# Tautological 1-form on the cotangent bundle

I'm trying to understand a little bit about symplectic geometry, in particular the tautological 1-form on the cotangent bundle. I'm following Ana Canas Da Silva's notes.

On page 10 she describes the coordinate free definitions and gives an exercise to find the expression in the local coordinates $$\sum_{i=1}^n \xi_i dx_i$$. I've tried to do this exercise but can't seem to be able to do it which is really annoying since everywhere I look it is said to be trivial and as a consequence never formally proved.

Let $(x^i)$ be local coordinates on our base manifold $M$ and let $(x^i, \xi_j)$ be the induced coordinates on the cotangent bundle $T^* M$. Let $\pi : T^*M \to M$ be the projection $(x^i, \xi_j) \mapsto (x^i)$. It induces a $C^\infty (M)$-linear map on $1$-forms, which I will write as $\pi^* : \Omega^1 (M) \to \Omega^1 (T^* M)$. In coordinates, this sends a $1$-form $\phi = \phi_i \, \mathrm{d} x^i$ (summation convention) to $(\phi_i \circ \pi) \, \mathrm{d} x^i$. As usual this induces a $\mathbb{R}$-linear map on the fibres, namely $\pi^*_{(x, \xi)} : T^*_x M \to T^*_{(x, \xi)} (T^* M)$, sending the covector $p$ to the covector $(p, 0)$. (We must be careful and distinguish between covectors and $1$-forms here, to avoid confusion.)

The tautological $1$-form on $T^* M$ is defined to be $\pi^*_{(x, \xi)} \xi$ at each point $(x, \xi)$ in $T^* M$. Why does this formula even make sense? Well, $\xi$ by definition is an element of $T_x^* M$, so it typechecks. Thus the point $(x, \xi)$ is mapped to the covector $(\xi, 0)$ in $T^*_{(x, \xi)} (T^* M)$, and so the tautological $1$-form in coordinates is given by $$\xi_i \, \mathrm{d} x^i$$ as claimed. (The coefficient of $\mathrm{d} \xi_j$ is $0$, of course.)

(Perhaps the reason no-one likes writing this out in full is because the tautological nature of the construction makes it quite confusing, unless one keeps track of the types of all the expressions involved.)

• Could someone explain the last part in a little more detail please? I'm not sure how to go from one goes from $(x,\xi )$ mapping to $(\xi,0)$ to writting the one from in coordinates. Commented Sep 12, 2019 at 17:43
• You might want to have a look at my answer. Commented Jan 13, 2023 at 21:53

EASY METHOD:

Suppose that $(p,\omega_p)\in Q \equiv T^*(M)$, where $p \in M, \omega_p \in T_p^*(M)$, represents a point in the cotangent bundle $Q$ whose fibres are the cotangent linear spaces at each point in the base manifold. We have the canonical fibre bundle projection,

\begin{align*} \pi: \quad Q &\to M \\ (p,\omega_p) &\mapsto p \ \end{align*}

which induces the pullback of 1-forms,

\begin{align*} \pi^*: \quad \Omega (M) &\to \Omega (Q) \\ \omega &\mapsto \pi^*(\omega)\equiv \theta. \ \end{align*}

In local coordinates, if we write $\omega = \sum_i \alpha_i \ \text{d}x^i$, then it is straightforward to compute the local representation of the pullback:

$$\theta = \pi^*(\sum_i \alpha_i \ \text{d}x^i) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)\ =" \sum_i \alpha_i \ \text{d}x^i$$

Note the quotation marks I used to emphasize that one should be careful about abuse of notation and use them smartly for convenience knowing that it should not be taken too literally.

ALTERNATIVE METHOD:

For a vector field $X \in \mathfrak{X}(Q)$, one could locally decompose it into a component parallel to the fibre $T_p^*(M)$ and another parallel to the base $M$ for each $p \in M$ :

$$X = \sum_j (X_{\parallel}^j\circ\pi)\frac{\partial}{\partial x^j} + \sum_i X_{\perp}^i \frac{\partial}{\partial \alpha^i}$$

where we have again considered $\omega = \sum_i \alpha_i \ \text{d}x^i$. By this decomposition, we mean that

$$\pi_*(X)= \sum_j X_{\parallel}^j \ \frac{\partial}{\partial x^j}$$

And therefore, we have that

\begin{align*} \theta_{(p,\omega_p)}(X) = \pi_{(p,\omega_p)}^*(\omega_p)(X) &= \omega_p (\pi_{*_{(p,\omega_p)}}(X)) \\ &= \sum_i \alpha_i(p)\ \text{d}x^i (\sum_j X_{\parallel}^j(p) \ \frac{\partial}{\partial x^j}) \\ &= \sum_i (\alpha_i\circ\pi)(p,\omega_p)(X_{\parallel}^i\circ\pi)(p,\omega_p)\ \end{align*}

Thus: $$\theta (X) = \sum_i (\alpha_i\circ\pi) (X_{\parallel}^i\circ\pi) = \sum_i (\alpha_i\circ\pi) \ \pi^*(\text{d}x^i)(X)$$

which is the same result that we convinced ourselves of before.

