I just found out about symplectic geometry in the context about this question on volume preservation in phase space.

It seems somewhat complicated and I am not sure what to do with the notation $\omega = dx\land dy$, or what a 2-form is.

Could someone explain how the mathematical objects in symplectic geometry are manifested within classical Hamiltonian mechanics? At least I'd like to understand the statement of the non-squeezing theorem and how it applies to phase space.

  • $\begingroup$ Read this en.wikipedia.org/wiki/Poisson_bracket $\endgroup$
    – OON
    Jul 10, 2020 at 11:24
  • $\begingroup$ I also would recommend Arnold's book "Mathematical Methods of Classical Mechanics" $\endgroup$
    – OON
    Jul 10, 2020 at 11:25
  • $\begingroup$ Can you clarify your present understanding, if any, of differential geometry? At least at the level of tangent vectors, covectors, and tensors. $\endgroup$
    – J. Murray
    Jul 13, 2020 at 14:58
  • $\begingroup$ @J.Murray, I have some understanding of it. I have an intuition of tangent vectors, I understand the definition of covectors, and I dont understand tensors. I've never written a proof about anything in differential geometry, but I think I get the idea behind why people study Riemannian manifolds, its role in general relativity etc. $\endgroup$
    – user56834
    Jul 13, 2020 at 16:49

3 Answers 3


It's easy to think of a $2N$-dimensional phase space simply as $\mathbb R^{2N}$, and generally this is what one does in elementary treatments. However, there are many, many phase spaces which cannot be put in this form. The phase space corresponding to a simple pendulum, for example, is not $\mathbb R^2$ but rather $S^1\times \mathbb R$ - a cylinder, not an infinite plane. This is a reflection of the fact that the angular coordinate is only meaningfully defined mod $2\pi$; the points $\theta$ and $\theta+2\pi$ are actually the same point, so we shouldn't model the angular coordinate as a point on a line, but rather a point on a circle.

This motivates us to try to define Hamiltonian dynamics on more general spaces than just $\mathbb R^{2N}$. As you know, a point in a $2N$-dimensional phase space can be labeled by a collection of $N$ position coordinates and $N$ momentum coordinates. If $F$ and $G$ are smooth functions of the $q$'s and $p$'s, then the Poisson bracket $\{F,G\}$ is given by

$$\{F,G\} = \sum_{i=1}^N \frac{\partial F}{\partial q_i} \frac{\partial G}{\partial p_i} - \frac{\partial G}{\partial q_i}\frac{\partial F}{\partial p_i}$$

Essentially all of Hamiltonian mechanics can be expressed in terms of the Poisson bracket$^\dagger$, so if we can generalize this to more interesting space than just $\mathbb R^{2N}$ then we are in business.

First, we need a bit of differential geometry, which I will quickly review. The space of smooth functions from $M$ to $\mathbb R$ is denoted $C^\infty(M)$. A tangent vector field on $M$ is a linear map from $C^\infty(M)$ to $C^\infty(M)$. If we have a coordinate system $(x^1,\ldots,x^N)$ for some patch of the manifold $U\subseteq M$, then we can express a vector field as

$$\mathbf X = X^\mu \frac{\partial}{\partial x^\mu}$$

The partial derivatives $\frac{\partial}{\partial x^\mu}$ constitute a basis for the space of tangent vectors at each point of $U$, and the functions $X^\mu$ are called the components of $\mathbf X$ in that basis.


Consider the manifold $\mathbb R^2$ equipped with coordinates $(x,y)$. An example of a vector field is $\mathbf X = x^2 \frac{\partial}{\partial x} + 2xy \frac{\partial}{\partial y}$. The $x$ and $y$ components of $\mathbf X$ are $x^2$ and $2xy$, respectively. If we let this vector act on an element of $C^\infty(\mathbb R^2)$ such $F(x,y)=x^3y^2$, then the result

$$\mathbf X(F) = x^2(3x^2y^2) + 2xy(2x^3y) = 3x^4y^2+4x^4y^2$$

is another element of $C^\infty(\mathbb R^2)$.

A covector field is an object which eats a vector field and spits out an element of $C^\infty(M)$. Just as the partial derivatives $\frac{\partial}{\partial x^\mu}$ constitute a basis for tangent vectors, we define the symbols $dx^\mu$ to constitute a basis for covectors, where $dx^\mu\left(\frac{\partial}{\partial x^\nu}\right) = \delta^\mu_\nu$. We can therefore express an arbitrary covector $\boldsymbol \omega$ as

$$\boldsymbol \omega = \omega_\mu dx^\mu$$

Note: It is useful to also allow vectors to eat covectors, by simply defining $\mathbf X(\boldsymbol\omega) := \boldsymbol\omega(\mathbf X)$.


