Yovani Villanueva
Some people think that the universe is locally like a box ($\mathbb{R}^{3}$), but actually it's a little bit complicated than the usual interpretation. Our world is so beautiful and a lot of mysterious things belong to it, however, in the deeper space can exist weird forms that got another physical properties so we aren't able to see. Althougt our intuition doesn't work very well int his situation, we can use all that we know in this locally and understand the rest through the pretty properties that we have in our close space. Differential geometry have done this possible and considering these transformations, this chapter pretends to serve as an introduction to global analysis through the study of the heat equation.
The Laplacian
Let $(M,g)$ be a n-dimensional Riemannian manifold.
Exercise 1: Show that for all $p \in M$, $g_{p}$ induces an isomorphism $\phi_{g}: T_{p}M \to T^{*}_{p}M$. between the tangent and cotangent spaces at the point $p$.
We recall that the Riemannian metric $g$ is a family of inner products subindexed by $p \in M$,
\begin{equation}
g_{p}: T_{p}M \times T_{p}M \to \mathbb{R},
\end{equation}
such that in local coordinates $g_{p} \in C^{\infty}(M)$. Let $\phi_{g}(X): T_{p}M \to \mathbb{R}$ be the element of $T_{p}^{*}M$ that, for tangent vectors, returns the element
\begin{equation}
\phi_{g}(X)(Y):=g_{p}(X,Y).
\end{equation}
$\phi_{g}$ is an isomorphism because
- If $\phi_{g}(X)=0$, $\phi_{g}(X)(X)=g_{p}(X,X)=0$, as the inner product has the property that $g_{p}(X,X)=0$ if and only if $X=0$, then $ker(\phi_{g})=0$ and $\phi_{g}$ is injective.
- Dim($T_{p}M$)=Dim($T^{*}_{p}M$), so $\phi_{g}$ is surjective $\square$
Definition
We define the \textit{gradient} of $f \in C^{\infty}(M)$, $\nabla:C^\infty(M) \to \Gamma(TM)$ as
\begin{equation*}
\nabla (f):=(\phi_{g})^{-1}(df)
\end{equation*}
Let $U$ be an open subset of $\mathbb{R}^{n}$ and $\psi : U \rightarrow M$ a chart of the manifold M, $\psi = (x_{1}, \dots, x_{n})$. We define
\begin{equation*}
g_{ij}:=g \left( \frac{\partial}{\partial x_{i}},\frac{\partial}{\partial x_{j}} \right),
\end{equation*}
where $\{\frac{\partial}{\partial x_{i}}|_p \}_{i \in [n]}$ is the base of $T_{p}M$ associated to the chart $\psi$ $\square$
Exercise 2: Show that the matrix $(g_{ij})$ is invertible.
We take again the property $g_{p}(X,X)=0$ if and only if $X=0$. Then we infer
- If $g(X,Y)=0$ for each $Y \in T_{p}M$ then $X=0$.
- If $g(X,Y)=0$ for each $X \in T_{p}M$ then $Y=0$.
Therefore $g_{p}$ is a symmetric bilinear form and then $(g_{ij})$ is invertible $\square$
Exercise 3: Deduce the following expression for the gradient in local coordinates
\begin{equation*}
\nabla(f)(x)=\sum_{j=1}^{n} \sum_{k=1}^{n} g^{jk}(x)\frac{\partial f}{\partial{x_{k}}}\frac{\partial}{\partial{x_{j}}}.
\end{equation*}
Let $X=\sum_{i=1}^{n}a_{i}(X)\frac{\partial}{\partial x_{i}}$ a derivation in local coordinates. The definition 1.1 can be expressed as $g(\nabla (f), X)=(\phi_{g})(\nabla (f))(X)=df(X)=X(f)$.
To simplify notation we denote $\partial_{i}=\frac{\partial}{\partial x_{i}}$, and $(g^{ij})$ as the inverse matrix of $(g_{ij})$. So, by the bilinearity of the inner product for each $g_{ij}$
\begin{equation*}
X(f)=\sum_{i=1}^{n}a_{i}\partial_{i}f=\sum_{ijk} (g_{ij}g^{jk})a_{i}\partial_{k}f=g \left(X,\sum_{j=1}^{n} \sum_{k=1}^{n} (g^{jk}\partial_{k}f) \partial_{j}\right) \square
\end{equation*}
{\bf Exercise 4:} If we consider $\mathbb{R}^{n}$ as a Riemannian manifold with the usual inner product and given $f \in C^{\infty}(\mathbb{R}^{n})$, demostrate that $\nabla(f)=(\partial_{1}(f), \cdots,\partial_{n}(f))$.
