While reading about the Ricci flow, I've ran into the following statement:
It is worth pointing out here that the Ricci tensor is invariant under uniform
scaling of the metric.
This quote was provided as part of a simple example – Ricci flow of an Einstein manifold of constant $\lambda \in \mathbb{R}$. See "Lectures on the Ricci Flow", P. Topping, for more context.
For some reason, that surprised me. So I've tried to prove it. I've obviously got it wrong, but I can't see where. Here's my go at it.
Let $(M,g)$ be a Riemannian manifold. Let $\tilde{g}=\lambda g$, be an alternative metric.
From the Koszul formula we have:
$$
\begin{align}
2\tilde{g}(\tilde{\nabla}_X Y ,Z)
&= X (\tilde{g}(Y,Z)) + Y (\tilde{g}(X,Z)) – Z (\tilde{g}(X,Y)) \\
&-\tilde{g}([Y,X],Z) – \tilde{g}([X,Z],Y) – \tilde{g}([Y,Z],X) \\
&= \lambda \{X (g(Y,Z)) + Y (g(X,Z)) – Z (g(X,Y)) \\
&-g([Y,X],Z) – g([X,Z],Y) – g([Y,Z],X) \} \\
&= \lambda 2g(\nabla_X Y ,Z)
\end{align}
$$
So:
$$
\lambda g (\tilde{\nabla}_X Y ,Z) = \lambda g (\nabla_X Y ,Z)
$$
By the uniqueness of the connection, we have $\tilde{\nabla} = \nabla$, and in particular:
$$ \tilde{R}(X,Y)Z = \nabla_X \nabla_Y Z – \nabla_Y \nabla_X Z -\nabla_{[X,Y]} Z = R(X,Y)Z $$
So far, it all sounds sort of sensible to me – using that identity with the uniform scaling of the sphere will lead to the desired change in sectional curvature.
But if all that was true, we'd have:
$$
\tilde{Ric}(v,w) = \sum_{i} \tilde{g}(R(e_i , v)w ,e_i) = \sum_{i} \lambda g(R(e_i , v)w ,e_i) = \lambda Ric (v,w)
$$
And not:
$$ Ric = \tilde{Ric} $$
What did I get wrong? Is there some geometric heuristic that you can offer as for why would Ricci curvature stays invariant under scaling?
Best Answer
Edit because the original answer was wrong: My original answer was wrong in its claim that the Ricci tensor would scale. By definition, we have $Ric(X,Y)= \text{tr}(Z \mapsto R(Z,X)Y)$ where $R$ is the Riemannian curvature tensor. So as the curvature tensor stays the same, so does the Ricci tensor. The mistake in the calculation above was that the formula $\text{Ric}(X,Y)= \sum_{i=1}^n g(R(E_i, X)Y, E_i)$ is only true if $\{E_i\}_{i=1}^n$ is an orthonormal basis of the tangent space with respect to $g$. Now if that is the case we have that $\{\frac{E_i}{\sqrt \lambda}\}_{i=1}^n$ is an orthonormal base with respect to $\tilde g$ and therefore $$\tilde{\text{Ric}}(v,w) = \sum \tilde g(R(\frac{E_i}{\sqrt \lambda}, v)w, \frac{E_i}{\sqrt \lambda}) = \sum \lambda g(R(\frac{E_i}{\sqrt \lambda}, v)w, \frac{E_i}{\sqrt \lambda}) = \sum g(R(E_i,v)w, E_i) = \text{Ric}(v,w).$$ My comment on the Ricci curvature staying the same under scaling stays true however.
Original, wrong answer: In order to calculate the Ricci curvature in a direction $v$ we always have to normalise the vector. If we denote by $r$ and $\tilde r$ the respective Ricci curvatures and choose $v$ to have unit length w.r.t. $\tilde g$ then we have
$$\tilde r(v)=\tilde{\text{Ric}}(\frac{v}{||v||_{\tilde g}},\frac{v}{||v||_{\tilde g}}) =\tilde{\text{Ric}}(v,v) = \lambda \text{Ric}(v,v) =\text{Ric}(\sqrt \lambda v, \sqrt \lambda v)= \text{Ric}(\frac{v}{||v||_{ g}},\frac{v}{||v||_{ g}}) =r(v) $$ because $||v||^2_g = \langle v, v \rangle_g = \frac{1}{\lambda} \langle v,v \rangle_{\tilde g}= \frac{1}{\lambda}$ so that $\sqrt \lambda v$ is the normalized version of $v$ w.r.t. $g$.
Hence the Ricci Curvature stays the same if we scale the metric even though the Ricci tensor $\text{Ric}(v,w)$ scales in the same way as the metric.