Question

Let $V$ be a real vector space. A bilinear form $\langle \rangle:V\times V\to {\mathbb{R}}$ induces a linear functional $\theta$ on the tensor product $V\otimes V$ given by sending the finite sum $\sum_i v_i\otimes w_i $ to $\sum_i \langle v_i,w_i\rangle$.

Is there a name for this induced linear functional?

In addition, if the bilinear form is symmetric, then this linear functional $\theta$ respects the natural involution on $V\otimes V$. That is $\theta(v\otimes w)=\theta(w\otimes v)$.

Answer 1

The correspondence you describe is part of the definition of the tensor product: $V \otimes W$ is defined to have the universal property that for any $U$, we have $$\operatorname{Hom}(V \otimes W, U) = \operatorname{Bil}(V \times W, U).$$ I wouldn't even give it a different name: the bilinear form is the same as the map out of the tensor product.

Answer 2

A bilinear form on $V$ (if non-degenerate) lets you identify $V$ with $V^\star$: $v \mapsto \langle v, \cdot \rangle$

In this case, your map is just a contraction of the identity map on $V^\star \otimes V$ (which, considering our identification, is the same as the identity on $V \otimes V$).

http://en.wikipedia.org/wiki/Tensor_contraction

Answer 3

There may be some interest in a variant: the natural map $V\otimes V^{\star}$ to $End(V)$ hits all finite-rank endomorphisms of $V$, and the bilinear map $V\times V^\star\to k$ "induces" trace on finite-rank endos. Thus, some name like "trace" is quite nearby to the literal question.

Answer 4

And symmetry of bilinear forms can be encoded in the tensor product: if we define the symmetric tensor product $V\otimes_s V$ to be the subspace of $V \otimes V$ spanned by the "symmetric tensors" $v\otimes_s w = (v\otimes w + w\otimes v)/2$, then there is a canonical bijection between linear forms on $V\otimes_s V$ and symmetric bilinear forms on $V\times V$.

Answer 5

Hello,

I don't know a specific name for that, but I would call it associated. I wouldn't call it induced, because the map $Bil(V) \to (V \otimes V)^*$ is one-to-one, since every linear $T \in (V\otimes V)^*$ induces a bilinear form on $V$ by sending $(v,w) \mapsto T(v\otimes w)$ and this clearly is the inverse.

Kind regards Konstantin