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Thread: Some help understanding the Tensor Product of vector spaces (Multilinear Algebra)

  1. #1
    Jun 2010

    Some help understanding the Tensor Product of vector spaces (Multilinear Algebra)

    I will begin the thread with the background knowledge needed to answer my question, and the question will follow afterwards.

    I'm currently getting through the beginning of Finite Dimensional Multilinear Algebra by Marvin Marcus (1973) and I'm curious about something (which I've seen in other textbooks as well).

    In this book, the Tensor Product is defined as a pair: a vector space P, and a mapping, Q, from the cartesian product of vector spaces (V1xV2x...xVn) to P with the following properties:

    1)The linear closure of the image of Q = P

    2) Q has the "Universal Property" that if a map L: V1x...xVn ---> U is defined, there is a unique map, H: P ---> U such that HQ = L.

    Now the author constructs the Tensor Space of a general vector space and proves that this Tensor Space is unique up to isomorphism.

    My question comes in the construction of the Tensor Space. The construction is as follows:

    Given a vector space, V=V1xV2x...xVn, construct the Free Vector Space, F(V) [Notes on Free Vector Space in the end of the post] over V. Using the canonical mapping, i, from a set to it's associated Free Vector Space, we now consider a subspace, M, of the Free Vector Space, and create the quotient space ofF(V)\M.

    Here is where my question arises.

    The subspace,M, of F(V) seems to be chosen out of thin air (to me) and I was wondering if I could get some background info on why this is the subspace that we mod out by for the Tensor Product.

    Here's the construction of the subspace [i is the canonical mapping from i:V-->F(V)]

    Let W1 be the set of elements in F(V)

    i(cv1+dv'1,v2,v3,...,vn) - ci(v1,v2,v3,...,vn) - di(v'1,v2,v3,

    Let Wt be the set (for each t, with t running from 1 until n)

    i(v1,...,cvt+dv't,...,vn) - ci(v1,...,vt, - di(v1,...,v't,

    Consider the linear closure of the union of all of these sets (W1,W2,W3, all the way through Wn), this is then M, the subspace we mod F(V) out by for the Tensor space.

    notes: Free Vector Space
    For any set, X, the Free Vector Space associated with X is the totality of functions defined on X which map to a subset of the field of scalers with the following property:
    f(x)=/= 0 for at most finitely many x in X.

    such that x-->f in F(V) such that f(x)=1 and f(v) = 0 for each v in X (that is not x).
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  2. #2
    MHF Contributor
    Opalg's Avatar
    Aug 2007
    Leeds, UK
    For simplicity, I'll stick to the case where you are just taking the tensor product of two spaces $\displaystyle V_1$ and $\displaystyle V_2$. It's usual to write $\displaystyle i(v_1,v_2)$ as $\displaystyle v_1\otimes v_2$, and it is part of the definition that the tensor product should be linear in each of its factors. So we need it to be true that $\displaystyle (cv_1+dv_1')\otimes v_2 = c(v_1\otimes v_2) + d(v_1'\otimes v_2)$ (with a similar property for the second coordinate). If you just take the free vector space over V then these properties will not hold. So you need to quotient F(V) by the subspace M generated by all these relations.
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