General Relativity, Wald, exercise 4b chapter 2

[Unconscious]

Suppose we have n vector fields $Y_{\left(1\right)},\ldots,Y_{\left(n\right)}$ such that at every point of the manifold they form a basis for the tangent space at that point .
I have to prove that:

$$\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}-\frac{\partial Y_\nu^{\left(\sigma\right)}}{\partial x^\mu}=C_{\alpha\beta}^\sigma Y_\mu^{\left(\alpha\right)}Y_\nu^{\left(\beta\right)}$$

where $C_{\alpha\beta}^\gamma$ are the coefficients of the expansion: $\left[Y_{\left(\alpha\right)},Y_{\left(\beta\right)}\right]=C_{\alpha\beta}^\gamma Y_{\left(\gamma\right)}$.



My proof attempt...

From a previous exercise that the book proposes, I was able to demonstrate that the components of the commutator vector field have this form:
$\left[Y_{\left(\alpha\right)},Y_{\left(\beta\right)}\right]^\mu=Y_{\left(\alpha\right)}^\nu\frac{\partial Y_{\left(\beta\right)}^\mu}{\partial x^\nu}-Y_{\left(\beta\right)}^\nu\frac{\partial Y_{\left(\alpha\right)}^\mu}{\partial x^\nu}$.
Putting this together with $\left[Y_{\left(\alpha\right)},Y_{\left(\beta\right)}\right]=C_{\alpha\beta}^\gamma Y_{\left(\gamma\right)}$, I can write that:

$\left(Y_{\left(\alpha\right)}^\nu\frac{\partial Y_{\left(\beta\right)}^\mu}{\partial x^\nu}-Y_{\left(\beta\right)}^\nu\frac{\partial Y_{\left(\alpha\right)}^\mu}{\partial x^\nu}\right)\frac{\partial}{\partial x^\mu}=C_{\alpha\beta}^\gamma Y_{\left(\gamma\right)}^\mu\frac{\partial}{\partial x^\mu}$.

Now, let this operator act on the dual vector $Y^{\left(\sigma\right)}$ of the dual basis $Y^{\left(1\right)},\ldots,Y^{\left(n\right)}$:

$\left(Y_{\left(\alpha\right)}^\nu\frac{\partial Y_{\left(\beta\right)}^\mu}{\partial x^\nu}-Y_{\left(\beta\right)}^\nu\frac{\partial Y_{\left(\alpha\right)}^\mu}{\partial x^\nu}\right)\frac{\partial\left(Y_\rho^{\left(\sigma\right)}dx^\rho\right)}{\partial x^\mu}=C_{\alpha\beta}^\gamma Y_{\left(\gamma\right)}^\mu\frac{\partial\left(Y_\rho^{\left(\sigma\right)}dx^\rho\right)}{\partial x^\mu}$

$\left(Y_{\left(\alpha\right)}^\nu\frac{\partial Y_{\left(\beta\right)}^\mu}{\partial x^\nu}-Y_{\left(\beta\right)}^\nu\frac{\partial Y_{\left(\alpha\right)}^\mu}{\partial x^\nu}\right)Y_\mu^{\left(\sigma\right)}=C_{\alpha\beta}^\gamma Y_{\left(\gamma\right)}^\mu Y_\mu^{\left(\sigma\right)}$

Taking account of:

1) $Y_\mu^{\left(\sigma\right)}\frac{\partial Y_{\left(\beta\right)}^\mu}{\partial x^\nu}=\frac{\partial\left(Y_\mu^{\left(\sigma\right)}Y_{\left(\beta\right)}^\mu\right)}{\partial x^\nu}-Y_{\left(\beta\right)}^\mu\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}=0-Y_{\left(\beta\right)}^\mu\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}$
2) analogously $Y_\mu^{\left(\sigma\right)}\frac{\partial Y_{\left(\alpha\right)}^\mu}{\partial x^\nu}=-Y_{\left(\alpha\right)}^\mu\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}$
3) $Y_{\left(\gamma\right)}^\mu Y_\mu^{\left(\sigma\right)}=\delta_\gamma^\sigma$

we can rewrite the expression as:

$\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}\left(Y_{\left(\alpha\right)}^\mu Y_{\left(\beta\right)}^\nu-Y_{\left(\alpha\right)}^\nu Y_{\left(\beta\right)}^\mu\right)=C_{\alpha\beta}^\sigma$.

