Geodesic flow preserves the volume (Liouville 's Theorem)

Question

I am reading Do Carmo's Riemannian Geometry. I got stuck on Problem 14 of Chapter 3. Here is the original problem with the hints:

(Liouville's Theorem) Prove that if $G$ is the geodesic field on $TM$ then $\mathrm{div}G=0$. Conclude from this that the geodesic flow preserves the volume of $TM$.

Hint: Let $p\in M$ and consider a system $(u_1, \ldots, u_n)$ of normal coordinates at $p$. Such coordinates are defined in a normal neighborhood $U$ of $p$ by considering an orthonormal basis $\{e_i\}$ of $T_pM$ and taking $(u_1, \ldots, u_n), q = \mathrm{exp}_p(\sum_iu_ie_i), i=1, \ldots, n$ as coordinates of $q$. In such a coordinate system, $\Gamma_{ij}^k(p)=0$, since the geodesics that pass through $p$ are given by linear equations. Therefore if $X=\sum x_i\frac{\partial}{\partial u_i}$, then $\mathrm{div}X(p)=\sum\frac{\partial x_i}{\partial u_i}$.

Now let $(u_i)$ be normal coordinates in a neighborhood $U\subset M$ around $p\in M$ and let $(u_i, v_j), v=\sum_j v_j\frac{\partial}{\partial u_j}, i, j = 1, \ldots, n$ be coordinates on $TM$ at $(q, v)$. Calculate the volume element of the natural metric of $TM$ at $(q, v), q\in U, v\in T_qM$, and show that it is the volume element of the product metric on $U\times U$ at the point $(q, q)$. Since the divergence of $G$ only depends on the volume element, and $G$ is horizontal, we can calculate $\mathrm{div}G$ in the product metric. Observe that in the coordinates $(u_i, v_j)$ we have $$G(u_i)=v_i, G(v_j)=-\sum_{ik}\Gamma^j_{ik}v_iv_k, k = 1,\ldots, n$$. Since the Christoffel sysbols of the product metric on $U\times U$ vanish at $(p, p)$, we can obtain finally, at $p$, $$\mathrm{div}G=\sum_i\frac{\partial v_i}{\partial u_i}-\sum_j\frac{\partial}{\partial v_j}\Big(\sum_{ik}\Gamma^j_{ik}v_iv_k\Big)=0.$$

Questions:

Where I got stuck is the bold part: Calculate the volume element of the natural metric of $TM$ at $(q, v), q\in U, v\in T_qM$, and show that it is the volume element of the product metric on $U\times U$ at the point $(q, q)$. I do not understand what "natural metric" is. If the natural metric is defined by $$\langle V, W\rangle_{(q, v)}=\langle d\pi(V), d\pi(W)\rangle_p + \langle \frac{Dv}{dt}(0), \frac{Dw}{ds}(0)\rangle_p$$ as is in Problem 2, then it seems too messy for me to calculate and simplify the volume element. Can any one help me on this? Thanks.

Answer 1

Assume that $TM$ has Sasaki metric $G$. Hence note that $M$ is $n$-dimensional totally geodesic submanifold in $TM$. In further $T_xM$ for all $x\in M$ is $n$-dimensional totally geodesic flat submanifold.

And $$ \pi : (TM,G)\rightarrow (M,g)$$ is a Riemannian submersion.

(1) Fix $(x,v)\in TM$. Consider tangent space of $TM$ at $(x,v)$.

Here $n$-dimensional subspace is horizontal lift of $T_xM$ by $\pi$, which is isometric to $(T_xM,g(x))$. And $n$-dimensional subspace orthogonal to that wrt $G$ is a fiber $(\pi^{-1}(x)=T_xM,g(x))$. Hence volume form $V(x,v)$ at $(x,v)$ wrt $G$ is a product of volume form $V(x)$ at $x$ wrt $g$.

Assume that $x$ is coordinate for $M$ Then we have a coordinate vector field $E_i$ Hence $$(x,v)\mapsto (x,v_iE_i)$$ is coordinate for $TM$.

