Confusion with Einstein tensor notation

[Loro]

Homework Statement

I'm confused about writing down the equation: $\Lambda \eta \Lambda^{-1} = \eta$ in the Einstein convention.

Homework Equations

The answer is: $\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho\sigma}$

However it's strange because there seems to be no distinction between $\Lambda$ and $\Lambda^{-1}$ if we write it this way.
However we know that:

$(\Lambda^{-1})^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}$

The Attempt at a Solution

If the equation was instead $\Lambda B \Lambda^{-1} = B$

Where $B$ is a tensor given in the form $B^{\mu}{}_{\nu}$ then it's clear to me how to write it:

$\Lambda^{\rho}{}_{\mu} B^{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = B^{\rho}{}_{\sigma}$

But $\eta$ is given in the form $\eta^{\mu\nu}$ and I don't understand how I can contract it with both $\Lambda^{\mu}{}_{\nu}$ and $\Lambda_{\nu}{}^{\mu}$ in order to arrive eventually at the result quoted in (2).

 

[Mandelbroth]

Homework Statement

I'm confused about writing down the equation: $\Lambda \eta \Lambda^{-1} = \eta$ in the Einstein convention.

Homework Equations

The answer is: $\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho\sigma}$

However it's strange because there seems to be no distinction between $\Lambda$ and $\Lambda^{-1}$ if we write it this way.
However we know that:

$(\Lambda^{-1})^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}$

The Attempt at a Solution

If the equation was instead $\Lambda B \Lambda^{-1} = B$

Where $B$ is a tensor given in the form $B^{\mu}{}_{\nu}$ then it's clear to me how to write it:

$\Lambda^{\rho}{}_{\mu} B^{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = B^{\rho}{}_{\sigma}$

But $\eta$ is given in the form $\eta^{\mu\nu}$ and I don't understand how I can contract it with both $\Lambda^{\mu}{}_{\nu}$ and $\Lambda_{\nu}{}^{\mu}$ in order to arrive eventually at the result quoted in (2).

Is there an actual question? :tongue:

So, your confusion is how (2) works?

 

[Loro]

Haha sorry :tongue:

I would like to know why (2) works, and possibly how I could arrive at it, starting from an expression that has both $\Lambda^{\mu}{}_{\nu}$ and $\Lambda_{\nu}{}^{\mu}$.

 

[Dick]

Homework Statement

I'm confused about writing down the equation: $\Lambda \eta \Lambda^{-1} = \eta$ in the Einstein convention.

Homework Equations

The answer is: $\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho\sigma}$

However it's strange because there seems to be no distinction between $\Lambda$ and $\Lambda^{-1}$ if we write it this way.
However we know that:

$(\Lambda^{-1})^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}$

The Attempt at a Solution

If the equation was instead $\Lambda B \Lambda^{-1} = B$

Where $B$ is a tensor given in the form $B^{\mu}{}_{\nu}$ then it's clear to me how to write it:

$\Lambda^{\rho}{}_{\mu} B^{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = B^{\rho}{}_{\sigma}$

But $\eta$ is given in the form $\eta^{\mu\nu}$ and I don't understand how I can contract it with both $\Lambda^{\mu}{}_{\nu}$ and $\Lambda_{\nu}{}^{\mu}$ in order to arrive eventually at the result quoted in (2).

Well, just raise the $\mu$ index and lower the $\rho$ index on the first $\Lambda$ in your form with the B tensor using the metric tensor.

 

[Loro]

Thanks,

Like that: ?

$\Lambda_{\rho}{}^{\mu} \eta_{\mu}{}_{\nu} \Lambda_{\sigma}{}^{\nu} = \eta_{\rho}{}_{\sigma}$

But then again both $\Lambda$'s are of the same form - this time they both seem to be inverses.