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Suppose $\mathbb{T}$ is a geometric theory, $\mathcal{E}$ is a topos, and $M$ is a model of $\mathbb{T}$ in $\mathcal{E}$. Is there any sort of elementary condition on $M$ and $\mathcal{E}$ (or, even better, on the geometric morphism $\mathcal{E}\to \mathbf{Set}$) which would allow us to recognize $\mathcal{E}$ as the classifying topos of $\mathbb{T}$ and $M$ as the generic $\mathbb{T}$-model therein?

I feel like this is a long shot, but I thought I would ask anyway.

Edit: Of course, such a condition could not be expressed in the internal logic of $\mathcal{E}$ (even including non-geometric logic), since then it would be preserved in all slices $\mathcal{E}/X$. This is one reason I feel it's a long shot; but the example of principal bundles mentioned in the comments suggests that it's not an entirely unreasonable question.

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  • $\begingroup$ Does anyone even know such a condition for the special case when this question is equivalent to asking "if we're given a principal G-bundle on a space X, can we tell if X = BG and the bundle is the canonical one?" ? $\endgroup$ – Dylan Wilson Dec 21 '11 at 5:30
  • $\begingroup$ (that question wasn't rhetorical- I'd be interested to know if there's an answer!) $\endgroup$ – Dylan Wilson Dec 21 '11 at 5:31
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    $\begingroup$ Dylan, if the total space of the bundle is weakly contractible. See for instance theorem 7.4 here: www-math.mit.edu/~mbehrens/18.906/prin.pdf $\endgroup$ – Urs Schreiber Dec 21 '11 at 9:04
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    $\begingroup$ Silly me! Now I somehow have more faith in Mike's question being answered... $\endgroup$ – Dylan Wilson Dec 21 '11 at 9:15
  • $\begingroup$ Too hard question for me, but I see that Olivia Caramello has worked so much (and still working about) classyfing topoi and theories. $\endgroup$ – Buschi Sergio Dec 21 '11 at 19:36
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This is several years late, but it may be helpful nonetheless.

As alluded to by Buschi, Olivia has given an explicit answer to this in Theorem 2.1.29 of her monograph Theories, Sites and Toposes:

Let $\mathbb{T}$ be a geometric theory, $\mathcal{E}$ a Grothendieck topos and $M$ a model of $\mathbb{T}$ in $\mathcal{E}$. Then $\mathcal{E}$ is a classifying topos of $\mathbb{T}$ and $M$ is a universal model (i.e. generic model) of $\mathbb{T}$ iff the following conditions are satisfied:

  1. The family $F$ of objects which can be built from the interpretations in $M$ of the sorts, function symbols and relation symbols over the signature of $\mathbb{T}$ by using geometric logic constructions (i.e. the objects given by the domains of the interpretations in $M$ of geometric formulae over the signature of $\mathbb{T}$) is separating for $\mathcal{E}$.
  2. The model $M$ is conservative for $\mathbb{T}$, that is for any geometric sequent $\sigma$ over the signature of $\mathbb{T}$, $\sigma$ is valid in $M$ if and only if it is provable in $\mathbb{T}$.
  3. Any arrow $k$ in $\mathcal{E}$ between objects $A$ and $B$ in the family $F$ of condition (1) is definable; that is, if $A$ (resp. $B$) is equal to the interpretation of a geometric formula $\phi(\vec{x})$ (resp. $\psi(\vec{y})$) over the signature of $\mathbb{T}$, there exists a $\mathbb{T}$-provably functional formula $\theta$ from
    $\phi(\vec{x})$ to $\psi(\vec{y})$ such that the interpretation of $\theta$ in $M$ is equal to the graph of $k$.

I'm not sure if you were looking necessary and sufficient conditions on $M$ and $\mathcal{E}$, or just merely sufficient conditions, but since this Theorem gives an 'iff' result, one might try and prove the sufficiency of certain (perhaps more intuitive) critiera on $M$ and $\mathcal{E}$ by checking against the conditions listed in this theorem, i.e. by proving results of the flavour: 'If $M$ and $\mathcal{E}$ satisfy condition $X$, then they satisfy the 3 conditions of this theorem.'

Extending this thought, I am curious to see how these conditions relate to the special case mentioned in Dylan's comment. In particular, how does weak contractibility relate to the conditions spelt out by Olivia? This is not obvious to me, but I haven't taken the time to properly work through the details.

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  • $\begingroup$ I suppose that's probably as good an answer as one could hope for, even though it looks like it might be rather hard to check in practice. $\endgroup$ – Mike Shulman Jul 6 at 5:52

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