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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2019
   • (edited Mar 15th 2019)
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   Author: Urs
   Format: MarkdownItexFelix Wellen has been thinking about the following, maybe somebody has further thoughts: Currently an open problem is to get an intrinsic construction of [[differential cohesion]], hence of the [[infinitesimal shape modality]] and/or the _smooth_ real line in some analogy to how Mike's real cohesion gets the [[shape modality]] from the topological real line constructed internally, in [arXiv:1509.07584](https://arxiv.org/abs/1509.07584). But if we take the base topos of the cohesion to be the Boolean $\not \not$-subtopos, so that $$ \sharp \; \simeq \;\not \not $$ (at least on 0-types?) then we may define infinitesimal disks to be the types $\mathbb{D}$ with $$ \sharp \mathbb{D} \simeq \ast $$ (as in "[[logical topology]]"). Then one could try to define $\Im$ as the localization at these $\mathbb{D}$. But also, the infinitesimal disks should know about the smooth real line. For instance the [[Kock-Lawvere axioms]] gives that for the first order 1d disk $\mathbb{D}^1(1)$ we have $[\mathbb{D}^1(1), \mathbb{D}^1(1)]^{\ast/} \simeq \mathbb{R} $, with the correct smooth real line on the right. So the collection of the $\mathbb{D}$-s, as above, should be enough to know the smooth $\mathbb{R}^1$, but I am not sure. If so, one could then declare the shape modality to be localization at this smooth $\mathbb{R}^1$. Maybe if all fits together, this construction could exhibit canonical [[differential cohesion]].

   Felix Wellen has been thinking about the following, maybe somebody has further thoughts:

   Currently an open problem is to get an intrinsic construction of differential cohesion, hence of the infinitesimal shape modality and/or the smooth real line in some analogy to how Mike’s real cohesion gets the shape modality from the topological real line constructed internally, in arXiv:1509.07584.

   But if we take the base topos of the cohesion to be the Boolean $\not \not$-subtopos, so that

   $$\sharp \; \simeq \;\not \not$$

   (at least on 0-types?) then we may define infinitesimal disks to be the types $\mathbb{D}$ with

   $$\sharp \mathbb{D} \simeq \ast$$

   (as in “logical topology”).

   Then one could try to define $\Im$ as the localization at these $\mathbb{D}$.

   But also, the infinitesimal disks should know about the smooth real line. For instance the Kock-Lawvere axioms gives that for the first order 1d disk $\mathbb{D}^1(1)$ we have $[\mathbb{D}^1(1), \mathbb{D}^1(1)]^{\ast/} \simeq \mathbb{R}$, with the correct smooth real line on the right.

   So the collection of the $\mathbb{D}$-s, as above, should be enough to know the smooth $\mathbb{R}^1$, but I am not sure. If so, one could then declare the shape modality to be localization at this smooth $\mathbb{R}^1$.

   Maybe if all fits together, this construction could exhibit canonical differential cohesion.

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMar 15th 2019
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   Author: Mike Shulman
   Format: MarkdownItexInteresting thought. Of course there are obstacles to overcome, e.g. it's not clear that there are only a small number of such $\mathbb{D}$, or how to identify one of them to call $\mathbb{D}^1(1)$.

   Interesting thought. Of course there are obstacles to overcome, e.g. it’s not clear that there are only a small number of such $\mathbb{D}$, or how to identify one of them to call $\mathbb{D}^1(1)$.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2019
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   Author: Urs
   Format: MarkdownItexSo Felix had this thought (if I may share this here) that the subobject of $Prop$ which is classified by $Prop \overset{\not \not}{\longrightarrow} Prop$ should be something like the universal formal disk, and that $\mathbb{R}$ should be the commutative part of its endomorphism ring. While I appreciate where he is headed, I don't see through this yet.

