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While reading Andrews and Hopper's book on Ricci flow, I found the following computation which I am not able to verify.

$M$ is a compact Riemannian manifold and $p \in M$ and $r>0$. We are interested in the function $\psi(x) = \phi\left(\frac{d_{g}(x,p)}{r}\right)$. Here $\phi$ is a smooth bump function $\phi: [0,\infty) \to \mathbb{R}$ with the following properties.

  1. $\phi = 1$ on $[0,1/2]$.
  2. $\phi = 0$ on $[1,\infty)$.
  3. $|\phi'| \leq 3$ on $[1/2,1]$.

Now we want to compute the derivative of $\psi$. Claim is that $|\nabla \psi| \leq \frac{1}{r} \sup |\phi'|$.

I assumed $r$ is small enough so that $B(p,r)$ lies in a normal neighborhood around $p$, then $d_{g}(p,x)= \sqrt{x_{1}^{2} + \dots + x_{n}^{2}}$ in normal coordinates around $p$. Then I see that $$ \frac{\partial \psi}{\partial x_{i}} = \frac{1}{r}\phi'\left( \frac{d_{g}(x,p)}{r}\right) \frac{x_{i}}{\sqrt{x_{1}^{2} + \dots + x_{n}^{2}}}. $$

Thus $$ |\nabla \psi|^{2} = g^{ij}(x)\frac{\partial \psi}{\partial x_{i}} \frac{\partial \psi}{\partial x_{j}} = \frac{1}{r^{2}}\phi'\left(\frac{d_{g}(x,p)}{r}\right)^{2}\frac{g^{ij}(x)x_{i}x_{j}}{x_{1}^{2} + \dots + x_{n}^{2}}. $$

I don't know how to go forward. I tried working in normal coordinates around $x$, that didn't seem to work either.

Also, I am not sure how to deal with the derivative of the distance function if $x$ is not in a normal neighborhood of $p$.

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    $\begingroup$ If $x$ is not in a normal neighborhood of $p$, $d$ might not be differentiable. $\endgroup$ – Arctic Char Aug 11 at 16:53
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Another more intuitive (and coordinate free) way to see this: Since $\psi (x) = \phi \left( \frac{d(x, p)}{r}\right)$, $$\nabla \psi = \phi' \left(\frac{d(x, p)}{r}\right) \cdot \frac{\nabla d}{r}$$

So it suffices to show that $|\nabla d|\le 1$. This follows from triangle inequality: Let $v\in T_xM$. Then $\gamma (t) = \exp_x (tv)$ is a curve on $M$ with $\gamma(0) = x$, $\gamma'(0) = v$. Then

\begin{align*} \langle \nabla d, v\rangle &= \frac{d}{dt} d(p, \gamma(t))\bigg|_{t=0} \\ &= \lim_{t\to 0} \frac{d(p, \gamma(t)) - d(p, x)}{t} \end{align*}

Since by triangle inequality, $$\left|\frac{d(p, \gamma(t)) - d(p, x)}{t}\right| \le \frac{d(x, \gamma(t))}{|t|} = \frac{|t|\| v\|}{|t|} = \|v\|$$

we have $$ |\langle \nabla d, v\rangle| \le \|v\|\Rightarrow |\nabla d| \le 1$$

(e.g. by picking $v = \nabla d$).

We use nothing but that the distance function $d(\cdot, p)$ is Lipschitz with Lipschitz constant $1$. This already implies that the gradient is $\le 1$.

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You've almost completed the computation. The last step is noting that in normal coordinates the radial vector field $\frac{\partial}{\partial r}=\frac{x_i}{\sqrt{x_1^2+\dots+x_n^2}}\frac{\partial}{\partial x_i}$ is a has unit magnitude (because it is the velocity of a unit speed geodesic). Thus, the second term in your expression $\frac{g^{ij}x_ix_j}{x_1^2+\dots+x_n^2}$ is equal to $1$.

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