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In Wikipedia - http://en.wikipedia.org/wiki/Tensor_product#Tensor_product_of_vector_spaces, tensor product space is explained as the following:

$$F(V\times W) = \left\{\sum_{i=1}^n \alpha_i e_{(v_i, w_i)} \ \Bigg| \ n\in\mathbb{N}, \alpha_i\in K, (v_i, w_i)\in V\times W \right\}$$

equivalence relation: \begin{align} e_{(v_1 + v_2, w)} &\sim e_{(v_1, w)} + e_{(v_2, w)}\\ e_{(v, w_1 + w_2)} &\sim e_{(v, w_1)} + e_{(v, w_2)}\\ ce_{(v, w)} &\sim e_{(cv, w)} \sim e_{(v, cw)} \end{align}

and $V \otimes W = F(V \times W) / R$

where $R$ is the space created by equivalence relations.

So, the first question is why is free vector space defined as a sum? I cannot get how vector space can be a sum. Second question is how can we take a quotient space when there is only $e_{(v_i,w_i)}$ and not $e_{(v_i,w_j)}$?

If there is anything I was mistaken, please inform me.

Thanks.

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If you don't like "formal sums", let $\mathcal{F}$ denote the set of all functions $V \times W \to K$. This is a vector space in an obvious way. Think of $e_{(v,w)}$ as just another name for the function $V \times W \to K$ that maps $(v,w)$ to $1$ and everyhing else to $0$. The addition and scalar multiplication appearing in the definition of $F(V \times W)$ are now those of $\mathcal{F}$ and $F(V \times W)$ is thus defined as a certain subspace of $\mathcal{F}$. FWIW, I recommend learning this material from a single textbook (any textbook) and not Wikipedia. –  leslie townes Apr 29 '12 at 6:56
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^ With the caveat that the functions $V\times W\to K$ are zero at all but finitely many elements of $V\times W$. To go between this functional interpretation and the constructive one, I find it helpful to think of it as a function that sends elements $(v,w)$ to their coefficient in the free vector space. (This is all prior to the quotient process to create the tensor product of course.) –  anon Apr 29 '12 at 7:42

3 Answers 3

up vote 3 down vote accepted

The free vector space construction is not just a single sum. Here's an intuitive picture of how I often think of the free vector space construction. We have a set $X$ - note that if this set has any additional structure defined on it, like a binary operation and some axioms, we decide to forget about it for the purpose of our construction. So, either naturally or artificially, we will assume that the elements of $X$ are not things that can truly be added together, like $X=\{\rm \square,~dog,~2\}$. Furthermore, we have a field $K$. We then pretend we can add things in $X$ together, and multiply them by scalars from $K$, and we end up with formal $K$-linear combinations of the form

$$\sum_{x\in X}\alpha_xx= \alpha_{\square} \square+\alpha_{\rm dog}\mathrm{dog}+\alpha_{2} 2.$$

There is a problem with this though. That numeral "$2$" there is supposed to have no structure to it, as part of our construction process, yet it also designates an element of $K$ too! We don't want to confuse ourselves, so let's put the elements of $X$ as subscripts of a generic "$e$." We then have linear combinations of the form

$$\sum_{x\in X}\alpha_x e_x=\alpha_\square e_\square +\alpha_{\rm dog} e_{\rm dog}+\alpha_2e_2.$$

Keep in mind the $\alpha$ coefficients are scalars from $K$. On top of this, we're well within our rights to now "pretend" that the set of all of these formal sums satisfies every single vector space axiom. We have by virtue of imagination created a new vector space out of $X$. (You will notice it is isomorphic to any other vector space of dimension three, or any other free vector space generated by a set of three elements. As I suggested earlier, the actual contents of $X$ are moot, they ultimately play the part of indexing a basis for our space.)

Remark. A vector space does not necessarily allow infinite sums. (It can be defined in finite-dimensional vector spaces, or on finite-dimensional subspaces of vector spaces, over a field that is also a complete metric space so that it allows infinite sums in the scalar field. Look up Hilbert or Banach spaces.) Thus in our formal sums, when we write $x\in X$ in the subscript, we run the risk of having an infinite sum if $X$ is itself infinite! To prevent this occurrence, we may instead write

$$\sum_{i=1}^n \alpha_i e_{x_{\Large i}} :~~ x_i\in X.$$

These are the actual forms our desired formal combinations will now have.

