I am reviewing the paper "On Harnack's Inequality and Entropy For Gaussian Curvature Flow" by Bennet Chow. On page 473 Lemma 3.1 (vii), Chow managed to find the evolution equation for Christoffel symbol. In particular, he says that we have the formula

\begin{align*}\partial_{t} \Gamma_{ij}^{k} = \frac{1}{2} g^{kl} (\nabla_{i}(\partial_{t}g_{jl}) + \nabla_{j}(\partial_{t}g_{il}) - \nabla_{l}(\partial_{t}g_{ij}))\end{align*}

Obviously, he used the local expression of Christoffel symbols to start with, but I just cannot get to simplifying the equation into that neat form. Can anyone please tell me how this could be done?

P.S. We are dealing with a solution of Gauss Curvature flow here. That is, suppose that $M^{n}$ is a compact, strictly convex, smooth manifold embedded in $\mathbb{R}^{n+1}$ by $X_{0}$ and $\{X_{t}\}$ is a solution of the GCF, given by

\begin{align*} X(p,t) & = -K^{\alpha} \nu, \quad p \in M \\ X(p,0) & = X_{0}(p) \end{align*}

where $\nu$ is the outward pointing normal and $K$ is the Gaussian curvature.


1 Answer 1


There are two key observations.

(1) Although a connection is not a tensor field, the difference between two connections is a tensor field. So if we let $\widehat \nabla$ denote the connection associated with the metric $g(0)$ and $\nabla$ the one associated with $g(t)$, then the following are well-defined $(1,2)$-tensor fields: $$\nabla - \widehat\nabla, \qquad \partial_t (\nabla - \widehat\nabla). $$ In any coordinate system, the latter has components $\partial_t \Gamma_{ij}^k$ (because the coefficients of $\widehat\nabla$ are independent of $t$). Thus the formula you're trying to prove is an equality between two invariantly defined $(1,2)$ tensor fields. (The right-hand side is a sum of contractions of $g^{-1}\otimes\nabla (\partial_t g)$.)

(2) Because you're trying to prove an equality between two well-defined tensor fields, you can verify it in any convenient coordinate system. Fix a time $t_0$ and a point $p\in M$, and let $(x^i)$ denote Riemannian normal coordinates for $g(t_0)$ centered at $p$. In these coordinates, when we evaluate at $t=t_0$ and $x=p$, we have $$ \partial_k g_{ij} = \partial_k g^{ij} = \Gamma_{ij}^k = 0, $$ and for any $2$-tensor field $T$, $$ \nabla_k T_{ij} = \partial_k T_{ij}. $$ So, to check the identity you're trying to prove at $t=t_0$ and $x=p$, just plug in these relations and simplify. Since this is true for every $t_0$ and every $p$, it's true everywhere.

  • $\begingroup$ First of all, it's an honor professor. I still have your book on my bookshelf. I was saving the Riemannian normal coordinates as my last resort but I guess it's just the simplest way. I will try this method. Thank you for your response! $\endgroup$
    – SW CHANG
    Aug 6, 2016 at 5:55

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