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Symplectic Basis of a Symplectic Vector Space

We construct a symplectic basis for a symplectic vector space using a method similar to the Gram-Schmidt algorithm for inner product spaces.

Author

Affiliation

Fei Ye

QCC-CUNY

1 Skew-Symmetric Bilinear Form

Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\) and \(\Omega: V \times V \to \mathbb{F}\) a bilinear map. The map \(\Omega\) is skew-symmetric if \(\Omega(v, w) = -\Omega(w, v)\) for all \(v, w \in V.\)

Theorem 1 Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\) and \(\Omega: V \times V \to \mathbb{F}\) a skew-symmetric bilinear map. Then there exists a basis \(\{u_1, \ldots, u_k, e_1, \ldots, e_n, f_1, \ldots, f_n\}\) of \(V\) such that \(\Omega(u_k, v)=0\) for all \(v\in V\) and \(i=1, \dots, k\), and \[ \Omega(e_i, e_j) = 0, \quad \Omega(f_i, f_j) = 0, \quad \Omega(e_i, f_j) = \delta_{ij} \] for all \(i, j = 1, \ldots, n\).

Proof. Let \(B = \{v_1, \ldots, v_{m}\}\) be a basis of \(V\). We can construct a basis \(\{u_1, \ldots, u_k, e_1, \ldots, e_n, f_1, \ldots, f_n\}\) of \(V\) as follows:

1. We may assume that \(v_1, \ldots, v_k\) satisfy that \(\Omega(v_i, v)=0\) for all \(v\in V\) and there is no other vector in \(B\) with this property. Let \(u_i = v_i\) for \(i=1,..., k\).

2. Now we assume that \(\Omega\) is nondegenerate on the subspace spanned by \(v_{k+1}, \ldots, v_m\), i.e. for each \(v_i\), there is a nonzero vector \(v\) such that \(\Omega(v_i, v)\ne 0\). Since \(\Omega\) is skew-symmetric, then \(\Omega(v_i, v_i) = 0\) for each \(v_i\). Let \(e_1 = v_{k+1}\). Then there is a vector \(v\) in the basis \(B\) such that \(\Omega(e_1, v)\neq 0\). We may assume that \(\Omega(e_1, v_{k+2})\ne 0\). Let \(f_1 = \frac{v_{k+2}}{\Omega(e_1, v_{k+1})}\). Then \(\Omega(e_1, f_1)=1.\)

3. Let \(e_2=v_{k+3} + \Omega(f_1, v_{k+3})e_1 + \Omega(v_{k+3}, e_1) f_1\). Then \(\Omega(e_1, e_2)=0\) and \(\Omega(f_1, e_2)=0\). We may assume that \(\Omega(e_2, v_{k+4})\ne 0\). Let \(f_2 = \frac{v_{k+4}}{\Omega(e_2, v_{k+4})} + \Omega(f_1, v_{k+3})e_1 + \Omega(v_{k+3}, e_1) f_1\). Then \(\Omega(e_2, f_2)=1\), \(\Omega(e_1, f_2)=0\), and \(\Omega(f_1, f_2)=0\).

4. Suppose that we have constructed \(e_1, \ldots, e_r\) and \(f_1, \ldots, f_r\) such that \(\Omega(e_j, e_k)=0\), \(\Omega(f_j, f_k)=0\), and \(\Omega(e_j, f_k)=\delta_{jk}\) for all \(j, k=1, \ldots, r\). Let \[e_{r+1} = v_{k+2r+1} + \sum\limits_{j=1}^{r} (\Omega(f_j, v_{k+2r+1})e_j + \Omega(v_{k+2r+1}, e_j)f_j). \] Assume that \(\Omega(e_{r+1}, v_{k+2r+2})\ne 0\). Let \[ f_{r+1} = \frac{v_{k+2r+2}}{\Omega(e_{r+1}, v_{k+2r+2})} + \sum\limits_{j=1}^{r} (\Omega(f_j, v_{k+2r+2})e_j + \Omega(v_{k+2r+2}, e_j)f_j). \] Then \(\Omega(e_{r+1}, f_{r+1})=1\), \(\Omega(e_j, f_{r+1})=\Omega(f_j, f_{r+1})=0\) for all \(j=1, \ldots, r\).

If \(m-k=2n\), then by induction, we have a basis \(\{u_1, \ldots, u_k, e_1, \ldots, e_n, f_1, \ldots, f_n\}\) of \(V\) with the desired properties.

If \(m-k=2n+1\), then by induction, we have a basis \(\{u_1, \ldots, u_k, e_1, \ldots, e_n, f_1, \ldots, f_n, e_m\}\) of \(V\). However, we have \(\Omega(e_m, v)=0\) for all \(v\in V\). Since \(\Omega\) is nondegenerate, then \(e_m=0\) which is a contradiction. Therefore, \(m-k=2n\) and \(\{u_1, \ldots, u_k, e_1, \ldots, e_n, f_1, \ldots, f_n\}\) is a desired basis of \(V\).

