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I am struggling to understand tensor products. I will first state what I think I understand and then ask questions.

Definition of Tensor Product:

http://en.m.wikipedia.org/wiki/Tensor_product

What I Think I Understand:

Consider two vector spaces $V$ and $W$. Let $V \times W$ be the Cartesian product of $V$ and $W$.
$F(V\times W)$ is a free vector space (no idea what this is).

$V \otimes W$ is a vector space. The vectors of this space are defined to be the equivalence classes of $F(V\times W)$ under the following equivalence relations \begin{align} &v, v_1, v_2 \in V; w, w_1, w_2 \in W; c \in K; \\ &(v_1,w) + (v_2,w) \sim (v_1 + v_2,w) \\ &(v,w_1) + (v,w_2) \sim (v,w_1+w_2) \\ &c(v,w) \sim (cv,w) \sim (v,cw) \end{align}

That's about all I understand.

My Question:

Is this the right way to approach understanding tensor products? If so, can you either fill in the gaps in my understanding or tell me where can I find more information about them (besides Wikipedia) so I can do so myself?

Stan Shunpike
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  • a free vector space is in some sense the most general, unconstrained vector space structure you can give to a set (in this case, to the set of pairs $(v,w)$ for $v\in V,w\in W$). then by imposing constraints (the given equivalence relation) you can derive from it the tensor product $V\otimes W$ – obataku Jan 17 '15 at 06:02
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    The free vector space construction is efficient but I don't find it very intuitive. I think it's more helpful to learn how to interpret a familiar object, like an inner product on a vector space $V$ as a tensor in its own right, namely as an element of $V^* \otimes V^*$. This example already contains a good deal of the intuition you'll need. – Travis Willse Jan 17 '15 at 06:18
  • It is helpful to come to grips with a free vector space (over $K$) on any set $S$. Denote it $F(S)$, or $F_K(S)$ if you want to show the dependence on $K$. Let each $s \in S$ correspond to a vector $e_s$ in $F(S)$. Your construction is the special case $S = V \times W$. And $V \otimes_K W$ is the quotient space $F(V \times W)/U$, where $U$ is the subspace of $F(V\times W)$ spanned by differences of things you want to be equal, such as $e_{(v+v',w)}-e_{(v,w)} - e_{(v',w)}$. Then, writing $v\otimes w$ for the coset $e_{(v,w)} \bmod U$, we get $(v+v')\otimes w = v\otimes w+ v'\otimes w$. – KCd Jan 17 '15 at 06:20
  • This explicit construction, alas, is not the right way to think about tensor products of vector spaces. It is only used to show a tensor product exists, where a tensor product of vector spaces has a certain universal property relating bilinear maps to linear maps. Just Google "tensor product mapping property." You need to realize that this stuff is hard for everyone, and there is no magic pill that will let you make sense of it without a certain amount of mental pain first. – KCd Jan 17 '15 at 06:27
  • Well, if you want to operate like a physicist then you could "define" a tensor product of two vector spaces by symbol pushing with expressions $v\otimes w$ satisfying certain rules of computation without ever being concerned about the consistency of the symbol pushing. To be fair to the physicists, their viewpoint is to some extent how tensor products were first created (late 19th century); the mathematician's attitude that a tensor product of spaces is a space satisfying a certain mapping property came much later (1930s). – KCd Jan 17 '15 at 06:33
  • Try one of these, perhaps: http://math.stackexchange.com/q/61916/1242, http://math.stackexchange.com/q/10282/1242 – Hans Lundmark Jan 17 '15 at 09:36
  • @KDc how can we define $F$ over any set? or is that why you mention the dependence on $K$...can you clarify this point? thx – Stan Shunpike Jan 17 '15 at 18:10
  • I found this helpful: http://math.stackexchange.com/questions/18315/free-vector-space-and-vector-space – Stan Shunpike Jan 19 '15 at 21:06

3 Answers3

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The major idea behind the tensor product $V\otimes W$ is that it allows us to study bilinear maps $\omega:V\times W\to Z$ as linear maps $\tilde\omega :V\otimes W\to Z$ on a space like $V\times W$ but with the bilinearity 'built-in', thus reducing the theory of bilinear maps to 'simple' linear algebra. In fact, this construction works for more generally multilinear maps $\omega : V_1\times\dots\times V_n\to Z$ as well ($\tilde\omega : V_1\otimes\dots\otimes V_n\to Z$).

obataku
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Saying something similar to what oldrinb said, the tensor product ( I am assuming here that you are tensoring vector spaces , though a lot of other Mathematical "objects" can be tensored) of two vector spaces is a vector space whose dimension is the product of the dimensions of the respective spaces (and there is a canonical construction of a basis {$v\otimes w$} for $ V\otimes W$) and so that, as oldrinb said, every multilinear map in the product vector space $V \times W$ is uniquely mapped into a linear map defined in the vector space $V \otimes W$ , i.e., there is a bijection between the set of bilinear maps from $V \times W$ into a third vector space $Z$ (over the same field) and the collection of linear maps from $V \otimes W$ into $Z$.

RikOsuave
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Here's a document, by Keith Conrad, that I found helpful: http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf - I have struggled understanding tensor products, just like you, and I still don't feel completely comfortable wth it. The point of the tensor product is that you can isolate the bilinearity of any bilinear function into a 'canonical' bilinear part and something linear - and because the bilinearity is a standard construction, the tensor product, you can more or less ignore it, because the interesting bits are in the linear "left-overs".

j4nd3r53n
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