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?

So, your confusion is how (2) works?

 

[Loro]

Haha sorry

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.