Soft Question: Can "invariants" always be defined without making a "specific" choice?

Question

I fear that I might know far too little category theory to formulate this question more precisely, that why it is only a soft question:

Give some mathematical object $A$, one can often assign to $A$ a numerical quantity $F(A) \in \mathbb R$ whose specific value may often shed light on important features of $A$ in return. Basic examples include:

1) The determinant of an endomorphism $f: V \to V$ on a finite-dimensional $\mathbb R$-vectorspace $V$,

2) The trace of an endomorphism $f: V \to V$ on a finite-dimensional $\mathbb R$-vectorspace $V$,

3) The Euler characterstic on a topological space $X$ admitting a finite $CW$-structure.

In each case, the quantity is usually defined at first with the aid of a specific choice. In the first two cases, one picks an arbitrary basis $B$ of $V$ and computes the corresponding quantities for the resulting square matrix. In the third case, one picks a finite $CW$-structure and computes the alternating sum of numbers of cells.

A major part of why these quantities turn out to be so interesting is of course that they are independent of the choices made (which is why they are also called invariants). In the first two cases, one usually learns this by observing that similar matrices have the same determinant and trace. Only later, after getting to know a little bit on dual spaces and exterior algebra, one finds out that there actually exist base-free definitions of both these quantities.

One usually learns about independence of the choice in the third case by being shown an equivalent, $CW$-free definition (with the aid of singular homology).

In all three cases, a numerical quantity of an object is

a) first defined with aid of a choice, then

b) shown to be independent of that choice, and finally,

c) is equivalently defined without the use of a choice.

My (vague) question now goes as follows:

If a numerical quantity of an object is defined using a specific choice and, within a context-dependent realm of possible choices, is a posteriori shown to be independent of that choice, can that same quantity always be defined without that choice ?

Answer 1

This isn't an explicit example of a "canonical" invariant without a canonical description, but I think its an important point that should be raised regarding these issues (and is too long for a comment).

Often we have multiple noncanonical choices in defining our invariant, and the existence of these choices is more important than insisting on a canonical choice free description.

An example is the fundamental group of a space. For a reasonable, connected space, the abstract group we get is independent of the basepoint chosen, but changing our basepoint by a given path induces a conjugation action of the fundamental group on itself. Thus, to get at this action, and all the basepoint changing phenomena, eg, covering spaces, we need to consider all the basepoint choices, and how they relate to one another.

However if we insist on basepoint independence, we get what remains after forcing the fundamental group's conjugation action on itself to be trivial, which gives us the first homology group of our space, that is, $\pi_1(X,,x_0)^{ab}\cong H_1(X)$.

Another instance of this is that one can recover a (finite) nonabelian group as the tensor automorphisms of a "forgetful" functor $F:Rep^G\rightarrow Vect$, but this choice of forgetful functor is noncanonical. However we can recover the abelianisation $G^{ab}$ of the group $G$ from its category of representations canonically i'm pretty sure, with no forgetful functor (basepoint) required, by considering the tensor automorphisms of the identity functor on the subcategory generated by invertible objects in $Rep^G$.

Answer 2

If you allow numerical invariants that aren’t real numbers, then “cardinal number” is an example. The cardinality of a set $A$ is the least ordinal $\alpha$ for which there exists a bijection between $\alpha$ and $A.$ But there is no canonical way to select a particular bijection (unless you make special set-theoretic assumptions like $V=L$), and there’s no way to bypass this.

Answer 3

Suppose we have some invariant in a set $B$ that is defined in terms of a choice of $a\in A$. Then without picking an $a\in A$ we may still construct the function $f:A\to B$ that maps $a\in A$ to the corresponding value of the invariant in $B$.

We can then prove that for all $a_1,a_2\in A$ we have $f(a_1)=f(a_2)$. This proof will involve considering some particular $a_1$ and $a_2$ in $A$, but I wouldn't say that it has made any choices since $a_1$ and $a_2$ both range over all of $A$.

I think this is as far as one can go without making a choice. To actually define the value of the invariant one has to say "pick $a\in A$ and consider $f(a)$". This is necessary because otherwise $A$ might be empty, and the invariant wouldn't be defined at all.

For example if your definition of "finite-dimensional vector space" is "has a finite basis" then your definition of determinant has to involve picking a finite basis. Otherwise the determinant would also be defined for infinite dimensional spaces. Even the definition of determinant involving exterior powers must involve picking a finite basis somewhere (it considers $\Lambda^{\mathrm{dim}(V)}(V)$, and you have to pick a finite basis to calculate $\mathrm{dim}(V)$).