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I'm having serious trouble to understand the definition of tensor products from Kostrikin's Linear Algebra and Geometry. Until now I've understood a tensor as a multilinear map from the cartesian product of $k$ copies of a vector space $V$ to the underlying field of scalars. However I was instructed to study the multilinear construction from Kostrikin's book, and I'm really not understanding what he's trying to do.

Well, i'll post how he constructs the tensor product, and then post my doubts. Any help or reference is very good.

He constructs the tensor product in three steps:

  1. Let $L_1,L_2\cdots L_p$ a family of vector spaces over the same field $\mathbb{K}$. Define the set $M$ of all functions with finite support on $L_1\times L_2\times\cdots\times L_p$ and with values in $\mathbb{K}$, in other words, all functions from this cartesian product to $\mathbb{K}$ which vanish at all except a finite number of points of the domain. One basis consists of the functions $\delta(l_1, l_2 \dots l_p)$ which is $1$ at the point $(l_1, l_2 ...l_p)$ and zero everywhere else. He also omits the $\delta$ symbol, so that we have $$M=\left\{\sum a_{l_1\cdots l_p}(l_1,l_2\cdots l_p) \mid a_{l_1\cdots l_p} \in \mathbb{K} \right\}$$
  2. Consider the subspace $M_0$ generated by the vectors on $M$ of the form: $$(l_1\cdots l'_j+l''_j\cdots l_p) - (l_1\cdots l'_j\cdots l_p) - (l_1\cdots l''_j\cdots l_p)$$ $$(l_1\cdots al_j\cdots l_p) - a(l_1\cdots l_j\cdots l_p) \quad \quad a\in \mathbb{K}$$
  3. The tensor product is then defined as $$L_1\otimes L_2 \otimes \cdots \otimes L_p = M / M_0$$ $$l_1\otimes l_2 \otimes \cdots \otimes l_p = (l_1, l_2\cdots l_p) + M_0$$ $$t: L_1 \times L_2 \times \cdots \times L_p \to L_1\otimes L_2 \otimes \cdots \otimes L_p, \quad t(l_1, l_2 \cdots l_p) = l_1 \otimes l_2 \otimes \cdots \otimes l_p$$

Well, I already did my best to understand this definition, but it just can't get it. First, he says that this definition is being made to be able to construct some universal multilienear application. That's fine, but why this definition allows us this? What's the intuition behind this definition?

Second, on step 1 he consider functions that vanishes at all points except a finite number. Well, why these delta functions form a basis ? I don't understand why, since for me it seems that for using those delta, we would need to know on which points the functions is not zero.

Third, why considering that subspace $M_0$ ? I just can't grasp what's the reason for this.

And finally fourth: why setting the tensor product as that quotient space ?

I really thought a lot already on this definition, trying to came out with something, however I couldn't have any idea. Can someone give some help?

Thanks in advance, and really sorry if this question is too long, too specific or too silly. I'm just very confused with this definition.

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1 Answer 1

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For simplicity, I will only talk about the construction of the tensor product of two $\mathbb{K}$-spaces, $L_1$ and $L_2$. The generalization to the tensor product of any finite number of spaces follows easily.

The intuition is this: Whatever the tensor product $L_1 \otimes L_2$ turns out to be, I want it to be generated by elements of the form $l_1 \otimes l_2$ with $l_1 \in L_1$ and $l_2 \in L_2$. I also want the canonical map $L_1 \times L_2 \to L_1 \otimes L_2$ with $(l_1,l_2) \mapsto l_1 \otimes l_2$ to be $\mathbb{K}$-bilinear, so that means I need the generators of $L_1 \otimes L_2$ to satisfy the following relations:

$$al_1 \otimes l_2 = a(l_1 \otimes l_2),$$ $$l_1 \otimes al_2 = a(l_1 \otimes l_2),$$ $$(l_1 + l_1')\otimes l_2 = l_1\otimes l_2 +l_1' \otimes l_2,$$ $$l_1 \otimes (l_2+l_2') = l_1 \otimes l_2 + l_1 \otimes l_2'.$$

And this is ALL that I want! Nothing more! I want $L_1 \otimes L_2$ to satisfy these properties, and I want absolutely no other relations (besides of course the relations that come from the definition of a vector space). The end product is a $\mathbb{K}$-space $L_1 \otimes L_2$ consisting of $\mathbb{K}$-linear combinations of the generators, and these generators satisfy the above relations.

The fancy language in the technical definition is just a means of formalizing this type of construction (the tensor product can be defined using its universal mapping property, but then one still has to prove that such a gadget actually exists). In fact, this type of construction has a name: We are defining $L_1 \otimes L_2$ by "generators and relations." Everything following this point in my answer is just formalization (keep the above intuition in mind).

