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    • CommentRowNumber1.
    • CommentAuthorMark Meckes
    • CommentTimeNov 9th 2011

    I’ve expanded the section on morphisms in Banach space, because the new page on isomorphism classes of Banach spaces refers to a different notion of isomorphism than what the Banach space page previously called the “usual” notion of isomorphism. (The issue is that what’s usual seems to be different for analysts and category theorists.)

    • CommentRowNumber2.
    • CommentAuthorAndrew Stacey
    • CommentTimeNov 9th 2011

    Good. I think that the Banach space page should be neutral in tone. It’s good to have both views on the page, but there shouldn’t be a particular preference displayed.

    Any existing bias should not be taken as an explicit preference - I think a lot of these pages were created around about when I was looking at the links between functional analysis and algebraic theories so it’s quite natural that there was a categorical bias.

    I’m just waiting for someone to write the page on category theory in the Baire sense …

    • CommentRowNumber3.
    • CommentAuthorMark Meckes
    • CommentTimeNov 9th 2011

    I tried to be neutral in tone; if anyone thinks I introduced a preference please correct it. I do think it’s reasonable, however, at least to start by defining analysts’ terminology. Someone refering to “isomorphic Banach spaces” is most likely an analyst, and therefore most likely using the term according to analysts’ conventions. If I get around to it, I’ll add something about the uniform and Lipschitz categories.

    Presumably someone will eventually write the page on Baire category, but I hope the “first/second category” terminology isn’t emphasized too much. None of the terminology surrounding Baire’s theorem seems intuitive to me, but “first/second category” has to be the worst.

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 9th 2011

    Right: categorists care about short linear maps to ensure that the category of Banach spaces has some “expected” completeness and cocompleteness properties, is monadic over suitable categories, and so on. Here isomorphisms preserve structure qua normed linear spaces. The coproduct is then given by taking the L 1L^1 completion of the direct sum (to put it roughly), and the product is given by taking the corresponding L L^\infty completion of the direct product.

    Analysts are more likely to consider general continuous linear maps, where isomorphisms preserve structure qua TVS. But here arbitrary coproducts and products do not exist in this category, if I’m not mistaken.

    I agree “usual notion of isomorphism” is probably an inapt expression. Depends on the community.

    • CommentRowNumber5.
    • CommentAuthorDavidRoberts
    • CommentTimeNov 9th 2011

    And they also care about unbounded linear maps and so on, which makes things even more interesting.

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTimeNov 10th 2011

    Analysts are more likely to consider general continuous linear maps

    And thus the categorist would argue that their category is actually the category of Banachable spaces rather than Banach spaces. (-:

    • CommentRowNumber7.
    • CommentAuthorTobyBartels
    • CommentTimeNov 10th 2011

    And yet sometimes analysts do care about the actual norm. It’s not so much that analysts use the category of Banachable TVSes as that analysts use the term ‘isomorphism’ in a specialised way that simply doesn’t mean ‹isomorphism in the relevant category›.

    • CommentRowNumber8.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 10th 2011

    I don’t think I agree with #7, although I’m still open to persuasion. As likely as caring about the actual norm, they care about norms only up to equivalence (i.e., up to producing the same uniformity).

    I find “the relevant category” a somewhat loaded expression. What may be relevant for many analysts is Banach spaces as constituting a full subcategory of TVS. If that is the relevant category, then ’isomorphism’ is being applied correctly in the sense of category theory.

    • CommentRowNumber9.
    • CommentAuthorTobyBartels
    • CommentTimeNov 10th 2011

    Sometimes functional analysts care about Banach spaces up to (surjective linear) isometry, while sometimes they only care about them up to (what they call) isomorphism. There’s only one category that allows you to think about both of these things at once, and that’s the isometric category. If one were to adopt language following strict category-theoretic principles, then (by working the isometric category) one could say ‘isomorphism’ to mean a surjective linear isometry and ‘isomorphism between underlying TVSes’ to mean (what functional analysts call) isomorphism. But the isomorphic category has no way to refer to isometries.

