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Problem of the Week #58 - July 8th, 2013

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Background Info: Let $\mathbb{V}$ be an $m$-dimensional vector space over the field $\mathbb{F}$ (where $\mathbb{F}$ can be either $\mathbb{R}$ or $\mathbb{C}$). If $\omega$ is a non-degenerate alternating bilinear form, then the pairing $(\mathbb{V},\omega)$ is called a symplectic vector space. A symplectic basis for $\mathbb{V}$ is a basis $v_1,\ldots,v_{2n}$ such that $\omega(v_i,v_j)=J_{i,j}$, which is the $(i,j)$-th entry of the $2n\times 2n$ matrix
\[J=\begin{pmatrix} \mathbf{0}_n & I_n\\ -I_n & \mathbf{0}_n\end{pmatrix}.\]A Lagrangian space $\mathbb{U}$ is a subspace of $\mathbb{V}$ of dimension $n$ such that $\omega$ is zero on $\mathbb{U}$; i.e. $\omega(u,w)=0$ for all $u,w\in\mathbb{U}$. A direct sum decomposition $\mathbb{V}=\mathbb{U}\oplus\mathbb{W}$ where $\mathbb{U}$ and $\mathbb{W}$ are Lagrangian subspaces is called a Lagrangian splitting, and $\mathbb{W}$ is called the Lagrangian complement of $\mathbb{U}$.

Problem:
  1. Let $\mathbb{U}$ be a Lagrangian subspace of $\mathbb{V}$. Show that there exists a Lagrangian complement of $\mathbb{U}$
  2. Let $\mathbb{V}=\mathbb{U}\oplus\mathbb{W}$ be a Lagrangian splitting and $x_1,\ldots,x_n$ any basis form $\mathbb{U}$. Show that there exists a unique basis $y_1,\ldots y_n$ of $\mathbb{W}$ such that $x_1,\ldots,x_n,y_1,\ldots,y_n$ is a symplectic basis for $\mathbb{V}$.

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Hints:
  1. Take $\mathbb{W}=J\mathbb{U}$ and show that it's a Lagrangian complement to $\mathbb{U}$.
  2. Define the annihilator of $\mathbb{W}$ by $\mathbb{W}^0 = \{f\in\mathbb{V}^{\ast}: f(e)=0\text{ for all $e\in\mathbb{W}$}\}$. Define $\phi_i\in\mathbb{W}^0$ by $\phi_i(w)=\omega(x_i,w)$ for $w\in\mathbb{W}$. Show that $\phi_1,\ldots,\phi_n$ forms a basis for $\mathbb{W}^0$.

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.

Proof:
  1. The following shows that the complement is not unique. Let $\mathbb{V}=\mathbb{F}^{2n}$ and let $\mathbb{U}\subset\mathbb{F}^{2n}$.
    Claim: $\mathbb{W}=J\mathbb{U}$ is a Lagrangian complement to $\mathbb{U}$.
    Proof of Claim: If $x,y\in\mathbb{W}$, then $x=Ju$, $y=Jv$ where $u,v\in\mathbb{U}$ or $\{u,v\}=0$ (here, $\{\cdot,\cdot\}$ denotes the Poisson bracket). But $\{x,y\}=\{Ju,Jv\}=\{u,v\}=0$, so $\mathbb{W}$ is Lagrangian. If $x\in\mathbb{U}\cap J\mathbb{U}$, the $x=Jy$ with $y\in\mathbb{U}$. So $x,Jx\in\mathbb{U}$ and so $\{x,Jx\}=-\|x\|^2=0$ or $x=0$. Thus, $\mathbb{U}\cap\mathbb{W}=\emptyset$. $\hspace{.25in}\blacksquare$
  2. Define $\phi_i\in\mathbb{W}^0$ by $\phi_i(w)=\omega(x_i,w)$ for $w\in\mathbb{W}$. If $\sum\alpha_i\phi_i=0$, then $\omega\left(\sum\alpha_ix_i,w\right)=0$ for all $w\in\mathbb{W}$ or $\omega\left(\sum\alpha_ix_i,\mathbb{W}\right)=0$. But because $\mathbb{V}=\mathbb{U}\oplus\mathbb{W}$ and $\omega(\mathbb{U},\mathbb{U})=0$, it follows that $\omega\left(\sum\alpha_ix_i,\mathbb{V}\right)=0$. This implies $\sum\alpha_ix_i=0$, because $\omega$ is nondegenerate, and this implies that $\alpha_i=0$ because the $x_i$'s are independent. Thus, $\phi_1,\ldots,\phi_n$ are independent, and so they form a basis for $\mathbb{W}^0$. Let $y_1,\ldots,y_n$ be the dual basis in $\mathbb{W}$; so $\omega(x_i,y_j)=\phi_i(y_j)=\delta_{ij}$. $\hspace{.25in}\blacksquare$
 
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