Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let's say that an $(\infty,1)$-topos is Boolean if for every object $X$, the lattice $Sub(X)$ of subobjects (i.e. $(-1)$-truncated morphisms into $X$) is a Boolean algebra. I think this is equivalent to asking that the subobject classifier be an internal Boolean algebra, and hence to asking that the underlying 1-topos of 0-truncated objects is Boolean in the classical sense.

In particular, the $(\infty,1)$-topos of sheaves on a topological space $X$ is Boolean if and only if the lattice of open sets in $X$ is a Boolean algebra, i.e. every open set is also closed. Thus, Booleanness is a sort of "zero-dimensionality" condition. Lurie shows in Higher Topos Theory that other sorts of "finite-dimensionality" conditions imply that an $(\infty,1)$-topos is hypercomplete. At the moment, however, I don't see whether Booleanness implies any of these other conditions.

Thus my question is: can a Boolean $(\infty,1)$-topos fail to be hypercomplete?

share|improve this question
    
Is there a good reference for the result that the ($\infty$, 1)-topos of sheaves on $X$ is Boolean iff $X$ satisfies "open=closed"? –  Noah S Jan 31 '13 at 3:43
add comment

1 Answer

If $G$ is a profinite group, then the topos of sets with a continuous $G$-action is Boolean, but the associated $\infty$-topos is usually not hypercomplete.

share|improve this answer
    
Thanks! Can you point me to an argument for why it is not hypercomplete? –  Mike Shulman Jan 31 '13 at 3:15
    
Ben Wieland showed me an argument that $B\mathbf{Z}_p$ is not hypercomplete; a sketch of his argument (which uses the Sullivan conjecture) is reproduced as Warning 7.2.2.31 in "Higher Topos Theory". And $\mathbf{Z}_p$ is about as tame as profinite groups get (the topos even has finite cohomological dimension). –  Jacob Lurie Jan 31 '13 at 5:47
add comment

Your Answer

 
discard

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.