## Introduction to Symplectic Topology

Hi everybody,

In my last post, I promised that I would talk about the Hamiltonian and Lagrangian formalism of classical mechanics. In order to do that, we need to first develop some topological ideas, specifically in the area of symplectic topology. (It appears that the nice folks over at The Everything Seminar has beaten me to it, but I had this mostly written, so I decided to post it anyway.) The outline of these notes comes from a mathematical physics course I took last year, while the content is drawn primarily from Ana Cannas da Silva’s Lectures on Symplectic Geometry; other sources will be mentioned as they come up.

First, let $V$ be a real vector space. A symplectic form on $V$ is a nondegenerate alternating bilinear map $\omega: V\times V\to \mathbb{R}$. One can do a process similar to Gram-Schmidt to show that $V$ has a basis $\{e_1, \ldots, e_n, f_1, \ldots, f_n\}$ such that $\omega(e_i, e_j) = \omega(f_i, f_j) = 0$ and $\omega(e_i, f_j) = \delta_{ij}$ for all $i,j$. This implies that $V$ must be even-dimensional, say $\dim V = 2n$. Then non-degeneracy condition is equivalent to the condition that $\omega^n$ is a non-vanishing alternating $2n$-form on $V$. Finally, $\omega$ induces an isomorphism of $V$ and its dual, $V^*$, given by $v\mapsto \omega(v, \cdot)$.

Before we define symplectic forms on manifolds, I’d like to review a result of differential topology that may not be familiar to some readers. Recall that a vector field $\xi$ on a manifold gives a map from the set of $n$-forms on $M$ to the set of $(n-1)$-forms, $\iota_\xi: \Omega^n(M) \to \Omega^{n-1}(M)$ that corresponds to plugging $\xi$ into the first slot of the form. More specifically, $\iota_\xi\alpha(Y_1, \ldots, Y_{n-1}) = \alpha(\xi, Y_1, \ldots, Y_{n-1})$ for $\alpha\in \Omega^n(M)$ and vector fields $Y_1,\ldots,Y_{n-1}$. One of the most important uses of this notion is in Cartan’s formula for the Lie derivative of an $n$-form: $Lie_\xi\alpha = d(\iota_\xi\alpha) + \iota_\xi(d\alpha)$, where $d$ is the exterior derivative. If you would like to see a proof of this relation, I recommend checking out John M. Lee’s Introduction to Smooth Manifolds, Proposition 18.13.

Now we’re ready to talk about some of the basics of symplectic topology. First, a symplectic form on a manifold $M$ is a CLOSED non-degenerate 2-form $\omega$. Immediately we see from the related ideas on vector spaces that if $M$ has a symplectic form, it must be even-dimensional, say dimension $2n$. Also, the non-degeneracy implies $\omega^n$ is a volume form (non-vanishing form of top degree) on $M$. Initially, it is not obvious why we demand that $\omega$ be closed. Hopefully we will soon see some of the nice results this produces. Finally, $\omega$ induces an isomorphism of the tangent bundle $TM$ to the cotangent bundle $T^*M$. Let’s explore this isomorphism a little closer.

Through $\omega$, each vector field $\xi$ on $M$ corresponds to unique 1-form $\iota_\xi\omega$. For any smooth function $f$, we define the symplectic gradient $\xi_f$ to be the vector field such that $\iota_{\xi_f} \omega = -df$. (The minus sign is completely by convention. The reason for it will soon become apparent.) It is interesting to note that, unlike the usual Euclidean gradient, the symplectic gradient points parallel to level surfaces of $f$ rather than orthogonal to them: $Lie_{\xi_f}f = \xi_ff = df(\xi_f) = -\omega(\xi_f,\xi_f) = 0$. Also, we can now see one of the benefits of the closedness of $\omega$: For any symplectic gradient $\xi_f$, we can use Cartan’s Formula to write $Lie_{\xi_f}\omega = d(\iota_{\xi_f}\omega) = d(-df) = 0$, so symplectic gradients preserve $\omega$.

The symplectic gradient allows us to define the Poisson bracket on $M$ associated to $\omega$, $\{\cdot, \cdot\}_\omega: C^\infty(M)\times C^\infty(M) \to C^\infty(M)$, by $\{f,g\}_\omega = \xi_fg$. Now, for those of you with physics background, this doesn’t look anything like the Poisson bracket that you’re used to, but we will compute it in nice coordinates later and get a very familiar result.