   So Felix had this thought (if I may share this here) that the subobject of $Prop$ which is classified by $Prop \overset{\not \not}{\longrightarrow} Prop$ should be something like the universal formal disk, and that $\mathbb{R}$ should be the commutative part of its endomorphism ring. While I appreciate where he is headed, I don’t see through this yet.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeMar 17th 2019
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   Author: Mike Shulman
   Format: MarkdownItexBTW, the equation $\sharp = \neg\neg$ only holds for (-1)-types (as it must, since $\neg\neg X$ is always a (-1)-type while $\sharp$ preserves $n$-types for all $n$). Also, I would expect that even in continuous $\infty$-groupoids, where there are no infinitesimals as such, there are still lots of (non-concrete) types $X$ with $\sharp X = \ast$.

   BTW, the equation $\sharp = \neg\neg$ only holds for (-1)-types (as it must, since $\neg\neg X$ is always a (-1)-type while $\sharp$ preserves $n$-types for all $n$). Also, I would expect that even in continuous $\infty$-groupoids, where there are no infinitesimals as such, there are still lots of (non-concrete) types $X$ with $\sharp X = \ast$.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMar 17th 2019
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   Author: Urs
   Format: MarkdownItex> BTW, the equation $\sharp = \neg\neg$ only holds for (-1)-types (as it must, since $\neg\neg X$ is always a (-1)-type while $\sharp$ preserves $n$-types for all $n$). Thanks, of course. I was misled by thinking about subtoposes labeled $J_\sharp = J_{\not \not}$, but of course that just means that $\sharp$ and $\not \not$ give the same Lawvere-Tierney operator. Still, I suppose it means we can _define_ $\sharp$ constructively in terms of $\not \not$, right? And all $\sharp$ corresponding to Boolean subtoposes should arise this way, I suppose. > I would expect that even in continuous $\infty$-groupoids, where there are no infinitesimals as such, there are still lots of (non-concrete) types $X$ with $\sharp X = \ast$. Hm, right. Maybe. Hm...

   BTW, the equation $\sharp = \neg\neg$ only holds for (-1)-types (as it must, since $\neg\neg X$ is always a (-1)-type while $\sharp$ preserves $n$-types for all $n$).

   Thanks, of course. I was misled by thinking about subtoposes labeled $J_\sharp = J_{\not \not}$, but of course that just means that $\sharp$ and $\not \not$ give the same Lawvere-Tierney operator.

   Still, I suppose it means we can define $\sharp$ constructively in terms of $\not \not$, right? And all $\sharp$ corresponding to Boolean subtoposes should arise this way, I suppose.

   I would expect that even in continuous $\infty$-groupoids, where there are no infinitesimals as such, there are still lots of (non-concrete) types $X$ with $\sharp X = \ast$.

   Hm, right. Maybe. Hm…

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeMar 17th 2019
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   Author: Mike Shulman
   Format: MarkdownItex> I suppose it means we can _define_ $\sharp$ constructively in terms of $\not \not$, right? Well, modulo hypercompletion. Assuming propositional resizing, we can nullify all the $\neg\neg$-closed propositions to yield a topological subtopos of double-negation sheaves, and $\sharp$ then coincides with this on all $n$-types for finite $n$ (8.13 in [BFP](https://arxiv.org/abs/1509.07584)). It's not immediately obvious that they coincide on untruncated types. But probably they do in the examples given that $\infty Gpd$ is hypercomplete? I haven't thought about this carefully. > And all $\sharp$ corresponding to Boolean subtoposes should arise this way, I suppose. Not all Boolean subtoposes are the double-negation sheaves, even in 1-topos theory. As a trivial case, the trivial subtopos is also Boolean.

   I suppose it means we can define $\sharp$ constructively in terms of $\not \not$, right?

   Well, modulo hypercompletion. Assuming propositional resizing, we can nullify all the $\neg\neg$-closed propositions to yield a topological subtopos of double-negation sheaves, and $\sharp$ then coincides with this on all $n$-types for finite $n$ (8.13 in BFP). It’s not immediately obvious that they coincide on untruncated types. But probably they do in the examples given that $\infty Gpd$ is hypercomplete? I haven’t thought about this carefully.

   And all $\sharp$ corresponding to Boolean subtoposes should arise this way, I suppose.

   Not all Boolean subtoposes are the double-negation sheaves, even in 1-topos theory. As a trivial case, the trivial subtopos is also Boolean.