Comparison Theorem in Riemannian geometry

 

Given $p\in M$, locally, there exists a diffemorphism of $T_M$ and $B_r(p)\subset$, this is the most important geometric mapping in Riemannian geometry, it is called the Gauss exponential mapping. It reveals enough geometric information near $p\in M$. In the following, we always assume that $M$ is complete.

Firstly, we need the definition of geodesic line. We only give the equation of it:
\begin{align}
\nabla_{\gamma'(t)}\gamma'(t)=0.
\end{align}
In local coordinate it can be written as
\begin{align}
x_k''(t)+\Gamma_{ij}^kx_i'(t)x_j'(t)=0.
\end{align}
This is a nonlinear ode, which can be solve locally with initial data.

The Gauss mapping is defined as
\begin{align}
exp:\ v\in T_pM\rightarrow M.
\end{align}

The most important property is the Gauss lemma.

\textbf{Gauss Lemma}
\begin{align}
\langle (dexp_p)_v(v),(dexp_p)_v(w)\rangle_{exp_p(v)}=\langle v,w\rangle_p.
\end{align}

Using the Gauss mapping and radial geodesic line, we get the most important coordinate in calculation, that is the normal coordinate. In fact, for any $a=(a_1,a_2,..,a_n)\in \varphi(U)$, we have
\begin{align}
x_ioexp_p(a)=a_i,\ \ i=1,...,n.
\end{align}
The radial geodesic line $exp_p(ta)$ has the parametric equation as
\begin{align}
x_i(t)=ta_i,
\end{align}
and plug it into the geodesic line equation, we get
\begin{align}
x_k''(t)+x'_i(t)x'_j(t)\Gamma_{ij}^k(ta)=0.
\end{align}
Let $t=0$, we get $a_ia_j\Gamma_{ij}^k(0)=0$. It follows that $\Gamma_{ij}^k(0)=0$ and $\frac{g_{ij}}{\partial x_k}(0)=0$, $g_{ij}(0)=delta_{ij}$.

Now we will turn to another topic, i.e., Jacobi field. After reading the book written by Weihuan Chen, and rethinking the process, we have some main idea to demonstrate.
(1) Firstly, what is called a Jacobi field, it comes from the geodesic line variation. It turn out that we call the vector field $J(t)$ along the geodesic line as Jacobi field if it satisfies the equation
\begin{align}
J''(t)-R(\gamma'(t),J)\gamma'(t)=0.
\end{align}
Recall that $\gamma(t)$ is a geodesic line.
It should notice that in the deduce process, the equation comes from the induced connection's compatibility, like
\begin{equation}
[\partial_t,\partial_s]=0,\nonumber
\end{equation}
and the definition of curvature operator, like
\begin{align}
R(X,Y)Z=\nabla_X \nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.\nonumber
\end{align}

Secondly, the Jocobi equation is a second order linear equation, it can be uniquely solved by change it to a linear first ode systems, which can be uniquely solved in the whole interval $[a,b]$ by Picard iteration. One thing should be emphasized that for linear ode, the Cauchy problem and the two ends point are different, the later one may causes nonuniqueness of solutions. It is a $2n$ dimension linear space. Note that let $f(t)=<J(t),\gamma'(t)>$, we get
\begin{align}
f''(t)=\langle J''(t),\gamma'(t)\rangle =\langle R(\gamma'(t),J(t))\gamma'(t),\gamma'(t)\rangle=0,
\end{align}
Then we have
\begin{align}
\langle J(t),\gamma'(t)\rangle =\lambda+\mu t,
\end{align}
if we use the $\gamma'(t)$ is the normal geodesic line and $t$ is the arc length parameter, it it easy to check that
\begin{align}
J(t)-\langle J(t),\gamma'(t)\rangle \gamma'(t)
\end{align}
still satisfies the Jacobi equation and it is a Jacobi field perpendicular to $\gamma'(t)$. Another existence result is the two ends ode, \textbf{if $q$ is not the conjugate point of $p$, then the Jacobi is uniquely determined by its value at $p$ and $q$.} It also should note that if $q$ is the conjugate to $p$, if and only if there exists nonzero Jacobi field along the geodesic line $\gamma(t)$ such that $J(t)$ are zero at $p$ and $q$ (At this time $J(t)$ is perpendicular to $\gamma'(t)$ along the geodesic line.