Under the basis constructed in the above theorem, the skew-symmetric bilinear form \(\Omega\) can be represented by a matrix of the form \[ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & I_n \\ 0 & -I_n & 0 \end{pmatrix}. \]

2 Symplectic Basis

Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\), and let \(\Omega: V \times V \to \mathbb{F}\) be a skew-symmetric bilinear map. We say that \(\Omega\) is symplectic if \(\Omega\) is nondegenerate, i.e. for all \(v\in V\), there is a \(w\in V\) such that \(\Omega(v, w)\ne 0\). A vector space equipped with a symplectic bilinear form is called a symplectic vector space.

As a corollary of Theorem 1 above, we have the following result.

Corollary 1 Let \(V\) be a symplectic space, and let \(\Omega: V \times V \to \mathbb{F}\) be a symplectic form. Then there exists a basis \(\{e_1, \ldots, e_n, f_1, \ldots, f_n\}\) of \(V\) such that \(\Omega(e_i, e_j) = 0\), \(\Omega(f_i, f_j) = 0\), and \(\Omega(e_i, f_j) = \delta_{ij}\) for all \(i, j = 1, \ldots, n\). In particular, \(\dim V = 2n\).

The basis in Corollary 1 is called a symplectic basis of \(V\).

Lemma 1 Let \(V\) be a symplectic space with a symplectic bilinear map \(\Omega: V \times V \to \mathbb{F}\), and let \(\{e_1, \ldots, e_n, f_1, \ldots, f_n\}\) be a symplectic space of \(V\). Then the dual space \(V^*\) is also a symplectic space. Moreover, the dual basis \(\{\theta_1, \ldots, \theta_n, \eta_1, \ldots, \eta_n\}\) of \(\{e_1, \ldots, e_n, f_1, \ldots, f_n\}\) is a symplectic basis of \(V^*\), where \(\theta_i = \Omega(e_i, -)\) and \(\eta_i = \Omega(f_i, -)\).

Proof. For any \(a_i, b_i \in \mathbb{F}\), let \(\alpha=\sum\limits_{i=1}^n a_i \theta_i + b_i \eta_i\). If \(\alpha=0\), then \(a_i = \alpha(f_i)=0\) and \(b_i = \alpha(e_i)=0\). Therefore, \(\{\theta_1, \ldots, \theta_n, \eta_1, \ldots, \eta_n\}\) forms a basis of \(V^*\). Let \(\Omega^*: V^*\times V^*\to \mathbb{F}\) be the bilinear map defined by \(\Omega^*(\theta_i, \eta_j) = \Omega(e_i, f_j) = \delta_{ij}\), \(\Omega^*(\theta_i, \theta_j)=\Omega(e_i, e_j)=0\), and \(\Omega^*(\eta_i, \eta_j)=\Omega(f_i, f_j)=0\). Then \(\Omega^*\) is a nondegenerate bilinear map on \(V^*\). Since \(\Omega\) is skew-symmetric, then \(\Omega^*\) is also skew-symmetric. It follows that \(V^*\) is a symplectic space with a symplectic bilinear map \(\Omega^*\).

Proposition 1 Let \(\Omega: V \times V \to \mathbb{F}\) be a skew-symmetric bilinear map. Let \(\tilde{\Omega}: V \to V^*\) be the map defined by \(\tilde{\Omega}(v) = \Omega(v, -)\). Then \(\tilde{\Omega}\) is an isomorphism of \(V\) onto \(V^*\) if and only if \(\Omega\) is a symplectic.

Proof. Suppose that \(\Omega\) is a symplectic.

Let \(v\in V\) be such that \(\tilde{\Omega}(v) = 0\). Then \(\Omega(v, w) = 0\) for all \(w\in V\). Since \(\Omega\) is nondegenerate, then \(v=0\). Therefore, \(\tilde{\Omega}\) is injective.

Let \(\alpha\in V^*\). Then \(\alpha = \sum\limits_{i=1}^n a_i \theta_i + b_i \eta_i\) for some \(a_i, b_i\in \mathbb{F}\). Let \(v = \sum\limits_{i=1}^n a_i e_i + b_i f_i\). Then \(\tilde{\Omega}(v) = \sum\limits_{i=1}^n a_i \Omega(e_i, -) + b_i \Omega(f_i, -) = \alpha\). Therefore, \(\tilde{\Omega}\) is surjective.

It follows that \(\tilde{\Omega}\) is an isomorphism of \(V\) onto \(V^*\).

Conversely, suppose that \(\tilde{\Omega}\) is an isomorphism of \(V\) onto \(V^*\). Let \(v\in V\) be such that \(\Omega(v, w) = 0\) for all \(w\in V\). Then \(\tilde{\Omega}(v) = 0\). Since \(\tilde{\Omega}\) is injective, then \(v=0\). Therefore, \(\Omega\) is nondegenerate. It follows that \(\Omega\) is a symplectic form on \(V\).