An example of a covector field on $\mathbb R^2$ is $\boldsymbol \omega = 3dx + 2y^2 dy$. If we let $\boldsymbol\omega$ act on the vector field $\mathbf X$ from the previous example, we obtain

$$\boldsymbol\omega(\mathbf X)= (3dx+2y^2dy)\left(x^2 \frac{\partial}{\partial x} + 2xy \frac{\partial}{\partial y}\right)$$

$$=3x^2 \ dx\left(\frac{\partial}{\partial x}\right) + 6xy \ dx\left(\frac{\partial}{\partial y}\right)+ 2y^2x^2\ dy\left(\frac{\partial}{\partial x}\right) + 4xy^3 dy\left(\frac{\partial}{\partial y}\right)$$ $$ = 3x^2 + 0 + 0 + 4xy^3$$

A $(p,q)$-tensor field is a linear map which eats $p$ covector fields and $q$ vector fields and spits out an element of $C^\infty(M)$. For example, a $(1,2)$-tensor field $\mathbf T$ eats one covector field and two vector fields:

$$\mathbf T(\boldsymbol\omega,\mathbf X,\mathbf Y) = \mathbf T\left(\omega_\mu dx^\mu, X^\nu \frac{\partial}{\partial x^\nu},y^\rho \frac{\partial}{\partial x^\rho}\right)$$ $$=\omega_\mu X^\nu Y^\rho \underbrace{\mathbf T\left(dx^\mu, \frac{\partial}{\partial x^\nu},\frac{\partial}{\partial x^\rho}\right)}_{\equiv T^\mu_{\ \ \nu\rho}} = \omega_\mu X^\nu Y^\rho T^\mu_{\ \ \nu\rho}$$

We can think of a $(1,2)$-tensor as a tensor product in the following way:

$$\mathbf T = T^\mu_{\ \ \nu\rho} \frac{\partial}{\partial x^\mu} \otimes dx^\nu \otimes dx^\rho$$

when we feed it a covector field $\boldsymbol \omega$ and two vector fields $\mathbf X$ and $\mathbf Y$, then we simply feed $\boldsymbol \omega$ to the first factor, $\mathbf X$ to the second factor, and $\mathbf Y$ to the third factor. In this light, it follows that a vector field can be thought of as a $(1,0)$-tensor field, a covector field can be thought of as a $(0,1)$-tensor field, and an element of $C^\infty(M)$ can be thought of as a $(0,0$-tensor field (i.e. a scalar field).


The object $\mathbf A = (2x+y)\frac{\partial}{\partial x} \otimes \frac{\partial}{\partial x}\otimes dy$ is a $(2,1)$-tensor. If we feed it two copies of $\boldsymbol \omega$ from the last example and one copy of $\mathbf X$ from the example before that, we will get

$$\mathbf A(\boldsymbol\omega,\boldsymbol\omega,\mathbf X)=(2x+y)(3)(3)(2xy) = 18xy(2x+y)$$

A differential $k$-form field is a completely antisymmetric $(0,k)$-tensor field. To make the antisymmetry manifest, we define the wedge product $\wedge$ to be the totally antisymmetrized tensor product. That is,

$$dx\wedge dy \equiv \frac{1}{2}(dx\otimes dy - dy \otimes dx)$$ $$dx\wedge dy \wedge dz \equiv\frac{1}{3!}(dx\otimes dy\otimes dz + dy \otimes dz \otimes dx + dz \otimes dx \otimes dy$$ $$ - dy \otimes dx \otimes dz - dx \otimes dz \otimes dy - dz \otimes dy \otimes dx)$$

so on and so forth. Note that $dx\wedge dx = 0$, due to the antisymmetry property.


