First order estimates of geodesic normal coordinates

Let $$(M^n,g)$$ be a complete Riemannian manifold with $$|Rm| \le 1$$. Can we find two positive constants $$C$$ and $$\epsilon$$, depending only on $$n$$, such that under the normal coordinates $$(g_{ij})$$ with respect to any point $$p \in M$$, we have $$|\partial_k g_{ij}(x)| \le C$$ for any $$|x| \le \epsilon$$?

As pointed out in the comment, if the injectivity radius at $$p$$ is small, then the estimates should be understood for the pull-back of $$g$$ to the tangent space, which is always well-defined by the curvature bound.

• I've deleted my original comment. My first statement, that a sufficiently thin flat cylinder is a counterexample, was correct. However, offhand, I don't see how to get the bound using exponential coordinates. You can get the $C^1$ bound on the metric using either harmonic or almost linear coordinates (as defined by Jost and Karcher). Commented Oct 27, 2020 at 14:51
• I deleted my answer. The comments by Deane Yang and Totoro were right and I'm not sure how to bound the non-radial directions. Commented Oct 27, 2020 at 15:01
• Let me add another comment about this: Exponential coordinates satisfy only an ODE, namely the Jacobi equation. So bounded curvature gives you a bound, in terms of the curvature, for only in the radial direction. To get a bound on the angular derivatives of the metric, you need to use the derivative of the Jacobi equation in the angular directions, so the covariant derivative of curvature appears. Commented Oct 27, 2020 at 18:24
• @DeaneYang You are right. I was wondering if there is an explicit counterexample. Commented Oct 27, 2020 at 19:47
• GabeK and DeaneYang, apparently one of you has not upvoted this question -- but the discussion suggests that you do find it surprisingly difficult, so probably worth an upvote.
– user44143
Commented Oct 28, 2020 at 3:02

The answer is 'no' for $$n=2$$ (and hence for all higher $$n$$). Here is how one can see this.

First, when $$n=2$$, recall that, by the Gauss Lemma, a metric $$g$$ in geodesic normal coordinates $$(x,y)$$ centered on $$p$$ takes the form $$g = \mathrm{d}x^2 + \mathrm{d}y^2 + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2,$$ where the function $$h$$ is arbitrary, subject to the condition that $$(x^2{+}y^2)h(x,y)+1>0$$.

Letting $$r^2 = x^2 + y^2$$ and letting $$R$$ be the radial vector field $$x\,\partial_x + y\,\partial_y$$, one computes the formula for the Gauss curvature of $$g$$ to be $$K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+2(5+3r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}.$$ Thus, in the geodesic disk of radius $$\epsilon>0$$ about $$p$$, i.e., where $$r^2=x^2 + y^2 \le\epsilon^2$$, we can keep $$|K|$$ as small as we like merely by imposing sufficiently small bounds on $$h$$, $$Rh$$ and $$RRh$$, i.e., $$h$$ and its first two radial derivatives. More precisely, for any $$M>0$$, there exists a $$\delta>0$$ such that, if $$|h|$$, $$|Rh|$$ and $$|RRh|$$ are bounded by $$\delta$$ when $$r\le\epsilon$$, then $$|K|\le M$$ when $$r\le \epsilon$$.

Let $$\rho(r)$$ be a smooth function that is identically zero near $$r=0$$ and $$r=\epsilon$$ and, say, positive, at $$r=\epsilon/2$$, but satisfies the condition that, for any constant $$\lambda$$ with $$|\lambda|\le 1$$, the function $$h(x,y) = \lambda\rho(r)$$ yields a $$K$$ that satisfies the bound $$|K|\le 1$$.

Let $$f(\theta)$$ be any $$2\pi$$-periodic smooth function bounded by $$1$$ and consider the smooth function $$h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta).$$ Then $$h$$ and its radial derivatives are bounded in such a way that the Gauss curvature $$K$$ for the corresponding metric $$g$$ will be bounded in absolute value by $$1$$, but the 'angular derivative' of $$h$$, i.e., $$xh_y-yh_x = \rho(r)f'(\theta)$$, need not be bounded. In particular, by choosing $$f$$ appropriately (bounded by $$1$$ but with very large first derivatives), we can be sure that the coefficients of $$g$$ in this coordinate system, i.e., $$g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y),$$ while bounded themselves, will have some very large first derivatives when $$r = \epsilon/2$$. In particular, there is no constant $$C>0$$ that would bound the first derivatives of these quantities independent of the choice of $$f$$.

• Could we take $h(r \cos \theta, r \sin \theta)= \sin^2(\pi r/\epsilon) \sin^2(n \theta)$?
– user44143
Commented Oct 28, 2020 at 15:36
• @MattF.: That $h$ won't be smooth when $n>0$ is large. That's why I wanted something like the $\rho(r)$ that vanishes identically near $r=0$. After all, $r^2\sin^2(n\theta)= r^{2-2n}\bigl(\mathrm{Im}((x+iy)^n\bigr)^2$. If you use my $\rho$, you could probably take $f(\theta) = \sin(n\theta)$ to get an example, though. Commented Oct 28, 2020 at 15:46