Motivation for the standard symplectic space

Question

I'm aware that for the vector space $V\oplus V^*$ that the product between $(v,\phi)$ and $(w,\psi)$ is given by $\psi(v)-\phi(v)$. What I'm struggling to see is why a person might want to define a product this way, other than to give an example of a symplectic product. Is there a particularly nice geometric or physics interpretation?

Answer 1

With $V=\Bbb R^n$, we note that on $T^*\Bbb R^n$ with standard coordinates $(q,p)$ this is the standard symplectic form $\omega=\sum dq_i\wedge dp_i$. Indeed, if we evaluate $\omega$ on the pair of tangent vectors $(v,\phi)$, $(w,\psi)$, we get $$\sum v_i\psi_i - w_i\phi_i = \psi(v)-\phi(w).$$ We think of using the standard orthonormal basis and dual basis here, so that $\psi(v)=\left(\sum\psi_i e^i\right)\left(\sum v_j e_j\right) = \sum \psi_i v_j e^i(e_j) = \sum \delta^i_j \psi_i v_j = \sum \psi_i v_i$.

Answer 2

Here's an algebraic reason one might consider that product canonical (at least up to a constant multiple).

The general linear group $GL(V)$ has a defining representation (i.e. left action) on $V$. It also has a canonical dual representation on $V^*$, given by $A\cdot\phi=(A^{-1})^*\phi$. Taking the direct sum of these represenations gives a canonical representation on $V\oplus V^*$, given by $$ A\cdot(v,\phi)=\left(Au,(A^{-1})^*\phi,\right) $$ One can show that there exists, up to a constant multiple, a unique symplectic product $\omega$ on $V\oplus V^*$ which is invariant under this representation: $$ \omega(A\cdot(v,\phi),A\cdot(w,\psi))=\omega((v,\phi),(w,\psi))\ \ \ \ \ \ \ \forall{A}\in GL(V),(v,\phi),(w,\psi)\in V\oplus V^* $$ This product is given by $\omega((v,\phi),(w,\psi))=c(\psi(v)-\phi(w))$ for $c\in\mathbb{R}\setminus\{0\}$.

Answer 3

I'll have to change your notation a little bit for this to make sense. Let $(V,\Omega)$ be a symplectic space, and let $L\subseteq V$ be a Lagrangian subspace. Then $L$ has a Lagrangian complement $L'$ in $V$ (e.g., complete a basis of $L$ for a Darboux basis of $V$ and take $L'$ to be the span of the added vectors). So $V = L\oplus L'$. Then $\Omega$ itself gives an isomorphism $L' \to L^*$ via $y \mapsto \Omega(\cdot,y)$. Hence we have an isomorphism $V \to L\oplus L^*$.

What does $\Omega$ look like when we use this isomorphism to see it as a symplectic form in $L\oplus L^*$? Up to a sign $\pm$, it's precisely the standard symplectic form in $L\oplus L^*$.

In short: since an isomorphism induced by $\Omega$ itself makes it look like the standard symplectic form, we see that once a choice of Lagrangian subspace has been made, one cannot fight nature any longer --- $\Omega$ wants to be standard.