The object $\boldsymbol\Omega = dx\wedge dy$ is a differential $2$-form field. If we feed it the vector fields $\mathbf X = 2x\frac{\partial}{\partial x} + y^2 \frac{\partial}{\partial y}$ and $\mathbf Y = x^2 \frac{\partial}{\partial x} + y^3 \frac{\partial}{\partial y}$, we get

$$\boldsymbol\Omega(\mathbf X,\mathbf Y) = dx\wedge dy \left(\left[2x\frac{\partial}{\partial x} + y^2 \frac{\partial}{\partial y}\right]\otimes\left[x^2 \frac{\partial}{\partial x} + y^3 \frac{\partial}{\partial y}\right]\right)$$ $$= \frac{1}{2}dx\otimes dy\left(\left[2x\frac{\partial}{\partial x} + y^2 \frac{\partial}{\partial y}\right]\otimes\left[x^2 \frac{\partial}{\partial x} + y^3 \frac{\partial}{\partial y}\right]\right)$$ $$ - \frac{1}{2}dy\otimes dx\left(\left[2x\frac{\partial}{\partial x} + y^2 \frac{\partial}{\partial y}\right]\otimes\left[x^2 \frac{\partial}{\partial x} + y^3 \frac{\partial}{\partial y}\right]\right)$$

$$ = \frac{1}{2}(2x)(y^3) - \frac{1}{2}(y^2)(x^2) = xy^3-\frac{1}{2}x^2y^2$$

This might look horrendous, but it's really not so bad. The object $dx\otimes dy$ eats two vector fields $\mathbf A$ and $\mathbf B$ and spits out $A^xB^y$; the object $dx\wedge dy$ eats $\mathbf A$ and $\mathbf B$ and spits out $\frac{1}{2}(A^xB^y-A^yB^x)$.

Let $\boldsymbol \beta=\beta_{\mu_1\ldots\mu_k} dx^1\wedge\ldots\wedge dx^k$ be a $k$-form field. Its exterior derivative is a $k+1$-form given by

$$d\boldsymbol\beta = \frac{\partial\beta_{\mu_1\ldots\mu_k}}{\partial x^\nu}dx^\nu \wedge (dx^1\wedge\ldots\wedge dx^k)$$

A $k$-form whose exterior derivative vanishes everywhere is called closed; a $k$-form which can be written as the exterior derivative of a $(k-1)$-form is called exact. It can be shown that every exact form is closed, but the reverse is not always true.


The function $F(x,y)=x^2y$ is a $0$-form. Its exterior derivative $$dF = \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y}dy = 2xy dx + x^2 dy$$

is a $1$-form. If I take a second exterior derivative, I get

$$d^2F = d(dF) = \frac{\partial}{\partial x}(2xy) dx\wedge dx + \frac{\partial}{\partial x}(x^2)dx\wedge dy + \frac{\partial}{\partial y}(2xy) dy\wedge dx +\frac{\partial}{\partial y}(x^2)dy\wedge dy$$ $$ = 0 + 2x dx\wedge dy + 2x dy\wedge dx + 0 $$ $$ = 2x(dx\wedge dy+dy\wedge dx) = 0$$

The Punchline:

Let $M$ be a $2N$-dimensional manifold, and let $\boldsymbol\Omega$ be a closed, non-degenerate $2$-form field on $M$.

In a sense, a $2$-form can be thought of as a kind of scalar product between two vector fields; the statement that $\boldsymbol \Omega$ is non-degenerate means that there are no non-zero vectors whose "scalar product" with every other vector is zero. In other words, if $\boldsymbol\Omega(\mathbf X,\mathbf Y)=0$ for all $\mathbf Y$, then $\mathbf X = 0$.

Such an $\boldsymbol\Omega$ provides a mapping between vectors and covectors. For any vector field $\mathbf X$, $\boldsymbol \Omega(\mathbf X,\bullet)$ is a covector (where $\bullet$ denotes an empty slot). Similarly, if $\boldsymbol\Omega^{-1}$ is the matrix inverse of $\Omega$ which is guaranteed to exist because $\boldsymbol\Omega$ is non-degenerate, then if $\boldsymbol\alpha$ is a covector, then $\boldsymbol\Omega^{-1}(\boldsymbol\alpha,\bullet)$ is a vector.

Such a $2$-form is called a symplectic form. It induces the structure of Hamiltonian mechanics on $M$ via the following definition. Let $F$ and $G$ be elements of $C^\infty(M)$. To each we can assign Hamiltonian vector fields $\mathbf X_F$ and $\mathbf X_G$ given by

$$\mathbf X_F=\boldsymbol \Omega^{-1}(dF,\bullet)$$ $$\mathbf X_G=\boldsymbol\Omega^{-1}(dG,\bullet)$$


$$\{F,G\} := \boldsymbol\Omega(\mathbf X_F,\mathbf X_G)$$

From this follows essentially all of Hamiltonian mechanics.


