📂Representation Theory

Symplectic Group

Background¹

An orthogonal matrix is a matrix that satisfies $Q^{\mathsf{T}}Q = I$, and this property is equivalent to $Q$ preserving the inner product. The set of matrices with this property forms a group (moreover a Lie group), called the orthogonal group and denoted as follows.

$$ \begin{align*} \operatorname{O}(n) &= \left\{ Q \in M_{n \times n}(\mathbb{R}) : Q^{\mathsf{T}}Q = I \right\} \\ &= \left\{ Q \in M_{n \times n}(\mathbb{R}) : \braket{Q\mathbf{x}, Q\mathbf{y}} = \braket{\mathbf{x}, \mathbf{y}} \quad \forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^{n} \right\} \end{align*} $$

The inner product is a kind of symmetric bilinear form, and, similarly to the orthogonal group, one can consider sets that preserve other bilinear operations. Let a symmetric bilinear form $[\cdot, \cdot]_{n,k}$ be given as follows. For $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n+k}$,

$$ [\mathbf{x}, \mathbf{y}]_{n,k} = x_{1}y_{1} + \cdots + x_{n}y_{n} - x_{n+1}y_{n+1} - \cdots - x_{n+k}y_{n+k} $$

The set of matrices preserving this operation is called the general orthogonal group.

$$ \begin{align*} \operatorname{O}(n, k) &= \left\{ Q \in M_{(n+k) \times (n+k)}(\mathbb{R}) : Q^{\mathsf{T}} \Lambda Q = \Lambda \right\} \\ &= \left\{ Q \in M_{(n+k) \times (n+k)}(\mathbb{R}) : [Q \mathbf{x}, Q \mathbf{y}]_{n, k} = [\mathbf{x}, \mathbf{y}]_{n, k} \right\} \end{align*} $$

Here $\Lambda = \begin{bmatrix} I_{n} & O \\ O & -I_{k} \end{bmatrix}$. The symplectic group is defined similarly to the above groups. Define the skew-symmetric bilinear form $\omega : \mathbb{R}^{2n} \times \mathbb{R}^{2n} \to \mathbb{R}$ as follows.

$$ \omega(\mathbf{x}, \mathbf{y}) = \sum_{i=1}^{n} (x_{i}y_{n+i} - x_{n+i}y_{i}) $$

The set of matrices preserving this form is defined to be the symplectic group.

Definition

Following the above notation, the symplectic group $\operatorname{Sp}(n, \mathbb{R})$ is defined as follows.

$$ \begin{align*} \operatorname{Sp}(n, \mathbb{R}) &= \left\{ Q \in M_{2n \times 2n}(\mathbb{R}) : Q^{\mathsf{T}} \Omega Q = \Omega \right\} \tag{1} \\ &= \left\{ Q \in M_{2n \times 2n}(\mathbb{R}) : \omega(Q \mathbf{x}, Q \mathbf{y}) = \omega(\mathbf{x}, \mathbf{y}), \quad \forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^{2n} \right\} \end{align*} $$

Explanation

It is also denoted $\operatorname{Sp}(2n, \mathbb{R})$. By the properties below, the two definitions above are equivalent.

If one expresses $\omega(\mathbf{x}, \mathbf{y})$ in terms of the inner product, one obtains $\braket{\mathbf{x}, \Omega \mathbf{y}}$. Here $\Omega = \begin{bmatrix} O & I_{n} \\ -I_{n} & O \end{bmatrix}$.

