Help pulling back the tautological one-form on the cotangent bundle $T^*M$ to the tangent bundle $TM$ using a Riemannian metric on $M$

differential-formsdifferential-geometrysymplectic-geometry

I am struggling with exercise 5 page 304 in Global Aspects of Classical Integrable Systems,Cushman and Bates

Let $g$ be a Riemannian metric on a smooth manifold $M$. In
local coordinates $x = (x_1,… x_n)$ the metric may be written as $g = \sum g_{ij} dx^i \otimes dx^j$

Let $v = (x, \nu) = (x^1,…, x^n, \nu^1,… \nu^n)$ be natural coordinates on TM. Show that the pullback
$\theta_g$ by the map $g^\#$ to $TM$ of the canonical 1-form $\theta$ on $T^*M$ may be written as
$\theta_g(\nu) = \sum g_{i j}\nu^i dx^j$

Setting the problem:

  • Take coordinates on $T^*M$ to be $m = (x, p)$

  • Canonical 1-form $\theta$ on $T^*M$
    \begin{equation}
    \begin{aligned}
    \theta\in \chi ^*(T^*M)\\
    \theta: \chi (T^*M) \rightarrow C^\infty(\mathbb{R})\\\theta_m: T_m(T^*M) \rightarrow \mathbb{R}
    \\
    \theta_{m=(x, p)} = \sum_i p_i dx^i
    \end{aligned}
    \end{equation}
    $\;\;\;\;\;$ * Let $w \in T_m(T^*M), \quad w = \sum_i^n w_i \dfrac{\partial}{\partial x^i} + \sum_{i=1}^{n} w'_{i} \dfrac{\partial}{\partial p^i}$
    $\;\;\;\;\;$ * Then $\theta_m(w) = \sum_i w_ip_i$.

  • The map $g(x)^\#$ on $T_xM$ is the isomorphism induced by $g(x)$ between $T_xM$ and $T^*_xM$
    \begin{equation}
    \begin{aligned}
    &g^\#(v) = g(x)^\#(\nu) \text{ is such that }
    [g(x)^\#(\nu)](\mu) = g_x(\nu, \mu), \quad \mu \in T_xM
    \end{aligned}
    \end{equation}

  • Pullback of a 1-form (Wikipedia):

Let $\phi:M \rightarrow N$ be a smooth map between smooth manifolds, and let $\alpha$ be a 1-form on $N$.
\newline Then the pullback of $\alpha$ by $\phi$ is the 1-form $\phi^*\alpha$ on $M$ defined by:
\begin{equation}
(\phi^*\alpha)_x(X) = \alpha_{\phi(x)}(d\phi_x(x))
\end{equation}
for $x \in M$ and $X \in T_xM$

  • Property of the tautological one-form from Wikipedia: The tautological one-form is the unique one-form that "cancels" the pullback. The tautological one-form $\theta$ is the only form with the property that $\beta^*\theta = \beta$, for every 1-form $\beta$ on $Q$

My confusion and my attempt
While $\theta$ is a one-form on $T^*M$, it seems to me that $g^\#$ maps to $T^*_xM$. (Indeed once we feed the metric a tangent vector we have decided on an attachment point $x$ of the manifold)

My solution was instead to define $g^\#:TM \rightarrow T^*M$
\begin{equation}
g^\#(x, \nu) = (x, g^2(x)^\#(\nu))
\end{equation}
where $[g^2(x)^\#(\nu)](\mu) = g(x)(\nu, \mu)$

  • Attempt component-wise
    \begin{equation}
    \begin{aligned}
    \theta_g(v) = [(g^\#)^*\theta] (v) &= \theta ( g^\#(v))\circ dg^\#_v \\
    &= \theta ( g^\#(x^i, \nu^i))\circ dg^\#_v\\
    &= \theta(\sum g_{ij}(x) \nu^i dx^j) \circ dg^\#_v\\
    &= (\sum g_{ij}(x) \nu^i dx^j) \circ dg^\#_v
    \end{aligned}
    \end{equation}
    Is this a valid method ? Can I show $dg^\#_v$ acts as the identity map on $\dfrac{\partial}{\partial x^i}$ components ?