Going back to the elementary treatment, assume that phase space is simply $\mathbb R^{2N}$, with coordinates $(x_1,\ldots,x_N,p_1,\ldots,p_N)$. The canonical symplectic form is

$$\mathbf \Omega = \sum_{i=1}^N dx_i\wedge dp_i$$ It is exact, because it is the exterior derivative of the canonical $1$-form $\boldsymbol \theta = -\sum_{i=1}^N p_i dx_i$, which means that it is closed. As a good exercise to see if you have understood what has been said here, you can let $N=2$, compute $\mathbf X_F$ and $\mathbf X_G$ and show that $\boldsymbol\Omega(\mathbf X_F,\mathbf X_G)$ reproduces the familiar Poisson bracket.

  • $\begingroup$ Extraordinary answer, +1 from me. $\endgroup$
    – Bcpicao
    Apr 13, 2023 at 17:40

To familiarize your self with the subject you may consult the aforementioned Arnold's book, or Nakahara "Geometry, Topology and Physics" https://www.academia.edu/29696440/GEOMETRY_TOPOLOGY_AND_PHYSICS_SECOND_EDITION_Nakahara, which is a very pedagogical introduction for physicists with the mathematical machinery.

In simple words, the symplectic geometry provides to extension of the Hamiltonian formalism for more general cases, where the manifold if not just simple $\mathbb{R}^{2n}$ formed by $(q, p)$. You have a $2n$-dimensional manifold with a 2-form $\omega$, such that:

  1. Is non-degenerate : $\omega^n \neq 0$
  2. Is closed : $d \omega = 0$

In the trivial case, when the manifold is cotangent bundle $T^{*} M$ it is simply familiar: $$ \omega = \sum_i dq_i \wedge d p_i $$ The nontrivial example is the symplectic form on a sphere $S^2$ ($\theta, \phi$ - angular variables): $$ \omega = \sin \theta \ d \theta \wedge d \phi $$ There are vector fields, which belong to the tangent bundle $T M$, which locally (or globally for the case of $\mathbb{R}^{2n}$) look like: $$ V = f(q, p) \frac{\partial}{\partial q} + g(q, p) \frac{\partial}{\partial p} $$ Action of such vector field on $\omega$ gives you some 1-form (eats a differential with a possible change of sign). One also defines a Lie derivative, which when acting on forms, can be calculated by the Cartan formula: $$ \mathcal{L}_V = i_V d + d i_V $$ If $\mathcal{L}_V \omega = 0$ than the field is said to preserve the symplectic form and is called the symplectic vector field. Looking at the Cartan formula and recalling that $d \omega = 0$, one may infer: $$ d i_V \omega = 0 $$ There is a Poincare's Lemma, which states, that locally any closed form is exact - may be represented as differential of some other form. In case of trivial topology this will hold globally. So : $$ d i_V \omega \Rightarrow i_V \omega = - d H $$ The $i_V \omega $ is 1-form, so the $H$ is a zero form - ordinary smooth function, which we will call Hamiltonian.

  • $\begingroup$ "In the trivial case, when the manifold is cotangent bundle T∗M it is simply familiar: $ω=\sum_i dq_i\land dp_i$". To be clear, this is not familiar to me. Does this just mean $\int ...\int dq_1dp_1...dq_ndp_n$? $\endgroup$
    – user56834
    Jul 10, 2020 at 16:47
  • $\begingroup$ @user56834 no, there is no integration, the wedge product has nothing to do with integration, I even would say more, the integration reduces the degree of form. It is familiar in sense, that it corresponds to the canonical Poisson bracket : $\{q_i, p_j\} = \delta_{i j}$ $\endgroup$ Jul 10, 2020 at 17:01
  1. If OP is already familiar with Poisson brackets then it seems that the central piece of information relevant to OP's question is the following theorem.

    Theorem: Let there be given a $2n$-dimensional manifold $M$. There is a canonical bijective correspondence between symplectic structures $\omega\in\Omega^2(M)$ and non-degenerate Poisson structures $\{\cdot,\cdot\}: C^{\infty}(M)\times C^{\infty}(M)\to C^{\infty}(M)$.

  2. Main example: A canonical Poisson bracket corresponds to $\omega=\sum_{j=1}^n\mathrm{d}p_j \wedge \mathrm{d}q^j$.

  3. Hamilton's equations can be expressed wrt. both structures. See also this related Phys.SE post.


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