MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is a good notation for the 'set' (or stack if you insist) of all principal G bundles over 'all' spaces for given G? BG is way over used. How about Bun(G)?

share|cite|improve this question
Within the context of stacks in algebraic geometry, $BG$ (or $pt/G$) is absolutely the notation that would be used. I suppose you don't like this because it is also the notation for the classifying space of $G$? From my perspective, I don't particularly like the notation $Bun(G)$, only because $Bun_G(X)$ is typically used to denote the moduli of $G$-bundles on the space $X$. But maybe others disagree. I'm struggling to think of a good alternative though. – Mike Skirvin Mar 6 '12 at 22:59
I certainly have seen $\operatorname{Bun}(G)$, but I agree with @Mike. Since from a stacky perspective $\operatorname{Bun}_G(X)$ is the space of maps $X \to $ this space that Jim is after, it's not bad to call the target space itself by $\operatorname{Bun}_G$. – Theo Johnson-Freyd Mar 7 '12 at 1:32
I have made a small change to the formatting. The issue is that Markdown considers the backwards single quote ` to be a special character. – Theo Johnson-Freyd Mar 7 '12 at 1:33
May I suggest that my learned colleagues' comments constitute genuine answers to the question asked, and might therefore be submitted as answers? – Tom Leinster Mar 7 '12 at 1:39
up vote 6 down vote accepted

In my comment to the question, I may have misinterpreted Jim's intent. Upon a second reading, I think Jim would like to consider the collection of all pairs consisting of a space $X$ and a $G$-bundle over $X$. If this is the correct interpretation, then I think the correct notation for this category is $\text{Spaces}_{/\mathrm{B}G}$. The meaning of this notation is: $\mathrm{B}G = \operatorname{Bun}_G$ is the stack of $G$-bundles, defined so that a map $X \to \mathrm{B}G$ is a $G$-bundle over $X$. The category $\text{Spaces}$ is whatever category of spaces you want to consider (topological spaces, manifolds, schemes, ...). The category of pairs $(X,$ $G$-bundle over $X)$ is the category of pairs $(X,f: X \to \mathrm{B}G)$, which is precisely the "comma" category of "spaces over $\mathrm{B}G$", and that's what's notated by the subscript.

share|cite|improve this answer
I can live with Spaces/BG since it is not previously used and will encourage the reader to understand what it denotes. – Jim Stasheff Mar 7 '12 at 13:15

In algebraic geometry, $BG$ only ever refers to a stack, viewed as a fibered category over the category of schemes, or affine schemes, depending on your conventions. The objects are principal $G$-bundles $P \to X$ of schemes (or if $G$ is a group sheaf, it is a map of sheaves). The morphisms are $G$-equivariant fiber squares.

More to the point, there is no alternative object in the algebro-geometric universe that takes the name $BG$, so there is no source of confusion there.

The name $\operatorname{Bun}_G$ is usually reserved for the enriched Hom-stack construction $\underline{\operatorname{Hom}}(-, BG)$. In other words, $\operatorname{Bun}_G(X)(T) = BG(X \times T)$.

share|cite|improve this answer
Perhaps no confusion within alg-geom, but we are not all alg-geometers. So your BG is a category and alg-geom has no use for a classifying space for G-bundles? or it has some other symbol? – Jim Stasheff Mar 7 '12 at 13:12
S. Carnahan's $BG$ is the classifying space in the algebro-geometric context. You might (I do) find it more intuitive to turn the category fibred in groupoids into a sheaf of groupoids/homotopy 1-types, using the canonical cleavage' construction. Then we are talking about $BG$ as a sheaf of homotopy types on affine schemes. For a scheme $X$, $\operatorname{Bun}_G(X)=\underline{\operatorname{Hom}}(X, BG)$ is another sheaf of homotopy types, and the global sections of $\pi_{0}\operatorname{Bun}_G(X)$ are the isomorphism classes of principal $G$-bundles over $X$, just like in topology. – Chris Brav Mar 7 '12 at 14:58
That does clarify the relation, but notice how many words it required. – Jim Stasheff Mar 8 '12 at 14:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.