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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 31st 2015
   • (edited Mar 31st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI keep trying to see to which degree one may nail down supergeometry via the yoga of adjoint modalities. My the starting point (maybe there is a better one, but for the time being that's the best I have come up with) is that the inclusion of commutative algebras into supercommutative algebras is reflective and coreflective, the reflector quotients out the ideal generated by the odd part, the coreflector picks the even subalgebra. Passing to sheaves over the sites of formal duals to algebras, this gives an adjoint modality which (and you may not like that now, but keep in mind that it's just notation which doesn't reall matter) I decided to denote $$ \rightrightarrows \; \dashv \; \rightsquigarrow $$ (The mnemonic is this: in Feynman diagrams $\rightsquigarrow$ is the symbol for the bosonic particles, so that denotes taking the bosonic subspace. Similarly a single fermion in a Feynman diagram appears as $\to$, so fermion bilinears look like $\rightrightarrows$ and that has to suffice to remind you of general even numbers of fermions.) The details (there are not many, it's straightforward) are at _[super formal smooth infinity-groupoid](http://ncatlab.org/nlab/show/super+formal+smooth+infinity-groupoid#Cohesion)_. Now, while nice, this falls short of _characterizing_ supercommutative algebras. For instance the inclusion of commutative algebras into commutative algebras equipped with $\mathbb{Z}/n\mathbb{Z}$-grading (for any $2 \leq n \lt \infty$, with no extra signs introduced when swapping factors) works just as well. (For the inclusion into $\mathbb{Z}$-graded algebras there are also left and right adjoints but they coincide, and so I declare that this case is excluded by demanding faithfulness/non-degeneracy of the model.) So the question is, which natural-looking further conditions could we impose on the above adjoints such as to narrow in a bit more on supercommutative algebras? Here is one observation: with the setup as described at _[super formal smooth infinity-groupoid](http://ncatlab.org/nlab/show/super+formal+smooth+infinity-groupoid#Cohesion)_ we also have the reduction modality $\Re$ which, on function rings, takes away the nilpotent ideal in an algebra, and its right adjoint, the de Rham stack functor $\Im$: $$ \Re \; \dashv \; \Im \,. $$ For both supercommutative algebras and for commutative algebras with grading we have inclusions of images of these functors $$ \array{ \rightrightarrows &\dashv& \rightsquigarrow \\ \bot && \bot \\ \rightsquigarrow &\dashv& \R \\ \vee && \vee \\ \Re &\dashv& \Im } $$ On the other hand, for supercommutative algebras but not for commutative algebras with grading, there is also an inclusion of images diagonally $$ \rightsquigarrow \Im \; \simeq \; \Im \,. $$ Because evaluating this via Yoneda on representables, then by adjunction this means that on these $$ \Re \rightrightarrows \; \simeq \; \Re $$ hence that the reduced part of the even part is the reduced part of the full algebra. But this says that the odd part of the algebra is nilpotent! This is something that is true for supercommutative algebras, but which need not be true for any old commutative algebras with grading. It is tempting to say that this is a kind of _[[Aufhebung]]_ , though it differs from what is currently discussed in that entry in that the diagonal inclusion is along nw-se instead of ne-sw. But I think that just means the entry should be generalized. In summary, while the Aufhebung-condition here gets closer on narrowing in the axioms on supercommutative algebras, it still does not quite characterize them. So far these axioms still allow in particular also (sheaves on formal duals to) cyclically-graded algebras whose non-0 graded parts are nilpotent. The first question seems to be: do we have further natural-looking axioms on the modalities that would narrow in further on genuine supergeometry? But possibly a second question to consider is: might there be room to declare that the further generality allowed by the axioms is something to consider instead of to discard. If there is a nice axiomatics that characterizes something a tad more general than supergeometry, maybe that's indication that this generalization is of interest? Not sure yet.

   I keep trying to see to which degree one may nail down supergeometry via the yoga of adjoint modalities.

   My the starting point (maybe there is a better one, but for the time being that’s the best I have come up with) is that the inclusion of commutative algebras into supercommutative algebras is reflective and coreflective, the reflector quotients out the ideal generated by the odd part, the coreflector picks the even subalgebra.