Thirdly, every Jocobi field can be generated by the geodesic variation, its proof need the uniqueness of the Jocobi equation. Especially, if the initial data $J(0)=0$, its geodesic variation can be written expressed by
\begin{align}
\phi(t,s)=exp_{p}\Big[t\big(\gamma'(0)+sJ'(0)\big)\Big].
\end{align}
This time, $J(t)$ can be written as
\begin{align}
J(t)=d\phi_{(t,0)}(\partial_s)=\Big(dexp_p\Big)_{t\gamma'(0)}\Big(tJ'(0)\Big)=t\Big(dexp_p\Big)_{t\gamma'(0)}\Big(J'(0)\Big).
\end{align}

This property is great! This formula gives a precise geometric interpretation of the Jocobi field. After the exponential mapping, the length of Jacobi field(with initial data is zeor) reflects the curvature of the manifold near $p\in M$. More precise formula are as follows.

Let $\gamma'(0)=V$ and $J'(0)=W$, $J'(0)=0$. By the above argument, we have
\begin{align}
J(t)=\Big(dexp_p\Big)_{tV}\Big(tW\Big).
\end{align}
Then
\begin{align}
|J(t)|^2=|W|^2t^2+\frac{1}{3}\langle R(V,W)V,W\rangle t^4+o(t^4).
\end{align}
The proof involves the Taylor expansion. Let $f(t)=|J(t)|^2=\langle J(t),J(t)\rangle$. Taking derivative repeatedly and using the Jocobi equation will derive the formula.In fact, $f(0)=0,f'(0)=0$, and
\begin{align}
f''(t)=2\langle J''(t),J(t)\rangle+2\langle J'(t),J'(t)\rangle\Longrightarrow f''(0)=2\langle W,W \rangle.
\end{align}
Differentiate the identity again, we have
\begin{align}
f'''(t)=2\langle J'''(t),J(t)\rangle+6\langle J''(t),J'(t)\rangle\Longrightarrow f'''(0)=0
\end{align}
and
\begin{align}
f''''(t)&=2\langle J''''(t),J(t)\rangle+8\langle J'''(t),J'(t)\rangle+6\langle J''(t),J''(t)\rangle\\
&\Longrightarrow f''''(0)=8\langle J'''(0),J'(0)\rangle\nonumber
\end{align}
Recall that
\begin{align}
J''(t)=R(\gamma'(t),J(t))\gamma'(t),
\end{align}
then
\begin{align}
\langle J'''(t),J'(t)\rangle=&\langle\nabla_{\gamma'(t)}\left(R(\gamma'(t),J(t))\gamma'(t)\right),J'(t)\rangle\\
=&\nabla_{\gamma'(t)}\Big( R(\gamma'(t),J(t))\gamma'(t),J'(t)\Big)-\langle R(\gamma'(t),J(t))\gamma'(t),J''(t)\rangle\nonumber\\
=&(\nabla_{\gamma'(t)}R)(\gamma'(t),J(t),\gamma'(t),J'(t))+R(\gamma'(t),J'(t),\gamma'(t),J'(t))\nonumber\\
&=(\nabla_{\gamma'(t)}R)(\gamma'(t),J(t),\gamma'(t),J'(t))+\langle R(\gamma'(t),J'(t))\gamma'(t)),J'(t)\rangle\nonumber
\end{align}
where we have used the definition of $\nabla R$. Note that $J(0)=0$, then
\begin{align*}
(\nabla R)(\gamma'(0),J(0),\gamma'(0),J'(0),\gamma'(0))=0,
\end{align*}
i.e.,
\begin{align*}
(\nabla_{\gamma'(t)} R)(\gamma'(t),J(t),\gamma'(t),J'(t))|_{t=0}=0.
\end{align*}
In fact, we can prove
\begin{align}
\nabla_{\gamma'(t)}(R(\gamma'(t),J(t))\gamma'(t))(0)=(R(\gamma'(t),J'(t))\gamma'(t))_{t=0}.
\end{align}

It follows that
\begin{align}
\langle J'''(0),J'(0)\rangle=\langle R(\gamma'(0),J'(0))\gamma'(0),J'(0)\rangle=\langle R(V,W)V,W\rangle.
\end{align}
Then the asymptotic expansion follows from Taylor' formula clearly.

