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{\small
\ni{Lectures by Robert Geroch}
\ni{{\TeX}ed and supplemented by R. Salgado}
}

\ni{\bf DIRECT SUM, TENSOR PRODUCT}
\bigskip

Let $H_1$ and $H_2$ be Hilbert Spaces.  There are two separate constructions
each of which combine these two Hilbert spaces to yield a third.

Denote by $K$ the collection of all pairs, $(\alpha_1,\alpha_2)$, with
$\alpha_1\in H_1$ and $\alpha_2\in H_2$.  Add and scalar-multiply in $K$
``component-wise'':
     $$(\alpha_1,\alpha_2)+(\alpha'_1,\alpha'_2)=
          (\alpha_1+\alpha'_1,\alpha_2+\alpha'_2)$$
     $$c(\alpha_1,\alpha_2)=(c\alpha'_1,c\alpha'_2)$$
One easily checks that this set $K$, with these operations of addition
and scalar-multiplication, is a vector-space.  Next, introduce on $K$
the following inner-product 
     $$\ip{(\alpha_1,\alpha_2)}{(\alpha'_1,\alpha'_2)}=
          \ip{\alpha_1}{\alpha'_1}+\ip{\alpha_2}{\alpha'_2}$$
One checks that this vector space $K$, with this inner-product,
satifies the condition for a Hilbert space.  (Each condition,
including completeness, follows directly from the corresponding
condition on $H_1$ and $H_2$.)  The Hilbert space so constructed
is called the {\bf direct sum}\marginpar{\bf direct sum}
of $H_1$ and $H_2$, written $H_1 \oplus H_2$.

As an illustration of this definition, let $\alpha_1, \alpha_2, \ldots$
be a basis for $H_1$, and $\beta_1,\beta_2,\ldots$ a basis for $H_2$.
Then a basis for $H_1 \oplus H_2$ consists of the elements
$(\alpha_1,0),(\alpha_2,0),\ldots$ and $(0,\beta_1),(0,\beta_2),\ldots$
In the case in which $H_1$ and $H_2$ are finite-dimensional,
the dimension of the direct sum is the sum of the dimensions of
$H_1$ and $H_2$.

\medskip
For the second construction, it is convenient to carry it out in terms of
bases.  Let $\alpha_i$ be a basis for $H_1$, and $\beta_j$ a basis
for $H_2$.  (Here, $i$ and $j$ might run over finite, or infinite, ranges.)
Consider the following collection of all pairs consisting of one basis
element of $H_1$ and one of $H_2$.  For later convenience, we will write
such a pair as $\alpha_i \otimes \beta_j$ (instead of the more usual
$(\alpha_i,\beta_j)$).  Denote by $L$ the Hilbert space with basis
consisting of all such pairs.  Thus, $L$ consists of all expressions of the
form $\sum_{i,j} c_{ij}\ \alpha_i \otimes \beta_j,$
where the $c_{ij}$ are complex numbers satisfying $\sum_{i,j} |c_{ij}|^2 < \oo$
(for, quite generally, every vector in a Hilbert space is a linear combination
of the basis vectors, where the sum of the squares of the absolute values
of the coefficients is finite).  The inner product of two vectors in $L$
is given by
     $$\ip{\sum_{i,j} c_{ij}\ \alpha_i \otimes \beta_j}
          {\sum_{i,j} c'_{ij}\ \alpha_i \otimes \beta_j}=
               \sum_{i,j} \overline{c_{ij}} c_{ij}$$
(for, quite generally, given two elements of a Hilbert space, each
expressed as a linear combination of basis elements, their inner product
is a sum of products of the coefficients, as on the right side above).
This Hilbert space is called the {\bf tensor product}
\marginpar{\bf tensor product} of $H_1$ and $H_2$,
written $H_1 \otimes H_2$.  In the finite dimensional case, the dimension
of the tensor product is the product of the dimensions of $H_1$ and $H_2$.

Let $\alpha=\sum c_i \alpha_i$ and $\beta=\sum d_j \beta_j$ be elements
of $H_1$ and $H_2$, repsectively.  We denote by $\alpha\otimes\beta$ that
element of $H_1 \otimes H_2$ given by 
$\sum_{i,j} c_i d_j\ \alpha_i \otimes \beta_j$ (noting that the coefficients
in this expression do indeed satisfy the condition for membership in 
$H_1 \otimes H_2$).  Note that, in the special case in which $\alpha$ and
$\beta$ are basis elements, this notation reduces to the earlier one.
The general element of $H_1 \otimes H_2$, then, is some linear combination
of these ``simple products'', those of the form $\alpha\otimes\beta$.
One easily checks the following facts:
     $$(c\alpha+\alpha')\otimes\beta=c(\alpha\otimes\beta)+(\alpha'\otimes\beta)$$
     $$\alpha\otimes(c\beta+\beta')=c(\alpha\otimes\beta)+\alpha\otimes\beta'$$
     $$\ip{\alpha\otimes\beta}{\alpha'\otimes\beta'}=
          \ip{\alpha}{\alpha'}\ip{\beta}{\beta'}.$$
Note that in particular that $\|\alpha\otimes\beta\| = \|\alpha\|\ \|\beta\|$.

In fact, one can define the tensor product ``directly'' (\ie, as was done
for the direct sum above, without reference to bases).  This definition
is more natural, more general, and more useful, but is technically
a little more complicated.  The idea is the following.
Fix Hilbert spaces $H_1$ and $H_2$. Introduce, for each $\alpha\in H_1$
and $\beta\in H_2$, a ``formal product of these elements'', written
$\alpha\otimes\beta$.  Next, consider the vector space of all finite
linear combinations of such formal products.  Now enforce, within this
vector space, linearity in each factor, \ie, enforce the first two relations
above. (More formally, one takes the quotient of the original vector space
by the subspace generated by these two relations.)  There results a (somewhat
smaller) vector space, in which linearity in each factor holds.  In this
vector space, introduce the inner product given by the third relation above.
The resulting vector-space-with-inner-product satisfies all of the conditions
for a Hilbert space except completeness.  (The failure of completeness
reflects the fact that we began with only finite linear combinations of
formal products.)  So, merely take the completion.  There results a
Hilbert space, which is just the tensor product, $H_1 \otimes H_2$,
introduced above in terms of bases.






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