I came across the following remark in the lecture:
Let $q$ be a quadratic form over the vector space $V$ and let $\widetilde{q}$ be its associated symmetric bilinear form. Then the mapping $f$ given by $q \rightarrow \widetilde{q}$ defines a bijection between the set of quadratic forms and the set of symmetric bilinear forms.
I can see why $f$ is injective, but I do not understand why $f$ is surjective; i.e. why we can identify each symmetric bilinear form with some $\widetilde{q}$. Could you please explain this to me?
Best Answer
Let $\mathcal{Q} = \mathcal{Q}(V)$ be the set of quadratic forms over the vector space $V$, and let $\mathcal{B} = \mathcal{B}(V)$ be the set of symmetric bilinear forms over the same space. Assuming that $2$ is invertible over your ground field $\mathbb{k}$, i.e. $\operatorname{char} \mathbb{k} \neq 2$, the map $\alpha: \mathcal{Q} \to \mathcal{B}$, from a quadratic form $q$ yields a symmetric bilinear form $\alpha(q)$ defined by $$ \alpha(q)(v, w) = \tfrac12 \bigl( q(v+w)-q(v)-q(w) \bigr). $$ Check that it's symmetric and bilinear! In the other direction, define $\beta: \mathcal{B} \to \mathcal{Q}$, takes a symmetric bilinear form $b$ and produces the quadratic form $\beta(b)$, defined by $$ \beta(b)(v) = b(v,v). $$ Check that it's a quadratic form. Now you have to verify that these are mutually inverse, i.e. that $\beta(\alpha(q)) = q$ for any $q \in \mathcal{Q}$ (shows that $\alpha$ is injective and $\beta$ is surjective) and that $\alpha(\beta(b)) = b$ for any $b \in \mathcal{b}$ (shows that $\alpha$ is surjective and $\beta$ is injective).