# Canonical symplectic structure.

I've got the next question, we have the canonical standard symplectic form, $\omega_{std}$, given in coordinate form $\omega_{std}=\sum_{i=1}^{n} dq_i \wedge dp_i$, and I want to show that it's non-degenrate.

I.e, for $x=(x_1,...,x_n) \ \forall y=(y_1,...,y_n)$ if $\omega_{std}(x,y)=0$ then $x=0$.

So if I understnad correctly I get:

$$0=\omega_{std}(x,y)= \sum dx_i \wedge dy_i$$

Somehow I want to pick such that it will imply that every $x_i$ is zero, any hints what will make it? It's easy right? :-(

Thanks.

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Your question is notationally confused. $\omega$ takes as inputs pairs of vectors of the form $v=(v_1,v_2,\ldots,v_{2n})$ where we can choose the basis so that $dx_i(v)=v_i$ and $dy_i(v)=v_{i+n}$. i.e. it is in $\Omega^2(\mathbb{R}^{2n})$ not $\Omega^2(\mathbb{R}^n)$. – Adam Apr 3 '12 at 18:04
I still don't see how to choose the $v'$ vector s.t $\omega(v,v')=0 \Rightarrow v=0$. – MathematicalPhysicist Apr 3 '12 at 18:29

To do it by bare hands, I'd argue as follows. Given $x$ with the property that $\omega_{std}(x,y) = 0$ for all $y$, express $x$ in the basis $\partial q_i, \partial p_j$. So $x = a^i \partial q_i + b^j \partial p_j$ (with the Einstein summation convention) for some choices of $a^i$ and $b^j$. Our goal is to show that each $a^i = b^j = 0$.

Well, set $y = \partial p_k$. Then we get \begin{align*} 0 &= \omega_{std}(x,\partial p_k)\\ &= \sum (dq^i \wedge dp^i)(x,\partial p_k)\\ &= \sum dq^i(x)dp^i(\partial p_k) - dq^i(\partial p_k) dp^i(x) \end{align*}

Now, the basis $\{dq^i, dp^j\}$ is dual to $\{\partial q_i, \partial p_j\}$, so this sum isn't too hard to evaluate. The second term of the sum is identically $0$ since $dq^i(\partial p_k) = 0$. The first is $0$ unless $i = k$ since $dp^i(\partial p_k) = 0$ for $i\neq k$ and equals $1$ when $i = k$.

So, the sum reduced to a single term: $$0 = dq^k(x)dp^k(\partial p_k) = dq^k(x)$$

Finally, to evaluate $dq^k(x)$, recall $x = a^i \partial q_i + b^j \partial p_j$. So \begin{align*} dq^k(x) &= dq^k(a^i \partial q_i + b^j \partial p_j)\\ &= a^i dq^k(\partial q_i) + b^j dq^k(\partial p_j)\\ &=a^k\end{align*}

where the last step uses the fact that you're working in a dual basis. Putting this altogether, we just showed $0 = a^k$. Doing this for all $k$, and then using $y = \partial q_k$ to deal with the $b^j$ shows $x = 0$.

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The matrix of the quadratic form is in block form $\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$ (where $I$ is the $n \times n$ identity matrix), which is nonsingular.

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