14.3 Symplectic geometry

14.3 Symplectic geometry

Concept links · terms present in this machine draft; source roles are unverified: vector space · linear map · dual space · orthogonal group

We saw in chapter 4 that given a basis ${\mathbf{e}}_j$ of a vector space $V,\mathrm{a}$ dual basis ${\mathbf{e}}_j^{\ast }$ of $V^{\ast }$ is given by taking ${\mathbf{e}}_j^{\ast }=v_j$ , where $v_j$ are the coordinate functions. If one instead is initially given the coordinate functions $v_j,$ a dual basis of $V=(V^{\ast }{)}^{\ast }$ can be constructed by taking as basis vectors the first-order linear diferential operators given by diferentiation with respect to the $v_j$ , in other words by taking

$${\mathbf{e}}_j=\frac{\partial }{\partial v_j}$$

Elements of $V$ are then identified with linear combinations of these operators. In efect, one is identifying vectors $v$ with the directional derivative along the vector

$$\mathbf{v}\leftrightarrow \mathbf{v}\cdot \nabla$$

We also saw in chapter 4 that an inner product $(\cdot ,\cdot )$ on $V$ provides an isomorphism of $V$ and $V^{\ast }$ by

$$v\in V\leftrightarrow l_v(\cdot )=(v,\cdot )\in V^{\ast }\text{\hspace{1em}(14.5)}$$

Such an inner product is the fundamental structure in Euclidean geometry, giving a notion of length of a vector and angle between two vectors, as well as a group, the orthogonal group of linear transformations preserving the inner product. It is a symmetric, non-degenerate bilinear form on $V$

A phase space $M$ does not usually come with a choice of inner product. Instead, we have seen that the Poisson bracket gives us not a symmetric bilinear form, but an antisymmetric bilinear form $\Omega ,$ defined on the dual space $\mathcal{M}$ . We will define an analog of an inner product, with symmetry replaced by antisymmetry:

Definition (Symplectic form)

A symplectic form $\omega$ on a vector space $V$ is a bilinear map

$$\omega :V\times V\to \mathbf{R}$$

such that

• $\omega$ is antisymmetric: $\omega (v,v^{\prime })=-\omega (v^{\prime },v)$

$\omega$ is nondegenerate: $ifv\ne 0$ , then $\mathbf{\omega }(v,\cdot )\in V^{\ast }$ is non-zero.

$A$ vector space $V$ with a symplectic form $\omega$ is called a symplectic vector space. The analog of Euclidean geometry, replacing the inner product by a symplectic form, is called symplectic geometry. In this sort of geometry, there is no notion of length (since antisymmetry implies $\omega (v,v)=0)$ . There is an analog of the orthogonal group, called the symplectic group, which consists of linear transformations preserving $\omega$, a group we will study in detail in chapter 16.

Just as an inner product gives an identification of V and $V^{\ast }$ , a symplectic form can be used in a similar way, giving an identification of M and $\mathcal{M}.$ . Using the symplectic form Ω on $\mathcal{M}$ , we can define an isomorphism by identifying basis vectors by

$$q_j\in \mathcal{M}\leftrightarrow \Omega (\cdot ,q_j)=-\Omega (q_j,\cdot )=-\frac{\partial }{\partial p_j}\in M$$ $$p_j\in \mathcal{M}\leftrightarrow \Omega (\cdot ,p_j)=-\Omega (p_j,\cdot )=\frac{\partial }{\partial q_j}\in M$$

and in general

$$u\in \mathcal{M}\leftrightarrow \Omega (\cdot ,u)=-\Omega (u,\cdot )\in M\text{\hspace{1em}(14.6)}$$

Note that unlike the inner product case, a choice of convention of minus sign must be made and is done here.

Recalling the discussion of bilinear forms from section 9.5, a bilinear form on a vector space $V$ can be identified with an element of $V^{\ast }\otimes V^{\ast }$ . Taking $V=M^{\ast }$ we have $V^{\ast }=(M^{\ast }{)}^{\ast }=M$ , and the bilinear form Ω on $M^{\ast }$ is an element of $M\otimes M$ given by

$$\Omega =\sum _{j=1}^d\left(\frac{\partial }{\partial q_j}\otimes \frac{\partial }{\partial p_j}-\frac{\partial }{\partial p_j}\otimes \frac{\partial }{\partial q_j}\right)$$

Under the identification 14.6 of M and $M^{\ast },\Omega \in M\otimes M$ corresponds to

$$\omega =\sum _{j=1}^d(q_j\otimes p_j-p_j\otimes q_j)\in M^{\ast }\otimes M^{\ast }\text{\hspace{1em}(14.7)}$$

Another version of the identification of M and $M$ is then given by

$$v\in M\to \omega (v,\cdot )\in \mathcal{M}$$

In the case of Euclidean geometry, one can show by Gram-Schmidt orthogonalization that a basis ${\mathbf{e}}_j$ can always be found that puts the inner product (which is a symmetric element of $V^{\ast }\otimes V^{\ast })$ in the standard form

$$\sum _{j=1}^n{v}_j\otimes v_j$$

in terms of basis elements of $V^{\ast }$ , the coordinate functions $v_j$ . There is an analogous theorem in symplectic geometry (for a proof, see for instance Proposition 1.1 of [8]), which says that a basis of a symplectic vector space $V$ can always be found so that the dual basis coordinate functions come in pairs $q_j,p_j$ , with the symplectic form $\omega$ the same one we have found based on the Poisson bracket, that given by equation 14.7. Note that one diference between Euclidean and symplectic geometry is that a symplectic vector space will always be even dimensional.

Digression. For those familiar with diferential manifolds, vector fields and diferential forms, the notion of a symplectic vector space can be extended to:

Definition (Symplectic manifold)

A symplectic manifold $M$ is a manifold with a diferential two-form $\omega (\cdot ,\cdot )$ (called a symplectic two-form) satisfying the conditions

• $\omega$ is non-degenerate $(i.e.,for$ a nowhere zero vector field $X,\omega (X,\cdot )$ is a nowhere zero one-form).

$d\omega =0$ , in which case $\omega$ is said to be closed.

The cotangent bundle $T^{\ast }N$ of a manifold $N$ $(i.e.$ , the space of pairs of a point on $N$ together with a linear function on the tangent space at that point) provides one class of symplectic manifolds, generalizing the linear case $N={\mathbf{R}}^d$ and corresponding physically to a particle moving on $N$ . A simple example that is neither linear nor a cotangent bundle is the sphere $M=S^2$ , with $\omega$ the area two-form. The Darboux theorem says that, by an appropriate choice of local coordinates $q_j,p_j$ on $M,$ symplectic two-forms $\omega$ can always be written in such local coordinates as

$$\omega =\sum _{j=1}^{d}d{q}_j\wedge d{p}_j$$

Unlike the linear case though, there will in general be no global choice of coordinates for which this true. Later on, our discussion of quantization will rely crucially on having a linear structure on phase space, so will not apply to general symplectic manifolds.

Note that there is no assumption here that M has a metric $(i.e.$ , it may not be a Riemannian manifold). A symplectic two-form $\omega$ is a structure on a manifold analogous to a metric but with opposite symmetry properties. Whereas a metric is a symmetric non-degenerate bilinear form on the tangent space at each point, a symplectic form is an antisymmetric non-degenerate bilinear form on the tangent space.

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