Not signed in (Sign In)

Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • CommentRowNumber1.
    • CommentAuthorparthapratim
    • CommentTimeApr 18th 2018

    I could not understand the proof of the fact that if a vector space VV over a field kk is dualisable in the monoidal category Vect k\mathbf{Vect}_k then it should be finite dimensional. I understand that the image of kk under the unit is a 11-dimensional subspace, but why should this imply the finite dimensionality of VV? I would be thankful if someone could provide me the complete proof or a pointer to it.

    With my regards,

    partha

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 18th 2018

    A vector space VV is a directed colimit of the cone consisting of its finite-dimensional subspaces and inclusions between them (every element of VV is contained in a finite-dimensional subspace). If VV is dualizable, then hom(V,)V *\hom(V, -) \cong V^\ast \otimes - and hence hom(V,)\hom(V, -) is a left adjoint and thus preserves that directed colimit. But if the canonical map Γ:colim f.d.subspaceF αhom(V,F α)hom(V,V)\Gamma: colim_{f.d.\; subspace\; F_\alpha} \hom(V, F_\alpha) \to \hom(V, V) is an isomorphism, then there is some α:VF α\alpha: V \to F_\alpha that maps to id Vid_V, i.e., id V:VVid_V: V \to V factors through a finite-dimensional subspace.

  1. While we’re here, I’ve been wondering about which definition of “finite” in a topos corresponds to dualisability. Which sets have dualisable free vector spaces?

    After thinking about it for a bit I realised I didn’t even know the definition of “vector space”. I think of vector spaces as places where you can take arbitrary finite linear combinations. But this definition itself involves the word “finite”. So the generalisation to toposes isn’t immediately clear. Exactly which internal sets do we allow sums over? Clearly if you allow sums over a set then you also allow sums over its subsets (since otherwise you could define the sum over that subset by summing over the original set a function that was 00 outside the subset). But except for that I have absolutely no intuition for which the correct definition is. Is there a standard choice?

  2. (Moved this discussion from the ’Scrap paper’ category, which can only be seen when people are logged in, to a different category).

    • CommentRowNumber5.
    • CommentAuthorparthapratim
    • CommentTimeApr 19th 2018

    Thanks to Todd for his help and the answer.

    Regarding the notion of finite from Oscar: given a natural number object (N,+,0)(N, +, 0) (= initial object in the category of internal algebras with a single nullary and a single unary and their homomorphisms) would calling an object AA finite if there exists a monomorphism from AA to NN but no isomorphism from AA to NN suffice?

    • CommentRowNumber6.
    • CommentAuthorDavidRoberts
    • CommentTimeApr 19th 2018

    I guess this is about free vector spaces on finite objects in toposes? If nn is a fibre of the generic natural number 𝒩\mathcal{N} \to \mathbb{N}, then I believe the free vector space over the field kk on nn is just k nk^n. Otherwise there’s a bit more work (especially if, for instance, the topos doesn’t have an NNO - at which point kk would probably be a finite field, whatever that means)

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTimeApr 19th 2018

    The standard biased definition of vector space, in which you can add two elements and zero elements, is the correct one constructively as well. The unbiased version that this generates allows you to sum (Bishop-)finite families of elements.

    Clearly if you allow sums over a set then you also allow sums over its subsets (since otherwise you could define the sum over that subset by summing over the original set a function that was 0 outside the subset).

    Actually that only works if the subset is decidable, otherwise you can’t define a function by cases to be “as given inside and 0 outside”. And a decidable subset of a finite set is still finite, so this doesn’t give you anything extra.

    I’m not actually sure offhand what sort of “finite” sets have dualizable free vector spaces constructively; it might be just finite ones, but sometimes this sort of “compactness” criterion gives you finitely indexed (K-finite) sets instead.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeApr 19th 2018

    The tricky thing is that constructively, not every vector space is free (has a basis). In particular, a finitely-generated subspace of a vector space is not necessarily finite-dimensional in the sense of having a finite basis. If we apply the argument in #2 to the cone of finitely-generated subspaces of VV, we conclude that every dualizable vector space is finitely generated. If we apply it instead to the cone of finite-dimensional vector spaces mapping to VV (not necessarily injectively), we conclude that VV is a retract of a finite-dimensional vector space, i.e. it is “finitely generated and projective”. Note that the latter is the correct abstract characterization of dualizable modules over general rings; constructively, vector spaces are often less different from modules over non-fields than they are classically.

    I don’t offhand see a reason that every finitely-generated projective vector space would be free. And I also don’t see any reason for the free vector space on even a finitely-indexed set to be projective (finitely indexed sets are not necessarily projective). So I suspect that only finite sets can be shown constructively to have dualizable free vector spaces, though I don’t at the moment know how to prove it. If a finitely-generated projective vector space is free, is it necessarily finite-dimensional (free on a finite set)?

