1550075568-C-Algebras_and_Finite-Dimensional_Approximations__Brown_

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60 3. Tensor Products

facts which help us rationalize the absence of formal proof in this section:
All of the results are simple, some hints and tricks are explained and we are
confident that the reader can fill the gaps with minimal effort. Of course,
one could consult an algebra book for the proofs, but this would be a waste
of time -it is just as easy to give the proofs as to look them up.
If X and Y are two vector spaces, then the algebraic tensor product of
X and Y is a new vector space which is both a useful tool and an interesting
object in its own right. I Though not common in the algebraic literature, we
will use X 0 Y to denote the algebraic tensor product.
The construction of X0Y goes as follows. Regard the Cartesian product
X x Y as a discrete space and consider the vector space of compactly (i.e.,
finitely) supported functions Cc(X x Y). Let X(x,y) E Cc(X x Y) be the
characteristic function over the point ( x, y) E X x Y and note that the
collection of all such characteristic functions is a basis for Cc(X x Y). Now
define a linear subspace K c Cc(X x Y) as the subspace spanned by elements
of the following four types:
(1) X(x1+x2,y) - X(x1,y) - X(x2,y)'
(2) X(x,y1 +y2) - X(x,y1) - X(x,Y2)'
(3) AX(x,y) - X(.Ax,y) and
(4) AX(x,y) - X(x,.Ay)·
Definition 3.1.1. Given vector spaces X and Y, their algebraic tensor
product is the quotient vector space
X 0 Y = Cc(X x Y)/K.
The image of an element X(x,y) E Cc(X x Y) under the canonic.al quotient
map Cc(X x Y) __, X 0 Y is called an elementary tensor and is denoted
x®y.

Note that X 0 Y is spanned by the elementary tensors (but they are not
a basis).
There are really only two things that one needs to know about tensor
products in order to handle most issues that arise. The first is the tensor
calculus and follows easily from the definition of the space K.

Proposition 3.1.2 (Tensor calculus). The following identities hold for all
vectors and scalars:
(1) (xI + x2) ® Y =XI® Y + X2 ® y and x ®(YI+ Y2) = x ®YI+ x ® Y2·


. (2) .A(x ® y) =(.Ax)® y = x ® (.Ay).


IAll of this can be done in much greater generality -for modules over rings -but for our
purposes the reader may assume that everything happens over IC.

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