The first thing I do is start out with my generators. So in part (1), I start with the $\mathbb{K}$-space $M$ with basis $L_1 \times L_2$. What is this? Formally, it consists of all functions $L_1 \times L_2 \to \mathbb{K}$ with finite support (and the vector space operations are pointwise). Why are those $\delta$'s actually a basis? Take an arbitrary element $m\in M$ (it's a finitely-supported function $L_1 \times L_2 \to \mathbb{K}$), and say it's nonzero at $(l_1,l_1'),\ldots,(l_n,l_n')$. Say $m(l_k,l_k')=a_k$. Then take a moment to convince yourself that the expression

$$ m = \sum_{k=1}^n a_k \delta(l_k,l_k')$$

is unique (remember the operations in $M$ are pointwise). This proves the $\delta$'s are a basis. I wanted $M$ to be the $\mathbb{K}$-space with basis $L_1\times L_2$, and $\delta(l_1,l_2)$ is how I think of $(l_1,l_2) \in L_1 \times L_2$ as living in $M$. Nobody ever thinks of these types of vector spaces (or modules) by their formal definition, we just think of them as all (formal) finite linear combinations of the generators. That is what is happening in (1) when he drops the $\delta$ notation.

Now that I have all my generators in order, I want to introduce the above relations ((2) in your definition). Ultimately, I want, for instance, $(al_1,l_2) = a(l_1,l_2)$ to be true. That is the same as wanting $(al_1,l_2) - a(l_1,l_2) = 0$ to be true. Hence, I want to kill the element $(al_1,l_2) - a(l_1,l_2)$ in $M$. How do you kill elements in vector spaces and modules? Mod out by them. So I take all of the elements that represent my relations, and consider the space $M/M_0$ where $M_0$ is the subspace generated by all of the elements representing my relations. Now the relation $(al_1,l_2) = a(l_1,l_2)$ really is true (modulo $M_0$) in the space $M/M_0$.

Now I set $L_1 \otimes L_2 = M/M_0$ and write $l_1 \otimes l_2$ for the image of $(l_1,l_2)$ in $L_1 \otimes L_2$. As mentioned above, this process completely formalizes my above "working definition" for the tensor product. As Atiyah and MacDonald say in their commutative algebra text, having seen this construction, you may safely forget it. The important part about the tensor product are its mapping properties.

And about those mapping properties... The tensor product, as you probably know, enjoys the following property: Any $\mathbb{K}$-bilinear map $L_1 \times L_2 \to V$ ($V$ a vector space) uniquely factors through the tensor product. To prove that our above definition satisfies this property, we use a "composition" of two universal mapping properties. The process of forming the vector space with a given basis has a universal mapping property (see free module, it's essentially "extending by linearity"), and quotients have their own universal mapping property. This last property is the following: If $V$ is a vector space and $V_0$ is a subspace, then any linear map $f: V \to W$ to another space $W$ which kills $V_0$ uniquely factors through the quotient $V/V_0$. Notice that we used both of these gadgets in our construction of $L_1 \otimes L_2$; if we put these two universal properties together, we obtain the universal mapping property of the tensor product.

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Oh my god! You've explained everything in detail. Thank you very much John Myers. There's just one point that I need clarification: the elements of the quotient space $M/M_0$ are equivalence classes right ? So, if $l_1 \otimes l_2 \in M / M_0$, this means that the tensor product of two vectors gives me an equivalence class of ordered pairs of vectors, in other words, an entire set of ordered pairs of vectors instead just one pair ? I just didn't get this part of using the equivalence relation, but everything else is much clearer now. Really thanks ! –  user1620696 Mar 14 '13 at 1:11
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Indeed, $l_1\otimes l_2$ by the above construction is an equivalence class. However, when dealing with quotients of vector spaces, say $V/W$, (or quotients of groups, modules, rings by ideals, etc.), it is sometimes psychologically more satisfying to think of the elements of the quotient literally as elements of $V$, but subject to whatever relations the quotient structure introduces. For instance, one shouldn't think of elements of the group $\mathbb{Z}/5\mathbb{Z}$ literally as sets of numbers, one should think of these elements as numbers that are subject to the the relation "modulo 5." –  John Myers Mar 14 '13 at 1:35
    
Also, this idea of defining some sort of structure by "generators and relations" is quite common. For instance, in group theory, a group presentation is essentially the same idea. If you study further in multilinear algebra, you'll see exterior and symmetric algebras; the parts making up these algebras are defined using generators and relations (in fact, by introducing more relations into the tensor product!). –  John Myers Mar 14 '13 at 1:39
    
Now I understood, we use the quotient just to impose relations on the elements of the space. I think I didn't get it before because I'm not yet used to this way of thinking used on algebra. Well, I'm just begining with multilinear algebra, but I'll get to exterior algebra. Can you recommend some book that also takes this approach, but giving more motivation for the definitions and so on ? Thanks again for your support. –  user1620696 Mar 14 '13 at 1:43
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One reference you might look at is math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf (it treats modules, not just vector spaces, so it may seem more abstract, but the basic construction is exactly the same). –  KCd Mar 16 '13 at 15:21

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