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 10th 2011
    • (edited Nov 10th 2011)

    Sorry, I’m not following. What’s the isometric category? Do you mean the category of Banach spaces and short linear maps?

    Edit: assuming that’s what you meant, I get the feeling that we’re probably not really in disagreement. But let me say you can’t refer to an isomorphism of norm 2 in the isometric category alone; you need that underlying functor U:Banach shortTVSU: Banach_{short} \to TVS as well, which is faithful but not full.

    In the end, there is no confusion. Analysts will work some of the time in Banach shortBanach_{short}, and sometimes in the full subcategory of TVS whose objects are what you called “Banachable” spaces. Isomorphisms in either category will give a notion to suit the purpose at hand.

    • CommentRowNumber11.
    • CommentAuthorYemon Choi
    • CommentTimeNov 10th 2011
    I seem to remember half-heartedly suggesting, or at least using, the notation Bang (the geometric category of Banach spaces, i.e. morphisms are "short") and the notation Bant (the topological category of Banach spaces).

    I don't know if the former category is the one I "should" be using, but the latter one is the one I actually *do* use more often than not...

    > As likely as caring about the actual norm, they care about norms only up to equivalence (i.e., up to producing the same uniformity).

    That seems close to my views/experience, although Mark would be better placed than me to assess Todd's interpretation.
    • CommentRowNumber12.
    • CommentAuthorYemon Choi
    • CommentTimeNov 10th 2011

    We are probably all in agreement that both categories are needed, right? Doing operator theory in Bang looks to me a bit Procrustean. Maybe Bant should be viewed as a category with some kind of enrichment that detects the extra structure in the “Hom-sets”?

    As it happens, if one tries to do homological algebra with Banach(able) algebras and Banach(able) modules – I know, why would you want to do that? – then one sometimes wants to perform “standard constructions with norm control”. So I really do want to say that certain nn-cocycles cobound with norm at most 3 n3^n, and it has never been clear to me whether the arguments would more naturally take place in Bang or in Bant.

    • CommentRowNumber13.
    • CommentAuthorTim_Porter
    • CommentTimeNov 10th 2011
    • (edited Nov 10th 2011)

    @13 Yemon, I, like you, can think of several good reasons for doing homological algebra with banach algebras!!!

    • CommentRowNumber14.
    • CommentAuthorAndrew Stacey
    • CommentTimeNov 10th 2011

    In the interests of completeness, and to see if we can get more people involved, I just asked about this on MO:

    http://mathoverflow.net/questions/80567/what-is-an-isomorphism-of-banach-spaces

    My motivation is just so that we don’t fix up this article only to find someone else complaining that we’ve left out their favourite type of morphism!

    • CommentRowNumber15.
    • CommentAuthorAndrew Stacey
    • CommentTimeNov 10th 2011

    With regard to Paul Siegel’s comment at http://mathoverflow.net/questions/80567/what-is-an-isomorphism-of-banach-spaces, that seems like a common enough concept: of having a subcategory of “special” morphisms. Something like a dagger category but not as strong. Is there such a thing?

    • CommentRowNumber16.
    • CommentAuthorMike Shulman
    • CommentTimeNov 10th 2011
    • (edited Nov 10th 2011)

    having a subcategory of “special” morphisms… Is there such a thing?

    Indeed! It doesn’t have a standard name yet, but I’ve been starting to see it all over the place. I first noticed it in the 2-categorical version, where Steve Lack and I called it an F-category. Here are some other important examples:

    • The Kleisli category of a monad, where the “special” morphisms are those arising from morphisms in the base category. This matters in computer science where the monad builds in “side effects” and the “special morphisms” are those with no side effects. There’s a paper by John Power, I think, where he makes the same observation in this context that we did for F-categories: they can be regarded as categories enriched over the category of injections.

