Can a valid metric always be inverted?

[Pacopag]

Homework Statement

I would like to know how to "invert" a tensor equation (see 2. Homework Equations ).

Homework Equations

x^{a} = h^{a}{ } _{b} y^{b}.
I just want to know how to get y-components in terms of x-components in general.

The Attempt at a Solution

I want to believe that h_{a}{ }^{b} h^{a}{ } _{c} = \delta^{b}{ } _{c}, but I'm not really sure that this is correct. Any help? Thanks.

 

[kdv]

Homework Statement

I would like to know how to "invert" a tensor equation (see 2. Homework Equations ).

Homework Equations

x^{a} = h^{a}{ } _{b} y^{b}.
I just want to know how to get y-components in terms of x-components in general.

The Attempt at a Solution

I want to believe that h_{a}{ }^{b} h^{a}{ } _{c} = \delta^{b}{ } _{c}, but I'm not really sure that this is correct. Any help? Thanks.

what is the definition of your h??

 

[Pacopag]

h_{a b}=g_{a b}+k_{a} N_{b} + N_{a} k_{b}

where k and N are both null vectors and N_{a} k^{a} = -1
Here, h is called the transverse metric (i.e. transverse the the null vector fields k and N.

 

[kdv]

h_{a b}=g_{a b}+k_{a} N_{b} + N_{a} k_{b}

where k and N are both null vectors and N_{a} k^{a} = -1
Here, h is called the transverse metric (i.e. transverse the the null vector fields k and N.

Then you can calculate explicitly h_a^b h^a_c, no?

 

[Pacopag]

I can expand it all out, but I don't know what to do with terms like g_{b}{}^{c} k_{c} N^{a}.
I mean, I can't do the typical contraction-like operation with the metric g.
Another thing: is g^{a}{}_{c} g_{b}{}^{c}=\delta^{a}{}_{b} ?
I'm pretty sure this is the case for the Minkowski metric, but what about a general metric?

 

[kdv]

h_{a b}=g_{a b}+k_{a} N_{b} + N_{a} k_{b}

where k and N are both null vectors and N_{a} k^{a} = -1
Here, h is called the transverse metric (i.e. transverse the the null vector fields k and N.

So h_{a }^b=g_{a}^b+k_{a} N^{b} + N_{a} k^{b}
and

h^{a }_c=g^{a}_c+k^{a} N_{c} + N^{a} k_{c}

So then you may contract them over "a"

 

[kdv]

I can expand it all out, but I don't know what to do with terms like g_{b}{}^{c} k_{c} N^{a}.
I mean, I can't do the typical contraction-like operation with the metric g.
Another thing: is g^{a}{}_{c} g_{b}{}^{c}=\delta^{a}{}_{b} ?
I'm pretty sure this is the case for the Minkowski metric, but what about a general metric?

g_{b}{}^{c} k_{c} N^{a} is just k_b N^a

 

[Pacopag]

In that case, I'm getting h^{a}{}_{c} h_{b}{}^{c}=g^{a}{}_{c} g_{b}{}^{c}+2(k_{b} N^{a} + N_{b} k^{a} +1)=g^{a}{}_{c} g_{b}{}^{c} + 2(h^{a}{}_{b}-g^{a}{}_{b}+1). This doesn't look like the kronecker delta I want in order to verify that I have the inverse of h.

 

[kdv]

Homework Statement

I would like to know how to "invert" a tensor equation (see 2. Homework Equations ).

Homework Equations

x^{a} = h^{a}{ } _{b} y^{b}.
I just want to know how to get y-components in terms of x-components in general.

The Attempt at a Solution

I want to believe that h_{a}{ }^{b} h^{a}{ } _{c} = \delta^{b}{ } _{c}, but I'm not really sure that this is correct. Any help? Thanks.

Sorry, I just went back to your first post. I am not sure why you want o invert h. why not simply apply g_a^c to both sides of your first equation?

 

[Pacopag]

g_{a}{}^{c} x^{a} =g_{a}{}^{c} h^{a}{ } _{b} y^{b}.
x^{c} = h^{c}{ } _{b} y^{b}.
I'm not getting it. Am I doing that wrong? I want to isolate the y-components in terms of the x-components. So my plan is to find the inverse of h, then apply it both sides of my first equation.

 

[Rainbow Child]

The vectors k^\alpha,\,N^\alpha are null eigenvectors of h_{\mu\nu}, i.e. h_{\mu\nu}\,N^\nu=0,\,h_{\mu\nu}\,k^\nu=0, thus h_{\mu\nu}\,h^{\nu\tau}=h_\mu{}^\tau.
Now let x^\alpha=h^\alpha{}_\beta\,y^\beta \quad (\ast) and contract with h^\gamma{}_\alpha in order to get

h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\alpha\,h^\alpha{}_\beta\,y^\beta\Rightarrow h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\beta\,y^\beta\overset{(\ast)}\Rightarrow h^\gamma{}_\alpha\,x^\alpha=x^\gamma \quad \forall x^\gamma

thus (\ast)\Rightarrow y^\alpha=x^\alpha.

 

[kdv]

The vectors k^\alpha,\,N^\alpha are null eigenvectors of h_{\mu\nu}, i.e. h_{\mu\nu}\,N^\nu=0,\,h_{\mu\nu}\,k^\nu=0, thus h_{\mu\nu}\,h^{\nu\tau}=h_\mu{}^\tau.
Now let x^\alpha=h^\alpha{}_\beta\,y^\beta \quad (\ast) and contract with h^\gamma{}_\alpha in order to get

h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\alpha\,h^\alpha{}_\beta\,y^\beta\Rightarrow h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\beta\,y^\beta\overset{(\ast)}\Rightarrow h^\gamma{}_\alpha\,x^\alpha=x^\gamma \quad \forall x^\gamma

thus (\ast)\Rightarrow y^\alpha=x^\alpha.

Ah! I was not told they were null eigenvectors of h!

 

[Pacopag]

Rainbow Child, your logic seems bulletproof, but the conclusion seems strange to me. If y^{a}=x^{a}, then doesn't that mean that h is just the identity (i.e. h^{a}{}_{b}=\delta^{a}{}_{b})? I don't think that this can be since k is arbitrary.

 

[Rainbow Child]

It should, if h^\alpha{}_\beta has an inverse. But I don't think it has!

 

[Pacopag]

Then that makes things much more difficult for me. I'll keep plugging away. Thank you kdv and Rainbow Child for your replies.

 

[Pacopag]

Does every valid metric have an inverse? If so, then I would expect that h has an inverse because its interpretation is that it is the metric on a hypersurface that is orthogonal to the congruence defined by k.