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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 $\mathrm{\Omega }:V\times V\to \mathbb{F}$ a bilinear map. The map $\mathrm{\Omega }$ is skew-symmetric if $\mathrm{\Omega }(v,w)=-\mathrm{\Omega }(w,v)$ for all $v,w\in V.$

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

Proof. Let $B=\{v_1,\dots ,v_m\}$ be a basis of $V$. We can construct a basis $\{u_1,\dots ,u_k,e_1,\dots ,e_n,f_1,\dots ,f_n\}$ of $V$ as follows:

1. We may assume that $v_1,\dots ,v_k$ satisfy that $\mathrm{\Omega }(v_i,v)=0$ 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 $\mathrm{\Omega }$ is nondegenerate on the subspace spanned by $v_{k+1},\dots ,v_m$, i.e. for each $v_i$, there is a nonzero vector $v$ such that $\mathrm{\Omega }(v_i,v)\ne 0$. Since $\mathrm{\Omega }$ is skew-symmetric, then $\mathrm{\Omega }(v_i,v_i)=0$ for each $v_i$. Let $e_1=v_{k+1}$. Then there is a vector $v$ in the basis $B$ such that $\mathrm{\Omega }(e_1,v)\ne 0$. We may assume that $\mathrm{\Omega }(e_1,v_{k+2})\ne 0$. Let $f_1=\frac{v_{k+2}}{\mathrm{\Omega }(e_1,v_{k+1})}$. Then $\mathrm{\Omega }(e_1,f_1)=1.$

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

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

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

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

Under the basis constructed in the above theorem, the skew-symmetric bilinear form $\mathrm{\Omega }$ can be represented by a matrix of the form $$\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& I_n\\ 0& -I_n& 0\end{array}\right).$$

2 Symplectic Basis

Let $V$ be a finite-dimensional vector space over a field $\mathbb{F}$, and let $\mathrm{\Omega }:V\times V\to \mathbb{F}$ be a skew-symmetric bilinear map. We say that $\mathrm{\Omega }$ is symplectic if $\mathrm{\Omega }$ is nondegenerate, i.e. for all $v\in V$, there is a $w\in V$ such that $\mathrm{\Omega }(v,w)\ne 0$. 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 $\mathrm{\Omega }:V\times V\to \mathbb{F}$ be a symplectic form. Then there exists a basis $\{e_1,\dots ,e_n,f_1,\dots ,f_n\}$ of $V$ such that $\mathrm{\Omega }(e_i,e_j)=0$, $\mathrm{\Omega }(f_i,f_j)=0$, and $\mathrm{\Omega }(e_i,f_j)={\delta }_{ij}$ for all $i,j=1,\dots ,n$. 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 $\mathrm{\Omega }:V\times V\to \mathbb{F}$, and let $\{e_1,\dots ,e_n,f_1,\dots ,f_n\}$ be a symplectic space of $V$. Then the dual space $V^{\ast }$ is also a symplectic space. Moreover, the dual basis $\{{\theta }_1,\dots ,{\theta }_n,{\eta }_1,\dots ,{\eta }_n\}$ of $\{e_1,\dots ,e_n,f_1,\dots ,f_n\}$ is a symplectic basis of $V^{\ast }$, where ${\theta }_i=\mathrm{\Omega }(e_i,-)$ and ${\eta }_i=\mathrm{\Omega }(f_i,-)$.

Proof. For any $a_i,b_i\in \mathbb{F}$, let $\alpha =\sum _{i=1}^n{a}_i{\theta }_i+b_i{\eta }_i$. If $\alpha =0$, then $a_i=\alpha (f_i)=0$ and $b_i=\alpha (e_i)=0$. Therefore, $\{{\theta }_1,\dots ,{\theta }_n,{\eta }_1,\dots ,{\eta }_n\}$ forms a basis of $V^{\ast }$. Let ${\mathrm{\Omega }}^{\ast }:V^{\ast }\times V^{\ast }\to \mathbb{F}$ be the bilinear map defined by ${\mathrm{\Omega }}^{\ast }({\theta }_i,{\eta }_j)=\mathrm{\Omega }(e_i,f_j)={\delta }_{ij}$, ${\mathrm{\Omega }}^{\ast }({\theta }_i,{\theta }_j)=\mathrm{\Omega }(e_i,e_j)=0$, and ${\mathrm{\Omega }}^{\ast }({\eta }_i,{\eta }_j)=\mathrm{\Omega }(f_i,f_j)=0$. Then ${\mathrm{\Omega }}^{\ast }$ is a nondegenerate bilinear map on $V^{\ast }$. Since $\mathrm{\Omega }$ is skew-symmetric, then ${\mathrm{\Omega }}^{\ast }$ is also skew-symmetric. It follows that $V^{\ast }$ is a symplectic space with a symplectic bilinear map ${\mathrm{\Omega }}^{\ast }$.

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

Proof. Suppose that $\mathrm{\Omega }$ is a symplectic.

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

Let $\alpha \in V^{\ast }$. Then $\alpha =\sum _{i=1}^n{a}_i{\theta }_i+b_i{\eta }_i$ for some $a_i,b_i\in \mathbb{F}$. Let $v=\sum _{i=1}^n{a}_i{e}_i+b_i{f}_i$. Then $\tilde{\mathrm{\Omega }}(v)=\sum _{i=1}^n{a}_i\mathrm{\Omega }(e_i,-)+b_i\mathrm{\Omega }(f_i,-)=\alpha$. Therefore, $\tilde{\mathrm{\Omega }}$ is surjective.

It follows that $\tilde{\mathrm{\Omega }}$ is an isomorphism of $V$ onto $V^{\ast }$.

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