# Inverting a tensor equation

1. Feb 18, 2008

### Pacopag

1. The problem statement, all variables and given/known data
I would like to know how to "invert" a tensor equation (see 2. Relevant equations).

2. Relevant 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.

3. 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.

Last edited: Feb 18, 2008
2. Feb 18, 2008

### kdv

what is the definition of your h??

3. Feb 18, 2008

### 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.

4. Feb 18, 2008

### kdv

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

5. Feb 18, 2008

### 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?

Last edited: Feb 18, 2008
6. Feb 18, 2008

### kdv

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"

7. Feb 18, 2008

### kdv

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

8. Feb 18, 2008

### 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.

Last edited: Feb 18, 2008
9. Feb 18, 2008

### kdv

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?

10. Feb 18, 2008

### 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.

Last edited: Feb 18, 2008
11. Feb 18, 2008

### 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 $(\ast)\Rightarrow y^\alpha=x^\alpha$.

12. Feb 18, 2008

### kdv

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

13. Feb 18, 2008

### 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.

Last edited: Feb 18, 2008
14. Feb 18, 2008

### Rainbow Child

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

15. Feb 18, 2008

### Pacopag

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

16. Feb 18, 2008

### 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.