1
$\begingroup$

According to this for $V$ a $2n$ (real) dimensional space any bilinear form $\omega: V \times V \to \mathbb{R}$ induces a linear map $\tilde{\omega}: V \to V^*$ via $$ \tilde{\omega}(v) := \omega(v, \bullet) $$ where, from what I understand $v \in V$ but then what is this $\bullet \,\,$? Can you give an example in the context of normal differential forms maybe? Or, if this $\omega$ is the symplectic form, say for simplicity in $\mathbb{R}^2$, $\omega = dx \wedge dy$, then what is $\tilde{\omega}$? Also is it correct to say that $\omega(x,y)=dx \wedge dy$?

$\endgroup$
1
$\begingroup$

The image of $\tilde{\omega}$ is an element of $V^*$. The notation $$\tilde{\omega}(v) := \omega(v, \bullet) $$ is just meaning that the element $\tilde{\omega}(v) \in V^*$ is defined by $$\tilde{\omega}(v)(u) := \omega(v, u) $$

$\endgroup$
0
$\begingroup$

What they mean is that $\omega(v, \bullet)$ is the linear map from $V$ to $\Bbb R$ given by $u \mapsto \omega(v, u)$. The notation $\omega(v, \bullet)$ therefore signifies a map from $V$ to $\Bbb R$, and "$\bullet$" means "the element from $V$ that you want to map into $\Bbb R$ goes here". That makes $\omega(v, \bullet)$ an element of $V^*$, so $\tilde\omega$ takes an element of $V$ and gives you an element of $V^*$.

$\endgroup$
0
$\begingroup$

We have that $dx(v)\wedge dy:V\to\Bbb R$ is just the linear map $u\mapsto dx\wedge dy(v,u)$

$\endgroup$
  • $\begingroup$ for your last question: $\omega(x,y)=dx\wedge dy$ makes not sense because $x,y$ are coordinates and $\omega$ eats a couple of vectors. $\endgroup$ – janmarqz Dec 21 '15 at 17:51
  • $\begingroup$ Could it not mean that $\omega$ eats the basis coordinates of the vector? $\endgroup$ – Marion Dec 21 '15 at 19:12
  • $\begingroup$ to calculate $\omega(v,u)$ of course one needs the components of the vectors $v,u$, which have $n$ in total and where $n$ is the dimension of the vector space $V$, but $x,y$ are only two coordinates of the space. $\endgroup$ – janmarqz Dec 21 '15 at 22:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.