Question of the Day
January 4th, 2008 by WaltIf you take the category with Banach spaces as objects, and continuous linear maps that have a closed image as morphisms, then do you get an abelian category?
If you take the category with Banach spaces as objects, and continuous linear maps that have a closed image as morphisms, then do you get an abelian category?
January 4th, 2008 at 1:10 am
Mh.. if i remember correctly to make an abelian category you need morphisms to be an abelian group (and you have a pre-additive category), all products (and you have an additive category), and good behaviour and good definition of exact sequences to have an abelian category.
But to me it looks like that with Banach spaces we cannot have arbitrary products, or we would get elements with infinite norm. Is this correct?
Regards!
January 4th, 2008 at 5:18 am
It is not even a category. Composing two morphisms with closed image does not necessarily give a morphism with closed image. Consider, for example, the obvious morphisms C([0,1]) -> C([0,1]) \oplus L¹([0,1]) -> L¹([0,1]).
January 4th, 2008 at 6:10 am
Suggest that terms be defined in all future questions,so that
non specialists can understand what is being asked.
January 4th, 2008 at 6:50 am
Re Maurizio’s question: you’re right, but a subterfuge that people often use is to consider instead the category of Banach spaces and bounded linear maps of norm less than or equal to 1. This has infinite products (with norm the sup of norms of the factors); in fact it’s complete and cocomplete, and has a number of nice properties.
It’s not quite abelian, since not all epis are quotients of their kernels. Consider for example maps with dense but not onto images — those are epis which aren’t quotient maps. I have a sneaking suspicion this observation was what motivated Walt’s question, at least in part. I don’t have a good answer myself.
One might object that the category Ban above isn’t even additive if we restrict to linear contractions, but the bit of subterfuge is that Ban with its usual function space objects hom(X, Y) is closed (symmetric monoidal) in the usual way (the necessary structure maps like evaluation
hom(X, Y) tensor_{Ban} X –> Y
still have norm less than or equal to 1), and one can define the Ab-valued hom to be the underlying additive group of the Banach space hom(X, Y). True, the underlying set of this additive group is not the same as the set of linear contractions X –> Y, and so one might still quibble that Ban is still not additive according to some textbook definitions, but I think one can counter that such definitions are unnecessarily restrictive in POV.
robert young: that’s a pretty onerous request imo. If I’m not mistaken, most or all the terms used (except possibly some in the present comment) would be familiar after two years or so of graduate study in mathematics at an American university — were there particular terms which were problematic? Also, Wikipedia will have answers to most such questions.
January 4th, 2008 at 11:03 am
I think the original question is not settled yet.
About Maurizio’s observation: By definition (Hilton-Stammbach, MacLane, Wikipedia, Planetmath all agree) only finite products exist in an abelian category. To finish your proof you should prove that in an abelian category also infinite products exist, and I don’t know whether it is true.
About thw’s observation: I must admit I haven’t understood the counterexample; as obvious morphisms I would take the injection on the first factor and the projection onto the second one, but then the composition would be the 0 map,
If I am not mistaken, I can prove that the composition of two continuous linear maps with closed image still has a closed image. Let f:U -> V and g:V -> W be two such functions, then g(V) is a banach space and the surjective map g:V -> g(V) is an open function by the open mapping theorem. Ker g is a closed subspace of V, and so is the direct sum Ker g \oplus f(U), then g(f(U))=g(Ker g \oplus f(U)) is closed in g(V) because Ker g \oplus f(U) is closed and saturated; but then g(f(U)) is closed in W too, because g(V) is closed.
Do you agree?
January 4th, 2008 at 11:27 am
Francesco, I think thw might have meant for the second morphism
C[0, 1] \oplus L^1[0, 1] –> L^1[0, 1]
to be the one which takes a pair (f, g) to i(f)+g, where i is the usual inclusion C[0, 1] –> L^1[0, 1]. This morphism clearly has a closed image, since it is surjective. I’m not seeing anything wrong with his example.
January 4th, 2008 at 12:30 pm
Thank you, Todd.
My wrong “pseudo-proof” fails because the sum of two closed subspaces need not be closed.
January 4th, 2008 at 2:39 pm
Further to Todd’s first comment: Ban_1 may be the more pleasant category but arguably a lot of natural questions and phenomena belong more naturally in Ban. Neither category is `exact’ in the sense of Barr, Bourn et al. and so looks very different from an abelian category, additivity or no additivity.
This reminds me of something I keep meaning to ask the categorists: is there some kind of fibred category or enriched category way to `stratify’ the morphisms of Ban in some sense according to their norm?
January 4th, 2008 at 5:00 pm
Robert Young: That is a reasonable request, except the question I’m asking is boringly technical. There’s a laundry list of nice properties that maps between vector spaces have. Banach spaces are a kind of infinite dimensional vector space, but the natural notion of maps between them don’t have nice properties. I was just musing on an idea to fix them up and make them nice again. I updated the post to link to Wikipedia’s page on “abelian category”, that that page itself is very technical.
January 4th, 2008 at 5:01 pm
thw: Why is the image of the first map closed?
Todd: Your musing on my motivation is correct.
January 4th, 2008 at 6:27 pm
Walt: to answer your question to thw, the map E -> E \oplus F; x \mapsto (x,0) always has closed image. Erm, this seems kinda obvious to me???
January 5th, 2008 at 1:01 am
Oh, I see. I was imagining the first map was something more complicated, like f -> (f,f).
January 5th, 2008 at 2:19 am
It would have closed image too, since it is an isomorphism onto its image, and the image being Banach makes it closed.
January 6th, 2008 at 3:47 am
What this question needs is a little mollification.
January 6th, 2008 at 11:34 pm
Todd, I now see why people work with the category you mention (Banach spaces with operators of norm at most 1). While thinking about this question, I’ve already reinvented that category three times.
January 7th, 2008 at 2:26 am
A : C([0,1]) -> C([0,1]) \oplus L¹([0,1]) is just the inclusion, while B: C([0,1]) -> C([0,1]) \oplus L¹([0,1]) is defined by B(f,g) := f + g. Then im A = ker pr_2 is closed and im B = L¹([0,1]) is also closed, but im BA = C([0,1]) \subset L¹([0,1]) is not closed (it is dense but it is not the whole set).
January 7th, 2008 at 6:09 pm
Walt: cool. (Uniform boundedness rules!)