$$ \begin{align*} \omega(\mathbf{x}, \mathbf{y}) &= \sum_{i=1}^{n} (x_{i}y_{n+i} - x_{n+i}y_{i}) \\ &= \Braket{\begin{bmatrix} x_{1} \\ \vdots \\ x_{n} \\ x_{n+1} \\ \vdots \\ x_{2n} \end{bmatrix}, \begin{bmatrix} y_{n+1} \\ \vdots \\ y_{2n} \\ -y_{1} \\ \vdots \\ -y_{n} \end{bmatrix}} \\ &= \Braket{\begin{bmatrix} x_{1} \\ \vdots \\ x_{n} \\ x_{n+1} \\ \vdots \\ x_{2n} \end{bmatrix}, \begin{bmatrix} 0 & \cdots & 0 & 1 & \cdots & 0 \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & 0 & \cdots & 1 \\ -1 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & -1 & 0 & \cdots & 0 \end{bmatrix} \begin{bmatrix} y_{1} \\ \vdots \\ y_{n} \\ y_{n+1} \\ \vdots \\ y_{2n} \end{bmatrix}} \\ &= \braket{\mathbf{x}, \Omega \mathbf{y}} \tag{2} \end{align*} $$

Subgroups

First, recall that if $A \in \operatorname{Sp}(n, \mathbb{R})$ then $A^{-1} \in \operatorname{Sp}(n, \mathbb{R})$.

$$ A^{\mathsf{T}} \Omega A = \Omega \iff \Omega = (A^{-1})^{\mathsf{T}} \Omega A^{-1} $$

By the subgroup test, to show $AB^{-1} \in \operatorname{Sp}(n, \mathbb{R})$ it suffices to assume $A, B \in \operatorname{Sp}(n, \mathbb{R})$. Assume $A, B \in \operatorname{Sp}(n, \mathbb{R})$. Then $A$ and $B^{-1}$ satisfy $(1)$, hence

$$ (AB^{-1})^{\mathsf{T}} \Omega (AB^{-1}) = (B^{-1})^{\mathsf{T}} A^{\mathsf{T}} \Omega A B^{-1} = (B^{-1})^{\mathsf{T}} \Omega B^{-1} = \Omega $$

Therefore $\operatorname{Sp}(n, \mathbb{R})$ is a subgroup of $\operatorname{GL}(n, \mathbb{R})$.

$$ \operatorname{Sp}(n, \mathbb{R}) \le \operatorname{GL}(n, \mathbb{R}) $$

Matrix Lie group

Since $\operatorname{Sp}(n, \mathbb{R})$ is a closed subgroup of $\operatorname{GL}(2n, \mathbb{R})$, it is a matrix Lie group. Define the function $f: \operatorname{GL}(2n, \mathbb{R}) \to M_{2n \times 2n}(\mathbb{R})$ as follows.

$$ f(Q) = Q^{\mathsf{T}} \Omega Q $$

Then $f$ is a continuous function. Recall that the preimage of a closed set under a continuous function is closed. The necessary and sufficient condition for $Q \in \operatorname{Sp}(n, \mathbb{R})$ is $Q^{\mathsf{T}} \Omega Q = \Omega$, and because $f^{-1}(\left\{ \Omega \right\}) = \operatorname{Sp}(n, \mathbb{R})$ is the preimage of the closed set $\left\{ \Omega \right\}$, $\operatorname{Sp}(n, \mathbb{R})$ is a closed subgroup of $\operatorname{GL}(2n, \mathbb{R})$ and therefore a matrix Lie group.

Complex matrices

The bilinear form $\omega$ defined on the real vector space can be defined unchanged on the complex vector space $\mathbb{C}^{2n}$.

$$ \omega : \mathbb{C}^{2n} \times \mathbb{C}^{2n} \to \mathbb{C} \\[0.5em] \omega(\mathbf{u}, \mathbf{v}) = \sum_{i} u_{i}v_{n+i} - u_{n+i}v_{i} $$

Note that it remains a skew-symmetric bilinear form, but because complex conjugation is not taken, it is not an inner product on the complex vector space. The complex symplectic group is defined as follows.