  3. The standard biased definition of vector space, in which you can add two elements and zero elements, is the correct one constructively as well. The unbiased version that this generates allows you to sum (Bishop-)finite families of elements.

    That’s nice. Taking the strictest possible notion of “sets you are allowed to sum over” gives you the most general possible definition of vector space. I suppose this means that the construction of the free vector space on the set AA is given by the formal sums of B-finite subsets of AA.

    Actually that only works if the subset is decidable, otherwise you can’t define a function by cases to be “as given inside and 0 outside”. And a decidable subset of a finite set is still finite, so this doesn’t give you anything extra.

    Ah yes, that makes sense. I worried a bit about that before I wrote it, but I managed to convince myself that it’s possible to define the complement (just pullback along the “false” map 121\to 2 rather than the “true” one). But the problem isn’t that you can’t define the complement; it’s that you can’t prove every element is either in the original set or the complement.

    I don’t offhand see a reason that every finitely-generated projective vector space would be free.

    Yeah, that sounds false. My understanding is that for any topological space XX we can recover the category of vector bundles over XX as the category of vector spaces over the Dedekind reals inside the topos of sheaves on XX. And a vector bundle is dualisable iff it’s locally finite-dimensional. So consider the “Möbius-strip” 1-dimensional real vector bundle over the circle. It’s not free, but it’s dualisable.

    In fact I think this gives us an example of a set that isn’t B-finite, but whose free module is dualisable. Again working in the topos of sheaves of S 1S^1, let AA be the sheaf of sections of the nontrivial double-cover of S 1S^1. Then AA is K-finite but not B-finite. But the free module on AA is dualisable, because as a bundle it’s locally 2-dimensional.

    (Actually I don’t quite understand this. The nLab says at finite object that

    In the category of sheaves Sh(X)\mathrm{Sh}(X) over a topological space, the decidable K-finite objects are those that are “locally finite;” i.e. there is an open cover of XX such that over each open in the cover, the sheaf is a locally constant function to NN. These are essentially the same as covering spaces of XX with finite fibres.

    but I don’t see how this works in the case of AA. The double-cover of S 1S^1 is definitely a covering space with finite fibres, but I don’t see which nn to take to get a surjection nAn\to A.)

    • CommentRowNumber10.
    • CommentAuthorTodd_Trimble
    • CommentTimeApr 19th 2018

    The Möbius strip is not a free module in the external sense of there being a global section as basis element, but might it be different internally, that at the level of stalks it is free? I seem to remember a statement somewhere that Kaplansky’s theorem, that projective modules over local rings are free, carries over in some degree to some toposes like sheaves over a compact space; see this excerpt from Johnstone’s Topos Theory.

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeApr 19th 2018

    The double-cover of S 1S^1 is definitely a covering space with finite fibres, but I don’t see which nn to take to get a surjection nAn\to A.

    The cover of S 1S^1 by two overlapping intervals has the property that over it the double cover of S 1S^1 is locally (indeed globally) constant at 22, which is what the characterization is asking for. I’m not sure what nn you are talking about.

    The Möbius strip is not a free module in the external sense of there being a global section as basis element, but might it be different internally, that at the level of stalks it is free?

    I think it depends on whether we formulate “VV is free” with a truncated \exists or with an untruncated Σ\Sigma. If “VV is free” means “there exists a set SS and a function f:SVf:S\to V such that every vVv\in V can be expressed uniquely as v=r 1f(s 1)++r nf(s n)v = r_1 f(s_1) + \cdots + r_n f(s_n)” and “there exists” is truncated, then it allows passing to a cover, and the Mobius strip is free over the cover of S 1S^1 by two overlapping intervals. But if it’s not truncated then we have to give a global SS and f:SVf:S\to V, and I don’t think that’s possible.

    • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeApr 19th 2018

    Similarly, the double cover DD of S 1S^1 is B-finite in the truncated internal sense of “there (merely) exists a natural number nn and an isomorphism [n]D[n]\cong D”.

Add your comments
  • Please log in or leave your comment as a "guest post". If commenting as a "guest", please include your name in the message as a courtesy. Note: only certain categories allow guest posts.
  • To produce a hyperlink to an nLab entry, simply put double square brackets around its name, e.g. [[category]]. To use (La)TeX mathematics in your post, make sure Markdown+Itex is selected below and put your mathematics between dollar signs as usual. Only a subset of the usual TeX math commands are accepted: see here for a list.

  • (Help)