It seems reveal the mystery. Differential Geometry is to differentiate it and differentiate it again and again.

Finally, after the argument as above, by the geodesic coordinate, we have $\frac{\partial}{\partial \theta_i}$($i=1,...,n-1$) are Jacobi field. This follows from the linearity of the map $(dexp)_{tV}(\cdot)$. As in the book of Hamilton's Ricci flow, let
\begin{align}
h_{ij}=\langle\frac{\partial}{\partial \theta_i},\frac{\partial}{\partial \theta_j}\rangle,
\end{align}
for $i,j=1,...,n-1$.
it is not hard to check that
\begin{align}
h_{ij}=-\Gamma_{ij}^n=\frac{1}{2}\frac{\partial g_{ij}}{\partial r},
\end{align}
and
\begin{align}
H=g^{ij}h_{ij}=\frac{1}{2}g^{ij}\frac{\partial g_{ij}}{\partial r}=\frac{\partial \log\sqrt{g}}{\partial r}=\frac{\partial \log J(r,\theta)}{\partial r}.
\end{align}

Now we go to the symmetric two linear form. It is note hard to see that if $X,Y$ are the Jacobi field along the geodesic line, then
\begin{align}
\langle X'(t),Y(t)\rangle-\langle X(t),Y'(t)\rangle\equiv constant
\end{align}
on $\gamma(t)$.

If $Y$ is any smooth vector field along the geodesic line, we define
\begin{align}
I(Y)=\int_0^l\langle Y'(t),Y'(t)\rangle+R(\gamma'(t),Y)\gamma'(t),Y\rangle dt.
\end{align}
It should note the this is related to the second derivative of the arc length variation with fixed two end point, i.e.,
\begin{align}
\frac{d^2L(t,s)}{dt^2}|_{t=0}=I(U^\perp,U\perp),
\end{align}
where $U$ is the transversal variation field.

Now polarized $I$, we have
\begin{align}
I(V,W)=\int_0^l\langle V'(t),W'(t)\rangle+\langle R(\gamma'(t),V)\gamma'(t),W\rangle dt.
\end{align}
Obviously, it is a symmetric two linear form. The fundamental lemma of two linear form it as follows,which will be using in the proof of comparison theorem.
Lemma：
If the exists no conjugate point along normal geodesic line $\gamma(t)$, and $J(t)$ is the Jacobi field along $\gamma(t)$. If $X(t)$ is a field along $\gamma(t)$ with $X(0)=J(0)$ and $X(l)=Y(l)$, then
\begin{align}
I(X,X)\geq I(J,J)
\end{align}
with inequality if and only if $X=J$.

Once we have this minimality of the index form of Jacobi field, we can prove the Rauch's comparison theorem. In Hongxi Wu's book, it gives the ode systems comparison which unified various comparison theorem, see also in Jianguo Cao's book. If $A(t)$ is a $n-1$ order square matrix of the coefficients of the Jacobi fields, then $A'(t)A(t)$ is symmetric matrix.

In fact, the Hessian comparison and Laplacian comparison theorem is related to index form as follows
\begin{align*}
D^2r(J(b),J(b))&=SSr-\nabla_{\nabla_SS}r\\
&=S\langle S,\nabla r\rangle-\langle \nabla_SS,\nabla r\rangle\\
&=S\langle S,T\rangle-\langle \nabla_SS,T\rangle\\
&=\langle S,\nabla_ST\rangle\\
&=\langle S,\nabla_TS\rangle\\
&=\langle J(b),J'(b)\rangle.
\end{align*}

[Rauch comparison theorem] Let $\gamma:[0,b]\rightarrow M$,$\tilde{\gamma}:[0,b]\rightarrow \tilde{M}$ are all normal geodesic line, and $dim(M)=dim(\tilde{M})$; Let $J,\tilde{J}$ are the Jacobi field along $\gamma,\tilde{\gamma}$ respectively, and $J(0)=0$, $\tilde{J}(0)=0$, $J'(0)\bot \gamma'(0)$,$\tilde{J}'(0)\bot \tilde{\gamma}'(0)$, $|J'(0)|=|\tilde{J}'(0)|$; $\gamma$ doesn't contain the conjugate point along $\gamma$. and $\tilde{K}(t)\leq K(t)$. Then
\begin{align}
|\tilde{J}(t)|\geq |J(t)|.
\end{align}

