Relating the Ricci curvatures of two Riemannian metrics

Question

Let $g,\bar{g}$ be two Riemannian metrics on a smooth $n$-manifold $M$ and consider the difference between the Levi-Civita connections of $g,\bar{g}$: $$W:=\nabla-\overline{\nabla}.$$

As indicated in Introduction to Riemannian Manifolds by John M. Lee, $W$ is a $(1,2)$-tensor field on $M$. And with a local frame $\{v_1,\ldots,v_n\}$, we see the components $W_{ij}^k$ of $W$ are given by $$\nabla_{v_i}v_j-\overline{\nabla}_{v_i}v_j=W_{ij}^k v_k,$$ which implies $$W_{ij}^k=\Gamma_{ij}^k-\overline{\Gamma}_{ij}^k$$ with $\overline{\Gamma}_{ij}^k$ denoting the connection coefficients of $\bar{g}$ w.r.t. the given frame.

In our later discussion, the Riemann curvature tensor $\mathrm{Riem}$ of $g$ will be a covariant $4$-tensor field on $M$ defined so that for vector fields $X,Y,Z$, and $W$ on $M$, we have $$\mathrm{Riem}(X,Y,Z,W)=g(-\nabla_X\nabla_Y Z+\nabla_Y\nabla_X Z+\nabla_{[X,Y]}Z,W).$$ And locally, some functions $R_{ijk\ell}$ together yield $$\mathrm{Riem}=R_{ijk\ell}v^i\otimes v^j\otimes v^k\otimes v^\ell,$$ where $v^i$ is the $i$-th covector field in the local coframe dual to $\{v_1,\ldots,v_n\}$.

Up next, the Ricci curvature $\mathrm{Ric}$ of $g$ is defined to be the trace of the Riemann curvature tensor over the second and fourth components. If we denote the components of $\mathrm{Ric}$ by $R_{ij}$, then the definition amounts to $$R_{ij}=g^{k\ell}R_{ikj\ell}.$$

Now I'd like to ask why the Ricci curvatures of $g,\bar{g}$ are related by $$R_{ij}=\bar{R}_{ij}+(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell).\tag{1}$$ I'm not so sure that $\overline{\nabla}_\ell W_{ij}^k$ symbolizes a component of $\overline{\nabla}W$ w.r.t. the given frame, but let's keep it that way for the moment. Anyway, to get the result, I began with a straightforward computation: $$\begin{align} R_{ij}&=g^{k\ell}R_{ikj\ell}\\ &=g^{k\ell}\mathrm{Riem}(v_i,v_k,v_j,v_\ell)\\ &=g^{k\ell}g\left(-\nabla_{v_i}(\Gamma_{kj}^\ell v_\ell)+\nabla_{v_k}(\Gamma_{ij}^\ell v_\ell)+\nabla_{[v_i,v_k]}v_j,v_\ell\right)\\ &=g^{k\ell}g\left(-\nabla_{v_i}((W_{kj}^\ell+\overline{\Gamma}_{kj}^\ell)v_\ell)+\nabla_{v_k}((W_{ij}^\ell+\overline{\Gamma}_{ij}^\ell)v_\ell)+c_{ik}^\ell\Gamma_{\ell j}^m v_m,v_\ell\right),\tag{2} \end{align}$$ where $c_{ij}^m$ are the functions defined by $$[v_i,v_j]=c_{ij}^m v_m.$$

As you can see, I was trying to bring in as many terms in (1) as possible, but I have one major concern: how do I get rid of $g$ and bring in $\bar{R}_{ij}$ in (2)?

Let's look at what we need to bring in $\bar{R}_{ij}$: $$\begin{align} \bar{R}_{ij}&=\bar{g}^{k\ell}\bar{R}_{ikj\ell}\\ &=\bar{g}^{k\ell}\overline{\mathrm{Riem}}(v_i,v_k,v_j,v_\ell)\\ &=\bar{g}^{k\ell}\bar{g}\left(-\overline{\nabla}_{v_i}(\overline{\Gamma}_{kj}^\ell v_\ell)+\overline{\nabla}_{v_k}(\overline{\Gamma}_{ij}^\ell v_\ell)+c_{ik}^\ell\overline{\Gamma}_{\ell j}^m v_m,v_\ell\right) \end{align}$$

