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