$

You are right. My mistake

 

[PeroK]

You are right. My mistake

In fact, this is quite important. One definition of a Lorentz Transformation is that it leaves the Minowski metric invariant. Your equation is not the general transformation rule, where the components of the metric tensor $\eta$ are transformed into different components in a new coordinate system. Instead, it's a statement that the Lorentz Transformation leaves the components unchanged.

 

[vanhees71]

Another utmost important point is to be concise in the Ricci (index) notation. It is important to keep both the vertical (co- versus contravariant tensor components) AND the horizontal position of the indices clean, i.e., you should write ${\Lambda^{\mu}}_{\nu}$ rather than $\Lambda_{\nu}^{\mu}$. The latter notation is not uniquely defining the object properly (in this case a Lorentz-transformation matrix).
In the former notation it's clear that in the matrix formalism $\mu$ (1st index) labels the rows and $\nu$ (2nd index) the columns. Then you equation reads
$$\eta_{\rho \sigma} {\Lambda^{\rho}}_{\mu} {\Lambda^{\sigma}}_{\nu}=\eta_{\mu \nu}.$$
and it's clear that by the definition of the matrix product, which follows the rule "row $\times$ column", must read
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta}.$$

 

[epovo]

Another utmost important point is to be concise in the Ricci (index) notation. It is important to keep both the vertical (co- versus contravariant tensor components) AND the horizontal position of the indices clean, i.e., you should write ${\Lambda^{\mu}}_{\nu}$ rather than $\Lambda_{\nu}^{\mu}$. The latter notation is not uniquely defining the object properly (in this case a Lorentz-transformation matrix).
In the former notation it's clear that in the matrix formalism $\mu$ (1st index) labels the rows and $\nu$ (2nd index) the columns. Then you equation reads
$$\eta_{\rho \sigma} {\Lambda^{\rho}}_{\mu} {\Lambda^{\sigma}}_{\nu}=\eta_{\mu \nu}.$$
and it's clear that by the definition of the matrix product, which follows the rule "row $\times$ column", must read
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta}.$$

This question of keeping the horizontal position of the indices has been baffling me for some time. My text book (Schutz) does it, but it never explains why. So I figured this was the use of those spaces, but since Schutz makes an occasional error, I was confused. This is the first time I see it spelled out. Thanks for this.

 

[vanhees71]

Indeed, there are tons of textbooks, which don't care about the horizontal placement of indices. Obviously the authors are too lazy to put the brackets in the LaTeX or, even better, to use the tensor package ;-).

 

[Ibix]

To be fair to Schutz, he does care about index placement. But as @epovo says, I don't think he explains why. This is part of why I feel he needed a better editor.

Carroll's lecture notes do explain, in chapter 2 I think, if you want to download and have a read.

An example of the importance of index ordering is something like the Riemann tensor, $R^a{}_{bcd}$. This completely captures curvature of spacetime, and one of the things it does is give you the change in a vector if you transport it around a small loop. So if you have a vector $V^b$ and two small displacement vectors $dx^c$ and $dy^d$ then the change in $V^b$ from parallel transporting around the parallelogram defined by the small displacements is $R^a{}_{bcd}V^bdx^cdy^d$. So you can see that each index has a particular function in this application: the first is the output vector, the second matches to the input vector, and the third and fourth match to the loop definition vectors. Mixing up the inputs (e.g. $R^a{}_{bcd}V^ddx^bdy^c$) will give you an answer, but not to the question you think you are asking.

The problem with index placement arises when you start raising and lowering indices. I mean, $R^a_{bcd}$ would be unambiguous if it were only ever that same index that was raised. But I can compute $g_{ea}g^{fc}R^a{}_{bcd}=R_{eb}{}^f{}_d$. And if I ignore index placement and write $R^f_{ebd}$, which tensor slot did I have raised again...?