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I Basis of a Tensor Product - Theorem 10.2 - Another Question

  1. Mar 18, 2016 #1
    I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

    I am focused on Section 10.1 Introduction to Tensor Products ... ...

    I need help with another aspect of the proof of Theorem 10.2 regarding the basis of a tensor product ... ...

    Theorem 10.2 reads as follows:


    A diagram involving the mappings [itex]\iota[/itex] and [itex]\gamma'[/itex] is as follows:


    My questions are as follows:

    Question 1

    How do we know that there exists a multilinear map [itex]\gamma' \ : \ X \longrightarrow Z'[/itex] ?

    Question 2

    What happens (what are the 'mechanics') under the mapping [itex]\gamma'[/itex] ... ... to the elements in X\X' ( that is [itex]X - X'[/itex])? How can we be sure that these elements end up in [itex]Z'[/itex] and not in Z\Z'? (see Figure 1 above)

    Hope someone can help ...


    Attached Files:

  2. jcsd
  3. Mar 20, 2016 #2


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    Re question 1:
    Let ##V^\dagger\equiv V_1\times,...,\times V_m## and let's use angle brackets [itex]\langle...\rangle[/itex] to enclose components of Cartesian product spaces. Let the components of [itex]\mathscr{B}_j[/itex] be [itex]v_{j1},...,v_{jn_j}[/itex] where [itex]n_j[/itex] is the dim of [itex]V_j[/itex].
    You haven't said what [itex]Z[/itex] is but let's assume it's the infinite-dimensional vector space over field [itex]F[/itex] with base set
    $$\mathscr{B}^Z\equiv \{\langle u_1,...,u_m\rangle\ |\ \forall k:\ u_k\in V_k\}$$
    More formally, [itex]Z[/itex] is the set of all functions from [itex]\mathscr{B}^Z[/itex] to [itex]F[/itex] for which each such function has finite support (ie is nonzero on only finitely many input values).
    [itex]Z'[/itex] is the subset of [itex]Z[/itex] containing only functions whose support lies in [itex]X'[/itex], and it is easily shown to be a subspace.
    Define the map [itex]\xi:X'\to Z'[/itex] that maps each Cartesian product of basis vectors [itex]\langle v_{1i_1},...,v_{mi_m}\rangle[/itex] to the function that returns zero for every input except [itex]\langle v_{1i_1},...,v_{mi_m}\rangle[/itex], for which it returns [itex]1_F[/itex]. Note that the image of [itex]\xi[/itex] is a basis for [itex]Z'[/itex].
    Then a map [itex]\gamma':V^\dagger\to Z'[/itex] is multilinear, and agrees with [itex]\xi[/itex] on [itex]X'[/itex], if and only if it satisfies:
    \gamma'\left(\left\langle \sum_{i_1}a_{1i_1}v_{1i_1},\, ...\, ,\sum_{i_m}a_{mi_m}v_{mi_m}\right\rangle\right)
    &=\sum_{i_1}\sum_{i_2}\ ...\ \sum_{i_m} \prod_{k=1}^m a_{ki_k}\gamma'\left(\left\langle
    v_{1i_1},\, ...\, ,v_{mi_m}\right\rangle\right)
    \sum_{i_1}\sum_{i_2}\ ...\ \sum_{i_m} \prod_{k=1}^m a_{ki_k}
    \xi\left(\langle v_{1i_1},...,v_{mi_m}\rangle\right)
    where the first equality implements multilinearity and the second implements the requirement to agree with [itex]\xi[/itex].
    Since [itex]X'[/itex] is a basis for [itex]V^\dagger[/itex], it follows that such a map [itex]\gamma'[/itex] exists. It is unique because the representation [itex]\sum_{i_k}a_{ki_k}v_{ki_k}[/itex] of the [itex]k[/itex]th coordinate of the input to [itex]\gamma'[/itex] is unique.
  4. Mar 20, 2016 #3

    Going through your post carefully shortly ...

    Cooperstein provides a definition of Z in the introduction to Section 10.1 including in the proof of Theorem 10.1 ...

    Relevant text from Cooperstein is as follows:


    Hope that helps ...


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