Equivalence between definitions of Tensor Products (Of Vector Spaces)

Asked Jul 08 '15 at 13:36

Active Jul 08 '15 at 14:53

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I've read the definitions in the book of Kostrikin and in the book of Sterling Berberian. The book of Berberian gives a definition for the product of two spacees, Kostrikin gives it more generally. The Berberian definition is:

$V\otimes W=\mathcal{L}(V^*,W)$, that is, the linear mappings from the dual of $V$ to $W$. Then he defines $v\otimes w\in V\otimes W$ such that for $f\in V^*$, $(v\otimes w)(f)=f(v)w$.

Let's denote this definition with $\otimes_1$.

The Kostrikin construction is given here, let's denote it with $\otimes_2$

The question is: Is there any way to extend the Berberian definition to $n$ spaces in a way that it coincides with Kostrikin's?

edited Apr 13 '17 at 12:21

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asked Jul 08 '15 at 13:36

LeviathanTheEsper

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One could always define $U \otimes_1 V \otimes_1 W := U \otimes_1 (V \otimes_1 W)$. With finite dimensional spaces, the two definitions of the tensor product are certainly isomorphic. – Ben Grossmann Jul 08 '15 at 14:56
So an inductive definition coincides with the other? Oh, thanks. – LeviathanTheEsper Jul 08 '15 at 20:44
Up to an isomorphism of vector spaces, certainly. However, taking the dual of infinite dimensional spaces can be problematic. – Ben Grossmann Jul 08 '15 at 21:22

Equivalence between definitions of Tensor Products (Of Vector Spaces)

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