At this point it gets tricky, because we are going to create the free vector space out of another vector space (in fact, out of a Cartesian product of vector spaces). At this point in our discussion, we must completely forget about the fact that $V$ and $W$ are vector spaces and have algebraic structure; temporarily to us they are just sets with no further facts about them available for use. As before, we have our set of formal linear combinations:

$$\left\{\sum_{i=1}^n \alpha_i e_{(v_i,w_i)}: n\in\Bbb N, (v_i,w_i)\in V\times W \right\} $$

Note: These $(v_i,w_i)$ are not the 2-tuples of basis vectors from $V$ and $W$. They are just a set of $n$ arbitrary vectors from $V\times W$. In fact we have designated no basis for $V$ or $W$. The $v_i$ and $w_i$, for each $i$, can be any two elements from $V$ and $W$ respectively.

Now, finally, after all of this, we can remember the vector space structure of $V\times W$ and create the relations given on Wikipedia, and quotient out by them. After the quotient, we can rename the $e_{(v,w)}$ objects (these were the basis vectors when we were still at the stage of a free vector space) as $v\otimes w$. (It gets tedious writing everything in subscripts!)

There are a few very important differences between $V\times W$ and $V\otimes W$ that need to be identified: even though the pure tensors $v\otimes w$ have two components, each taken from $V$ and $W$ resp., our tensor product will have sums of pure tensors (so, e.g., $u\otimes v+x\otimes y$) that cannot always be written as pure tensors, because the two components of pure tensors are independently linear. For example, in $V\times W$, we can split $(v+x,w+y)$ into $(v,w)+(x,y)$ (you cannot split the first component without also having to split the second), whereas in the tensor product we have to split each individually:

$$\begin{array}{c c c} (v+x)\otimes(w+y) && =v\otimes(w+y) & +x\otimes(w+y) \\ && =v\otimes w +v\otimes y & +x\otimes w+x\otimes y. \end{array}$$

Finally, scalar multiplication of $u\otimes v$ does not affect both componenets at once; only one or the other, resulting in $c(u\otimes v)=(cu)\otimes v=u\otimes(cv)$. In the Cartesian product, scalar multiplication affects both components simultaneously, where $c(u,v)=(cu,cv)$. Hope this helps!

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+1 for the "dog" thing. XD I like your answer. –  Patrick Da Silva Apr 29 '12 at 7:43

What they mean is that you take the free vector space on $V\times W$. Which means you take an indexed set $\{e_{(v,w)}\}_{(v,w)\in V\times W}$ and you consider all formal (finite) sums $\displaystyle \sum_{(v,w)\in V\times W}\alpha_{(v,w)}e_{(v,w)}$--these are the ELEMENTS of the free vector space, not the vector space itself.

So, if you had the vector space $\mathbb{Z}_3$ and $\mathbb{Z}_3^2$ a typical element of $\mathbb{Z}_3(\mathbb{Z}_3\times\mathbb{Z}_3^2)$ is $2e_{(1,(0,1))}+1e_{(2,(1,2))}$

As for your second question, I could see how that could be slightly confusing. Ignore their notation for a second and look at mine. There is no "dependence" on the indices in the sum, as it sort of appears in their sum (even though really there isn't)--just as there is no "dependence" in the indices of the relation you are quotienting by.

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The free vector space is not defined as a sum but rather as a set of sums. Namely, all linear combinations of elements of the basis $\{e_{(v,w)}\}$ with coefficients in $K$.

This is not much different from what you call a linear span, i.e. the vector space you get if you start with a set of vectors. For example, if we are in $\mathbb R^3$, the vector space spanned by $B = \{\left ( \begin{array}{c} 1 \\ 0 \\ 0\end{array} \right ), \left ( \begin{array}{c} 0 \\ 1 \\ 0\end{array} \right ) \}$ is $\mathbb R^2$ and you can denote it by $\{ \sum_{i=1}^2 r_i e_i \mid e_i \in B, r_i \in \mathbb R \}$.

As for your second question: the quotient relation is defined in this way because we want the tensor product to be bilinear. Note that we can define any equivalence relation on any set we like. The set we are taking the quotient over in this case is all linear combinations of elements in $\{(v,w) \mid v \in V, w \in W \}$.

Please let me know if this answers your questions, I am not sure I understand your second question and I'm happy to elaborate.

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