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I'm trying to figure out exactly what the tensor product of vector spaces is. This is what I understand so far:

If $V, W$ are vector spaces over a field $R$ then the free vector space $C(V\times W)$ is a vector space which has an infinite basis (one element for each pair $(v,w)$ such that $v\in V, w\in W$. Then let the subgroup $Z$ of $C(V\times W)$ be generated by elements of the form:

1) $(v, w_1 + w_2)-(v,w_1)-(v,w_2)$

2) $(v_1+v_2,w)-(v_1,w)-(v_2,w)$

3) $(av,w)-a(v,w)$

4) $(v,aw)-a(v,w)$

Where $a\in R$, $v \in V$, $w \in W$. The tensor product $V\otimes W$ is the quotient group $C(V\times W)/Z$.

Apparently this group now obeys the rules $(v, w_1 + w_2)-(v,w_1)-(v,w_2)=0$, and the other corresponding rules from the above, and this follows from the definition of the quotient. I haven't seen this explained anywhere and it's not immediately apparent to me at any rate. Thanks for any replies!

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It's literally immediately from the definition of quotient. If $V/W$ is a quotient space then for $x\in V$ we have $\bar{x}=0$ in $V/W$ iff $x\in W$. –  Seth May 9 '14 at 23:34
What do you want to be explained exactly? You've explained how the tensor product is formed and why it obeys bilinearity. What does confuse you? –  math.n00b May 9 '14 at 23:34
I saw that in an earlier post you mentioned that your background is mostly in linear algebra and you are doing this for a project. Quotient spaces are not really emphasized in a first course in linear algebra. They are more natural to study in the context of groups/rings/modules. But you may want to review the definition of quotient space. –  Seth May 9 '14 at 23:40

1 Answer 1

It's literally immediately from the definition of quotient. If $V/W$ is a quotient space then for $x\in V$ we have $\bar{x}=0$ in $V/W$ iff $x\in W$.

So $(v,w_1+w_2)−(v,w_1)−(v,w_2)=0$ in $V\otimes W=C(V×W)/Z$ since $(v,w_1+w_2)−(v,w_1)−(v,w_2)\in Z$.

If you are still confused you might want to review the definition of quotient space.

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Thanks for the fast reply! I mean I guess my confusion is that as far as I know $(v,w_1+w_2)-(v,w_1)-(v,w_2)$ isn't an element in $V\otimes W$, it's an element in $Z$. I thought we're meant to get the linearity relation between the equivalence classes in the quotient group. –  James Machin May 9 '14 at 23:57
Or is the class [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] =0 and then you seperate it into the 3 different classes and get the result? –  James Machin May 10 '14 at 0:26
Right, really what they are saying is [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] =0. And from definition of quotient [(v,w_1 + w_2)-(v,w_1)-(v,w_2)] = [(v,w_1 + w_2)]-[(v,w_1)]-[(v,w_2)] –  Seth May 10 '14 at 0:36
When I said "in $V\otimes W$" what I meant was to pretend that I had written the equivalence class. I was being a little sloppy with notation I guess. –  Seth May 10 '14 at 0:37

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