Taking the canonical base of $\mathbb{R}^{n}$ ($T_{p}\mathbb{R}^{n} \simeq \mathbb{R}^{n}$) and the equation (3)
\begin{equation*}
\nabla(f)=\sum_{j=1}^{n} \sum_{k=1}^{n} \delta_{jk}\partial_{k} f \partial_{j}=\sum_{k=1}^{n}(\partial_{k}f) \partial_{k}=(\partial_{1}(f), \cdots,\partial_{n}(f)) \; \square
\end{equation*}
Exercise 5: Recall the definitions of volume form and Lie
derivatives of $p$--forms.
A \textit{volume form} is a no-where-vanishing differential n-form in $M$.
Definition
The \textit{inner product} of differential r-forms \cite{IC} is defined, according to \cite{IC}, by $i_{X}: \bigwedge^{r} T_{p}M^{*} \rightarrow \bigwedge^{r-1} T_{p}M^{*}$ such that
\begin{equation*}
i_{X}v (Y_{1},\dots, Y_{r-1}):= v(X, Y_{1},\dots, Y_{r-1})
\end{equation*}
\noindent and the Lie derivative is defined by
\begin{equation}
L_{X} := i_{X} \circ d + d \circ i_{X}
\end{equation}
Definition
We define the divergence of $X$ with respect to $\omega$ by the equation:
\begin{equation*}
L_X(\omega)=(div(X))\omega \; \square
\end{equation*}
Observe that $div(X) \in C^{\infty}(M)$.
Exercise 6: Prove that $(div X)\omega=d(i_X\omega )$.
Since $dw=0$ and applying this in the equation (2) we get
\begin{equation*}
L_{X}(w)=(i_{X} \circ d + d \circ i_{X})(w)=(i_{X} \circ d)(w) + (d \circ i_{X})(w)
\end{equation*}
\begin{equation*}
=i_{X}(dw) + d (i_{X}w)=i_{X}(0) + d (i_{X}w)=d (i_{X}w) \; \square
\end{equation*}
Definition
We say that the manifold M is orientable if there exists a covering $\{(U_{i}, \; \psi_{i})\}_{i\in I}$ of charts ($\phi_{i}:U_{i} \subseteq \mathbb{R}^{n} \rightarrow M$) such that the Jacobians of the transition maps $\psi_{i} \circ \psi_{j}^{-1}$ have positive determinants. This definition is equivalent to the existence of a volume form \cite{IC}.
Exercise 7: Let $\{ (U_{i}, \psi_{i}) \}_{i \in I}$ be a compatible charts covering, then the locally defined form $\omega:=\sqrt{det((g_{ij}))} \; dx_1 \wedge \cdots \wedge dx_n$ is in fact a global form.
Let $(x_{1}, \dots, x_{n})=\psi_{i}:U \subseteq \mathbb{R}^{n} \rightarrow M$ and $(x^{'}_{1}, \dots, x^{'}_{n})=\psi_{j}:V \subseteq \mathbb{R}^{n} \rightarrow M$. If we take the following two volume forms $\omega_{1}:=\sqrt{det((g_{ij}))} \; dx_1 \wedge \cdots \wedge dx_n$ and $\omega_{2}:=\sqrt{det((g^{'}_{ij}))} \; dx^{'}_1 \wedge \cdots \wedge dx^{'}_n$, then we need to proof that $\omega_{1}|_{U \cap V}=\omega_{2}|_{U \cap V}$.
