## Lecture 11. The Laplace-Beltrami operator on a Riemannian manifold

In this lecture we extend the previous results in the framework of smooth manifolds. The main idea to extend those results is that, similar computations may be performed in local coordinates charts and then we use a partition of unity.

Lemma: Let $\mathbb{M}$ be a paracompact manifold. Let $(U_i)_{i \in I}$ be a locally finite covering of $\mathbb{M}$ such that each $\bar{U}_i$ is compact. Then, there exists a system $(\phi_i)_{i \in I}$ of smooth functions on $\mathbb{M}$ such that:

• Each $\phi_i$ has a compact support contained in $U_i$,
• $\phi_i \ge 0$, $\sum_{i \in I} \phi_i =1$.

We recall that on a topological space $\mathbb{M}$, a covering $(U_i)_{i \in I}$ is said to be locally finite if each $m\in \mathbb{M}$ has a neighborhood that intersects only finitely many of the sets $U_i$‘s. The space $\mathbb{M}$ is said to be paracompact if for each covering $(U_i)_{i \in I}$ of $\mathbb{M}$, there is a locally finite covering of $\mathbb{M}$ which is a refinement of $(U_i)_{i \in I}$.

From now on, in this lecture $\mathbb{M}$ will be a smooth manifold with dimension $n$.

Definition: A Riemannian structure $g$ on $\mathbb{M}$ is a smooth, symmetric and positive $(0,2)$ tensor on $\mathbb{M}$.

In other words, a Riemannian structure $g$ induces for each $x \in \mathbb{M}$ an inner product $g_x$ on the tangent space $\mathbf{T}_x \mathbb{M}$ and the dependence $x \rightarrow g_x$ is required to be smooth.

Unlike the case of $\mathbb{R}^n$, in general we may not define a Riemannian structure on a manifold by using global frames. For instance on the two-dimensional sphere $\mathbb{S}^2$, it is impossible to find smooth vector fields $(V_1,V_2)$ such that for every $x \in \mathbb{S}^2$, $(V_1(x),V_2(x))$ is a basis of $\mathbf{T}_x \mathbb{S}^2$. However, of course, we may always deal with local orthonormal frames: That is, if $\mathbb{M}$ is a smooth Riemannian manifold (i.e. a smooth manifold endowed with a Riemannian structure), for every $x$ in $\mathbb{M}$, we can find an open set $x \in U \subset \mathbb{M}$ and smooth vector fields $(V_1,\cdots,V_n)$ on $U$ such that for every $y \in U$, $(V_1(y),\cdots, V_n(y))$ is an orthonormal basis of the tangent space $\mathbf{T}_y \mathbb{M}$ for the inner product $g_y$.

From now on we consider a smooth Riemannian manifold $(\mathbb{M},g)$. It is possible to find a locally finite covering of $\mathbb{M}$ by local coordinate charts $(U_i,\phi_i)_{i \in I}$ and smooth vector fields $(V_1^i,\cdots,V_n^i)$ on $U_i$ such that for every $x \in U_i$, $(V^i_1(x),\cdots, V^i_n(x))$ is an orthonormal basis of the tangent space $\mathbf{T}_x \mathbb{M}$ for the inner product $g_x$. Let $(\psi_i)_{i \in I}$ be a partition of unity subordinated to this covering.

Our first goal is to define the canonical Riemannian measure on $\mathbb{M}$. The vector fields $(V_1^i,\cdots,V_n^i)$ induce smooth vector fields $(\tilde{V}_1^i,\cdots,{V}_n^i)$ on $\phi_i(U_i)$. Without loss of generality, we may assume that on $\phi_i(U_i)$, $\mathbf{det}(\tilde{V}_1^i,\cdots,\tilde{V}_n^i) >0$. Consider on $\phi_i(U_i)$ the Borel measure with density $d\mu_i = \frac{1}{\mathbf{det} (\tilde{V}_1^i,\cdots,\tilde{V}_n^i)} d x$, where $dx$ is the Lebesgue measure on $\mathbb{R}^n$. If $f:\mathbb{M} \rightarrow \mathbb{R}$ is a non negative Borel function with a compact support included in $U_i$, it is natural to define $\mu(f)= \int_{\phi_i(U_i)} f \circ \phi_i^{-1} d\mu_i.$ We observe that if the support of $f$ is included in $U_i \cap U_j$, $\int_{\phi_i(U_i)} f \circ \phi_i^{-1} d\mu_i= \int_{\phi_j(U_j)} f \circ \phi_j^{-1} d\mu_j,$ so that $\mu$ is well defined. Now, for a general non negative Borel function $f:\mathbb{M} \rightarrow \mathbb{R}$, we define $\mu (f)= \sum_{i \in I} \mu(\psi_i f),$ where $(\psi_i)_{i \in I}$ is the partition of unity subordinated to the covering $(U_i)_{i \in I}$. This defines a Borel measure $\mu$ on $\mathbb{M}$ which is called the Riemannian measure.