$$ \begin{align*} \operatorname{Sp}(n, \mathbb{C}) &= \left\{ Q \in M_{2n \times 2n}(\mathbb{C}) : Q^{\ast} \Omega Q = \Omega \right\} \\ &= \left\{ Q \in M_{2n \times 2n}(\mathbb{C}) : \omega (Q \mathbf{u}, Q \mathbf{v}) = \omega (\mathbf{u}, \mathbf{v}), \quad \forall \mathbf{u}, \mathbf{v} \in \mathbb{C}^{2n} \right\} \end{align*} $$

Here $A^{\ast}$ is the conjugate transpose of $A$. The complex symplectic group has the same properties as the real symplectic group. On the other hand, among complex matrices there is a separate set that preserves the inner product, namely the unitary group. Define the set $\operatorname{Sp}(n)$ as follows; it is the set of matrices that preserve both the inner product on the complex vector space and the bilinear form $\omega$, and it is called the compact symplectic group.

$$ \operatorname{Sp}(n) = \operatorname{Sp}(n, \mathbb{C}) \cap \operatorname{U}(2n) $$

Properties

(a) $\Omega^{-1} = -\Omega$.

(b) For $Q \in \operatorname{Sp}(n, \mathbb{R})$, $\det Q = \pm 1$.

(c) A necessary and sufficient condition for $Q \in \operatorname{Sp}(n, \mathbb{R})$ is $ Q^{\mathsf{T}} \Omega Q = \Omega$.

$$ \omega(Q \mathbf{x}, Q \mathbf{y}) = \omega(\mathbf{x}, \mathbf{y}) \iff Q^{\mathsf{T}} \Omega Q = \Omega $$

Proof

(a)

Since computations with block matrices are carried out like matrix multiplication,

$$ \Omega (-\Omega) = \begin{bmatrix} O & I_{n} \\ -I_{n} & O \end{bmatrix} \begin{bmatrix} O & -I_{n} \\ I_{n} & O \end{bmatrix} = \begin{bmatrix} I_{n} & O \\ O & I_{n} \end{bmatrix} $$

$$ \implies \Omega^{-1} = -\Omega $$

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(b)

By property (c) we have $Q^{\mathsf{T}} \Omega Q = \Omega$, hence by the properties of the determinant we obtain the following.

$$ \begin{align*} && \det(Q^{\mathsf{T}} \Omega Q) &= \det \Omega \\ \implies && \det Q^{\mathsf{T}} \det \Omega \det Q &= \det \Omega \\ \implies && (\det Q)^{2} &= 1 \\ \implies && \det Q &= \pm 1 \\ \end{align*} $$

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(c)

This can be shown without much difficulty as follows.

$$ \begin{align*} && Q \in \operatorname{Sp}(n, \mathbb{R}) \\ \iff && \omega(Q \mathbf{x}, Q \mathbf{y}) &= \omega(\mathbf{x}, \mathbf{y}) \\ \iff && \braket{Q \mathbf{x}, \Omega Q \mathbf{y}} &= \braket{\mathbf{x}, \Omega\mathbf{y}} \\ \iff && (Q \mathbf{x})^{\mathsf{T}} \Omega Q \mathbf{y} &= \mathbf{x}^{\mathsf{T}} \Omega\mathbf{y} \\ \iff && \mathbf{x}^{\mathsf{T}} Q^{\mathsf{T}} \Omega Q \mathbf{y} &= \mathbf{x}^{\mathsf{T}} \Omega\mathbf{y} \\ \iff && \mathbf{x}^{\mathsf{T}} \left( Q^{\mathsf{T}} \Omega Q - \Omega \right) \mathbf{y} &= 0 \quad\forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^{2n} \\ \iff && \left( Q^{\mathsf{T}} \Omega Q - \Omega \right) \mathbf{y} &= \mathbf{0} \quad\forall \mathbf{y} \in \mathbb{R}^{2n} \\ \iff && \left( Q^{\mathsf{T}} \Omega Q - \Omega \right) &= O \\ \iff && Q^{\mathsf{T}} \Omega Q &= \Omega \\ \end{align*} $$

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