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1. • CommentRowNumber1.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 7th 2015
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexPutting together $$ SuperSmooth\infty Grpd \coloneqq Sh_\infty(CartSp \rtimes SuperPoint) $$ and [[complex analytic ∞-groupoid]], presumably there's nothing to stop $$ Super \mathbb{C} Analytic \infty Grpd \coloneqq Sh_\infty(CplxMfd \rtimes SuperPoint) $$ to generalise [[complex supermanifold]].

   Putting together

   and complex analytic ∞-groupoid, presumably there’s nothing to stop

   to generalise complex supermanifold.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 7th 2015
   • PermaLink
   Author: Urs
   Format: MarkdownItexYes, that should work without problem.

   Yes, that should work without problem.

3. • CommentRowNumber3.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 13th 2015
   • (edited May 13th 2015)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexWould we still say that shape is $loc_{\mathbb{R}}$ and that $R$ is $loc_{\mathbb{R}^{0|1}}$ in this complex setting?

   Would we still say that shape is and that is in this complex setting?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMay 13th 2015
   • PermaLink
   Author: Urs
   Format: MarkdownItexSo in the complex setting shape would not be localization at $\mathbb{R}$. It might be that it is now localization at $\mathbb{C}$, but I haven't really thought this through. But $Rh \simeq loc_{\mathbb{R}^{0|1}}$ would still hold, and I would still write it this way, yes.

   So in the complex setting shape would not be localization at . It might be that it is now localization at , but I haven’t really thought this through.

   But would still hold, and I would still write it this way, yes.

5. • CommentRowNumber5.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 13th 2015
   • (edited May 13th 2015)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexAh, so I wonder how things should be said in a model independent way. We could have $loc_{\mathbb{A}^1}$ and $loc_{\mathbb{R}^{0|1}}$, but isn't it possible that in some other nontrivial model of solid cohesion that we wouldn't have $Rh$ as $loc_{\mathbb{R}^{0|1}}$?

   Ah, so I wonder how things should be said in a model independent way.

   We could have and , but isn’t it possible that in some other nontrivial model of solid cohesion that we wouldn’t have as ?

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeMay 13th 2015
   • (edited May 13th 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexYes, there could be models of solid cohesion where $Rh$ is not $loc_{\mathbb{R}^{0\vert 1}}$. We discussed this before, that it seems unclear and in fact unlikely that solid cohesion as presently set up is sufficient to fully characterize supergeometry. But slight variants of this localization business will work more generally. For instance I suppose it follows readily that in complex analytic cohesion the shape modality is given by localization at the collection of all all [[Stein spaces]]. Similarly, reduction will maybe not in general be localization at $Spec(\mathbb{R}[\epsilon]/(\epsilon^2))$, but should fairly generally be localization at the collection at all infinitesimally thickened points.

   Yes, there could be models of solid cohesion where is not . We discussed this before, that it seems unclear and in fact unlikely that solid cohesion as presently set up is sufficient to fully characterize supergeometry.

   But slight variants of this localization business will work more generally. For instance I suppose it follows readily that in complex analytic cohesion the shape modality is given by localization at the collection of all all Stein spaces.

   Similarly, reduction will maybe not in general be localization at , but should fairly generally be localization at the collection at all infinitesimally thickened points.