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I have a tangled web of ideas about natural transformations, vector spaces, equivalence classes, local coordinates, etc. in my head that I'm trying to unravel. So here are some of the questions I thought of:

  • Vector spaces: Why is it that in linear algebra we always calculate with basis but when we think of a definition we always try to make sure it is basis independent? What kinds of symmetries are we trying to preserve by this?

  • Categories: The obvious condition for natural transformations isn't so obvious to me and again I think the definition is the way it is because there is again some kind of symmetry that a natural transformation is preserving. What are the symmetries that a natural transformation is preserving?

  • Equivalence classes: Almost the same thing as in the vector space scenario. We do things with representatives but we make sure our definitions are true regardless of this choice. Ok, this one is kinda silly because we are trying to preserve $\cong$ so there isn't anything too complicated happening here.

  • Differential geometry: We again calculate with local coordinates but our definitions shouldn't depend on them. What symmetries are we trying to preserve in differential geometry?

So it seems to me we always break symmetry, whatever that means, when we want to work with something concrete but we try to make sure our calculations are preserved by these symmetries. Feel free to correct me and add your answers to some of my questions.

Edit: Dmitri gives a good answer for the definition of natural transformation in terms of the exponential and its adjoint.

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  • $\begingroup$ Once we make some choice (basis, representative etc) in a contstruction or a definition, we'll have to prove that what we are doing doesn't depend on that choice. For example, Poincare (I think) define homology in terms of triangulation of the space, and the constructions are quite geometric and natural. However, it's hard to prove that this definition doesn't depend on the triangulation. So now homology is usually introduced using singular chains, which are "invariant" (no choice is made). $\endgroup$
    – Evgeny Shinder
    Commented Dec 9, 2009 at 22:20
  • $\begingroup$ My original ideas were motivated by group actions and orbits. Often various constructions are done by choosing a representative from each orbit and then checking to see if the construction doesn't depend on that choice. So I wanted to see how far I could stretch this analogy with group actions and orbits. I think this analogy works well in the linear algebra case because the orbit consists of the images of the standard basis under all invertible linear maps and we are trying to make sure our construction just depend on the orbit of the standard basis and nothing else. $\endgroup$
    – user577
    Commented Dec 10, 2009 at 0:47

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Your examples about vector spaces and differential geometry do not make any sense to me. One does not need coordinates or bases to prove statements in linear algebra and differential geometry. Personally, I always use coordinate- and basis-free proofs. For me the reason to avoid coordinates and bases is that we lose geometric intuition whenever we use them. See my manifesto on this matter here: When to pick a basis?

One way to make the definition of natural transformation more natural is to consider the category A with exactly two objects and one non-trivial morphism between them. Then the set of morphisms of an arbitrary category C is the set of functors Fun(A, C). If we want the set of functors Fun(C, D) between two categories C and D to be a category, then the set of morphisms of this category is Fun(A, Fun(C, D)). But it is natural to assume that we have the standard adjunction Fun(A, Fun(C, D)) = Fun(A × C, D). Unraveling the definition of functor from A × C to D yields precisely the usual axioms for natural transformations.

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  • $\begingroup$ I don't see how you can calculate a bound for a derivative or compute curvature and other local information without using local coordinates. $\endgroup$
    – user577
    Commented Dec 8, 2009 at 20:41
  • $\begingroup$ I like your answer for the definition of natural transformation as motivated by the adjoint property for exponentials. $\endgroup$
    – user577
    Commented Dec 9, 2009 at 3:25
  • $\begingroup$ What kind of a bound for a derivative do you mean? As for curvature, it can be defined without coordinates and I know many computations of curvature without coordinates (e.g., the curvature of a connection on a trivial vector bundle). Of course, if the connection itself is specified in coordinates, then there might be no way to compute the curvature without using coordinates. $\endgroup$
    – Dmitri Pavlov
    Commented Dec 10, 2009 at 13:09
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I don't agree with your "always." It is certainly possible to write many proofs in linear algebra or involving an equivalence relation that don't require the choice of a basis or a representative of an equivalence class, and there are lots of good reasons to do this:

  • Sometimes there's just not a good choice available. For example when talking about the Lebesgue integral we quotient by equivalence almost anywhere, and for a generic Lebesgue-integrable function it's far from clear what the "right" choice of representative in an equivalence class is.

  • Something that's independent of choices is likely to generalize. For example, the basis-independent part of linear algebra is the part that makes sense in a much more general context: direct sums, duals, tensor products, and so forth.

  • Related to the previous point, a choice-independent proof tends to be more conceptual than a choice-dependent proof.

  • If you prove that something's true regardless of what choices you make, you are free to make choices that take advantage of that! For example, you know that the trace can always be computed as the sum of the diagonals regardless of the basis, so one way to compute the trace of a particular linear transformation is to pick a basis where it's obvious what the diagonal entries are (and this isn't always the eigenvector basis).

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  • $\begingroup$ A choice independent proof is often more geometric, in linear algebra at least, but I wouldn't say it is any more conceptual. $\endgroup$
    – user577
    Commented Dec 7, 2009 at 5:57
  • $\begingroup$ I never understood the quotienting in the Lebesgue case and it always seemed silly to me and I think that's because it is not motivated by any kind of symmetry consideration. It is simply because the integral is insensitive to measure 0 variations and since a lot of the defintions involve the integral we just mod out all the way in the beginning so we don't have to worry about mentioning "up to variation on sets of measure 0". $\endgroup$
    – user577
    Commented Dec 7, 2009 at 6:01
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    $\begingroup$ You quotient so that the L^p seminorms become a norm. $\endgroup$
    – Michael
    Commented Dec 7, 2009 at 14:29
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    $\begingroup$ I think you have too specific a view of what "symmetry consideration" means. Before quotienting, the set of Lebesgue-integrable functions don't form a metric space because distinct functions may be at distance 0 from each other. If one thinks of metric spaces as categories, then what this means is that distinct objects are allowed to be isomorphic, which is perfectly fine unless all you want to do is apply standard theorems about metric spaces, in which case you should pass to the skeletal category, i.e. quotient out by isomorphisms. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 7, 2009 at 14:33
  • $\begingroup$ I have not once stumbled upon any theorem that made any real use of identifying functions because they only differ on sets of measure 0. It's a non-issue and that's because there is no "deep" reason for identifying functions that only differ on sets of measure 0. Contrast with the case in linear algebra where many constructions and calculations are sensitive to choice of bases but the theory is set up to avoid such sensitivities. $\endgroup$
    – user577
    Commented Dec 7, 2009 at 21:10

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