# Antisymmetric but non-symplectic

1. Mar 27, 2009

### Cantab Morgan

I understand that a symplectic scalar product is bilinear and antisymmetric. But is that the only such scalar product? In other words, is it possible to have a bilinear and antisymmetric scalar product that is not symplectic?

2. Mar 28, 2009

### yyat

A symplectic form is by definition an anti-symmetric, nondegenerate, bilinear form.
I am not sure what you mean by "scalar product" in this case, but if it should be nondegenerate then the answer to your question is no.

3. Mar 28, 2009

### Cantab Morgan

Thanks for your reply, yyat! By scalar product I mean a map from pairs of points (vectors) onto a scalar.

For example, the traditional dot product is a Euclidean scalar product that's such a map. In two dimensions, we'd have

$$\left( \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} x' \\ y' \end{array} \right) \right) = xx' + yy'$$

That's a concrete example of a scalar product that's symmetric and bilinear, but it's not the only one. For example, the Lorentz scalar product is also symmetric and bilinear.

$$\left< \left( \begin{array}{c} t \\ x \end{array} \right) , \left( \begin{array}{c} t' \\ x' \end{array} \right) \right> = tt' - xx'$$

Anyway, my understanding is that the symplectic scalar product $$\omega(,)$$ is given by

$$\omega\left( \left( \begin{array}{c} x \\ y \end{array} \right) , \left( \begin{array}{c} x' \\ y' \end{array} \right) \right) = xy' - x'y$$

I'm just curious as to whether that's the only possible bilinear antisymmetric scalar product, or whether there could be other flavors, just as there are different flavors of the symmetric scalar product.