Not signed in (Sign In)

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

Site Tag Cloud

2-category 2-category-theory abelian-categories adjoint algebra algebraic algebraic-geometry algebraic-topology analysis analytic-geometry arithmetic arithmetic-geometry book bundles calculus categorical categories category category-theory chern-weil-theory cohesion cohesive-homotopy-type-theory cohomology colimits combinatorics complex complex-geometry computable-mathematics computer-science constructive cosmology deformation-theory descent diagrams differential differential-cohomology differential-equations differential-geometry digraphs duality elliptic-cohomology enriched fibration foundation foundations functional-analysis functor gauge-theory gebra geometric-quantization geometry graph graphs gravity grothendieck group group-theory harmonic-analysis higher higher-algebra higher-category-theory higher-differential-geometry higher-geometry higher-lie-theory higher-topos-theory homological homological-algebra homotopy homotopy-theory homotopy-type-theory index-theory integration integration-theory k-theory lie-theory limit limits linear linear-algebra locale localization logic mathematics measure-theory modal modal-logic model model-category-theory monad monads monoidal monoidal-category-theory morphism motives motivic-cohomology nlab noncommutative noncommutative-geometry number-theory of operads operator operator-algebra order-theory pages pasting philosophy physics pro-object probability probability-theory quantization quantum quantum-field quantum-field-theory quantum-mechanics quantum-physics quantum-theory question representation representation-theory riemannian-geometry scheme schemes set set-theory sheaf simplicial space spin-geometry stable-homotopy-theory stack string string-theory subobject superalgebra supergeometry svg symplectic-geometry synthetic-differential-geometry terminology theory topology topos topos-theory tqft type type-theory universal variational-calculus

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.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010
    Hello everyone,

    A friend and myself have been (slowly) making our way through some papers by Freed discussing extended Chern-Simons theory in 3-dimensions. In one of his papers, he shows that the 2-Hilbert space associated to the circle (by the 3-dim theory) is equal to the category of equivariant vector bundles over the gauge group (which we assume is finite) G, E(S^1) = Vect_G(G). Now, he asks for the reader to explicitly construct E(S^1) for an abelian group. Can anyone give us some hints/explain how to do this? Thanks.
    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010

    An equivariant vector bundle on XX with action GG is a morphism X//GVectX//G \to Vect from the action groupoid of GG acting on XX to Vect.

    A morphism between two is hence a natural transformation between the two.

    A very little bit on this is at nLab:equivariant bundles.

    (If you want to take the smooth/topological structure into account both X//GX//G and VectVect are to be regarded as stacks on suitable test spaces, but for extracting the part of the answer that you are interested in, you can pretty much ignore this, becuase I guess you already know what a smooth/topological action on a smooth/topological vector bundle is).

    • CommentRowNumber3.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010
    So, the action groupoid of G (for G abelian) acting on itself by conjugation, G//G, has objects given by all elements of G and for any two different objects in G//G there are no morphisms between them (since a morphism s between objects g and h is an element s in G such that s * g = h, which since G is abelian says that g = h) unless you look at the morphisms between each object. In this case, since G is abelian, every element of G gives a morphism?

    Thus, I have two categories, so how do we explicitly construct E(S^1); that is, how do we write down all the morphs from G//G to Vect?
    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010
    • (edited Apr 12th 2010)

    So, the action groupoid of G (for G abelian) acting on itself by conjugation, G//G, has objects given by all elements of G and for any two different objects in G//G there are no morphisms between them (since a morphism s between objects g and h is an element s in G such that s * g = h, which since G is abelian says that g = h) unless you look at the morphisms between each object. In this case, since G is abelian, every element of G gives a morphism?

    Yes, for GG abelian, we have G// AdG gG*//G= gGBGG//_{Ad} G \simeq \coprod_{g \in G} *//G = \coprod_{g \in G} \mathbf{B}G.

    two categories, so how do we explicitly construct E(S^1); that is, how do we write down all the morphs from G//G to Vect?

    As I said, it’s the natural transformations between the functors G//GVectG//G \to Vect.

