Basis of a Tensor Product - Theorem 10.2 - Another Question

[Math Amateur]

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 \iota and \gamma' is as follows:

My questions are as follows:Question 1

How do we know that there exists a multilinear map \gamma' \ : \ X \longrightarrow Z' ?
Question 2What happens (what are the 'mechanics') under the mapping \gamma' ... ... to the elements in X\X' ( that is X - X')? How can we be sure that these elements end up in Z' and not in Z\Z'? (see Figure 1 above)
Hope someone can help ...

Peter

 

[andrewkirk]

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

 

[Math Amateur]

Andrew,

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 ...

Peter