Symplectic Manifolds, First Look

A symplectic manifold is a smooth even-dimensional manifold equipped with a closed nondegenerate 2-form. It is the coordinate-independent model of classical phase space.

More precisely, a symplectic manifold is a pair $(M,\omega)$ where $M$ is a smooth manifold and $\omega$ is a smooth 2-form such that:

• $\omega_p$ is nondegenerate on every tangent space $T_pM$;
• $\omega$ is closed:

$$d\omega=0.$$

Nondegenerate means that, at each point $p$,

$$\omega_p(u,v)=0 \quad \text{for all }v\in T_pM \quad \Longrightarrow \quad u=0.$$

Thus every tangent space $T_pM$ is a symplectic vector space. The manifold part says that those linear phase spaces vary smoothly from point to point.

Helpful background. Symplectic Vector Spaces gives the pointwise linear algebra; Differential Forms supplies the global two-form language.

Geometry of Hamiltonian motion

On each small region of phase space, positions and momenta obey the familiar Hamilton equations. A symplectic manifold explains how these local descriptions fit together, and why the same bracket relations survive a canonical change of coordinates.

Local Model

In canonical coordinates

$$(q^1,\ldots,q^n,p_1,\ldots,p_n),$$

the standard symplectic form is

$$\omega = \sum_{i=1}^n dq^i\wedge dp_i.$$

At a point, tangent vectors can be written as infinitesimal variations

$$u=(\delta q,\delta p), \qquad v=(\Delta q,\Delta p).$$

Then

$$\omega_p(u,v) = \sum_i \left( \delta q^i\Delta p_i - \delta p_i\Delta q^i \right),$$

which is exactly the symplectic-vector-space pairing.

The word “local” matters. On a general manifold, one may need several coordinate charts to cover phase space. The symplectic form is the geometric object that stays meaningful when the coordinates change.

Cotangent-Bundle Phase Space

For a configuration manifold $Q$, the natural phase space is the cotangent bundle

$$T^*Q.$$

A point of $T^*Q$ is a position $q\in Q$ together with a covector $p\in T_q^*Q$. In local coordinates this gives the familiar pair $(q^i,p_i)$.

There is a canonical 1-form on $T^*Q$,

$$\theta = \sum_i p_i\,dq^i.$$

With the convention used here, the canonical symplectic form is

$$\omega = \sum_i dq^i\wedge dp_i = - d\theta.$$

Some texts define the canonical symplectic form as $d\theta$ and then reverse a sign elsewhere. The important rule is to keep the sign convention consistent with Hamilton’s equations and the Poisson bracket.

This cotangent-bundle construction is the geometric reason canonical momenta are covectors. Momentum is not just “mass times velocity” in arbitrary coordinates; it is the object that pairs naturally with a virtual displacement.

Hamiltonian Vector Fields

Let $H:M\to\mathbb R$ be a smooth Hamiltonian function. The symplectic form turns the differential $dH$ into a vector field $X_H$ by the rule

$$\omega(X_H,Y)=dH(Y) \qquad \text{for every vector field }Y.$$

Nondegeneracy guarantees that $X_H$ exists and is unique. This is the symplectic analogue of using a metric to turn a differential into a gradient vector, but the resulting vector field is not a gradient flow. It is Hamiltonian flow.

In canonical coordinates, write

$$X_H = \sum_i \left( A^i\frac{\partial}{\partial q^i} + B_i\frac{\partial}{\partial p_i} \right).$$

Using

$$\omega = \sum_i dq^i\wedge dp_i,$$

the equation $\omega(X_H,Y)=dH(Y)$ gives

$$A^i = \frac{\partial H}{\partial p_i}, \qquad B_i = - \frac{\partial H}{\partial q^i}.$$

Therefore

$$X_H = \sum_i \left( \frac{\partial H}{\partial p_i} \frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i} \frac{\partial}{\partial p_i} \right).$$

Integral curves of this vector field satisfy Hamilton’s equations:

$$\dot q^i = \frac{\partial H}{\partial p_i}, \qquad \dot p_i = - \frac{\partial H}{\partial q^i}.$$

Poisson Bracket from the Form

The Poisson bracket is the observable algebra induced by the symplectic form. With the convention above,

$$\{f,g\} = X_g(f) = \omega(X_f,X_g).$$

In canonical coordinates this becomes

$$\{f,g\} = \sum_i \left( \frac{\partial f}{\partial q^i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q^i} \right).$$

This is the same bracket used in the canonical-coordinate page. The symplectic-manifold viewpoint explains why the formula is not tied to one special coordinate chart.

Darboux Coordinates

Darboux’s theorem says that every symplectic manifold is locally equivalent to the standard model: near any point, there are coordinates

$$(q^1,\ldots,q^n,p_1,\ldots,p_n)$$

such that

$$\omega = \sum_i dq^i\wedge dp_i.$$

This is a striking difference from Riemannian geometry. A Riemannian metric can have local curvature invariants. A symplectic form has no local curvature invariant of that kind; locally it always looks standard.

Global information can still be highly nontrivial. Topology, boundary conditions, periodic variables, singularities, and global coordinate obstructions can all affect quantization and semiclassical formulas. Darboux’s theorem is local, does not provide global canonical coordinates.

