# Curvature for $C^{1,\alpha}$-metrics

According to Gromov's Metric Structures for Riemannian and Non-Riemannian Spaces every limit $V_0$ of sequences in the class of manifolds $V$ with $|K(V)| \leq 1$ and $\mathrm{InjRad}(V) \geq \rho > 0$ is a Riemannian $C^{1,1}$-manifold, which according to Gromov's ad hoc definition satisfies the following (among other things, see p. 387):

i. There exists a $C^{1,1}$-atlas (differentiable w/ Lipschitz derivative), and the metric tensor is Lipschitz in these coordinates (using harmonic coordinates I think one should obtain that $g$ is $C^{1,1}$ as well).

ii. The squared distance function is locally $C^{1,1}$.

iii. There exists a bounded measurable quadratic form $B_S$ (one for each hyperplane in $T_v V_0$) satisfying the following tube formula (see p. 372):

For every hypersurface $W\subset V_0$ through $v$ with shape operator $A_0$ and $T_v W = S$, $$\frac{d}{dt} A^\ast_t \bigg\vert_{t=0} = - (A_0^\ast)^2 + B_S,$$ where $A_t^\ast$ denotes the pullback of the shape operator $A_t$ of the hypersurface $W_t$ which is obtained from $W$ by equidistant translation.

The interesting thing about $B_S$ is that it can be used to define sectional curvature by setting $$K(\tau_1 \land \tau_2) = - g(B_S(\tau_1),\tau_2),$$ where $\tau_1 \bot S$ and $\tau_2 \in S$.

Question 1: How is B_S obtained? How does one see that it is measurable? I'm assuming that using harmonic coordinates one sees that $A$ is Lipschitz and thus differentiable almost everywhere and then obtains $B_S$ in terms of the derivative of $A$ using the tube formula as definition for $B_S$...

Question 2: Can one obtain a similar definition for sectional curvature if the metric tensor is only $C^{1,\alpha}$ (differentiable with H\"older continuous derivative)?

• I do not see a proper reason to close it ("no longer relevant" is not an option anymore), so the best way to go could be for you to post your edit as an answer, possibly in a slightly more detailed form, and accept it. – Benoît Kloeckner Feb 27 '14 at 21:07
• Could you post the answers (if they are really in Peters's paper which is not how I recall the paper)? – Igor Belegradek Feb 28 '14 at 4:47

## 1 Answer

The following sort of answers the question in case of both sided curvature bounds:

In Convergence of Riemannian Manifold Peters cites a result due to Nikolaev asserting that if a manifold $M$ with a metric tensor $g$ with bounded curvature in the Alexandrov sense admits a domain on which harmonic coordinates are defined, then the metric components $g_{ij}$ are actually of Sobolev class $H^2(\Omega)$. Since harmonic coordinates on approximating manifolds "converge" to harmonic coordnates on the limit manifold, one obtains that the limit metric is of class $H^2(M)$. In particular, its second derivatives and hence the Riemannian curvature tensor and defined outside a Riemannian zero-set.

Since the form $B_S$ (which is, at least in classical surface theory, referred to as Weingarten map) may be defined using $g$ and its first and second derivatives (see Gromov's book or article On Curvature and Sign) one sees how to obtain $B_S$ in the limits space.

As I pointed out, this sort of answers the question. I'm sure that Gromov had a more-geometric-less-analytic method of obtaining the quadratic form. So the question of what Gromov's proof looks like is still open. In fact, I seriously doubt that Gromov has ever had this written down.

• Nikolaev's result assumes two sided curvature bound (in the Alexandrov sense). I thought in Question 2 you wanted a definition of curvature of any $C^{1,\alpha}$ metric, not necessarily of bounded Alexandov curvature. See also Otsu's survey for generalizations: library.msri.org/books/Book30/files/otsu.pdf – Igor Belegradek Feb 28 '14 at 16:20
• Forgive me Igor! "My" manifolds are always compact. I'm so spoiled by it that I keep forgetting to bring it up. On the other hand, I guess I'll leave the question as it is. And thank you for the link! – user35946 Feb 28 '14 at 16:37
• How is compactness relevant? Are you suggesting that any compact $C^1,\alpha$ manifold has bounded Alexandrov curvature? – Igor Belegradek Feb 28 '14 at 16:49
• No you're right, of course they are not. (I think there's even a counterexample in Peters' article...) – user35946 Feb 28 '14 at 17:01