If $g_{ij}:=g(E_i,E_j)$, then $$ V(x)=\sqrt{{\rm det}\ g_{ij} }\ dx^1\cdots dx^n $$

$$ V(x,v)={\rm det}\ g_{ij}\ dx^1\cdots dx^n dv^1\cdots dv^n $$

(2) If $c(t)=x(t)$ is a geodesic in $M$ with $$c'=x_i'(t)E_i(x(t)) $$ then $$ x_i''(t) +x_i' x_j' \Gamma_{jk}^i=0$$

Hence trajectory $(c(t),x_i'(t) E_i )$ has a tangent $$(x_i'E_i,x_j''E_j)=(x_i'E_i,-x_j'x_k'\Gamma_{jk}^iE_i ) $$

So if $x$ is normal coordinate, \begin{align*}&\ \ di_{(x_i'E_i,x_j''E_j)} V(x,v)\\&= d \bigg( {\rm det}\ g_{ab}\ x_i' \bigg)(-1)^i\ dx^1\cdots \widehat{dx^i}\cdots dx^n dv^1\cdots dv^n \\&- d \bigg( {\rm det}\ g_{ab}\ x_j'x_k'\Gamma_{jk}^l\bigg)(-1)^{n+l}\ dx^1\cdots dx^n dv^1\cdots \widehat{dv^l}\cdots dv^n\\&=0\end{align*}

Hence geodesic field has divergence $0$.

Answer 2

Step 1 : When $v_i'(0)=w_i$, then $\frac{d}{dt} {\rm det}\ [v_1\cdots v_n]= \sum_i\ {\rm det}\ [v_1\cdots w_i\cdots v_n]$

Step 2 : When $f$ is a coordinate for $M$, then we have a coordinate $F$ for $TM$ : $F(x,y) = (f(x) ,\sum_i\ y^i E_i (f(x)) )$ where $f_{x_i}=E_i$ is a coordinate field for $M$.

Hence $$F_{x_i} = (E_i,y^k\Gamma_{ik}^s E_s)$$ where $\Gamma_{ik}^s$ is a Christoeffel symbol for $M$.

$$ F_{y_i} = (0, E_i) $$

When $g$ is a metric on $M$ and $g_{TM}$ is a metric for $TM$, then $$ g_{TM} (F_{y_i},F_{y_j} ) =g_{ij} $$

$$ g_{TM} (F_{y_i},F_{x_j} ) = \bigg((0,E_i),(E_j,y^k \Gamma_{jk}^s E_s) \bigg) =g_{is} y^k \Gamma_{jk}^s $$

$$ g_{TM} (F_{x_i},F_{x_j} ) = g_{ij} + y^k y^l \Gamma_{ki}^s \Gamma_{lj}^t g_{st} $$

Hence we set

$$ H = {\rm det}\ \left( \begin{array}{cc} g_{ij} & g_{is} y^k \Gamma_{jk}^s \\ g_{js} y^k \Gamma_{ik}^s & g_{ij} +y^ky^l \Gamma_{ki}^s \Gamma_{lj}^t g_{st} \\ \end{array} \right) $$ which is related with volume form on $TM$, as we already know.

Step 3 : When $X$ is a geodesic field with its flow $(\gamma,\gamma ')$, then \begin{align*} X & = \frac{d}{dt} (\gamma, \gamma') \\&= \bigg( \gamma'(t)=_{set} \sum_i \ a_i(t) E_i(t) , 0 \bigg) \end{align*}

Since $ \gamma$ is a geodesic, then $$ a_i' + a_s a_j\Gamma_{js}^i =0 $$

Hence $$ X = a_i \bigg(F_{x_i} - y^k\Gamma_{ki}^s F_{y_s} \bigg) $$

Step 4 : Here ${\rm div}\ X =0 $ iff $XH=0$.

When $f$ is a normal coordinate, then we have $\frac{\partial }{\partial x_i} g_{jk}=0$ so that $XH=0$.