2
$\begingroup$

This is an exercise in AC da Silva's Lectures onn Symplectic Geometry; I am having trouble showing the following.

$(X,g)$ is a geodesically complete manifold, and $f: X \times X \to \mathbb{R}$ is given by $f(x,y) = - \frac{1}{2} d(x,y)^2$ where $d$ is the Riemannian distance function (i.e. the infimum of arc-lengths of piecewise smooth curves joining $x$ to $y$).

I need to show that under the identification of $TX$ and $T^*X$ by the metric $g$, the symplectomorphism generated by $f$ coincides with the map $TX \to TX$ given by $(x,v) \to (\exp(x,v)(1), \frac{d}{dt} \exp(x,v) (1))$.

This amounts to solving the equations $g_x(v, \cdot ) = d_x f (\cdot )$ and $g_y (w, \cdot) = -d_y f(\cdot)$ for $(y, w)$ with fixed $(x,v)$, where $d_x f, d_y f$ are components of $df_{(x,y)}$ where $T^*_{(x,y)} (X \times X) \simeq T_x^* X \times T_y ^* X$.

I am having trouble in trying to solve the first equation, $g_x (v, \cdot) = d_x f(\cdot)$ for $y$, where both are members of $T_x^* X$. (The answer should be $y = \exp(x,v)=1$.) In particular, when $g$ (and hence $d$) is not concretely given, how should I go about finding $df$ at a fixed point $(x,y)$? Thank you for any help.

$\endgroup$
1
$\begingroup$

Given $(x,v) \in TX$, let $\exp_x(v): \mathbb R \to X$, $t \mapsto \exp_x(tv)$ be the unique geodesic with initial conditions

$$\begin{cases} \exp_x (0)= x \\ \frac{d}{dt}\big|_{t=0}\exp_x (tv) = v \end{cases}$$ Consider the map

$$\begin{aligned}\tilde g_x : T_xM &\to T_x^*M \\ v &\mapsto g_x (v, \cdot) \end{aligned}$$ We need to solve

$$\begin{cases} \tilde g_x (v) = \xi_i = d_xf \\ \tilde g_x(w) = \eta_i = - d_yf \end{cases} \iff \begin{cases} g_x (v, \cdot) = d_xf (\cdot) &(\star) \\ g_x(w, \cdot) =- d_yf (\cdot) &(\star \star) \end{cases} $$

First we solve $(\star)$ for $y \in X$. Since $X$ is geodesically convex there exists $u \in T_xM$ such that $\exp_x (u) = y$.

Claim: For $u,v \in T_x M$ $$\frac{d}{dt}\bigg|_{t=0} \left(-\frac{1}{2} d(\exp_x(tv), \exp_x (u))^2\right) = g_x(u,v)$$

Proof: Let $s \in \mathbb R$ be sufficiently small so that $\exp_x(su)$ is contained in a geodesic ball centered at $x$ and $\exp_x(u) = y \in \partial_sB(x)$ (see Figure 1). Then we have that $d(\exp_x(su) , \exp_x(u)) = |1-s||u|$. Indeed, let $\gamma :\mathbb R \to X$ be the geodesic starting at $x$ with velocity $u$, then

$$\begin{aligned} L\left(\gamma|_{[s,1]}\right) = \int_s^1 |\dot\gamma (t)| dt = |u|\int_s^1 dt = |1-s||u| \end{aligned}$$

enter image description here

Since the radial geodesic $\exp_x(su)$ is the unique minimizing geodesic from $\exp_x(u)$ to $\exp_x(su)$ it follows

$$d(\exp_x(su) , \exp_x(u)) = |1-s||u|$$

Decomposing $v = \alpha u + z$, where $z$ is the tangent vector to the geodesic sphere $\partial_s B(x)$. By Gauss Lemma this is an orthogonal decomposition, plus there is a curve $\beta : \mathbb R \to X$ starting at $x$ with velocity $z$ such that $f (\beta(t),y)$ is constant.

Now define $\hat f : X \to \mathbb R$, by $x \mapsto f(x, \exp_x(u))$ and $l : \mathbb \to X $, by $t \mapsto \exp_x(tv)$ then

$$\begin{aligned} \frac{d}{dt}\bigg|_{t=0}(f \circ l)(t) & = df_0(l(0))\cdot l'(0) = df_0(x)(v) \\&= df_0 (x)(\alpha u + z)\\&=\alpha df_0(x)(u) + \underbrace{df_0(x)(z)}_{0}\\&= \alpha \frac{d}{ds}\bigg|_{s=0} \left(-\frac{1}{2}d(\exp_x(su), \exp_x(u))\right) \\&= \alpha \frac{d}{ds}\bigg|_{s=0} \left(-\frac{1}{2} |1-s|^2|u|\right) \\&= \alpha |u|^2\\&=g_x(u,v) \end{aligned}$$ as we wanted.

Evaluating $\tilde g_x (v) = g_x(v , \cdot)$ and $d_xf (\cdot)$ at $v$, by the claim we get

$$|v|^2 = d_xf (v) = \frac{d}{dt}\bigg|_{t=0} \left(-\frac{1}{2} d(\exp_x(tv), \exp_x (u))^2\right) = g_x (u,v)$$

For any $v' \perp v$ we have that

$$0 = d_xf (v') = g_x (u,v')$$ Therefore, $u=v$ and $y = \exp_x (v)$. To solve $(\star \star)$ let $W = \frac{d}{dt}\Big|_{t=1} \exp_x (tv)$ and fix any $w' \perp w$. Again by the Gauss Lemma $d_f (w') =0$. Thus $w = k W$. Since geodesics have constant velocity $|W|^2 = |v|^2$. Hence the left hand side of $(\star \star)$ is

$$g_x(w,W) = k |W|^2 = k |v|^2$$ while the right hand side is

$$\begin{aligned} -d_yf (W) &= \frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}d(x , \exp_x (1+s) (v))^2\right) \\&= \frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}d(\exp_x (0v), \exp_x (1+s) (v))^2\right) \\&\underset{(*)}= v\frac{d}{ds}\bigg|_{s=0} \left(\frac{1}{2}(1+s)^2|v|^2\right)\\&= |v|^2 \end{aligned}$$ where we used the same first argument of the claim in $(*)$. Thus we have

$$k |v|^2 = |v|^2 \implies k =1$$ and $w = W$ as required.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.