Functors Between Real and Complex Vector Spaces
(Unfinished)
In Chapter 12, R.G. describes three possible transformations “from real to complex vector spaces and back”. Inspired by Chapter 17 (Functors), I have examined these transformations from the functor point of view.
We can distinguish not two, but three categories here, the first and third of which are equivalent:
: Complex vector spaces with their usual morphisms and composition law.
: Real vector spaces with their usual morphisms and composition law.
: The category of pairs
, where
is a real vector space and
a vector space morphism
such that
(with
the identity morphism on
). A morphism
in this category is any linear mapping from
to
subject to the condition:
. These morphisms are composed as mappings. It is simple to check that this does make up a category.
Equivalence of
and 
The equivalence of these two categories stems from the procedures described in Chapter 12, procedures that we will describe as isofunctors.
Functor
from
to 
Action on objects:
If is a complex vector space,
is such that:
is the real vector space obtained from
by “forgetting” multiplication by imaginaries, that is the one based on the same set as
with the same addition, the multiplication by a scalar being simply restricted to real scalars.
is the linear mapping
that sends a vector
to
, the scalar multiplication here being that in the complex vector space
. One immediately checks that we do have
.
Action on morphisms:
A morphism between complex vector spaces is also, as a mapping, a morphism between the corresponding real vector space obtained by “forgetting” multiplication by imaginaries; thus, for a -morphism
, we can define
as the same mapping. It is simple to check that the condition on
-morphisms
is satisfied.
Functor
from
to 
The second construction described by R.G. takes a pair with
real vector space and
morphism
such that
to a complex vector space. We can make this into a functor
from
to
by defining what happens to morphisms: they are left as they are, as mappings; the above condition on morphisms in
ensures that the resulting mapping is a
-morphism.
These two functors are isofunctors
It is easy to check (R.G. should at least mention it – criticism) that these two operations are the inverse one of the other, which makes an isofunctor with
its inverse, and
and
equivalent categories.
Taking free objects between these categories yields nothing new, as we saw in the post on isofunctors.
The main interest in this category is that it allows us to work with real vector spaces only, avoiding in particular the need to consider complex and real versions of a given vector space, with the scalar multiplication in the former that «knows how» to multiply by complex numbers, while it does not in the latter. In the rest of this post I will talk of complex vector spaces and these «real plus» vector spaces as if they were the same thing.
Forgetful functor
from complex to real vector spaces
One can make a forgetful functor from
to
simply by forgetting the
part. The functor leaves morphisms unchanged. This is the first construction described by R.G. in Chapter 12, from
to
.
A sort of inverse construction of 
The functor forgets the
, which cannot be retrieved. It can, however, be reinvented, though in general there is a necessary arbitrariness in the process.
Let be a real vector space. Suppose that it is possible to find in
two subspaces,
and
, that are complementary and isomorphic. There will in general be an infinite number of ways of choosing the pair
, and, this pair being chosen, an infinite number of ways of choosing and isomorphism
. (One can, for instance, arbitrarily choose bases on
and
, and create
from an arbitrary bijection between these bases.)
is a mapping between two specific subspaces of
. We can, however, create a full mapping
by posing:
It is easy to check that this is linear, and that
. Hence
is an object of
. Applying the
functor defined above, we obtain a complex vector space.
The necessary and sufficient condition for it to be possible to find in two complementary and isomorphic subspaces is that
be either finite and even-dimensional, or infinite-dimensional (exercise 84).
Providing this condition is satisfied, this procedure involves the arbitrary step of choosing the subspaces, and the second arbitrary step of choosing the isomophism between them. For this reason, one cannot make it into a functor; for a functor must consist of a rule for mapping objects and morphisms. We don’t have an inverse functor for .
This arbitrariness is present even in the simplest non-trivial case, that of a two-dimensional real vector space. This is another formulation of the fact that in a complex vector space there is no non-arbitrary notion of “real” and “imaginary” vectors (as noted in Chapter 12).
Free objects following the forgetful functor from complex to real vector spaces
We do, however, have a functor from to
, but one that implies doubling the initial vector space. It uses the third construction described in Chapter 12.
It is simpler to talk of this construction with rather than
; we have seen that these categories are equivalent. Given an object
in
, that is, a real vector space, we take
as the real vector space direct sum
. We define the mapping
by
. It is easy to check that
is linear, and that
. Hence
is an object in
. We thus have a rule
that maps objects of
to objects of
. Now let us consider a morphism
in
, between real vector spaces
and
. Let
and
. We can form the mapping
. It is linear, and furthermore
; hence
is a morphism in category
, between
and
. We write
. We can check that this rule
preserves composition and identities. It is thus a functor from
to
.
Now let us examine if this cannot be viewed as the — or better, a — free object functor following functor
.
Let be a real vector space, and
, that is
and
.
and
are objects in
. Let
be the
-morphism
defined by
. We wish to check that
is a free object on
following
and that the transformation of morphisms follows suit.
Let be another object of
and
a
-morphism from
to
, that is a linear mapping
.
We wish to find a -morphism
such that
. But
is just the mapping
itself. For
,
. So
must map any
to
. Since
must also be a
-morphism, we must have
.
In other words, necessarily, our must be such that
.
Now we must check that thus defined is indeed a
-morphism
and that
(for up to now we have shown what
must be; we must check that the result is indeed adequate). This checking is unproblematic.
Thus our is a free object on the real vector space
following the functor
.
But it’s not quite over! We must also check that, taking