# Sheaf-theoretically characterize a Riemannian structure?

A smooth structure on a topological manifold can be characterized as a sheaf of local rings, see for example the discussion here.

Q: Is there a way to characterize a Riemannian structure on a smooth manifold by a sheaf of functions?

A most likely horrible guess to clarify the type of answers I'm thinking about: define a Riemannian manifold to be a locally ringed space that locally looks like the sheaf $(\mathbb R^n, \mathcal H_g)$ where $g$ is some non degenerate symmetric positive definite matrix and $\mathcal H_g$ is the sheaf (is it even a sheaf?) that assigns to open subsets harmonic functions solving the Laplace equation given by $g$.

Please forgive my ignorance in the above, this is not my field. Just had to do a little Riemannian geometry today and was thinking whether there's a sheaf-theoretic/functor of points way to think about things.

• I think the sheaf of locally Lipschitz functions with Lipschitz constant $1$ will do the trick, as it will tell you the metric (in the metric spaces sense), which determines the metric (in the Riemmanian metric sense). – Will Sawin Jun 3 '18 at 5:59
• @WillSawin thanks! my analysis is rusty - is there a textbook reference for how the algebra of lipschitz functions lets you recover the metric? – zzz Jun 3 '18 at 6:38
• Lipschitz functions do not form an algebra. Maybe one in fact needs also the map from the sheaf of Lipschitz functions to the sheaf of continuous functions. Using maximal ideals in the sheaf of continuous functions, one can define the evaluation map. Then one simply defines the distance between two points as the maximum difference between their evaluations in any connected open set containing both points. – Will Sawin Jun 3 '18 at 6:47
• This is very similar to (maybe almost a duplicate of) this question: mathoverflow.net/questions/56833/… – Qfwfq Jun 3 '18 at 13:36
• @Qfwfq indeed sorry did not see that question, but as there’s some good discussion in this thread now maybe we keep this open rather than closing it as duplicate? – zzz Jun 3 '18 at 17:42

Suppose that $M$ is a smooth manifold and $g_0, g_1$ are Riemann metrics on $M$. $\newcommand{\eH}{\mathscr{H}}$ Denote by $\eH_{g_i}$, $i=0,1$ and the sheaf of $g_i$-harmonic functions. More precisely for any open set $U\subset M$

$$\eH_{g_i}(U)=\big\{\; f\in C^\infty(U):\;\;\Delta_{g_i} u=0\;\big\},$$

where $\Delta_{g_i}$ denotes the scalar Laplacian of the metric $g_i$.

Suppose that $\eH_{g_0}(U)=\eH_{g_1}(U)$, for any open set $U\subset M$.

• If $\dim M\geq 3$, then there exists $c\in (0,\infty)$ such that $g_1=c g_0$.

• If $\dim M=2$, then there exists a smooth function $f: M\to (0,\infty)$ such that $g_1=fg_0$, i.e., the metrics $g_0$ and $g_1$ live in the same conformal class.

The strong unique continuation property of harmonic functions shows that this statement is really a statement about the stalks of the sheaves $\eH_{g_i}$. Note that these are sheaves of vector spaces, not rings. In dimension $\geq 3$ these sheaves determine the metric up to a multiplicative positive constant.

• Don't harmonic functions form a sheaf? I thought that the vanishing of the Laplacian was a local condition – Denis Nardin Jun 3 '18 at 11:37
• I tried to be cautious in my statement. – Liviu Nicolaescu Jun 3 '18 at 12:31
• Does it ever suffice to consider only the sections over the open set $M$ (just the global harmonic functions, and not the entire sheaf)? Or maybe there are problems with extension of harmonic functions? (I know that some harmonic functions cannot be extended, but that wouldn't rule out a stronger result as far as I can tell...) – Lorenzo Najt Jun 5 '18 at 12:10
• If $M$ is compact and connected, then the only harmonic functions are constant. The same happens for other noncompact Riemann manifolds. – Liviu Nicolaescu Jun 5 '18 at 12:18
• Given an elliptic partial differential operator $L:C^\infty(E_0)\to C^\infty(E_1)$, $E_0,E_1$ smooth vector bundles over the compact manifold $M$, one can form the sheaf $\ker L$, $$\ker L(U)=\{ s\in C^\infty(E_0|_U):\;\;Ls=0\}$$ one can show that the Euler characteristic of the cohomology of $\ker L$ is equal to the index of $L$. In fact $H^0(M,\ker L)= \ker L(M)$ – Liviu Nicolaescu Jan 28 at 10:05