Firstly, since $\partial_{i}=\sum_{k=1}^{n}\frac{\partial x^{'}_{k}}{\partial x_{i}} \frac{\partial }{\partial x^{'}_{k}}$ we have
\begin{equation*}
\sqrt{det((g_{ij}))}=\sqrt{det((g(\partial_{i},\partial_{j})))}=\sqrt{det\left( \left( \sum_{k,l=1}^{n}\frac{\partial x^{'}_{k}}{\partial x_{i}}\frac{\partial x^{'}_{l}}{\partial x_{j}} g\left( \partial^{'}_{k}, \partial^{'}_{l}\right) \right)\right)}
\end{equation*}
\begin{equation*}
=\sqrt{det\left( \left( \frac{\partial x^{'}_{k}}{\partial x_{i}} \right)^{T} \left( \frac{\partial x^{'}_{l}}{\partial x_{j}}\right) (g(\partial^{'}_{k},\partial^{'}_{l})) \right)} = det(J)\sqrt{det((g^{'}_{kl}))}
\end{equation*}
where $J=\left( \frac{\partial x^{'}_{k}}{\partial x_{i}} \right)$ is the derivative of the change of coordinates. Also, we can realise that $dx^{'}_{i}=\sum_{k=1}^{n}\frac{\partial x^{'}_{k}}{\partial x_{i}} dx_{k}$, so
\begin{equation*}
dx^{'}_1 \wedge \cdots \wedge dx^{'}_n = \sum_{j_{1},\dots,j_{n}} \frac{\partial x^{'}_{1}}{\partial x_{j_{1}}} \dots \frac{\partial x^{'}_{n}}{\partial x_{j_{n}}} dx_{j_1} \wedge \cdots \wedge dx_{j_n}
\end{equation*}
\begin{equation*}
= \sum_{j_{1} < j_{2} < \dots < j_{n}} \frac{\partial x^{'}_{1}}{\partial x_{j_{1}}} \dots \frac{\partial x^{'}_{n}}{\partial x_{j_{n}}} dx_{j_1} \wedge \cdots \wedge dx_{j_n} = det(J)dx_1 \wedge \cdots \wedge dx_n
\end{equation*}
$\square$
Exercise 8: Suppose that $M$ is an oriented Riemaniann
manifold. Let $X$ be a vector field of $M$. If $X=\sum_{i=1}^n a_i(x) \partial_{i}$ in local coordinates, prove that
\begin{equation*}
div(X)=g^{-1/2} \partial_{j}(g^{1/2}).
\end{equation*}
Hint: Exercise 6 could help.
From \cite[Pag 150]{IC}
\begin{equation*}
(div(X))(w)=L_{X}(w)=d(i_{X}(w))=d(i_{X}(\sqrt{g} \; dx_1 \wedge \cdots \wedge dx_n))
\end{equation*}
\begin{equation*}
=d \left( \sum_{j=1}^{n} (-1)^{j-1} a_{i}(x) \sqrt{g} \; dx_1 \wedge \cdots \wedge \widehat{dx_{j}} \wedge \dots \wedge dx_n \right)
\end{equation*}
\begin{equation*}
= \sum_{j=1}^{n} (-1)^{j-1} d(a_{i}(x) \sqrt{g}) \wedge dx_1 \wedge \cdots \wedge \widehat{dx_{j}} \wedge \dots \wedge dx_n
\end{equation*}
\begin{equation}
= \sum_{j=1}^{n} \frac{\partial (a_{i}(x) \sqrt{g})}{\partial x_{j}} \; dx_1 \wedge \cdots \wedge dx_n = \left( \frac{1}{\sqrt{g}} \sum_{j=1}^{n} \frac{\partial (a_{i}(x) \sqrt{g})}{\partial x_{j}} \right) (w)
\end{equation}
$\square$
Exercise 9: Consider $M:=\mathbb{R}^{n}$ as a Riemaniann manifold.
Show that $div(X)=\sum_{i=1}^n \partial_{i}(a_i(x))$.
Since $\sqrt{g}=\sqrt(det((\hat{g}_{ij})))=1$ choosing the canonical base of $\mathbb{R}^{n}$ with the usual inner product and orientation, the exercise 8 shows that
\begin{equation*}
div(X)=\sum_{j=1}^{n} \frac{\partial (a_{i}(x))}{\partial x_{j}}
\end{equation*}
Stokes's Theorem
If $M$ is an open and oriented manifold that has a smooth boundary $\partial M$, let $\phi$ a differential k-form in $M$ and $In:\partial M \rightarrow M$ the inclusion of the boundary to the manifold., then
\begin{equation}
\int _{M} d\phi = \int _{\partial M} IN^{*}\phi
\end{equation}
$\square$
Exercise 10: Use Stokes theorem to prove $\int_M (div X)\omega=\int i_X(\omega)$.
From the exercise 6, $(div(X))w=d(i_{X}w)$, the divergence is the exterior derivative of the form $w(X)$, therefore using the Stoke's theorem
\begin{equation*}
\int _{M} d(i_{x}w) = \int _{M} (div(X))w = \int _{\partial M} i_{x}w
\end{equation*}
$\square$
Exercise 11: Suppose that $(M,[\omega])$ is an oriented manifold with boundary. Prove that $\partial M$ has an orientation inherited
from the orientation of $M$.
Let $p \in \partial M$ an element of the boundary, we consider $T_{p}M$,the tangent space of the boundary at $p$, $T_{p}\partial M$, and $E=\{N \in T_{p}M ; a^{n} \leq 0 \; and \; \sum_{i=1}^{n-1}a^{i}\partial_{i} + a^{n}\partial_{n} = N \}$.
