Inverse of tensor using Einstein notation

[bdforbes]

If I have a (1,1) tensor, eg a Lorentz transform, how do I write its inverse? For example:

x'^\mu=\Lambda^\mu_\nu x^\nu

Would I multiply on the left by:

(\Lambda^{-1})^\nu_\mu?

It seems to make sense, but I'm not 100% sure. I'd prefer to not use anything from matrix multiplication, if that's possible.

 

[Phrak]

You could put the primes on the primed coordinates as many do. After all, primed coordinates deserve primes. x is a vector. It is the same vector in any coordinate system. It doesn't deserve primes without reference to a coordinate system, where priming the coordinates takes care of that.

x ^{\mu '} = \Lambda ^{\mu '}{} _{\nu} \ x ^{\nu}

And fully decorated,

x ^{\mu '} \hat{e}_{\mu '} = \Lambda ^{\mu '}{} _{\nu} \hat{e}_{\mu '} \hat{e} ^{\nu} \ x ^{\nu} \hat{e} _{\nu}

it still looks nice--except for the oversized primes that the latex font wants to use.

 

[bdforbes]

So using the second equation you gave Phrak, how would I invert the Lorentz transform? I'm not used to using the tensor basis vectors, it was only covered briefly in my general relativity course.

 

[Phrak]

I'm not sure if you want to know how to write the inverse tensor or express it.

As to the first, it depends on the convension of the author.

You might see the inverse matrix written as
\Lambda^{\mu}{}_{\nu '}
or
\Lambda_{\nu '}\:{}^{\mu}

You will see the inverse of a Lorentz transform written with the same symbol, lambda, but they are not the same matrix.

Keeping this notation in mind,

x^{\sigma} = \Lambda^{\sigma}{}_{\mu '}\Lambda^{\mu '}{}_{\nu} x^{\nu}

must be true for all vectors, for a transform and its inverse, so

\Lambda^{\sigma}{}_{\mu '} \Lambda^{\mu '}{}_{\nu} = \delta^{\sigma}_{\nu}

 

[bdforbes]

So does that mean x^sigma and x^nu are the same vector, in the same coordinate system?

 

[cristo]

So does that mean x^sigma and x^nu are the same vector, in the same coordinate system?

Yes.