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I'm following this lecture in symplectic geometry and I'm trying to show the result stated at 31 minutes that the canonical 1-form on the cotangent bundle $M = T^*X$ is well defined regardless of which coordinates we choose, that is:

$$\alpha = \xi_j dx^j = \xi'_j dx'^j$$

From what I understand, we have each $M \ni p = (x,\xi)$ where $x \in X$ and $\xi \in T_x^*X$ so each "point" on $M$ is actually a tuple of a point in $X$ and a 1-form on $X$.

It starts to get confusing from here, but from what I understand $\alpha$ is a valid 1-form on $M$ because although $\xi_j$ are forms they are also coordinate maps on $M$ for a given chart, so even though it looks like we are multiplying forms nonsensically it's all well defined if we think of this construction on the manifold $M$, so we indeed have a differential form written as functions in front of the exterior derivative of some coordinate maps.

If we change coordinate chart from $(x^1, \ldots, x^n, \xi^1, \ldots, \xi^n)$ to $(x'^1, \ldots, x'^n, \xi'^1, \ldots, \xi'^n)$ then I know my forms will translate as $\displaystyle dx^j = \frac{\partial x^j}{\partial x'^i} dx'^i$ but I'm having trouble showing that the $\xi_j$ will transform as we want.

I was thinking that since each $\xi_j$ is a 1-form on $X$ I can write them as $\displaystyle \xi_j = a^j_i dx^i$ and then use the coordinate transformation rules to get $\displaystyle \xi_j = a^j_i \frac{\partial x^i}{\partial x'^k} dx'^k :=a'^j_kdx'^k$

This leads to $\displaystyle \alpha = \xi_j dx^j = a'^j_kdx'^k \frac{\partial x^j}{\partial x'^i} dx'^i$

Ideally I was hoping for the appearance of a term like $\displaystyle \frac{\partial x^j}{\partial x'^i}\frac{\partial x'^i}{\partial x^k} = \delta^j_k$ which would cancel out, but expanding out $a'^j_k$ gives me a second $\displaystyle \frac{\partial x^i}{\partial x'^k}$ which doesn't lead to anything.

Later I found this question which provides a solution that I don't quite understand, what is the meaning of $\xi_i(dx^j)$ of a form being evaluated on a form?

I'm looking for explanation of the answer in the linked post and what the error was in my attempted proof, it's all well if my approach doesn't go anywhere useful, but I am unsure why I get something that looks so wrong/ugly just by applying what I think are simple and correct rules.

Theo Diamantakis
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1 Answers1

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The $\xi_j$ are definitely not forms. They are functions defined on an open subset of $M = T^*X$ mapping into $\Bbb{R}$. Just to be clear, let me introduce the following notation. Let $\pi:T^*X \to X$ be the canonical projection (map each covector to its base point). Now, given a chart $(U, x)$ on the manifold $X$ (i.e $x:U \to x[U] \subset \Bbb{R}^n$ is the chart map, and we set $x^i := \text{pr}^i \circ x$), we obtain a chart for the cotangent bundle as follows: on $T^*U$, we get a coordinate chart $(x^1 \circ \pi, \dots, x^n \circ \pi, \xi_1, \dots, \xi_n)$, defined as follows: $\xi_i:T^*U \to \Bbb{R}$ \begin{align} \xi_i(\lambda) := \lambda\left(\dfrac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)} \right) \in \Bbb{R} \end{align} Now, observe what each object is.

  • $x^i \circ \pi$ is a function $T^*U \to \Bbb{R}$ (people often abuse notation slightly and write simply $x^i$ when it really should be $x^i \circ \pi = \pi^*(x^i)$, where the RHS is the pull-back of a function).

  • Next, $\lambda \in T^*U$ is a covector, which means $\lambda \in T_{\pi(\lambda)}^*X$ lies in this specific cotangent space.

  • Next, $\frac{\partial}{\partial x^i}|_{\pi(\lambda)} \in T_{\pi(\lambda)}X$ is a tangent vector in this specific tangent space, so the evaluation of the covector on this tangent vector yields a number.

  • Finally, $\xi_i$ is a function $T^*U \to \Bbb{R}$, so it makes sense to feed it a covector. Note that essentially what $\xi_i$ is doing is telling us what the $i^{th}$ component of $\lambda$ is with respect to the basis $\{dx^1|_{\pi(\lambda)}, \dots dx^n|_{\pi(\lambda)}\}$ of the cotangent space. In other words, \begin{align} \lambda &= \xi_i(\lambda) \cdot dx^i|_{\pi(\lambda)} \end{align} (this should be (hopefully) somewhat familiar from linear algebra).

Now, for the sake of precision, let me write the Tautological form as: \begin{align} \alpha := \xi_i \, d(x^i \circ \pi) = \xi_i \, d(\pi^*x^i) = \xi_i \, \pi^*(dx^i) \end{align} This currently is a form defined on $T^*U$ (because $\xi_i$ and $x^i \circ \pi$ are functions on $T^*U$, while $d(x^i \circ \pi)$ is a 1-form on $T^*U$, so their product is still a form on $T^*U$).

The objective is to show that this formula yields a globally well-defined $1$-form on the whole manifold $M=T^*X$. So, let's take another chart $(V,z)$ on the base manifold $X$, and then we "lift it" to a chart $(T^*V, z^1 \circ \pi, \dots, z^n \circ \pi, \zeta_1, \dots \zeta_n)$ (excuse me not using primes for the other coordinates, because I'll definitely make mistakes lol). To complete the proof, we really need to understand how the $\zeta_j$ are related to the $\xi_i$. This is simple: given any covector $\lambda \in (T^*U)\cap (T^*V) = T^*(U \cap V)$, we have by definition: \begin{align} \xi_i(\lambda) &:= \lambda\left(\dfrac{\partial}{\partial x^i}\bigg|_{\pi(\lambda)} \right) \\ &= \lambda\left(\dfrac{\partial z^j}{\partial x^i}\bigg|_{\pi(\lambda)}\cdot \dfrac{\partial}{\partial z^j}\bigg|_{\pi(\lambda)} \right) \\ &= \dfrac{\partial z^j}{\partial x^i}\bigg|_{\pi(\lambda)}\cdot\zeta_j(\lambda), \end{align} where in the last line I used $\Bbb{R}$-linearity of the covector $\lambda$, along with the definition of $\zeta_j$. If we write this as an equality of functions on $T^*(U\cap V)$, we get \begin{align} \xi_i &= \zeta_j \cdot \left(\dfrac{\partial z^j}{\partial x^i} \circ \pi\right) = \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i} \right) \end{align} Now, finally proving the well-definition is simple: \begin{align} \xi_i \cdot \pi^*(dx^i) &= \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i} \right) \, \pi^*(dx^i) \\ &= \zeta_j \cdot \pi^*\left(\dfrac{\partial z^j}{\partial x^i}\, dx^i \right) \\ &= \zeta_j \cdot \pi^*(dz^j) \end{align}

Remarks.

Typically, this final computation is presented with the following abuse of notation (usually for good reason, since with a bit of practice, it get very cumbersome to keep track of the $\pi$): \begin{align} \xi_i\, dx^i &= \zeta_j \cdot \dfrac{\partial z^j}{\partial x^i} \, dx^i = \zeta_j\, dz^j. \end{align} (so $x^i$ can mean either a coordinate function on the base manifold $X$ or its pullback to the bundle $T^*X$).

Also, you should take note that there is a completely chart-free definition of $\alpha$, which shouldn't be too hard to find (but also, try to construct it by yourself if you can).

peek-a-boo
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