There exists an orientation of the boundary at $p$ defined as: $B^{'}=\{ e_{2},\dots,e_{n} \}$ is a positively oriented basis of $T_{p}\partial M$ if $B_{N}=\{ N, e_{2},\dots,e_{n} \}$ is positively oriented in $(T_{p}M, [w|_{p}])$ for each $N \in E$. This orientation is induced by $E$ and $[\omega]$.
This definition doesn't depend on either $B^{'}$ or $N$, if we take another basis $\hat{B}^{'}$ and $\hat{N} \in E$ then
\begin{equation*}
M_{B_{N}}^{\hat{B}_{\hat{N}}}=\left(
\begin{array}{cc}
a_{1} & 0 \cdots 0 \\
a_{2} & \\
\vdots & M_{B^{'}}^{\hat{B}^{'}} \\
a_{n} & \\
\end{array}
\right)
\end{equation*}
where $a_{1}>0$. So $B_{N}$ has the same orientation of ${\hat{B}_{\hat{N}}}$ iff the orientations of ${B^{'}}$ and ${\hat{B}^{'}}$ are consistent.
Remark that
\begin{equation*}
i_{N}\omega|_{T_{p}\partial M}(e_{2}, \dots, e_{n})=\omega(N,e_{2},\dots,e_{n})>0 \; \forall N \in E
\end{equation*}
So $i_{N}\omega|_{T_{p}\partial M}$ is a volume element in $T_{p}\partial M$ and is consistent with the induced orientation. $\square$
The following exercises provide an interpretation of the
divergence on Riemannian manifolds.
Exercise 12: Suppose that $M$ is an oriented Riemannian
manifold with boundary. Show that if $\omega=dvol_g(x)$ is the
volume form defined in exercise 8, then for all $X\in \Gamma (TM)$
and $x \in
\partial M$
$i_X(\omega)(x)=g(X,n)(x) dvol(x),$
where $n$ is the unitary vector orthogonal to $T_x \partial M$
such that $n$ joined to any compatible oriented base of $T_x
\partial M$ conforms an compatible oriented base of $T_x M$.
Let $X^{'}=X-g(X,N)N \in \Gamma (TM)$, then
\begin{equation*}
i_{X}\omega=i_{X^{'}}\omega(v_{2}, \dots, v_{n})+g(X,N)i_{N}\omega (v_{2}, \dots, v_{n})
\end{equation*}
So $i_{N}\omega|_{\partial M} = dvol_{S}$. Let $\{ v_{2}, \dots v_{n} \}$ be vectors of $T_{p}\partial M$, then $\{X^{'}, v_{2}, \dots v_{n} \}$ aren't linearly independent and $(i_{X^{'}}\omega(v_{2}, \dots, v_{n}))=0$. $\square$
In the next exercises we will make use of the following theory on
measure theory.
Definition
A family $\{ E_{r} \}_{r>0}$ of Borel subsets of $\mathbb{R}^{n}$ is said to \textit{shrink nicely} if
- $E_{r} \subset B(r,x)$ for each r.
- There is a constant $\alpha>0$, independent of r, such that $m(E_{r})>\alpha m(B(r,x))$.
Theorem
Let $f \in C^\infty(M)$. We will work in some coordinates $x_1,\cdots, x_n$ of an open $U \subset M$ diffeomorphic to $\mathbb{R}^n$. For
all $r>0$ we define the function $A_r(f)$ in $C^\infty(M)$ by:
$A_r(f)(x)=\frac{\int_{E_r}f(y)dvol_g(y)}{\int_{E_r} dvol_g(y)}$
where $E_r$ is a family that shrinks nicely to $x$.
Exercise 13: Let $X \in \Gamma (TM)$. Prove that $\lim_{r
\to 0} A_r(div_g(X)) =div_g(X)(x)$.
By the divergence theorem
\begin{equation*}
\int_{M}div(w)=\int_{\partial M} i^{*}w
\end{equation*}
Using the theorem 3.22 from \cite{GF}, as $div(X) \in L_{1}$ the result is reached.
$\square$
Definition
Now we define the Laplacian as:
\begin{equation}
\Delta_{g}f(x)=\lim_{\epsilon \rightarrow 0} \frac{\int_{B_{\epsilon}(x)} n \cdot \nabla f(g) \;\; dVol_{g}(y)}{Vol(B_{\epsilon}(x))}.
\end{equation}
Bibliography
- {IC}Chavel, I. Riemannian Geometry. Second Edition. 2006.
- {GF}Folland, G. Real analysis: modern techniques and their applications. Ed. 2. Wiley Interscience. 1999.
- {MS}Spivak, M. Calculus on Manifolds. Brandeis University. 1965.
- {FW}Warner, F. Foundations of Differentiable Manifolds and Lie Groups. University of Pennsylvania.
