16.1 The symplectic group

16.1 The symplectic group

Concept links · terms present in this machine draft; source roles are unverified: orthogonal group · dual representation

Recall that the orthogonal group can be defined as the group of linear transformations preserving an inner product, which is a symmetric bilinear form. We now want to study the analog of the orthogonal group that comes from replacing the inner product by the antisymmetric bilinear form Ω that determines the symplectic geometry of phase space. We will define:

Definition (Symplectic group)

The symplectic group $Sp(2d,\mathbf{R})$ is the subgroup of linear transformations g of $\mathcal{M}={\mathbf{R}}^{2d}$ that satisfy

$$\Omega (g{v}_1,g{v}_2)=\Omega (v_1,v_2)$$

for $v_1,v_2\in \mathcal{M}$

While this definition uses the dual phase space $\mathcal{M}$ and Ω, it would have been equivalent to have made the definition using M and $\omega ,$ since these transformations preserve the isomorphism between M and $\mathcal{M}$ given by Ω (see equation 14.6). For an action on $\mathcal{M}$

$$u\in \mathcal{M}\to gu\in \mathcal{M}$$

the action on elements of M (such elements correspond to linear functions $\Omega (u,\cdot )$ on $\mathcal{M})$ is given by

$$\Omega (u,\cdot )\in M\to g\cdot \Omega (u,\cdot )=\Omega (u,g^{-1}(\cdot ))=\Omega (gu,\cdot )\in M\text{\hspace{1em}(16.1)}$$

Here the first equality uses the definition of the dual representation (see 4.2) to get a representation on linear functions on $M$ given a representation on $\mathcal{M}$ and the second uses the invariance of Ω.

16.1.1 The symplectic group for d=1

Concept links · terms present in this machine draft; source roles are unverified: linear map · Lie algebra · unitary representation · irreducible representation · group action

In order to study symplectic groups as groups of matrices, we’ll begin with the case $d=1$ and the group $Sp(2,\mathbf{R})$ . We can write Ω as

$$\Omega \left(\left(\genfrac{}{}{0ex}{}{c_q}{c_p}\right),\left(\genfrac{}{}{0ex}{}{c_q^{\prime }}{c_p^{\prime }}\right)\right)=c_q{c}_p^{\prime }-c_p{c}_q^{\prime }=\left(\begin{array}{cc}{c}_q& c_p\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\genfrac{}{}{0ex}{}{c_q^{\prime }}{c_p^{\prime }}\right)\text{\hspace{1em}(16.2)}$$

A linear transformation $g$ of M will be given by

$$\left(\genfrac{}{}{0ex}{}{c_q}{c_p}\right)\to \left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\left(\genfrac{}{}{0ex}{}{c_q}{c_p}\right)\text{\hspace{1em}(16.3)}$$

The condition for Ω to be invariant under such a transformation is

$${\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)}^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\text{\hspace{1em}(16.4)}$$

or

$$\left(\begin{array}{cc}0& \alpha \delta -\beta \gamma \\ -\alpha \delta +\beta \gamma & 0\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$$

so

$$det\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)=\alpha \delta -\beta \gamma =1$$

This says that we can have any linear transformation with unit determinant. In other words, we find that $Sp(2,\mathbf{R})=SL(2,\mathbf{R})$ . This isomorphism with a special linear group occurs only for $d=1$

Now turning to the Lie algebra, for group elements $g\in GL(2,\mathbf{R})$ near the identity, $g$ can be written in the form $g=e^{tL}$ where $L$ is in the Lie algebra $\mathfrak{gl}(2,\mathbf{R})$ . The condition that $g$ acts on $\mathcal{M}$ preserving Ω implies that (diferentiating 16.4)

$$\frac{d}{dt}\left((e^{tL}{)}^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)e^{tL}\right)=(e^{tL}{)}^T\left(L^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)L\right)e^{tL}=0$$

Setting $t=0$ , the condition on $L$ is

$$L^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)L=0\text{\hspace{1em}(16.5)}$$

This requires that L must be of the form

$$L=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)\text{\hspace{1em}(16.6)}$$

which is what one expects: $L$ is in the Lie algebra sl(2, $R$) of 2 by 2 real matrices with zero trace.

