Symmetric bilinear form vs. quadratic form

It has just struck me that symmetric bilinear forms and quadratic forms are different objects, corresponding to symmetric tensors and symmetric algebra respectively of certain tensor algebra (in this post “tensor algebra” will take the narrow meaning of 2nd tensor power of a vector space). I first came to this observation when I was told in IB Linear Algebra that to recover a symmetric bilinear form from a quadratic form, one requires the characteristic of the ground field to be not 2. This oddity can be explained by the “averaging” map

$\frac{1}{2}(\varphi(u, v) + \varphi(v, u))$

which requires 2 to be a unit. This explanation is stilted (at least to me) and troubled me since it hinges on this explicit construction and does not tell us a priori why so. It occurred to me recently when I was going through some old notes that the formula is precisely the symmetrization map of the tensor algebra. It then follows easily that symmetric bilinear forms are images of the “averaging” map and thus is the subalgebra of symmetric tensors. I’ll elaborate briefly on this idea in this post.

Let $k$ be a field and $V$ be a finite-dimensional $k$ -vector space. The space of bilinear forms on $V$ can be identified with $T^2(V^*)$ , the 2-tensors by noting that

$\operatorname{Hom}(V \otimes V, k) \cong \operatorname{Hom}(V, \operatorname{Hom}(V, k)) = \operatorname{Hom}(V, V^*) \cong V^* \otimes V^*$

where the first isomorphism comes from Hom-tensor adjunction and the last one is an easy exercise. It is also possible to directly construct a bilinear map

$\begin{align*} V^* \times V^* &\to \operatorname{Bilin}(V, V) \\ (\varepsilon, \eta) &\mapsto ((u, v) \mapsto \varepsilon(u) \eta(v)) \end{align*}$

In this way, symmetric tensors in $T^2(V^*)$ are precisely those invariant under action of $S_2$ , i.e. permutation, and thus can be identified with symmetric bilinear forms.

Now if the characteristic of $k$ is not divisible by $2!$ , the symmetrization map

$\frac{1}{2!}\operatorname{Sym}: T^2(V^*) \to T^2(V^*)$

induces an isomorphism of $S^2(V^*)$ , the symmetric algebra, and the subalgebra of symmetric 2-tensors. To identify quadratic forms with the symmetric algebra, we can construct a similar bilinear map from $V^* \times V^*$ to the space of quadratic forms and resort to universal property of symmetric multilinear map. And in this way we have shown that symmetric forms and quadratic forms are isomorphic.

One more thing we get for free from this construction is that the space of bilinear forms is the direct sum of symmetric and skew-symmetric forms: since the symmetrization map (call it $\pi$ ) is a projection, we have

$T^2(V^*) \cong \operatorname{ker}(\pi) \oplus \operatorname{Im}(\pi)$

where the former is the ideal generated by elements of the form $\varepsilon \otimes \eta - \eta \otimes \varepsilon$ that we quotient by and consists of skew-symmetric forms, and the latter can be identified by symmetric forms.

Just a word to end the post: as long as the characteristic of the field is not 2, it is possible to generalise this construction (and thus equivalence) to *-algebras. This is related to epsilon-quadratic form.

References

D. Dummit & R. Foote, Abstract Algebra, §11.5

References

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