# Trace of hessian

So this is a result used in Peter Topping's,Lectures on Ricci Flow. What is a quick of showing

$$\text{tr}\nabla_{X,\cdot}^2h(\cdot,W)=-(\nabla\delta h)(X,W)$$

where $\delta A=-\text{tr}_{12}\nabla A$, is the divergence operator and $h=\partial_tg$.

• Are you sure you've transcribed this correctly? Looks to me like the right hand side has an extra index/slot. – Anthony Carapetis Nov 17 '15 at 4:08
• Yes, its transcribed correctly. Peter uses the convention that $\nabla_XA(Y,Z,...)=\nabla A(X,Y,Z,...)$ for any tensor field $A$. – Enigma Nov 17 '15 at 4:17
• Are you sure $\delta$ is the Laplacian and not the codifferential? – Anthony Carapetis Nov 17 '15 at 4:22
• Sorry, Peter actually wrote that $\delta$ is the divergence operator. – Enigma Nov 17 '15 at 4:23
• In any case $\delta$ is defined above, so the name is of no particular concern. – Enigma Nov 17 '15 at 4:38

## 1 Answer

This is just raising/lowering the indices in the trace - the particular value of $h$ is irrelevant. In abstract index notation we have

$${\rm tr} \nabla_{X,\cdot}^2 h(\cdot,W) = \nabla_X \nabla_j h^j{}_W = \nabla_X \nabla^j h_{jW} = \nabla_X (-\delta h)(W)=-(\nabla\delta h)(X,W).$$

More abstractly, just note that the RHS is $-\nabla (-{\rm tr}_{12} \nabla h)$ which is ${\rm tr}_{23} \nabla^2 h$ because contractions commute with the covariant derivative, and we have to change the slots we are contracting over because the $\nabla$ adds an extra slot at the front.

• Could you explain your notation a little bit. What is $\nabla_j$ and $\nabla^j$, and what do you mean by taking $W$ as a subscript?. Also why is $\nabla_jh^j_W=\nabla^jh_{jW}$. – Enigma Nov 17 '15 at 5:55
• Both sides are bilinear in $X,W$ so it suffices to prove it for the case when they're coordinate vector fields $X = \partial_i$, $W = \partial_j$. $\nabla_j$ is just $\nabla_{\partial_j}$; so when I use $W$ as an index I mean $j$. Both sides of your last equation are just $g^{jk} \nabla_j h_{kW}$ written in short form - on one we raise the $j$, on the other we raise the $k$ and rename it to $j$. You don't really need to do this shuffling at all - see my edit. – Anthony Carapetis Nov 17 '15 at 5:57
• Thanks. I see how you get the result now. – Enigma Nov 17 '15 at 6:01