Linear Algebra, Canonical Isomorphism.

Question

I cannot understand the difference between canonical and non-canonical isomorphisms between vector spaces. Could someone explain the definition and give some examples?

Answer 1

A canonical morphism between two vector spaces (or other algebraic structures) is one that is obtained without involving arbitrary choices, or better yet: given by the very constructions involved and thus in a way obvious.

For example, if $V$ is a vector space and $U$ is a sub-space, then we define the quotient space $V/U$ in the usual manner, using co-sets $v+U$ as elements. This very definition gives us without any further thought, arbitrary choice, or other specification a homomorphism from $V$ to $V/U$, the canonical projection $V\to V/U$, $v\mapsto v+U$. In the same situation, the very fact that $U$ is a subspace of $V$ gives us the canonical inclusion homomorphism $U\to V$, $u\mapsto u$. And if in such or similar situations, such a canonical homomorphism turns out to be an isomorphism (e.g., between $V/U$ and $(V/W)\bigm/(U/W)$ if additionally $W$ is a subspace of $U$), we of course call that a canonical isomorphism.

More generally, if an object is given by a universal property, i.e., as an object together with certain homomorphisms such that in certain diagrams with similar homomorphisms, there exists one and only one homomorphism making the combined diagram commutative, the the homomorphisms mentioned above (as "together with certain homomorphisms") are referred to as canonical. Concretely, this gives us canonical projections to the factor objects of a direct product, canonical embeddings of the summands of a direct sum, and at least some of the examples from the preceding paragraph (subspace and quotient space) can be viewed as being of this kind.

On the other hand, we know abstractly that $\Bbb R$ and $\Bbb C$ can be viewed as vector spaces over the field $\Bbb Q$. As such, they have the same dimensions (not only are both infinite-dimensional, but it's in fact the same infinity), hence isomorphisms between them do exist (just pick bases and map one to the other). However, none of these suggests itself as being "obvious"; in fact, no-one can even describe any such isomorphism explicitly (i.e., without invoking the Axiom of Choice).

Another important specific example of a homomorphism called canonical because it involves no choice etc., is this: If $V$ is a vector space over a field $k$, we call the vector space of linear maps $V\to k$ its dual space $V^*$. This is again a vector space, hence has a dual space $V^{**}$, the bi-dual space of $V$. So elements of $V^{**}$ map linear maps $V\to k$ to elements of $k$. For each $v\in V$, we have the "evaluate at $v$"-map $V^*\to k$ that maps a linear map $\phi\colon V\to k$ to the element $\phi(v)\in k$. Mapping $v$ to the "evaluate at $v$"-map is a homomorphism $V\to V^{**}$, $v\mapsto (\phi\mapsto \phi(v))$. If $V$ is finite-dimensional, this turns out to be an isomorphism (simply because $V$, $V^*$, $V^{**}$ have the same dimension). Do you see how we put no arbitrary choices into this (e.g., we did not have to pick a basis for $V$)? That's why this homomorphism $V\to V^{**}$ is often called canonical as well. Note that on the other hand, there is no non-trivial canonical homomorphism $V\to V^*$, not even if both $V$ and $V^*$ are just one-dimensional!