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(This is a close retelling of Wald, problem 2.4b. Not for homework; just curiosity and an increasingly alarming suspicion that I've never actually understood anything.)

Let $Y_1 ... Y_n$ be a collection of smooth vector fields on an $n$-dimensional manifold $M$ such that at every point they form a tangent vector space. Let $Y^{*1} ... Y^{*n}$ be the corresponding dual basis. Show that the components $(Y^{\gamma *})_\mu$ of $Y^{\gamma *}$ in any coordinate basis satisfy

$$\frac{\partial (Y^{\gamma *})_\mu}{\partial x^\nu} - \frac{\partial (Y^{\gamma *})_\nu}{\partial x^\mu} = \sum_{\alpha,\beta} C^{\gamma}_{\;\;\alpha\beta}(Y^{\alpha *})_\mu (Y^{\beta *})_\nu$$

The hint is that you should contract both sides with $(Y_\sigma)^\mu (Y_\rho)^\nu$. Also, presumably relevant is that the commutator of $Y_\alpha$ and $Y_\beta$ can be expressed $[Y_\alpha,Y_\beta] = \sum C^{\gamma}_{\;\alpha\beta} Y_\gamma$ for some $C^{\gamma}_{\;\;\alpha\beta}$.

Now, my brain is encountering a parse error on this problem, particularly the left-hand side. Thoughts and questions:

  1. This is how the prompt sounds to me: we introduce some coordinates with an associated coordinate basis $\{v_\mu\}$ such that $Y_\gamma = \sum (Y_\gamma)^\mu v_\mu$ with components $(Y_\gamma)^\mu$ in that coordinate basis. Then we take the corresponding dual basis $\{v^{\mu *}\}$ and express every $Y^{\gamma *}$ as $Y^{\gamma *} = \sum (Y^{\gamma *})_\mu v^{\mu *}$. And these $(Y^{\gamma *})_\mu$ components are what we have in the problem. But... what happens now? I don't understand what it means to contract with the aforementioned $(Y_\sigma)^\mu (Y_\rho)^\nu$ when that component is in a partial derivative, as it is in the LHS. Since these are tangent spaces, should the derivative itself be taken as a basis vector in the coordinate basis?

  2. When trying to contract the RHS, I get this:

$$\sum_{\alpha,\beta} C^{\gamma}_{\;\;\alpha\beta}\sum_\mu (Y^{\alpha *})_\mu (Y_\sigma)^\mu \sum_\nu(Y^{\beta *})_\nu (Y_\rho)^\nu$$

Now, for any coordinate basis vector $\{v_\nu\}$ and dual vector from the corresponding dual basis $\{v^{\mu *}\}$, we have $v^{\mu *} (v_\nu) = \delta^{\mu}_{\;\nu}$, and thus

$$Y^{\mu *} (Y_\nu) = \delta^{\mu}_{\;\nu} = \sum_\kappa (Y^{\mu *})_\kappa v^{\kappa *} \sum_\eta (Y_{\nu})^\eta v_{\eta} = \sum_\kappa (Y^{\mu *})_\kappa (Y_{\nu})^\kappa$$

So the RHS just reduces to

$$\sum_{\alpha,\beta} C^{\gamma}_{\;\;\alpha\beta} \delta^{\alpha}_{\;\sigma} \delta^{\beta}_{\;\rho} = C^{\gamma}_{\;\;\sigma\rho}$$

Likewise, from here it's not clear to me how to proceed. A former physics professor advised "relax and let the math carry you", but that doesn't seem to be working.

So... Is there something here that jumps out as a flaw or gap in my understanding? I assume there is a methodical way of doing this, but I can't grasp it. Would somebody be so kind as to give a pointer on how to keep going?

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    $\begingroup$ "[...] increasingly alarming suspicion that I've never actually understood anything" We've all been there, many times in my case! $\endgroup$ – Danu Feb 2 '16 at 9:42
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I think you have all the right pieces to answer the question, here are a few hints that should be of some use.

You say that you picked coordinates $ \{v^{\mu} \}$. It seems to me that they should instead be called $ \{ x^{\mu} \}$, as that is what you're taking partial derivatives with respect to. As you correctly pointed out, you are working with components of vectors and dual vectors. These are defined in a particular chart, and they should depend on a particular coordinate.

In other words a slightly more pedantic way of writing your expression for the (co)vectors is $Y^{\gamma\ast} = \sum (Y^{\gamma \ast})_{\mu} (x) x^{\mu} $. Here $(Y^{\gamma \ast})_{\mu} (x) $ can be viewed as $n$ functions of the coordinate $x$.

Now to address each point: For the RHS, try using the chain rule backwards (something like $ \partial (u) v = \partial (u v) - u \partial(v) $ )after hitting the expression with the vectors.

Also, you mentioned the commutators of two vectors. Remember how the commutator acts on the components of these vectors, and keep only the component. See if it is reminiscent of anything you have in your computation.

Hopefully that helps you get started!

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  • $\begingroup$ Thank you, you da best! The reverse product rule thing had vaguely occurred to me, but didn't congeal until you said it. It makes a lot of sense if you think of $Y^{\gamma *}$ as functions of $x$. Wald had said as much, but I hadn't realized the full significance. $\endgroup$ – 0x00019913 Feb 3 '16 at 3:02

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