Question on index notation and metric tensor

Question

I found this expression in my SR notes:

$$ (\Lambda^{-1})^{\lambda}_{\ \ \ \sigma} = g^{\lambda\mu}~\Lambda^{\rho}_{\ \ \ \mu} ~g_{\rho\sigma} = \Lambda_\sigma^{\ \ \ \lambda}$$

I know where it comes from, so I don't need a proof, but:

1. First off, I thought that when doing matrix multiplication the dummy indices had to be next to each other, as in $\mathbf{A} \cdot \mathbf{B} = A_{ij}B_{jk} $ where $j$ is a column for $A$ and a row for $B$, while here $\mu$ is column for both $g$ and $\Lambda$....

2. Why is this expression true? Maybe I am missing something, but what is the action of the $g$'s on $\Lambda$?

3) Is $\Lambda_\sigma^{\ \ \ \lambda}$ the transpose?

Answer 1

For doing matrix multiplication, yes, we usually put the dummy indicies together, but this is (assuming one is not working over $\mathbb{H}$) convention really.

Don't forget that this is shorthand Einstein convention.

When you write

$$ \mathbf{A} \cdot \mathbf{B} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} $$

you are writing a shorthand version in Einstein convention. What this is for a single matrix element is

$$ (\mathbf{A} \cdot \mathbf{B})_{i j} = A_{ij} B_{jk} = \sum_{j=1}^n A_{ij} B_{jk} = \sum_{j=1}^n B_{jk} A_{ij} $$

assuming that your matrices are $n\times n$ matrices, since real-valued (or complex-valued) numbers commute.

However, if we always remember to obey the Einstein convention, for tensor calculus, it doesn't really matter$^{\dagger}$, since our metric in Special Relativity is symmetric.

So I could write

$$ g^{\mu \nu} = \Lambda^{\mu}_{\space \space \alpha} \Lambda^{\nu}_{\space \space \beta} g^{\alpha \beta} = \Lambda^{\mu}_{\space \space \alpha} g^{\alpha \beta} \Lambda^{\nu}_{\space \space \beta} = g^{\alpha \beta} \Lambda^{\mu}_{\space \space \alpha} \Lambda^{\nu}_{\space \space \beta} $$

say, since we know that

$$ g^{\mu \nu} = g^{\nu \mu} $$

as long as we're careful in summing over the indices as indicated, these are all the same.

Now, as for you're second question, I don't exactly know what you're asking, since you say that you understand where the expression comes from. The action of the metrix tensor is to raise and lower indices.

So for the first of your matric tensors in the second expression, $ g^{\lambda \mu} $, say, we see that it is contracted with the $\mu$ index of your Lorentz transformation tensor and it raises this index to a $\lambda$.

Since $ g^{\lambda \mu}$ is symmetric as I say, maybe you would see this more clearly if written as

$$ g^{\lambda \mu} \Lambda^{\rho}_{\space \space \mu} = g^{ \mu \lambda} \Lambda^{\rho}_{\space \space \mu} = \Lambda^{\rho \lambda} $$

Sometimes it's easier to see if the contracted indices are 'on the same side', the left in this case , and sometimes if the free index, $\lambda$ here, is on the 'same side as it will end up', again, the left in this case.

As for your final question, well, I've never thought about that before, but for $\Lambda$ a Lorentz transformation we have

$$ \Lambda \in \mbox{SO}(3,1) $$

in which case I would say yes, the inverse is just the transpose for orthogonal matrices

$$ \Lambda \Lambda^{T} = I $$

so

$$ \Lambda^{T} = \Lambda^{-1} $$

I hope that helps!

$\dagger$ I should put in a disclamer, when you're doing Supersymmetry with Weyl fermions your 'metric' is antisymmetric, so in that case it does matter, beause

$$ \epsilon^{\mu \nu} = - \epsilon^{\nu \mu} $$

and there is a distinction about summing 'North-East', $\nearrow$ and 'South-East' $\searrow$.

I realise that this doesn't apply to you here, since you're doing Special Rel., but I just didn't want you to think that this is always the case.