    • CommentRowNumber5.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010
    Yes, of course! Sorry for just going blank and not reading the rest of your first response.
    • CommentRowNumber6.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010
    To ease computations a bit, we restrict to G = Z/2Z:

    Let F: Z/2Z // Z/2Z ---> Vect be the functor which sends 0 to the v.s. V_1 and 1 to the v.s. V_2 along with Aut(0) to End(V_1) and Aut(1) to End(V_2).
    Let G: Z/2Z // Z/2Z ---> Vect be the functor which sends 0 to the v.s. W_1 and 1 to the v.s. W_2 along with Aut(0) to End(W_1) and Aut(1) to End(W_2).

    Now, a nat. trans. a: F ---> G takes 0 and gives a morph. F(0) ---> G(0), i.e., it takes 0 and gives an element a_0 in Hom(V_1,W_1) such that for any f in Aut(0) we have a_0 o F(f) = G(f) o a_0 (the commutative diagram in the link you gave above for nat. trans.). And similarly for 1 in Obj(Z/2Z // Z/2Z).

    So, does this imply that F(f) = G(f)? If so, does this imply that V_1 is equal (isom.) to W_1? Similarly for V_2 and W_2?
    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010

    Please below your comment box, choose the radio bottun “Mardown + itex”. Then include all laztex in dollar signs, as usual. Then I can read this more quickly.

    • CommentRowNumber8.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010
    • (edited Apr 12th 2010)

    To ease computations a bit, we restrict to G=Z 2G = Z_2:

    Let F:Z 2//Z 2VectF: Z_2 // Z_2 \rightarrow Vect be the functor which sends 0Obj(Z 2)0\in Obj(Z_2) to the v.s. V 1V_1 and 1Obj(Z 2)1\in Obj(Z_2) to the v.s. V 2V_2 along with Aut(0)Aut(0) to End(V 1)End(V_1) and Aut(1)Aut(1) to End(V 2)End(V_2). Let G:Z 2//Z 2VectG: Z_2 // Z_2 \rightarrow Vect be the functor which sends 0Obj(Z 2)0\in Obj(Z_2) to the v.s. W 1W_1 and 1Obj(Z 2)1\in Obj(Z_2) to the v.s. W 2W_2 along with Aut(0)Aut(0) to End(W 1)End(W_1) and Aut(1)Aut(1) to End(W 2)End(W_2).

    Now, a nat. trans. α:FG\alpha: F \rightarrow G takes 00 and gives a morph. F(0)G(0)F(0)\rightarrow G(0), i.e., it takes 00 and gives an element α 0Hom(V 1,W 1)\alpha_0 \in Hom(V_1,W_1) such that for any fAut(0)f \in Aut(0) we have α 0F(f)=G(f)α 0\alpha_0 \circ F(f) = G(f) \circ \alpha_0 (the commutative diagram in the link you gave above for nat. trans.). And similarly for 1Obj(Z 2)1 \in Obj(Z_2).

    So, does this imply that F(f)=G(f)F(f) = G(f)? If so, does this imply that V 1V_1 is equal (isom.) to W 1W_1? Similarly for V 2V_2 and W 2W_2? Otherwise, we have no idea how to classify the set of Z 2Z_2-equivariant vector bundles over Z 2Z_2, other than saying that it is the cat. of all functors and nat. transformations as given above.

    • CommentRowNumber9.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010
    • (edited Apr 12th 2010)

    So, does this imply

    No. Think of the example where FF is a direct sum of GG with some other equivariant vector bundle, and α\alpha the projection.

    In fact, maybe it helps to see how we are now dealin here with nothing but representation theory. Notice that we have

    Rep(G)Func(*//G,Vect) Rep(G) \simeq Func(*//G, Vect)

    the category of linear representations of a group is equivalently the functor category of functors from *//G=BG*//G = \mathbf{B}G to VectVect.

    • CommentRowNumber10.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    So, are you saying that the category of GG-equivariant bundles over GG, which is just Func(//G,Vect)Func(\star//G, Vect), is isomorphic to the cat. of linear reps. of GG, Rep(G)Rep(G), whose objects are reps of GG and whose morphs are intertwiners?

    • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010

    No, but something close to that in the special case that you were giving me: if the action of GG on XX is trivial, as in the case of the adjoint action of an abelian GG on itself for X=GX = G, then an equivariant vector bundle on XX is clearly just a vector bundle with a representation of GG, or equivalently a bundle of such representations. If then XX is the point, this is just a single representation.

