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This is essentially a follow-up on this question here. Specifically, I'm studying the construction of the tensor product of two vector spaces as indicated by the OP's 2nd definition. Namely, that the tensor product of two vector spaces $V$ and $W$ is the quotient space $V \times W / E$ where $E$ is the subspace of $V \times W$ generated by linear combinations of the form

$$ (v_1 + v_2, w) - (v_1, w) - (v_2, w) $$ $$ (v, w_1 + w_2) - (v, w_1) - (v, w_2) $$ $$ (av, w) - a(v,w) $$ $$ (v,aw) - a(v,w) $$

We also denote an element of this quotient space by $v \otimes w$. Now, I'm trying to understand precisely why it follows from this construction that, for example, $(v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w$. I believe it follows from the following argument:

First, we note that

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

By definition of what it means to be a quotient space, this last equation means that

$$ [ (v_1 + v_2, w)] = [(v_1, w) + (v_2, w)] $$

where $[ \cdot ]$ denotes the induced equivalence class. But, according to the definition of the $\otimes$ operator, we can express this equality as

$$ (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w $$

Is this argument valid and is this the appropriate way to think about these things?

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Be careful! This is not a quotient of $V \times W$. You want to form the free vector space on the set $V \times W$. –  Dylan Moreland Sep 4 '11 at 22:19
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Here's one thing: It's not the space $V \times W$ but rather the space (freely) generated by the set of symbols $[v,w]$ with $v \in V$ and $w \in W$. You'll probably have heard that $\dim{V \otimes W} = \dim{V} \cdot \dim{W}$, while $\dim{V \times W} = \dim{V} + \dim{W}$, so it can't be $V \times W$. –  t.b. Sep 4 '11 at 22:20
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1 Answer 1

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You have a couple of misconceptions here.

First, the tensor product is not defined as a quotient of the vector space $V\times W$. Rather, you consider a vector space $Z$ whose basis elements are the elements of $V\times W$. So, for example, if $V=W=\mathbb{R}$, you would have one basis element for $(1,0)$, another basis element for $(2,0)$, another basis element for $(3,0)$, etc. That is why in that post they are written with double brackets, so as not to confuse $Z$ with $V\times W$.

Note that this is way bigger than $V\times W$. The vector space $V\times W$ has dimension $\dim(V)+\dim(W)$. The vector space $Z$ has dimension $|V\times W|$, which is, usually, much larger! For $V=W=\mathbb{R}$, $V\times W$ has dimension $2$, but $Z$ has dimension $\mathfrak{c}=2^{\aleph_0}$.

So you have one basis element for each element of $V\times W$; you should think of $V\times W$ as the index set for the basis. The basis element $[[v,w]]$ is the basis element that corresponds to the element $(v,w)$ of $V\times W$.

Then $E$ is the subspace of $Z$ generated be all the relations you write down; so, in my example above, you would have the vector $2[[1,0]]-[[2,0]]$ in $E$, etc.

Now, the image of the basis vector $[[v,w]]$ in the quotient is denoted by $v\otimes w$. So in general it's not every vector of $Z/E$ that can be written as $v\otimes w$: these are only the images of the basis of $Z$. So you know that these elements generate $Z/E$, but they need not be all of $Z/E$ (in general, they won't be). The elements of $Z/E$ are linear combinations of these "pure tensors" $v\otimes w$.

So, why does it follow from the construction that $(v_1+v_2)\otimes w = (v_1\otimes w) + (v_2\otimes w)$?

This equality is saying that the equivalence class of $[[v_1+v_2,w]]$ is the same as the equivalence class of $[[v_1,w]]+[[v_2,w]]$ in the quotient. By definition of quotient, this is the same as saying that the vector $$[[v_1+v_2,w]] - [[v_1,w]] - [[v_2,w]]$$ of $Z$ lies in the subspace $E$. But it lies in the subspace $E$ because it is one of the generating elements of $E$. So the equality holds in $Z/E$.

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Thanks for the elaboration; I guess the major thing I was missing is that we consider the free vector space generated by all $(v,w) \in V \times W$ and then quotient that space with $E$. –  ItsNotObvious Sep 5 '11 at 0:16
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