# Help with the definition of a bilinear form $\omega$

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 $\omega(x,y)=dx \wedge dy$?

## 3 Answers

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 $$\tilde{\omega}(v)(u) := \omega(v, u)$$

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 $V^*$.

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

• 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. – janmarqz Dec 21 '15 at 17:51
• Could it not mean that $\omega$ eats the basis coordinates of the vector? – Marion Dec 21 '15 at 19:12
• 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. – janmarqz Dec 21 '15 at 22:01