Not signed in

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

Site Tag Cloud

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

• 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 $X$ with action $G$ is a morphism $X//G \to Vect$ from the action groupoid of $G$ acting on $X$ 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 $X//G$ and $Vect$ 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 $G$ abelian, we have $G//_{Ad} G \simeq \coprod_{g \in G} *//G = \coprod_{g \in G} \mathbf{B}G$.

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 $G//G \to Vect$.

• 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 $G = Z_2$:

Let $F: Z_2 // Z_2 \rightarrow Vect$ be the functor which sends $0\in Obj(Z_2)$ to the v.s. $V_1$ and $1\in Obj(Z_2)$ to the v.s. $V_2$ along with $Aut(0)$ to $End(V_1)$ and $Aut(1)$ to $End(V_2)$. Let $G: Z_2 // Z_2 \rightarrow Vect$ be the functor which sends $0\in Obj(Z_2)$ to the v.s. $W_1$ and $1\in Obj(Z_2)$ 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. $\alpha: F \rightarrow G$ takes $0$ and gives a morph. $F(0)\rightarrow G(0)$, i.e., it takes $0$ and gives an element $\alpha_0 \in Hom(V_1,W_1)$ such that for any $f \in Aut(0)$ we have $\alpha_0 \circ F(f) = G(f) \circ \alpha_0$ (the commutative diagram in the link you gave above for nat. trans.). And similarly for $1 \in Obj(Z_2)$.

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 $W_2$? Otherwise, we have no idea how to classify the set of $Z_2$-equivariant vector bundles over $Z_2$, 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 $F$ is a direct sum of $G$ with some other equivariant vector bundle, and $\alpha$ the projection.

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

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

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

• CommentRowNumber10.
• CommentAuthorrdkw10
• CommentTimeApr 12th 2010

So, are you saying that the category of $G$-equivariant bundles over $G$, which is just $Func(\star//G, Vect)$, is isomorphic to the cat. of linear reps. of $G$, $Rep(G)$, whose objects are reps of $G$ 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 $G$ on $X$ is trivial, as in the case of the adjoint action of an abelian $G$ on itself for $X = G$, then an equivariant vector bundle on $X$ is clearly just a vector bundle with a representation of $G$, or equivalently a bundle of such representations. If then $X$ is the point, this is just a single representation.

If you know what an action groupoid $X//G$ 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 $Func(X//G, Vect)$ 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 $G$ on $X$ is trivial, as in the case of the adjoint action of an abelian $G$ on itself for $X=G$, then an equivariant vector bundle on $X$ is >clearly just a vector bundle with a representation of $G$, or equivalently a bundle of such representations. If then $X$ is the point, this is just a single >representation.

So, over the points of $X$, we have a representation of $G$. And so, if $X=Z_2$, 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 $Vect_{Z_2}(Z_2)$ 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 $Z_2$, 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 $X$, we have a representation of $G$

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

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

Yes.

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

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeApr 12th 2010

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

• CommentRowNumber17.
• CommentAuthorrdkw10
• CommentTimeApr 12th 2010

In a paper by Freed, he says that $Vect(Z_2)$ is the category of vector bundles over $Z_2$, and we are wondering if there is anyway to simplify this as we did before for $Vect_{Z_2}(Z_2)$ 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 $Vect(\mathbb{Z}_2)$ is the category of vector bundles over the set of two elements. Right?

If so then, as I said, we have $Vect(\mathbb{Z}_2) \simeq Vect \times Vect$: 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!