The same idea allows to construct the Laplace-Beltrami operator on $\mathbb{M}$. If $f:\mathbb{M} \rightarrow \mathbb{R}$ is a smooth function on $U_i$, we define $L^i f=\sum_{k=1}^n (V_k^i)^2 f+V_0^i f,$ where $V_0^i$ is the smooth vector field on the open set $U_i$ constructed as in the linear case. Let us now observe that, on $U_i \cap U_j$, we have $L^i f=L^jf$. This leads to the following definition of the Laplace-Beltrami operator on $\mathbb{M}$: If $f: \mathbb{M} \rightarrow \mathbb{R}$ is a smooth function, $Lf= \sum_{i \in I} L^i(\psi_i f)$, where $(\psi_i)_{i \in I}$ is the partition of unity subordinated to the covering $(U_i)_{i \in I}$.

Exercise: Show that the Laplace-Beltrami operator $L$ is symmetric with respect to the Riemannian measure $\mu$.

Diffusion operators on manifolds are intrinsically defined as follows:

Definition: Let $\mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R})$ be the set of smooth functions $\mathbb{M} \rightarrow \mathbb{R}$ and $\mathcal{C} (\mathbb{M}, \mathbb{R})$ be the set of continuous functions $\mathbb{M} \rightarrow \mathbb{R}$. A diffusion operator $L$ is an operator
$L: \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R}) \rightarrow \mathcal{C} (\mathbb{M}, \mathbb{R})$ such that:

• $L$ is linear;
• $L$ is a local operator; That is, if $f,g \in \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R})$ coincide on a neighnorhood of $x$, then $Lf(x)=Lg(x)$;
• If $f \in \mathcal{C}^{\infty} (\mathbb{M}, \mathbb{R})$ has a local minimum at $x$, $Lf (x) \ge 0$.

And it is easily seen, that the Laplace-Beltrami is a diffusion operator. It is moreover elliptic in the sense that if $(\phi,U)$ is a local coordinate chart, then the operator $L$ read in this chart is an elliptic operator on $\phi(U)$.

As usual, we associate to $L$ the differential bilinear form $\Gamma(f,g) =\frac{1}{2}(L(fg)-fLg-gLf).$ The bilinear form $\Gamma$ is related to the notion of Riemannian gradient.

Definition: Let $f:\mathbb{M} \rightarrow \mathbb{R}$ be a smooth function. There is a unique smooth vector field on $\mathbb{M}$ which is denoted by $\nabla f$ and that is called the Riemannian gradient that satisfies for every $x \in \mathbb{M}$ and $u \in \mathbf{T}_x \mathbb{M}$, $df_x(u)=g_x( \nabla f (x), u)$, where $df$ is the differential of $f$.

If $U$ is an open set of $\mathbb{M}$, and $V_1,\cdots, V_n$ are smooth vector fields on $U$ such that for $x \in U$, $(V_1(x),\cdots, V_n(x))$ is an orthonormal frame of $\mathbf{T}_x \mathbb{M}$, it is readily checked that $\nabla f (x) =\sum_{i=1}^n V_if(x) V_i (x), \quad x \in U.$

The bilinear form $\Gamma$ is related to the Riemannian gradient by the following formula:

Lemma: Let $f,h :\mathbb{M} \rightarrow \mathbb{R}$ be smooth functions. We have $\Gamma (f,h)=g(\nabla f, \nabla h).$

Proof: Let $x \in \mathbb{M}$. Let $U$ be an open neighborhood of $x$ and $V_1,\cdots, V_n$ smooth vector fields on $U$ such that for $y \in U$, $(V_1(y),\cdots, V_n(y))$ is an orthonormal frame of $\mathbf{T}_y \mathbb{M}$. On $U$, we have $\Gamma(f,h) =\frac{1}{2}(L(fh)-fLh-hLf)=\sum_{i=1}^n V_i f V_i f=g(\nabla f, \nabla h),$ so that in particular $\Gamma (f,h)(x)=g(\nabla f, \nabla h)(x).$ Since $x$ is arbitrary, the proof is complete $\square$

The last result we wish to extend to manifolds is the relation between completeness and essential self-adjointness of the Laplace-Beltrami operator.

The Riemannian distance on a Riemannian manifold is defined exactly as in $\mathbb{R}^n$.

Given an absolutely continuous curve $\gamma: [0,1] \rightarrow \mathbb{M}$, we define its Riemannian length by $L_g (\gamma)=\int_0^1 \sqrt{g_{\gamma(s)} (\gamma'(s),\gamma'(s) )} ds.$ If $x,y \in \mathbb{M}$, let us denote by $\mathcal{C}(x,y)$ the set of absolutely continuous curves $\gamma: [0,1] \rightarrow \mathbb{M}$ such that $\gamma(0)=x, \gamma(1)=y.$ The Riemannian distance between $x$ and $y$ is then defined by $d(x,y)=\inf_{\gamma \in \mathcal{C}(x,y)} L_g(\gamma).$ The Hopf-Rinow theorem also holds on manifolds:

Theorem:(Hopf-Rinow theorem on manifolds) The metric space $(\mathbb{M},d)$ is complete if and only the compact sets are the closed and bounded sets.

We then have the following expected theorem:

Theorem: If the metric space $(\mathbb{M},d)$ is complete, then the Laplace-Beltrami operator $L$ is essentially self-adjoint on the space of smooth and compactly supported functions $f:\mathbb{M} \rightarrow \mathbb{R}$.

This entry was posted in Curvature dimension inequalities. Bookmark the permalink.

### 2 Responses to Lecture 11. The Laplace-Beltrami operator on a Riemannian manifold

1. symphonyofpi says:

In the last equation in the definition of Riemannian gradient, u has been mistakenly replaced by y.