    • The category of sets in a material set theory, where the “special” morphisms are literal subset inclusions (not arbitrary injections). A generalized version of this called a “directed structural system of inclusions” plays an important role in this paper

    • More generally, any category of subobjects (or quotients) of some given object, with morphisms that don’t necessarily respect the inclusions, where the “special” morphisms are those that do. (The previous example is where the given object is the class of all sets.) For instance, Peter May’s Galois theory example of an “evil” category on the categories mailing list: subfields of an extension field, but with field homomorphisms that don’t necessarily respect the inclusion.

    I’ve thought of calling these “\mathcal{M}-categories” since they can be identified with categories enriched over the category of monomorphisms in Set (a category which one might call \mathcal{M}), but if anyone has a better name, I’m all ears.

    • CommentRowNumber17.
    • CommentAuthorMark Meckes
    • CommentTimeNov 11th 2011

    Re Toby #7: It’s worth keeping in mind that, as with many (most?) other uses of “isomorphism” in mathematics, analysts’ usual usage of the term predates the invention of category theory.

    Re Toby #9: I’m not convinced that there’s any utility to having a single category which encompasses those two different notions, but I’m open to persuasion.

    • CommentRowNumber18.
    • CommentAuthorFinnLawler
    • CommentTimeNov 11th 2011

    Mike wrote:

    There's a paper by John Power, I think, where he makes the same observation in this context that we did for F-categories: they can be regarded as categories enriched over the category of injections.

    It's Premonoidal categories as categories with algebraic structure, Theoretical Comp. Sci. 278, 2002.

    • CommentRowNumber19.
    • CommentAuthorTobyBartels
    • CommentTimeNov 12th 2011

    Of course one doesn’t have to spend one’s whole life in a single category, but for me, having a category (or at least a groupoid, or most generally an ω\omega-groupoid) in mind is necessary to fully define what a Banach space (or whatever concept you’re thinking about) is. This is why people are inspired to say ‘Banachable [topological vector] space’ for things that are considered the same when a mere linear homomorphism exists between them; when you see the definition of Banach space as an abstract structure, you wouldn’t naturally think that two such things are the same unless there was a surjective linear isometry between them: a bijection which (and whose inverse which) preserves all of the structure in that definition.

    I keep thinking of group theory, and how sometimes group theorists are interested in arbitrary functions between groups. They even have a term for an isomorphism class in the category of groups and functions (‘order’). Yet nobody has any doubt about what ‘the category of groups’ means or when two groups should be considered the same (although they may be equivalent for certain purposes without being the same, as having the same order, being Morita-equivalent, etc).

    Of course, we should not steal the term ‘isomorphism’. The functional analysts have decided what that means in the context of Banach spaces, and we shouldn’t denigrate it just because the same term was adopted in category theory for something else.

    • CommentRowNumber20.
    • CommentAuthorMike Shulman
    • CommentTimeNov 13th 2011

    I really think the \mathcal{M}-category viewpoint has a lot to recommend it. Any \mathcal{M}-category has two kinds of isomorphism, which we might in general call “tight” and “loose”, as Steve and I did for \mathcal{F}-categories. But in lots of examples these two already have specific names.

    • In a \dagger-category, like Hilbert spaces, the tight isomorphisms are called unitary.

    • In the \mathcal{M}-category of material-sets, the tight isomorphisms are called equality and the loose ones bijections.

    • In the orbit category of a group GG, whose objects are the GG-sets G/HG/H, whose loose morphisms are GG-maps, and whose tight morphisms are those that commute with the quotient maps from GG, a loose isomorphism G/HG/KG/H \cong G/K means that HH and KK are conjugate, whereas a tight one means they are equal (as subgroups of GG).

    So in the \mathcal{M}-category of Banach spaces, there’s no problem with saying that the tight isomorphisms are called isometries while the loose isomorphisms are just isomorphisms.

    • CommentRowNumber21.
    • CommentAuthorTobyBartels
    • CommentTimeNov 13th 2011

    I created M-category.

