Question about the role of Quotient Spaces as Tensor Producs

Question

According to Steven Roman's Advanced Linear Algebra, the very first objects that is required to construct the Tensor Product lies on a free vector space $\mathfrak{F}_{(\mathfrak{U}\times\mathfrak{V})}$ which elements are in the form: $$\sum^{N}_{i=1} r_{i}(u_{i},v_{i})$$

And a subspace $\mathfrak{S}$ of $\mathfrak{F}_{(\mathfrak{U}\times\mathfrak{V})}$ which are spanned by vectors of this form:

$$r(u,w)+s(v,w)-(ru+sv,w)$$

$$r(u,v)+s(u,w)-(u,rv+sw)$$

Then he writes:

where $r,s \in$ Field $\mathbb{F}$ and $u,v$ and $w$ are in the appropriate vector spaces. Note that vectors are precisely what we must identify as the zero vector in order to enforce bilinearity.

Well, my question is, what suppose to mean identify in this context? If we want to identify vectors we construct quotient spaces?

Furthermore, later in the text he constructs the tensor product as:

$$\mathfrak{V}\otimes\mathfrak{W} := \frac{\mathfrak{F}_{(\mathfrak{U}\times\mathfrak{V})}}{\mathfrak{S}}$$

Answer 1

The idea is the following: first, form the free vector space generated by $\mathfrak{U}\times\mathfrak{V}$; free, in the sense that there are no relations among the elements, which are just finite formal sums of the form $\sum r_i(u,v)$ floating around in the vector space. So, for example, the elements $r(u,v)+s(u,w)$ and $(u,rv+sw)$ are different. But, we would "like" them to be the same, because it seems natural to "distribute" the $u$ out of each term and multiply the $r$ and $s$ "in". This is where the identification comes in. How do we declare vectors like these to be the same?

One way is to take the vector subspace $\frak G$, generated by the vectors of the form you have displayed in your question and declare two vectors $x,y$ in the free product to be "the same" (they are $identified$) if and only if $x-y\in \frak G$. That is, $x\sim y\Leftrightarrow x-y\in \frak G.$ Now, collect all the vectors identified by the (equivalence) relation $\sim$ into sets, which will turn out to be disjoint (This nice property is the reason we chose to make our identification in this way), so each set can be considered ONE vector in a NEW space, $\frac{\mathfrak{F}_{(\mathfrak{U}\times\mathfrak{V})}}{\mathfrak{S}}$, which is called the quotient space, and as we have seen, is a collection of equivalence classes, which we defined by $\sim$. It also turns out that $\frac{\mathfrak{F}_{(\mathfrak{U}\times\mathfrak{V})}}{\mathfrak{S}}$ is itself a vector space, but now all the relations we want to hold, that did not hold in the free vector space, now do. This is the tensor product.

Although the ideas of quotient spaces obtained from equivalence relations is certainly important, this particular construction, in my opinion, is not very important. Once you know there is such a thing as the tensor product of two vector spaces, you go right to the universal property it satisfies, and never look back.

If this answer seems trivial to you, or is not what you are after, please let me know and I will gladly delete it.

Answer 2

In this context, by identify he means to impose a relation on the elements of the free space so that they satisfy bilinearity.

Notice that the "bilinearity condition" for all $u, w \in W$ and $v \in V$ in the product $W \times V$ is stating that:

$(u+w, v) = (u, v)+(w, v)$

Now if we move the entire RHS to the left then we obtain

(1) $(u+w, v)-(u, v)-(w, v)=0$

Recall that if you quotient a vector space $V$ by a subspace $U$, you are declaring all elements of the subspace $U$ to belong to the zero element of the quotient space.

Thus if we consider the set of all elements of the form $(u+w, v)-(u, v)-(w, v)$, then look at the subspace generated by them, and then quotient the larger space by that subspace, we have determined that all of these elements belong to the zero element of the quotient, and imposed the relation $(u+w, v)-(u, v)-(w, v) =0$ on our vector space as we wanted in (1).