It seems to be a wise decision to express (2) as $$R_{ij}=g^{k\ell}g\left(-\nabla_{v_i}(W_{kj}^\ell v_\ell)-\overline{\Gamma}_{kj}^\ell W_{i\ell}^m v_m\mathbf{-\overline{\nabla}_{v_i}(\overline{\Gamma}_{kj}^\ell v_\ell)}+\nabla_{v_k}(W_{ij}^\ell v_\ell)+\overline{\Gamma}_{ij}^\ell W_{k\ell}^m v_m\mathbf{+\overline{\nabla}_{v_k}(\overline{\Gamma}_{ij}^\ell v_\ell)}+c_{ik}^\ell W_{\ell j}^m v_m\mathbf{+c_{ik}^\ell\overline{\Gamma}_{\ell j}^m v_m},v_\ell\right).\tag{3}$$ As indicated by the boldfaced terms, we are pretty close to an extra $\bar{R}_{ij}$, but the question remains: how do I get rid of $g$? That is, how do I get from $g^{k\ell}g(\cdot,\cdot)$ to $\bar{g}^{k\ell}\bar{g}(\cdot,\cdot)$? Thank you.

Edit. Now I'm pretty sure that $\overline{\nabla}_\ell W_{ij}^k$ symbolizes a component of $\overline{\nabla}W$ w.r.t. the given frame, because I have successfully derived the coordinate representation of $\mathrm{Ric}$ in terms of the Christoffel symbols using (1).

Answer 1

The definition I use for the Riemann curvature tensor is $$ R(X,Y)Z = \nabla_X\nabla_YZ - \nabla_Y\nabla_XZ - \nabla_{[X,Y]}Z,$$ which is a section of the tangent bundle. Notice that the metric $g$ does not appear in this formula. Only the Levi-Civita connection is needed. Your version of the Riemann curvature tensor is $$R(X,Y,Z,W) = g(R(X,Y,)Z, W),$$ which does use the metric $g$. When doing calculations, I recommend using the first version.

If we now choose any basis $(e_1, \dots, e_n)$ of the tangent space and let $(\omega^1, \dots, \omega^n)$ be the dual basis, then we can define the Ricci curvature (up to sign) to be $$ R(X,Y) = \langle \omega^i, R(X,e_i)Y\rangle. $$ Again, note that the metric does not appear in this formula. I suggest trying to do the calculation this way. You should, of course, verify that this definition agrees with the one you're using. Also, this might help you figure out how to get rid of the metric in your calculation.

The calculation in coordinates goes like this: \begin{align*} \nabla_k(\nabla_l\partial_j) &= \nabla_k(\Gamma_{jl}^p\partial_p)\\ &= \partial_k\Gamma_{jl}^p\partial_p + \Gamma_{jl}^p\Gamma_{kp}^q\partial_q\\ &= (\partial_k\Gamma_{jl}^p + \Gamma_{jl}^q\Gamma_{kq}^p)\partial_p \end{align*} Therefore, \begin{align*} R_{ijkl} &= g(\partial_i,(\nabla_k\nabla_l-\nabla_l\nabla_k)\partial_j)\\ &= g(\partial_i,(\partial_k\Gamma_{jl}^p + \Gamma_{jl}^q\Gamma_{kq}^p - \partial_l\Gamma_{jk}^p - \Gamma_{jk}^q\Gamma_{lq}^p)\partial_p )\\ &= g_{ip}(\partial_k\Gamma_{jl}^p + \Gamma_{jl}^q\Gamma_{kq}^p - \partial_l\Gamma_{jk}^p - \Gamma_{jk}^q\Gamma_{lq}^p). \end{align*} You can now define the Ricci tensor (up to sign) in one of two ways. Either $$ R_{ik} = g^{jl}R_{ijkl} $$ or $$ R_{jl} = g^{ik}R_{ijkl}. $$ It is clear that the second way leads to an easier calculation and a nicer formula than the first, yielding $$ R_{jl} = \partial_k\Gamma^k_{jl} - \partial_l\Gamma_{jk}^k + \Gamma_{jl}^q\Gamma_{kq}^k - \Gamma_{jk}^q\Gamma^k_{lq}. $$ At this point, the rest is straightforward. You could try to do the calculation using the first formula for $R_{ik}$, but the calculation looks a lot more painful to me.

Answer 2

Finally, I succeeded in deriving the relation, but without @Deane's timely help, I wouldn't have gone this far. I really owe him a lot. Thanks for everything.