Symplectomorphisms and Canonical Transformations

A diffeomorphism $\Phi:M\to M$ is a symplectomorphism if it preserves the symplectic form:

$$\Phi^*\omega=\omega.$$

In canonical mechanics, symplectomorphisms are the geometric version of canonical transformations. They preserve Poisson brackets:

$$\{f\circ\Phi,g\circ\Phi\} = \{f,g\}\circ\Phi.$$

Linear symplectomorphisms are exactly the symplectic matrices discussed in Symplectic Vector Spaces. Nonlinear canonical transformations are their manifold-level extension.

Hamiltonian Flow Preserves the Form

Hamiltonian time evolution is not just any flow on phase space. It preserves $\omega$.

Using Cartan’s formula for the Lie derivative,

$$\mathcal L_{X_H}\omega = d(\iota_{X_H}\omega) + \iota_{X_H}(d\omega).$$

With the convention $\iota_{X_H}\omega=dH$ and the symplectic condition $d\omega=0$,

$$\mathcal L_{X_H}\omega = d(dH)+0 = 0.$$

Thus Hamiltonian flow preserves the symplectic form. This is the geometric core of Liouville’s theorem and of the statement that time evolution is a canonical transformation.

Phase-Space Volume

On a $2n$-dimensional symplectic manifold, the form

$$\frac{1}{n!}\omega^n$$

is a natural volume form. Since Hamiltonian flow preserves $\omega$, it also preserves this volume form.

In ordinary mechanics this is Liouville’s theorem. In semiclassical quantum mechanics it underlies the idea that phase-space volume measured in cells of size set by $\hbar$ estimates numbers of states. That estimate is subtle and convention-dependent, but it begins with the symplectic volume, not with an arbitrary coordinate volume.

Common Mistakes

• Treating $\omega$ as a metric or an inner product.
• Forgetting the closedness condition $d\omega=0$ in the definition.
• Assuming Darboux coordinates are global coordinates.
• Confusing a symplectomorphism with any volume-preserving map.
• Losing a sign when moving between $\omega$, $X_H$, and the Poisson bracket.
• Thinking Hamiltonian flow preserves energy only; it also preserves the symplectic form.
• Equating classical symplectic geometry with quantum Hilbert-space geometry. They are linked by quantization and semiclassical limits, but they are not the same structure.

Exercises

1. In canonical coordinates, derive Hamilton’s equations from $\omega(X_H,Y)=dH(Y)$.

Solution

Write

$$X_H = A^i\frac{\partial}{\partial q^i} + B_i\frac{\partial}{\partial p_i}.$$

For an arbitrary vector

$$Y = C^i\frac{\partial}{\partial q^i} + D_i\frac{\partial}{\partial p_i},$$

the symplectic form gives

$$\omega(X_H,Y) = A^iD_i-B_iC^i.$$

The differential is

$$dH(Y) = \frac{\partial H}{\partial q^i}C^i + \frac{\partial H}{\partial p_i}D_i.$$

Since this must hold for all $C^i$ and $D_i$,

$$A^i=\frac{\partial H}{\partial p_i}, \qquad B_i=-\frac{\partial H}{\partial q^i}.$$

The integral curves therefore satisfy Hamilton’s equations.

2. Use Cartan’s formula to show that Hamiltonian flow preserves $\omega$.

Solution

Cartan’s formula gives

$$\mathcal L_{X_H}\omega = d(\iota_{X_H}\omega) + \iota_{X_H}(d\omega).$$

For a Hamiltonian vector field, $\iota_{X_H}\omega=dH$. For a symplectic form, $d\omega=0$. Therefore

$$\mathcal L_{X_H}\omega = d(dH)+0 = 0.$$

The infinitesimal change of $\omega$ along the flow is zero, so the flow preserves $\omega$.

3. Explain why Darboux’s theorem does not imply that all symplectic manifolds are globally the same.

Solution

Darboux’s theorem is local: it provides canonical coordinates in a neighborhood of any point. It says nothing by itself about how different neighborhoods patch together globally. A manifold may have periodic coordinates, nontrivial topology, boundaries, singular regions, or global obstructions to one coordinate chart. Those global features can affect quantization and semiclassical formulas.

4. For one canonical pair, let $\theta=p\,dq$. Compute $-d\theta$.

Solution

Using the product rule for the exterior derivative,

$$d\theta = d(p\,dq) = dp\wedge dq +p\,d(dq).$$

Since $d(dq)=0$,

$$d\theta = dp\wedge dq.$$

Therefore

$$-d\theta = -dp\wedge dq = dq\wedge dp.$$

This matches the convention $\omega=dq\wedge dp$ used on this page.

5. Why is volume preservation weaker than symplectic preservation?

Solution

The volume form $\omega^n/n!$ records total oriented phase-space volume. The symplectic form $\omega$ records the pairing of every conjugate direction. A map can preserve total volume while mixing or rescaling conjugate directions in a way that changes $\omega$. In one canonical pair these notions coincide for linear maps, but in several degrees of freedom symplectic preservation is stronger.

References

• V. I. Arnold, Mathematical Methods of Classical Mechanics, 2nd ed., Springer, 1989.
• A. Cannas da Silva, Lectures on Symplectic Geometry, Lecture Notes in Mathematics 1764, Springer, 2001; author version revised 2006.