The homogeneous degree two polynomials in $p$ and $q$ form a three dimensional sub-Lie algebra of the Lie algebra of functions on phase space, since the non-zero Poisson bracket relations on a basis $\frac{q^2}{2},\frac{p^2}{2},qp$ are

$$\{\frac{q^2}{2},\frac{p^2}{2}\}=qp\{qp,p^2\}=2p^2\{qp,q^2\}=-2q^2$$

We have

Theorem 16.1

The Lie algebra of degree two homogeneous polynomials on $M={\mathbf{R}}^2$ is isomorphic to the Lie algebra sp $(2,\mathbf{R})=\mathfrak{sl}(2,\mathbf{R})$ , with the isomorphism given explicitly by

$$-aqp+\frac{b{q}^2}{2}-\frac{c{p}^2}{2}=\frac{1}{2}\left(\begin{array}{cc}q& p\end{array}\right)L\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\genfrac{}{}{0ex}{}{q}{p}\right)\leftrightarrow L=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)\text{\hspace{1em}(16.7)}$$

Proof

One can identify basis elements as follows:

$$\frac{q^2}{2}\leftrightarrow E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)-\frac{p^2}{2}\leftrightarrow F=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)-qp\leftrightarrow G=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\text{\hspace{1em}(16.8)}$$

The commutation relations amongst these matrices are

$$[E,F]=G[G,E]=2E[G,F]=-2F$$

which are the same as the Poisson bracket relations between the corresponding quadratic polynomials. □

The moment map for the $SL(2,\mathbf{R})$ action on $\mathcal{M}={\mathbf{R}}^2$ of equation 16.3 is given by

$${\mu }_L=-aqp+\frac{b{q}^2}{2}-\frac{c{p}^2}{2}\text{\hspace{1em}(16.9)}$$

To check this, first compute using the definition of the Poisson bracket

$$-X_{{\mu }_L}F(q,p)=\{{\mu }_L,F\}=(bq-ap)\frac{\partial F}{\partial p}+(aq+cp)\frac{\partial F}{\partial q}$$

Elements $e^{tL}\in SL(2,\mathbf{R})$ act on functions on $M$ by

$$e^{tL}\cdot F(q(m),p(m))=F(q(e^{-tL}\cdot m),p(e^{-tL}\cdot m))$$

where (for $m\in M$ written as column vectors) $e^{-tL}\cdot m$ is multiplication by the matrix $e^{-tL}$ . On linear functions $l\in \mathcal{M}$ written as column vectors, the same group action takes l to $e^{tL}l$ and acts on basis vectors $q,p$ of $\mathcal{M}$ by

$$\left(\genfrac{}{}{0ex}{}{q}{p}\right)\to (e^{tL}{)}^T(\genfrac{}{}{0ex}{}{q}{p})$$

The vector field $X_L$ is then given by

$$\begin{array}{rl}-X_{L}F(q,p)& =\frac{d}{dt}F((q(e^{-tL}\cdot m),p(e^{-tL}\cdot m){)}_{t=0}\\ & =\left(\frac{\partial F}{\partial q},\frac{\partial F}{\partial p}\right)\cdot \frac{d}{dt}(e^{tL}{)}^T{\left(\genfrac{}{}{0ex}{}{q}{p}\right)}_{t=0}\\ & =\left(\frac{\partial F}{\partial q},\frac{\partial F}{\partial p}\right)\cdot L^T\left(\genfrac{}{}{0ex}{}{q}{p}\right)\\ & =\left(\frac{\partial F}{\partial q},\frac{\partial F}{\partial p}\right)\cdot \left(\genfrac{}{}{0ex}{}{aq+cp}{bq-ap}\right)\end{array}$$

and one sees that $X_L=X_{{\mu }_L}$ as required. The isomorphism of the theorem is the statement that ${\mu }_L$ has the Lie algebra homomorphism property characterizing moment maps:

$$\{{\mu }_L,{\mu }_{L^{\prime }}\}={\mu }_{[L,L^{\prime }]}$$

Two important subgroups of $SL(2,\mathbf{R})$ are

• The subgroup of elements one gets by exponentiating G, which is isomorphic to the multiplicative group of positive real numbers

$$e^{tG}=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{-t}\end{array}\right)$$

Here one can explicitly see that this group has elements going of to infinity.

• Exponentiating the Lie algebra element $E-F$ gives rotations of the plane

$$e^{\theta (E-F)}=\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right)$$

Note that the Lie algebra element being exponentiated here is

$$E-F\leftrightarrow \frac{1}{2}(p^2+q^2)$$

the function studied in section 15.2, which we will later re-encounter as the Hamiltonian function for the harmonic oscillator in chapter 22.

The group $SL(2,\mathbf{R})$ is non-compact and its representation theory is quite unlike the case of $SU(2)$ . In particular, all of its non-trivial irreducible unitary representations are infinite dimensional, forming an important topic in mathematics, but one that is beyond our scope. We will be studying just one such irreducible representation (the one provided by the quantum mechanical state space), and it is a representation only of a double cover of $SL(2,\mathbf{R})$ , not of $SL(2,\mathbf{R})$ itself.