At this point, I contract on $\alpha$ and $\beta$ arriving at:

$\frac{\partial Y_\mu^{\left(\sigma\right)}}{\partial x^\nu}\left(Y_{\left(\alpha\right)}^\mu Y_\gamma^{\left(\alpha\right)}Y_{\left(\beta\right)}^\nu Y_\rho^{\left(\beta\right)}-Y_{\left(\alpha\right)}^\nu Y_\gamma^{\left(\alpha\right)}Y_{\left(\beta\right)}^\mu Y_\rho^{\left(\beta\right)}\right)=C_{\alpha\beta}^\sigma Y_\gamma^{\left(\alpha\right)}Y_\rho^{\left(\beta\right)}$

To conclude, I would need to show (for example) that $Y_{\left(\alpha\right)}^\mu Y_\gamma^{\left(\alpha\right)}=\delta_\gamma^\mu$, but I can't see how...

From duality I can only deduce this on coordinates:

$Y^{\left(\alpha\right)}Y_{\left(\beta\right)}=\delta_\beta^\alpha$

$Y_\mu^{\left(\alpha\right)}Y_{\left(\beta\right)}^\nu dx^\mu\frac{\partial}{\partial x^\nu}=\delta_\beta^\alpha$

$Y_\mu^{\left(\alpha\right)}Y_{\left(\beta\right)}^\mu=\delta_\beta^\alpha$

something that I used in the steps above, but it seems to me to be different from what is needed in the last step to conclude.

 

[PeterDonis]

Moderator's note: Thread moved to advanced physics homework help.

 

[ergospherical]

Remember that in something like $Y_{(\mu)}^{a}$, the $(\mu)$ part labels the particular vector in the basis whereas the $a$ is an index. Note that you have used Greek indices for both the vector-labels and the indices, whereas I will make it especially clear by using Latin indices (a,b,...) for the indices. The defining relation is:

$g_{ab} Y_{(\mu)}^{a} Y_{(\nu)}^{b} = \eta_{(\mu)(\nu)}$

From this, you can derive the following: $\eta_{(\mu)(\nu)} Y_a^{(\mu)} Y_b^{(\nu)} = g_{ab}$. To show that it's true, contract both sides with $Y_{(\rho)}^{b}$ to obtain

$\eta_{(\mu) (\nu)} Y_a^{(\mu)} Y_b^{(\nu)} Y_{(\rho)}^b = \eta_{(\mu)(\nu)} Y_a^{(\mu)} \delta_{(\rho)}^{(\nu)} = \eta_{(\mu)(\rho)} Y_a^{(\mu)} = (Y_{(\rho)})_a = g_{ab} Y_{(\rho)}^b$

Raising the index $b$ gets you to the equation you want: $Y_a^{(\mu)} Y_{(\mu)}^b = \delta_a^b$

 

[Unconscious]

The defining relation is:

$g_{ab} Y_{(\mu)}^{a} Y_{(\nu)}^{b} = \eta_{(\mu)(\nu)}$

Sorry for my ignorance... what is $\eta_{(\mu)(\nu)}$?

 

[ergospherical]

You can drop those brackets - I just mean the components of the matrix $\mathrm{diag}(-1,1,1,1)$, with the brackets included for consistency with the left hand side. (It's also probably more common to just drop the brackets everywhere and rely on just Greek/Latin, i.e. $Y_{\mu}^a$)

 

[Unconscious]

Ok, clear, thank you :)