I'm reading the book "The Ricci Flow: An Introduction" and I'm at the part where the authors prove the Bianchi identities using the diffeomorphism invariance of the curvature. I'm stuck on some computations from the paragraph below:

Consider the scalar curvature operator $g \mapsto R_g$ and it's linearization $DR_g$ defined by

$$D R_{g}(h)=-g^{i j} g^{k \ell}\left(\nabla_{i} \nabla_{j} h_{k \ell}-\nabla_{i} \nabla_{k} h_{j \ell}+R_{i k} h_{j \ell}\right) \ \ \ \ \ (1)$$

for any $2$ tensor $h$. Substituting

$$h_{i j}=\left(\mathcal{L}_{X} g\right)_{i j}=\nabla_{i} X_{j}+\nabla_{j} X_{i}$$

(where $X$ is an arbitrary vector field) and commuting covariant derivatives yields

$$\begin{align} D R_{g}\left(\mathcal{L}_{X} g\right) &=-2 \Delta \nabla_{i} X^{i}-2 R_{i j} \nabla^{i} X^{j}+\nabla^{i} \nabla_{j} \nabla_{i} X^{j}+\nabla_{i} \nabla_{j} \nabla^{j} X^{i} \ \ (2)\\ &=2 X^{i} \nabla^{j} R_{i j} \ \ (3) \end{align}$$

I don't understand how to go from $(1)$ to $(2)$. Commuting derivatives, we get (where I'm using the obvious notation $\nabla_{j, k} = \nabla_j \nabla_k$):

$$\nabla_{j, k} X_{\ell} - \nabla_{k, j} X_{\ell} = R_{jks}^{\ell} X^{s}$$

and with some work we can substitute $h = \mathcal{L}_{X} g$ into $(1)$ and obtain:

$$DR_g(h) = -g^{ij}g^{kl}\left( \nabla_{i} \left(R_{jks}^{\ell} X^{s} \right)- \nabla_{i}\left(R_{j{\ell}s}^{k} X^{s}\right) +R_{i k} h_{j \ell}\right) $$

but I still can't get from here to $(2)$. Nor can I see how $(3)$ follows from $(2)$. I've been stuck at this for a while now and would really appreciate some help.


You don't have to commute derivatives to go from (1) to (2); just directly plug in the choice of $h_{ij}$. That is, \begin{align*} g^{ij}g^{kl}\nabla_i\nabla_j h_{kl} & = \Delta g^{kl} h_{kl} = 2\Delta\nabla_i X^i, \\ g^{ij}g^{kl}\nabla_i\nabla_k h_{jl} & = \nabla^i\nabla^k h_{ik} = \nabla^i\nabla^k\nabla_i X_k + \nabla^i\nabla^k\nabla_k X_i, \\ g^{ij}g^{kl} R_{ik}h_{jl} & = R^{ik}h_{ik} = 2R^{ik}\nabla_iX_k ; \end{align*} the first and third lines here use the symmetry of $h_{ij}$.

To go from (2) to (3), first use the definition of the Ricci curvature to write $$ \nabla_i\nabla_j\nabla^j X^i = \nabla_j\nabla_i\nabla^j X^i - R_{ij}\nabla^j X^i + R_{ij}\nabla^i X^j = \nabla_j\nabla_i\nabla^j X^i . $$ Use the definition of the Ricci curvature again to write $$ \nabla_j\nabla_i\nabla^j X^i = \Delta\nabla_i X^i + \nabla_j(R_{ij}X^i) . $$ Plugging these into (2) in order yields \begin{align*} DR_g(\mathcal{L}_Xg) & = -2\Delta\nabla_iX^i + 2\nabla^i\nabla_j\nabla_i X^j - 2R_{ij}\nabla^iX^j \\ & = 2\nabla^j(R_{ij}X^i) - 2R_{ij}\nabla^i X^j \\ & = 2X^i\nabla^jR_{ij} , \end{align*} where the last equality follows from the product rule. This gives (3).

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  • $\begingroup$ Thanks! Are you assuming an orthonormal frame in these computations? $\endgroup$ – Matheus Andrade Apr 10 at 0:13
  • $\begingroup$ I am writing in Abstract index notation. However, there is no loss in thinking of this as a computation using an orthonormal frame. $\endgroup$ – Jeffrey Case Apr 10 at 0:51
  • $\begingroup$ Thanks! I'll accept your answer shortly. Could you explain why there is no loss of generality in carrying out the computations in an orthonormal frame? I mean, I know it's always possible to find such a frame, but I still feel like that's kinda cheating. $\endgroup$ – Matheus Andrade Apr 10 at 1:04
  • $\begingroup$ I would also really appreciate it if you elaborated on the last part of your answer. Putting everything in your answer back into (3), the final answer I get is $$-R_{ij}\nabla_{i}X^{j} + 2 \nabla_{i}(R_{ij} X^{j})$$ instead of what we want. $\endgroup$ – Matheus Andrade Apr 10 at 1:32
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    $\begingroup$ You missed a factor of two somewhere in going from (2) to (3). I've written in the details here. As for your question about the use of an orthonormal frame, again I am not doing this in coordinates, but rather using abstract indices. This means, for example, that $R^{ik}\nabla_i X_k$ is shorthand for $g^{ik}R_{il}\nabla_k X^l$. This is also being done in your Equation (2) to make sense of the symbol $\nabla^j$. That being said, since everything written is tensorial, there no loss in generality in doing the computation with respect to an orthonormal frame. $\endgroup$ – Jeffrey Case Apr 10 at 16:52

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