• I know that's an old answer...anyway, could I ask you why $\pi^*(dx^i) = dx^i$ in your third equation? Commented Jan 26, 2019 at 19:11
• @LoScrondo It's not really. Note that the equality is within quotes emphasising that it is a notational abuse. Commented Jan 27, 2019 at 19:00

I find that most of the confusion comes from using the same names on different mathematical objects, so first I'll try to clarify this by explicitly defining things with distinct names.

The $$n$$ coordinate functions of some chart on the base manifold, $$x^i : X \to \mathbb{R}$$

Any covector from a cotangent space at $$x \in X$$, $$\omega \in \text{T}^*_xX$$, has $$n$$ components in this chart, $$\omega_i \in \mathbb{R}$$, such that $$\omega = \omega_i \text{d}x^i$$. (It's implied in this expansion that the basis 1-form $$\text{d}x^i$$ is evaluated at $$x$$ to provide a basis covector).

The $$2n$$ coordinate functions of some chart on the tangent bundle $$\text{T}^*X$$ are divided into 2 sets and denoted differently:

First, the $$n$$ functions associated with the coordinates of the points of $$X$$ are denoted $$\bar{x}^i:\text{T}^*X \to \mathbb{R}$$

Second, the $$n$$ functions associated with the coordinates of the covectors of $$X$$ are denoted $$\xi_i:\text{T}^*X \to \mathbb{R}$$

In general, a chart on the cotangent bundle doesn't have to be related to the chosen chart on the base manifold, but we can naturally choose it such that for any $$p=(x,\omega) \in \text{T}^*X$$, \begin{aligned} \bar{x}^i(p) &= x^i(\pi(p)) \\ \xi_i(p) &= \omega_i \end{aligned}

Given this choice we have a simple component-wise representation of the pullback by $$\pi$$: $$\pi^*\text{d}x^i = \text{d}\bar{x}^i$$

The text defines the tautological 1-form pointwise as $$\alpha_p = \pi^*_p\omega$$, where $$\omega$$ is the covector associated with $$p$$. Then expanding this in coordinate bases we have:

\begin{aligned} \alpha_p &= \pi^*(\omega_i\text{d}x^i) \\ &= \omega_i\pi^*\text{d}x^i \\ &= \xi_i(p)\text{d}\bar{x}^i \end{aligned}

Which is the pointwise representation of $$\alpha = \xi_i\text{d}\bar{x}^i$$

Here we can explicitly see that $$\alpha$$ it is expressed purely in the coordinate functions of the cotangent bundle, which wasn't so clear when $$x^i$$ was used for both the base manifold and the cotangent bundle. I also found it needlessly ambiguous that arbitrary covectors of $$X$$ and the covector coordinate functions of $$\text{T}^*X$$ were both denoted $$\xi$$.

If anything I hope this can supplement the other answers which are very detailed and very well written, but have this notational ambiguity.

Actually, we have the following. Let $$\pi: T^*M \to M$$ be the canonical projection. Let $$(x,\xi) \in T^*M$$. Let $$(x_i)_{i=1}^n$$ be local coordinates around $$x$$ on our base manifold $$M$$, corresponding to a chart $$(U,\phi)$$. Then $$\xi$$, as a covector in $$T^*_xM$$, can be expressed as $$\sum_{i=1}^n \xi_i {dx_i}_{|x}$$ and the tautological 1-form evaluated at $$(x,\xi)$$, $$\theta_{(x,\xi)}$$, is a covector in $$T^*_{x,\xi}(T^*M)$$ equal by definition to $$\pi^*_{x,\xi} \xi$$, which by linearity is, $$\sum_{i=1}^n \xi_i \pi^*_{x,\xi} ({dx_i}_{|x})$$. So, writing by abuse of notation the one-form $$\begin{cases} T^*U &\to T^*(T^*U) \\ (x,\xi) &\mapsto \pi^*_{x,\xi} ({dx_i}_{|x}) \end{cases}$$ as "$$dx_i$$", the 1-form $$\theta$$ can be written locally on $$T^*U$$ as $$\sum_{i=1}^n \xi_i dx_i$$.

NB: Recall that $$\pi^*_{x,\xi} \xi = (\pi^* \xi)_{|x,\xi}$$ is defined by the rule $$(\pi^* \xi)_{|x,\xi}(v) := \xi_x((T_{x,\xi}\pi) v)$$ where $$v \in T_{x,\xi}(T^*M)$$.