Before the proof of Rauch's comparison theorem, we need a technique lemma.
Lemma：
Suppose the assumption of the theorem holds. Let $J$,$\tilde{J}(t)$ are normal Jacobi field along $\gamma,\tilde{\gamma}$ respectively, such that $J(0)=0$, $\tilde{J}(0)=0$ and such that for some $\beta\in[0,b]$, $|J(b)|=|\tilde{J}(b)|$, then
\begin{align}
\langle J'(b),J(b)\rangle\leq \langle \tilde{J}'(b),\tilde{J}(b)\rangle.
\end{align}

Recall that
\begin{align*}
I(J,J)&=\int_0^b\langle J'(t),J'(t)\rangle+\langle R(\gamma'(t),J)\gamma'(t),J\rangle dt\\
&=\int_0^b\langle J'(t),J'(t)\rangle+\langle J''(t),J\rangle dt\\
&=\langle J'(t),J(t)\rangle\Big|_{t=0}^{t=b}\\
&=\langle J'(b),J(b)\rangle,
\end{align*}
Then it is not hard to see that the lemma can be prove by the Jacobi comparison theorem.

Now we give the skeleton of the proof. Consider $g(t)=\frac{|\tilde{J}(t)|^2}{|J(t)|^2}$ for $t\in(0,b]$, using L.Hospital's rule twice, we have $g(t)\rightarrow 1$ as $t$ tends to $0_+$. Differentiate $g(t)$, for $t=\beta\in(0,b]$, we get
\begin{align}
g'(\beta)=\frac{\frac{\langle \tilde{J}'(\beta),\tilde{J}(\beta)\rangle}{|\tilde{J}(\beta)|^2}-\frac{\langle J'(\beta),J(\beta)\rangle}{|J(\beta)|^2}}{|\tilde{J}(\beta)|^2|J(\beta)|^4}.
\end{align}
We only need to show
\begin{align}
\frac{\langle \tilde{J}'(\beta),\tilde{J}(\beta)\rangle}{|\tilde{J}(\beta)|^2}\geq\frac{\langle J'(\beta),J(\beta)\rangle}{|J(\beta)|^2}.
\end{align}
Let $J_1(t)=\frac{J(t)}{|J(\beta)|}$, $\tilde{J}_1(t)=\frac{\tilde{J}(t)}{|\tilde{J}(\beta)|}$, by the previous lemma, we have
\begin{align}
\langle J_1'(b),J_1(b)\rangle\leq \langle \tilde{J}_1'(b),\tilde{J}_1(b)\rangle,
\end{align}
which is exactly the inequality we need.

One more thing, we should note that the Hodge Laplace act on form although is the some as the Berltrami-Laplace, but it is completely different from the covariant Laplace for $1$ forms, such as $df$.

For $r=dist(x,p)$, we have $|\nabla r|=1$ near $p\in M$, by the Bochner's formula, we get
\begin{align}
0=|Hess(r)|^2+\langle\nabla \Delta r,\nabla r\rangle+Ric(\nabla r,\nabla r).
\end{align}
Note that
\begin{align}
\Delta r=\sum_{i=1}^{n-1}\sum_{j=1}^{n-1}g^{ij}h_{ij}=H=\frac{\partial \log J(r,\theta)}{\partial r},
\end{align}
which is a very useful geometric formula in comparison theorem.

By Cauchy's inequality and the geodesic polar coordinate, we have
\begin{align}
(n-1)|Hess(r)|^2\geq(\Delta r)^2 ,
\end{align}
we obtain
\begin{align}
0\geq\frac{\partial}{\partial r}\left(\Delta r\right)+(\Delta r)^2+Ric(\nabla r,\nabla r).
\end{align}
If $Ric\geq (n-1)K$, we have
\begin{align}
0\geq\frac{\partial}{\partial r}\left(\Delta r\right)+(\Delta r)^2+(n-1)K.
\end{align}