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What's the easiest way to show that the covariant derivative $\nabla U^{\mu}$ is linearly independent to $U^{\mu}$, which is a vector?

I mean I'm assuming they are since I'm proving the second Bianchi identity and my proof requires this to be the case; and I'm quite confident that the Bianchi identity is true!

So I start with the Jacobi identity


I then substitute in the definition of the Riemann tensor, giving me

$$\nabla_a(R^d_{\space\space\space ebc}U^e)+\text{cyclic}=0$$

which gives

$$\nabla_a(R^d_{\space\space\space ebc})U^e+R^d_{\space\space\space ebc} \nabla_aU^e+(\text{cyclic in }a,b,c)\space\space=0.$$

So we have 6 expressions, from which if the second Bianchi identity is true, the vector and the covariant derivate of the vector must be linearly independent.

Or am I just OK to say that as the covariant derivative is (1,1) tensor and a vector is a (1,0) tensor, the result must follow straight away?

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It sounds like you have the stress-energy tensor of a perfect fluid. Is that the case? –  Muphrid May 24 '14 at 16:05
Could you show what you tried to prove this yourself so far, please? –  Neuneck May 24 '14 at 16:10
What you're asking if the definition of the covariant derivative or close to it, depending on how you define it. –  auxsvr May 24 '14 at 16:47
The covariant derivative $U^\mu{}_{;\nu}$ is a rank two tensor, whereas $U^\mu$ is a rank one tensor. Are you sure this is what you mean? –  Robin Ekman May 24 '14 at 16:57
@Neuneck see edit –  Phibert May 24 '14 at 17:34

3 Answers 3

In general, it is false that $U$ and $\nabla_XU$ are linearly independent, however it does not matter here, since the wanted proof does not need such a general statement.

If $M$ is your differentiable manifold with dimension $n$, consider a point $p\in M$ and a geodesically normal coordinate system centered on $p$. In that coordinate system $\Gamma^a_{bc}(p)=0$. Therfore, if a vector field $U$ has constant components in that coordinate system, exactly at $p$, one has $\nabla_aU^e=0$. There are $n$ linearly independent such vector fields. Inserting them in the identity you are considering, you have $U^e(p) \nabla_a(R^d_{\space\space\space ebc})|_p+0+$(cyclic in a,b,c)$\space\space=0$. As the number of linearly independent fields $U$ is $n$, it means that $$\nabla_a(R^d_{\space\space\space ebc})|_p+\mbox{(cyclic in a,b,c)}\space\space=0\:.$$ Since we have used covariant derivatives, the result does not depend on the adopted coordinate system. Since $p$ is arbitrary, the found identity holds everywhere on $M$.

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Just reading your first statement saying that it's false; there's a difference between $\nabla_X U$ and $\nabla_a U$ so in my questions, surely they are linearly independent? If it's false, then you're saying me can make a matrix my multiplying a vector by a scalar, so to speak? –  Phibert May 24 '14 at 18:59
Nope, $\nabla_a U= \nabla_X U$ for $X= \partial/\partial x^a$. –  Valter Moretti May 24 '14 at 19:00
Oh sure, in that case you're right. But since when are we considering basis vectors? Can we just do that? –  Phibert May 24 '14 at 19:01
And anyway, I'm struggling to see how a (1,1) tensor and a (1,0) tensor aren't clearly linearly independent? Even though this was my original question! –  Phibert May 24 '14 at 19:06
But if we take it down to the very elementary linear algebra, how can we make a (1,1)-tensor from a (1,0) tensor by scalar multiplication and addition? Unless I've forgotten the definition of linear independence from calculus? –  Phibert May 24 '14 at 19:11

While V. Moretti's proof is entirely correct and the standard one taught (at least it is the first one I learned, and it's the one given in the text of Misner, Thorne and Wheeler, it is also the one given by Weinberg), the Bianchi identity can be proven in purely algebraic manner without appealing to existence of special coordinates. This is Section 14.5 and exercise 14.17 in MTW.

Let $\newcommand{\be}{\mathbf{e}} \be_\mu$ be a set of basis vector fields. We can view them as vector-valued 0-forms. Then $d\be_\mu$ are vector-valued 1-forms. They must have an expansion like $$d\be_\nu = \be_\mu \omega^\mu{}_\nu$$ where $\omega^\mu{}_\nu$ are a set of 1-forms. They are the connection 1-forms. This is Cartan's first structure equation. Cartan's second structure equation is $$R^{\mu\nu} = d\omega^{\mu\nu} - \omega^\mu{}_\alpha \wedge \omega^{\nu\alpha} $$ and relates the connection forms to the curvature 2-forms $R^{\mu\nu}$ (this is not the Ricci tensor). The Riemann tensor $\mathcal R$ is related to this by $$\mathcal R = \frac{1}{2}\be_\mu\wedge \be_\nu\, R^{\mu\nu}.$$ You can see that this is an anti-symmetric 2-tensor (a 2-form) taking anti-symmetric 2-tensors as values, which is the familiar property that the Riemann tensor is antisymmetric in both pairs of indices. The Bianchi identity in this formulation is $$d\mathcal R = 0$$ and it is fairly straight-forward to show it using Cartan's two structure equations, and the properties of $d$, in particular that $d^2 = 0$ and that $$d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^p \alpha \wedge d\beta$$ when $\alpha$ is a $p$-form.

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I think your main problem was that you missed a term when going from the Jacobi identity to the next line. The commutator is

$$[\nabla_a,[\nabla_b,\nabla_c]]U^d = \nabla_a\left([\nabla_b,\nabla_c]U^d\right)-[\nabla_b,\nabla_c]\left(\nabla_a U^d\right) $$

The first term on the RHS is what you wrote down in the second line. The second term will give you two more terms depending on the Riemann tensor, one of which cancels the $\nabla_a U^e$ term you were worried about. You should also find a term that looks like $R^e_{\;abc}\nabla_e U^d + \mbox{cyclic}.$ Here you need to apply the first Bianchi identity $R^e_{\;[abc]}=0$ in order to derive second Bianchi identity (i.e. the one involving the covariant derivative). This step is essential since this first Bianchi identity holds only when the torsion vanishes. If you decided to work with torsion present, you would need to keep track of the places where the torsion appears when going from the Jacobi identity expression to the expression in terms of curvature tensors, and also when applying the first Bianchi identity.

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