  • I have also been attempting a solution using the "tautological one form cancels pullback" identity
    My "solution" has been to fix $\nu$ while varying $x$ thus defining the one-form:
    $$g^\#_\nu: M \rightarrow T^*M
    $$
    $$g^\#_\nu(x) \text{ is such that: }$$
    $$g^\#_\nu(x)(\mu) = g_x(\nu, \mu)
    $$

Pulling $\theta$ back by $g^\#_\nu$ :
$$\theta_g = (g^\#_\nu)^*\theta = g^\#_\nu = \iota_\nu g = \sum g_{ij} \nu^i dx^j
$$

There are obviously several problems with this solution, $\theta_g$ acts in $M$ instead of $TM$ and it is odd to fix a covector $\nu$ in this way.

I am unsure how to define $g^\#$ to get spaces to properly match up, I am unsure how to deal with the fact the tautological form on $T^*M$ "looks" like a one-form on $M$

Any help with this exercise, or suggestions on simpler exercises to tackle that migh help would be very welcome.

Best Answer

First a small remark: the standard notation for the map induced by $g$ from $TM\to T^*M$ (for which you have given a correct definition) is $g^{\flat}$ rather than $g^{\sharp}$ or $g^{\#}$; we define $g^{\sharp}:=(g^{\flat})^{-1}:T^*M\to TM$.

Next, I hope you know that the coordinate $x^i$ is being used 3 times with 3 different meanings (once as coordinates on the base manifold, once as projection to base coordinates on $TM$ and once as the projection to base coordinates on $T^*M$). For the sake of clarity, I shall explicitly use a different notation: $(x^1,\dots, x^n)$ be the coordinates on $M$, then the corresponding "base coordinates" on $TM$ and $T^*M$ are $x^i\circ \pi_{TM}=(\pi_{TM})^*x^i$ and $x^i\circ \pi_{T^*M}=(\pi_{T^*M})^*x^i$.

Now, the tautological one-form is $\theta=p_i\,d(x^i\circ \pi_{T^*M})$. So, if we pull-back using $g^{\flat}$, then using the fact that pull-back commutes with multiplication and exterior derivatives, and that pullback of functions is just composition, we have \begin{align} \theta_g&:=(g^{\flat})^*\theta\\ &=(g^{\flat})^*p_i \cdot (g^{\flat})^*[d(x^i\circ \pi_{T^*M})]\\ &=(p_i\circ g^{\flat})\cdot d(x^i\circ \pi_{T^*M}\circ g^{\flat})\\ &=(p_i\circ g^{\flat})\cdot d(x^i\circ \pi_{TM}) \end{align} Now, all that remains is to calculate $p_i\circ g^{\flat}$. If we consider a tangent vector $\xi\in T_aM$, then we can expand it as $\xi=\nu^j(\xi)\cdot \frac{\partial}{\partial x^j}(a)$, and thus due to fiber-wise linearity, \begin{align} (p_i\circ g^{\flat})(\xi)&=p_i\left[\nu^j(\xi)\cdot g^{\flat}\left(\frac{\partial}{\partial x^j}(a)\right)\right]\\ &=p_i\left[\nu^j(\xi)g_{jk}(a)\cdot dx^k(a)\right]\\ &:=\nu^j(\xi)\cdot g_{ji}(a)\\ &=[(g_{ij}\circ \pi_{TM})\cdot \nu^j](\xi) \end{align} Therefore, \begin{align} \theta_g&=[g_{ij}\circ \pi_{TM}]\cdot \nu^j\,d(x^i\circ \pi_{TM}), \end{align} or if we suppress the projection $\pi_{TM}$ (which usually causes no harm once we get used to this sort of calculation) then we can write this as (using symmetry of $g_{ij}$) \begin{align} \theta_g&=g_{ij}\nu^i\,dx^j. \end{align}

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