    If you know what an action groupoid X//GX//G is, what the category Vect is, what a functor and a natural transformation is, then you have enough information to unwind the definitions to your heart’s content of what the functor category Func(X//G,Vect)Func(X//G, Vect) is. try to play around a bit with it until you feel comfortable and see the light.

    • CommentRowNumber12.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    if the action of GG on XX is trivial, as in the case of the adjoint action of an abelian GG on itself for X=GX=G, then an equivariant vector bundle on XX is >clearly just a vector bundle with a representation of GG, or equivalently a bundle of such representations. If then XX is the point, this is just a single >representation.

    So, over the points of XX, we have a representation of GG. And so, if X=Z 2X=Z_2, then we have two points and hence two reps. Hence, our bundle is with two fibres (or reps) over the base space of two points. Thus, the category Vect Z 2(Z 2)Vect_{Z_2}(Z_2) is the cat. with objects different forms of the previously mentioned bundles and morphs are bundle maps which obey the extra structure on the fibres. However, since there is exactly two irreps of Z 2Z_2, does this limit the different kinds of bundles we can have; i.e., there are four possible bundles (bundle A whose fibres are both the trivial irrep, bundle B whose fibres are both the signature irrep, …)?

    Thanks you so much for your help, and hopefully after you answer this you will have exhausted all of our questions (for today at least ;) ).

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010
    • (edited Apr 12th 2010)

    So, over the points of XX, we have a representation of GG

    If the action of GG on XX is trivial, yes. Just beware that this degenerate case is typically not what one is interested in when discussing equivariant bundles.

    And so, if X=Z 2X=Z_2, then we have two points and hence two reps.

    Yes.

    Thus, the category Vect Z 2(Z 2)Vect_{Z_2}(Z_2) is the cat. with objects different forms of the previously mentioned bundles and morphs are bundle maps which obey the extra structure on the fibres.

    If you mean what I think you mean, then: yes.

    However, since there is exactly two irreps of Z 2Z_2, does this limit the different kinds of bundles we can have;

    Well, sure, it means that every such bundle is obtained from just a handful by forming direct sums of these.

    • CommentRowNumber14.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    Dear Urs,

    Thank you so much, from the both of us!!! Now, we can move along.

    • CommentRowNumber15.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    Is there any way to explicitly construct Vect(Z 2)Vect(Z_2) rather than just saying it is the category of vector bundles over Z 2Z_2; i.e., when we restrict to Z 2Z_2-equivariant vector bundles over Z 2Z_2 (with conjugate action of Z 2Z_2 on Z 2Z_2) we then require the vector bundles to have a rep of Z 2Z_2 on the fibres? Is there any such equivalent statement here, or is the category of vector bundles over Z 2Z_2 as far as we can go?

    • CommentRowNumber16.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010

    I am not sure what you are after here. Vect( 2)=Vect×VectVect(\mathbb{Z}_2) = Vect \times Vect is simply the category whose objects are pairs of vector spaces, and whose morphisms are pairs of linear maps. Similarly Vect 2( 2)Vect_{\mathbb{Z}_2}(\mathbb{Z}_2) is simply the category whose objects are pairs of 2\mathbb{Z}_2-representations (which are vector bundles with an involution).

    • CommentRowNumber17.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    In a paper by Freed, he says that Vect(Z 2)Vect(Z_2) is the category of vector bundles over Z 2Z_2, and we are wondering if there is anyway to simplify this as we did before for Vect Z 2(Z 2)Vect_{Z_2}(Z_2) in the previous comments.

    • CommentRowNumber18.
    • CommentAuthorUrs
    • CommentTimeApr 12th 2010
    • (edited Apr 12th 2010)

    Okay, but how much simpler can it get?? :-)

    Unless I am immensely misunderstanding what you are trying to tell me, the category Vect( 2)Vect(\mathbb{Z}_2) is the category of vector bundles over the set of two elements. Right?

    If so then, as I said, we have Vect( 2)Vect×VectVect(\mathbb{Z}_2) \simeq Vect \times Vect: objects are pairs of vector spaces, morphisms are pairs of linear maps.

    If you mean something else, you need to give me some more information.

    • CommentRowNumber19.
    • CommentAuthorrdkw10
    • CommentTimeApr 12th 2010

    Ok, as we thought, there is no simplification. Thank you!