    • CommentRowNumber22.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 13th 2011

    Thanks, Toby. Always nice when an nForum discussion leads to a useful nLab entry. :-)

    • CommentRowNumber23.
    • CommentAuthorRodMcGuire
    • CommentTimeNov 13th 2011

    ummm, M-category never says what “tight” and “loose” mean.

    • CommentRowNumber24.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 13th 2011

    Rod, if x,yx, y are objects of an MM-category, then hom(x,y)hom(x, y) is by definition a pair of sets (T,L)(T, L) where TLT\subset L. An element of the first component TT is called a tight morphism from xx to yy. We can think of the hom-set of the underlying ordinary category as just LL, so every morphism of the ordinary category is loose by default, but some morphisms, the ones that land in TT, are also tight.

    • CommentRowNumber25.
    • CommentAuthorMike Shulman
    • CommentTimeNov 13th 2011

    Thanks, Toby – I hadn’t created it yet because I wasn’t sure about what to call it. But \mathcal{M}-category is certainly fine. The page looks great! I corrected the description of the \mathcal{M}-category \mathcal{M}, and modified the Idea section so it doesn’t suggest that \mathcal{F}-categories are an “incomplete” categorification of \mathcal{M}-categories (which I don’t think they are – there are just also other possible categorifications, which might have other applications).

    Re 24: Yes – but just so we don’t get confused, we should remember that the “underlying ordinary category” in the usual sense of enriched category theory actually has TT as its hom-sets.

    • CommentRowNumber26.
    • CommentAuthorTodd_Trimble
    • CommentTimeNov 13th 2011

    Oh, you’re right Mike! It’s because the monoidal unit is (1,1)(1, 1), and hom((1,1),(T,L))Thom((1, 1), (T, L)) \cong T.

    • CommentRowNumber27.
    • CommentAuthorTobyBartels
    • CommentTimeNov 14th 2011

    we should remember that the “underlying ordinary category” in the usual sense of enriched category theory actually has TT as its hom-sets.

    And this only proves that the category theorists were right about BanBan all along! (^_^)

    • CommentRowNumber28.
    • CommentAuthorTobyBartels
    • CommentTimeNov 14th 2011

    I corrected the description of the \mathcal{M}-category \mathcal{M}

    I don’t know why I could define \mathcal{M} correctly as a category but not as an \mathcal{M}-category! Thanks.

    • CommentRowNumber29.
    • CommentAuthorMike Shulman
    • CommentTimeNov 14th 2011

    Toby: In #16 I originally wrote “sets-and-injections” in describing \mathcal{M}, then corrected it. I guess for some reason that is an easy mistake to make!

    • CommentRowNumber30.
    • CommentAuthorTobyBartels
    • CommentTimeNov 14th 2011

    It’s the old issue of whether the category of foos has foos as objects or morphisms. I’m a partisan of the former position (which is the same as making the distinction between a group and its delooping), but the latter is obviously inviting sometimes!

    • CommentRowNumber31.
    • CommentAuthorDavid_Corfield
    • CommentTimeApr 17th 2021

    Added a reference

    diff, v54, current

  1. The image of linear operators between Banach spaces need not have closed image, and so the usual definition of cokernels need not yield a Banach space, as quotients by non-closed subspaces of a Banach space can be non-complete. However, the category of Banach spaces still admits coequalizers by taking the quotient by the closure of the image of fgf-g instead.

    Javier Villar

    diff, v56, current

    • CommentRowNumber33.
    • CommentAuthorUrs
    • CommentTimeMay 13th 2022

    just to highlight that the above comment refers to an edit in this paragraph.

    diff, v57, current

  2. linking to absolute convergence

    Anonymous

    diff, v60, current

    • CommentRowNumber35.
    • CommentAuthorTobyBartels
    • CommentTimeMar 11th 2024

    If the underlying set of a Banach space, treated as an object in a closed category, is the closed unit ball, then how to do we describe this as this set with extra structure. We can do this very concretely!

    diff, v65, current