Let's start from (3) and rewrite it as $$\begin{align} R_{ij}=&g^{k\ell}g\left(\underbrace{-\overline{\nabla}_{v_i}(\overline{\Gamma}_{kj}^\ell v_\ell)+\overline{\nabla}_{v_k}(\overline{\Gamma}_{ij}^\ell v_\ell)+c_{ik}^\ell\overline{\Gamma}_{\ell j}^m v_m}_\bullet,v_\ell\right)\\ &+g^{k\ell}g\left(-\nabla_{v_i}(W_{kj}^\ell v_\ell)-\overline{\Gamma}_{kj}^\ell W_{i\ell}^m v_m+\nabla_{v_k}(W_{ij}^\ell v_\ell)+\overline{\Gamma}_{ij}^\ell W_{k\ell}^m v_m+c_{ik}^\ell W_{\ell j}^m v_m,v_\ell\right). \end{align}$$ The first term on the right is none other than $\bar{R}_{ij}$! Indeed, if you follow @Deane's suggestion and observe that $$g^{k\ell}g(\bullet,v_\ell)=\delta_m^k\bullet^m=\bar{g}^{k\ell}\bar{g}(\bullet,v_\ell),$$ you will end up with $$R_{ij}=\bar{R}_{ij}+g^{k\ell}g\left(-\nabla_{v_i}(W_{kj}^\ell v_\ell)-\overline{\Gamma}_{kj}^\ell W_{i\ell}^m v_m+\nabla_{v_k}(W_{ij}^\ell v_\ell)+\overline{\Gamma}_{ij}^\ell W_{k\ell}^m v_m+c_{ik}^\ell W_{\ell j}^m v_m,v_\ell\right)!$$ And the proof will be complete if we show that $$\begin{align} &g^{k\ell}g\left(-\nabla_{v_i}(W_{kj}^\ell v_\ell)-\overline{\Gamma}_{kj}^\ell W_{i\ell}^m v_m+\nabla_{v_k}(W_{ij}^\ell v_\ell)+\overline{\Gamma}_{ij}^\ell W_{k\ell}^m v_m+c_{ik}^\ell W_{\ell j}^m v_m,v_\ell\right)\\ &=(\overline{\nabla}_k W_{ij}^k-\overline{\nabla}_j W_{ki}^k)+(W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell).\tag{4} \end{align}$$ The RHS of (4) is, by definition, $$\begin{align} &v_k(W_{ij}^k)+W_{ij}^\ell\overline{\Gamma}_{k\ell}^k-W_{\ell j}^k\overline{\Gamma}_{ki}^\ell-W_{i\ell}^k\overline{\Gamma}_{kj}^\ell\color{red}{-v_j(W_{ik}^k)}-W_{ik}^\ell\overline{\Gamma}_{j\ell}^k\\ &+W_{\ell k}^k\overline{\Gamma}_{ji}^\ell+W_{i\ell}^k\overline{\Gamma}_{jk}^\ell+W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell.\tag{5} \end{align}$$ For the record, sometimes I write $W_{ij}^k$ in place of $W_{ji}^k$ since the components of $W$ can be proved to be symmetric w.r.t. their lower indices.

Up next, we expand the LHS of (4) to get $$-\overline{\Gamma}_{kj}^\ell W_{i\ell}^k+\overline{\Gamma}_{ij}^\ell W_{k\ell}^k+c_{ik}^\ell W_{\ell j}^k\color{red}{-v_i(W_{kj}^k)}-W_{kj}^\ell\Gamma_{i\ell}^k+v_k(W_{ij}^k)+W_{ij}^\ell\Gamma_{k\ell}^k.$$ Since $-v_i(W_{kj}^k)$ differs from $-v_j(W_{ik}^k)$, continuing the process may lead to a dead end. Luckily, we can switch the roles of $i$ and $j$ by observing that $$\begin{align} R_{ij}&=R_{ji}\\ &=\bar{R}_{ji}-\ldots\\ &=\bar{R}_{ij}-\overline{\Gamma}_{ki}^\ell W_{j\ell}^k+\overline{\Gamma}_{ji}^\ell W_{k\ell}^k+c_{jk}^\ell W_{\ell i}^k-v_j(W_{ik}^k)-W_{ki}^\ell\Gamma_{j\ell}^k+v_k(W_{ij}^k)+W_{ji}^\ell\Gamma_{k\ell}^k. \end{align}$$ After comparing it to (5), we know the job can be done by showing that $$\begin{align} c_{jk}^\ell W_{\ell i}^k-W_{ki}^\ell\Gamma_{j\ell}^k+W_{ji}^\ell\Gamma_{k\ell}^k=&W_{ij}^\ell\overline{\Gamma}_{k\ell}^k-W_{i\ell}^k\overline{\Gamma}_{kj}^\ell-W_{ik}^\ell\overline{\Gamma}_{j\ell}^k\\ &+W_{i\ell}^k\overline{\Gamma}_{jk}^\ell+W_{k\ell}^k W_{ij}^\ell-W_{j\ell}^k W_{ik}^\ell. \end{align}$$ And that is simply an exercise of $c_{jk}^\ell=\overline{\Gamma}_{jk}^\ell-\overline{\Gamma}_{kj}^\ell$ (Homework!) and $\Gamma_{j\ell}^k=W_{j\ell}^k+\overline{\Gamma}_{j\ell}^k$. Done!