16.1.2 The symplectic group for arbitrary d

Concept links · terms present in this machine draft; source roles are unverified: linear map · orthogonal group · Lie algebra · group action

For general d, the symplectic group $Sp(2d,\mathbf{R})$ is the group of linear transformations g of M that leave Ω (see 14.4) invariant, $i$.e., satisfy

$$\Omega \left(g\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q}{{\mathbf{c}}_p}\right),g\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right)\right)=\Omega \left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q}{{\mathbf{c}}_p}\right),\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right)\right)$$

where ${\mathbf{c}}_q,{\mathbf{c}}_p$ are d dimensional vectors. By essentially the same calculation as in the $d=1$ case, we find the d dimensional generalization of equation 16.4. This says that $Sp(2d,\mathbf{R})$ is the group of real 2d by 2d matrices $g$ satisfying

$$g^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)g=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\text{\hspace{1em}(16.10)}$$

where 0 is the d by $d$ zero matrix, 1 the d by $d$ unit matrix.

Again by a similar argument to the $d=1$ case where the Lie algebra sp(2, $R$) was determined by the condition 16.5, sp $(2d,\mathbf{R})$ is the Lie algebra of 2d by 2d matrices $L$ satisfying

$$L^T\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)L=0\text{\hspace{1em}(16.11)}$$

Such matrices will be those with the block-diagonal form

$$L=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)\text{\hspace{1em}(16.12)}$$

where $A,B,C$ are d by $d$ real matrices, with $B$ and C symmetric, $i$.e.,

$$B=B^T,C=C^T$$

Note that, replacing the block antisymmetric matrix by the unit matrix, in 16.10 one recovers the definition of an orthogonal matrix, in 16.11 the definition of the Lie algebra of the orthogonal group.

The generalization of 16.7 is

Theorem 16.2

The Lie algebra sp $(2d,\mathbf{R})$ is isomorphic to the Lie algebra of order two homogeneous polynomials on $M={\mathbf{R}}^{2d}$ by the isomorphism (using a vector notation for the coeficient functions $q_1,\cdot \cdot \cdot ,q_d,p_1,\cdot \cdot \cdot ,p_d)$

$$L\leftrightarrow {\mu }_L$$

where

$$\begin{array}{rl}{\mu }_L& =\frac{1}{2}\left(\begin{array}{cc}\mathbf{q}& \mathbf{p}\end{array}\right)L\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\genfrac{}{}{0ex}{}{\mathbf{q}}{\mathbf{p}}\right)\\ & =\frac{1}{2}\left(\begin{array}{cc}\mathbf{q}& \mathbf{p}\end{array}\right)\left(\begin{array}{cc}B& -A\\ -A^T& -C\end{array}\right)\left(\genfrac{}{}{0ex}{}{\mathbf{q}}{\mathbf{p}}\right)\\ & =\frac{1}{2}(\mathbf{q}\cdot B\mathbf{q}-2\mathbf{q}\cdot A\mathbf{p}-\mathbf{p}\cdot C\mathbf{p})\end{array}\text{\hspace{1em}(16.13)}$$

We will postpone the proof of this theorem until section 16.2, since it is easier to first study Poisson brackets between order two and order one polynomials, then use this to prove the theorem about Poisson brackets between order two polynomials. As in $d=1$ , the function ${\mu }_L$ is the moment map function for $L.$

The Lie algebra $\mathfrak{sp}(2d,\mathbf{R})$ has a subalgebra $\mathfrak{gl}(d,\mathbf{R})$ consisting of matrices of the form

$$L=\left(\begin{array}{cc}A& 0\\ 0& -A^T\end{array}\right)$$

${}_{\mathrm{or}},$ in terms of quadratic functions, the functions

$$-\mathbf{q}\cdot A\mathbf{p}=-\mathbf{p}\cdot A^T\mathbf{q}\text{\hspace{1em}(16.14)}$$

where $A$ is any real d by $d$ matrix. This shows that one way to get symplectic transformations is to take any linear transformation of the position coordinates, together with the dual linear transformation (see definition 4.2) on momentum coordinates. In this way, any linear group acting on position space gives a subgroup of the symplectic transformations of phase space.

An example of this is the group $SO(d)$ of spatial rotations, with Lie algebra so $(d)\subset \mathfrak{gl}(d,\mathbf{R})$ , the antisymmetric d by $d$ matrices, for which $-A^T=A$ . The special case $d=3$ was an example already worked out earlier, in section $15.4,$ where ${\mu }_L$ gives the standard expression for the angular momentum as a function of the $q_j,p_j$ coordinates on phase space.

Another important special case comes from taking $A=0,B=1,C=-1$ in equation 16.12, which by equation 16.13 gives

$${\mu }_L=\frac{1}{2}(|\mathbf{q}{|}^2+|\mathbf{p}{|}^2)$$

This generalizes the case of $d=1$ described earlier, and will be the Hamiltonian function for a d dimensional harmonic oscillator. Note that exponentiating L gives a symplectic action on phase space that mixes position and momentum coordinates, so this an example that cannot be understood just in terms of a group action on configuration space.

Chapter contents · Next section: 16.2 The symplectic group and automorphisms of the Heisenberg group · 中文版本

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