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 ?

  • 2
    $\begingroup$ See also: mathoverflow.net/questions/244131/nuances-regarding-naturality $\endgroup$ – Oscar Cunningham Mar 19 '19 at 7:33
  • $\begingroup$ very helpful and informative. I am more or less aware of all the nuances involved when dealing with isomorphisms between vector spaces because of the strong distinction between the finite-dimensional and the infinite-dimensional case. that's why I chose to stick to invariants that are only common in the first case (although there are "natural" generalizations of both the trace and the determinant, but lets not go there) $\endgroup$ – H1ghfiv3 Mar 19 '19 at 8:00
  • 2
    $\begingroup$ One solution to some of these problems involving arbitrary choices is to make all possible choices (mathoverflow.net/questions/50025/…). This applies to your example (3). Instead of picking a particular triangulation of a space, we take the singular simplicial complex in which the $n$-simplices are all functions from $\Delta_n$ to the space. $\endgroup$ – Oscar Cunningham Mar 19 '19 at 9:35

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.

| cite | improve this answer | |
  • $\begingroup$ Looking closer, it doesn't seem that the definition of cardinality, unlike the examples I gave, really chooses a specific bijection. It only checks whether there is some bijection between $A$ and an ordinal $\alpha$. $\endgroup$ – H1ghfiv3 Mar 18 '19 at 12:47

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 $Rep^G$.

| cite | improve this answer | |
  • $\begingroup$ In terms of functoriality, it is of course very important to stick to choices. To my knowledge, this is also part of the reason why the category of based topological spaces has been introduced in the first place. Moreover, I agree with you that in many instances, trying to find a "coordinate-free" definition may deviate from the actual purpose. This is also why I'm asking this only in the case of numerical invariants (to which fundamental groups don't belong to). $\endgroup$ – H1ghfiv3 Mar 18 '19 at 11:04

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 $\mathrm{dim}(V)$).

| cite | improve this answer | |
  • $\begingroup$ for finite-dimensional vector spaces $V$, couldn't I define "dim(V)" as the largest integer $n$ such that $\Lambda^n V \neq 0$ and thereby weasel around picking a basis ? $\endgroup$ – H1ghfiv3 Mar 19 '19 at 8:26
  • $\begingroup$ @BerniWaterman How do you know such an $n$ exists? $\endgroup$ – Oscar Cunningham Mar 19 '19 at 8:34
  • $\begingroup$ I think I see where you are getting at: it is true that for finite-dimensional vector spaces, $\Lambda^k V$ contains only one element for $k > \dim(V)$. However, in order to determine that, one needs to understand how a basis of $\Lambda^k V$ is built from a basis of $V$, right? $\endgroup$ – H1ghfiv3 Mar 19 '19 at 8:42
  • $\begingroup$ In programming terms, one could say something like list(rang(f))[0]. $\endgroup$ – J.G. Mar 19 '19 at 8:48
  • $\begingroup$ @J.G. Then the question would be that of how list chooses which order to arrange the elements. $\endgroup$ – Oscar Cunningham Mar 19 '19 at 9:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.