Let’s take a look at this Poisson bracket. First, by unraveling the definitions we see it is anti-symmetric: $\{f,g\}_\omega = \xi_fg = dg(\xi_f) = \omega(\xi_f, \xi_g) = -\omega(\xi_g,\xi_f) = -\{g,f\}_\omega$. (Here, the minus sign in the definition of the symplectic gradient was cancelled out by the alternating property of $\omega$; the minus sign was included in that definition precisely so that $\{f,g\}_\omega = \omega(\xi_f,\xi_g)$.) Also, we can see that the previous consequence of the closedness of $\omega$ leads to the relation: $\xi_{\{f,g\}} = [\xi_f, \xi_g]$, where the bracket is the normal Lie bracket on vector fields. Therefore, the closedness of $\omega$ leads to a natural Lie algebra structure on the space of smooth functions.

Let’s take a look at some examples:

First, say $M = \mathbb{R}^{2n}$ and say $(x_1, \ldots, x_n)$ are the first $n$ standard coordinate functions and $(p_1, \ldots, p_n)$ are the last. The standard symplectic form on $\mathbb{R}^{2n}$ is $\omega = \sum_i dx_i \wedge dp_i$. If $f$ is a smooth function on $\mathbb{R}^{2n}$, then $df = \sum_i \frac{\partial f}{\partial x_i}dx_i + \frac{\partial f}{\partial p_i}dp_i$. The symplectic gradient of $x_i$ is $\frac{\partial}{\partial p_i}$ and the symplectic gradient of $p_i$ is $-\frac{\partial}{\partial x_i}$. Thus, $\xi_f = \sum_i \frac{\partial f}{\partial x_i} \frac{\partial}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial}{\partial x_i}$. Therefore, we can compute the Poisson bracket of two functions in terms of local coordinates: $\{f,g\} = \sum_i \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial x_i}$. This is the usual formula for the Poisson bracket that one sees in classical mechanics.

As another example, there is an important class of manifolds that always has a symplectic structure: cotangent bundles. Let $M$ be an $n$-dimensional manifold. Say $U$ is a chart on $M$ with coordinate functions $x^1,\ldots, x^n$. Then a cotangent vector $\lambda$ over a point $x\in U$ has a unique decomposition $\lambda = \sum_i p_i dx^i_x$. This induces a map $T^*U \to \mathbb{R}^{2n}: (x,\lambda) \mapsto (x^1, \ldots, x^n, p_1, \ldots, p_n)$. You probably recognize this as the standard construction of coordinate charts of $T^*M$. We can put the symplectic structure $\sum_i dx^i \wedge dp_i$ on $T^*U$, and it turns out that this form is coordinate invariant and so can be defined on all of $T^*M$. This is called the “canonical” symplectic form on the cotangent bundle. Notice that in this case, the symplectic form is exact. The 1-form given by the local coordinate formula $\sum_i p_idx^i$ is generally denoted by $\alpha$ and is called the “tautological” 1-form on the cotangent bundle. $\omega = -d\alpha$. There is a coordinate-invariant way of defining $\alpha$ and $\omega$; for details, see da Silva’s book.

Before I wrap up, I’d like to give just a word or two on symmetries of symplectic manifolds. A symplectomorphism $\phi$ of a symplectic manifold $(M, \omega)$ is a diffeomorphism of $M$ that preserves the symplectic form; i.e. $\phi^*\omega = \omega$. An infinitesimal symplectomorphism of $M$ is a vector field $\xi$ such that $Lie_\xi \omega = 0$. The name of this term comes from the fact that the 1-parameter family of diffeomorphisms generated by $\xi$ will be symplectomorphisms, entirely analogous to the situation in Riemannian geometry where Killing fields generate families of isometries. Infinitesimal symplectomorphisms will become very important in the study of Hamiltonian and Lagrangian mechanics, as they will lead to conserved quantities.

That’s all I have for now. Until next time.

-Joe

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### 5 Responses to Introduction to Symplectic Topology

1. Eitan says:

Maybe a dumb question, what does omega^n mean? Also, in paragraph 3 you meant that Y_i is in Gamma(TM) or that i_xi alpha is in Omega^{n-1}(M). Also, is this how you solved that question on the Lie groups homework?

2. jwwalsh says:

By $\omega^n$, I mean $n$ copies of $\omega$ wedged together. Yes, I did mean that $Y_i$ was in $\Gamma(TM)$ (they are vector fields.) Thanks for catching my mistake; it has been corrected.

3. Eitan says:

Oh right, because they’re 2-forms…I keep thinking that wedging a form with itself gives 0 but that’s only for odd forms (right?)

4. jwwalsh says:

Precisely.