Ricci curvature: step in proof of a paper by Hamilton

Question

In Hamilton's paper "The Ricci Curvature Equation" (in Seminar on Nonlinear Partial Differential Equations, here), I can do all of Lemma 4.2 except for the following relation: $$ -g^{ik}g^{j\ell}h_{pk}\partial_jF^p_{i\ell}=-\frac{1}{2}\Delta E+g^{ik}g^{j\ell}(\partial_jh_{kq})F^q_{i\ell} $$ where $g$ and $h$ are metrics, $\Delta$ is the Laplacian w.r.t. $g$, and $E=g^{ij}h_{ij}$.

From what I understand the relation follows from directly calculating the LHS: $$ -g^{ik}g^{j\ell}h_{pk}\partial_jF^p_{i\ell} $$ where $$ F^p_{i\ell}=\frac{1}{2}h^{sp}(\partial_ih_{\ell s}+\partial_\ell h_{is}-\partial_sh_{i\ell}), $$ however what I'm currently trying isn't working.

Edit: A correct answer has been given.

Answer 1

This $\partial_i$ is used by Hamilton to denote the Levi-Civita connection with respect to the metric $g$, or with respect to $\Gamma_{i}{}^{k}_{j}$

Notice that there is a typo in the original paper in the fifth line on p.52 where it is said that "$\partial_i$ is covariant differentiation with respect to $F_{i}{}^{k}_{j}$".

Let me change the notation to make the things clearer. I use $\nabla^{(g)}$ and $\nabla^{(h)}$ for the Levi-Civita connections of the metrics $g$ and $h$ respectively. In other words, I use $\nabla^{(g)}_i$ instead of Hamilton's $\partial_i$ to avoid a possible confusion with the partial derivatives.

The difference of the connections $F_{i}{}^{k}_{j}$ in then expressed by the identity: $$ \nabla^{(g)}_i \omega_j = \nabla^{(h)}_i \omega_j + F_{i}{}^{k}_{j} \omega_k $$ where $\omega_i$ is an arbitrary $1$-form (covector).

Now we can calculate the Laplacian w.r.t. $g$ of the function $E = g^{i j} h_{i j}$. $$ \begin{align} \Delta^{(g)} E & = g^{i j} \nabla^{(g)}_i \nabla^{(g)}_j (g^{k l} h_{k l}) \\ & = g^{i j} g^{k l} \nabla^{(g)}_i \nabla^{(g)}_j h_{k l} \\ & = g^{i j} g^{k l} \nabla^{(g)}_i \Big( \nabla^{(h)}_j h_{k l} + F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p} \Big) \ \end{align} $$

Now we observe that $\nabla^{(h)}_j h_{k l} = 0$ and recover the identity $$ \nabla^{(g)}_j h_{k l} = F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p} $$

Continuing this process we obtain

$$ \begin{align} \Delta^{(g)} E & = g^{i j} g^{k l} \nabla^{(g)}_i \Big( F_{j}{}^{p}_{k} h_{p l} + F_{j}{}^{p}_{l} h_{k p} \Big) \\ & = g^{i j} g^{k l} \Big( h_{p l} \nabla^{(g)}_i F_{j}{}^{p}_{k} + F_{j}{}^{p}_{k} \nabla^{(g)}_i h_{p l} + h_{k p} \nabla^{(g)}_i F_{j}{}^{p}_{l} + F_{j}{}^{p}_{l} \nabla^{(g)}_i h_{k p} \Big) \\ & = 2 \, g^{i j} g^{k l} h_{p l} \nabla^{(g)}_i F_{j}{}^{p}_{k} + 2\, g^{i j} g^{k l} F_{j}{}^{p}_{k} \nabla^{(g)}_i h_{p l} \end{align} $$ which is equivalent to the equation in the question. This also confirms the identity in @Avitus's answer: $$ \begin{align} \Delta^{(g)} E & = 2 \, g^{i j} g^{k l} \nabla^{(g)}_i \Big( h_{p l} F_{j}{}^{p}_{k} \Big) \end{align} $$

Answer 2

Hint: considering the r.h.s. of the equation in the OP we can write

$$g^{ik}g^{jl}(\partial_j h_{kq})F^{q}_{il}=g^{ik}g^{jl}\partial_j\left( h_{kq}F^{q}_{il}\right)-g^{ik}g^{jl}h_{kq}(\partial_j F^{q}_{il}); $$

but

$$-g^{ik}g^{jl}h_{kq}(\partial_j F^{q}_{il})=-g^{ik}g^{jl}h_{qk}(\partial_j F^{q}_{il})=-g^{ik}g^{jl}h_{pk}(\partial_j F^{p}_{il}),$$

i.e. the l.h.s. of the given equation in the OP, by symmetry of the tensor $h$ and changing names to repeated indices.

It remains to prove that

$$g^{ik}g^{jl}\partial_j\left( h_{kq}F^{q}_{il}\right)=\frac{1}{2}\Delta(E), $$

by using the definition of the Laplace-Beltrami operator $\Delta$ w.r.t. $g$.