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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 $\theta(v\otimes w)=\theta(w\otimes v)$.

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I would simply identify your bilinear map and the induced map on the tensor product, and simply say they are the same thing. The universal product of the tensor product tells us that I would not run into much trouble. –  Mariano Suárez-Alvarez Aug 28 '11 at 17:57

5 Answers 5

up vote 5 down vote accepted

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.

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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

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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 $V\times V$.

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Don't divide by 2 if you want your remark to make sense over all fields (i.e., including those of characteristic 2). –  KConrad Aug 28 '11 at 20:53

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

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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.

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