   Passing to sheaves over the sites of formal duals to algebras, this gives an adjoint modality which (and you may not like that now, but keep in mind that it’s just notation which doesn’t reall matter) I decided to denote

   $$\rightrightarrows \; \dashv \; \rightsquigarrow$$

   (The mnemonic is this: in Feynman diagrams $\rightsquigarrow$ is the symbol for the bosonic particles, so that denotes taking the bosonic subspace. Similarly a single fermion in a Feynman diagram appears as $\to$, so fermion bilinears look like $\rightrightarrows$ and that has to suffice to remind you of general even numbers of fermions.)

   The details (there are not many, it’s straightforward) are at super formal smooth infinity-groupoid.

   Now, while nice, this falls short of characterizing supercommutative algebras.

   For instance the inclusion of commutative algebras into commutative algebras equipped with $\mathbb{Z}/n\mathbb{Z}$-grading (for any $2 \leq n \lt \infty$, with no extra signs introduced when swapping factors) works just as well. (For the inclusion into $\mathbb{Z}$-graded algebras there are also left and right adjoints but they coincide, and so I declare that this case is excluded by demanding faithfulness/non-degeneracy of the model.)

   So the question is, which natural-looking further conditions could we impose on the above adjoints such as to narrow in a bit more on supercommutative algebras?

   Here is one observation:

   with the setup as described at super formal smooth infinity-groupoid we also have the reduction modality $\Re$ which, on function rings, takes away the nilpotent ideal in an algebra, and its right adjoint, the de Rham stack functor $\Im$:

   $$\Re \; \dashv \; \Im \,.$$

   For both supercommutative algebras and for commutative algebras with grading we have inclusions of images of these functors

   $$\array{ \rightrightarrows &\dashv& \rightsquigarrow \\ \bot && \bot \\ \rightsquigarrow &\dashv& \R \\ \vee && \vee \\ \Re &\dashv& \Im }$$

   On the other hand, for supercommutative algebras but not for commutative algebras with grading, there is also an inclusion of images diagonally

   $$\rightsquigarrow \Im \; \simeq \; \Im \,.$$

   Because evaluating this via Yoneda on representables, then by adjunction this means that on these

   $$\Re \rightrightarrows \; \simeq \; \Re$$

   hence that the reduced part of the even part is the reduced part of the full algebra.

   But this says that the odd part of the algebra is nilpotent! This is something that is true for supercommutative algebras, but which need not be true for any old commutative algebras with grading.

   It is tempting to say that this is a kind of Aufhebung , though it differs from what is currently discussed in that entry in that the diagonal inclusion is along nw-se instead of ne-sw. But I think that just means the entry should be generalized.

   In summary, while the Aufhebung-condition here gets closer on narrowing in the axioms on supercommutative algebras, it still does not quite characterize them. So far these axioms still allow in particular also (sheaves on formal duals to) cyclically-graded algebras whose non-0 graded parts are nilpotent.

   The first question seems to be: do we have further natural-looking axioms on the modalities that would narrow in further on genuine supergeometry?

   But possibly a second question to consider is: might there be room to declare that the further generality allowed by the axioms is something to consider instead of to discard. If there is a nice axiomatics that characterizes something a tad more general than supergeometry, maybe that’s indication that this generalization is of interest?

   Not sure yet.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMar 31st 2015
   • (edited Mar 31st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexOne stand-alone aspect of this question might be the following: in general commutative geometry behaves very different from genuine non-commutative geometry. But what if the non-commutativity is restricted to the nilpotent ideal? Has this been considered anywhere in some generality in the literature?

   One stand-alone aspect of this question might be the following:

   in general commutative geometry behaves very different from genuine non-commutative geometry. But what if the non-commutativity is restricted to the nilpotent ideal?

   Has this been considered anywhere in some generality in the literature?

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 31st 2015
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItex>But what if the non-commutativity is restricted to the nilpotent ideal? Is this at all like what appears in Connes' spectral triple approach to SM (compactified NC space, non-compact ordinary space)?

   But what if the non-commutativity is restricted to the nilpotent ideal?

   Is this at all like what appears in Connes’ spectral triple approach to SM (compactified NC space, non-compact ordinary space)?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 31st 2015
   • (edited Mar 31st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexIn the Connes-Lott model the noncommutative fiber spaces have ungraded matrix algebra as their algebras of functions.

   In the Connes-Lott model the noncommutative fiber spaces have ungraded matrix algebra as their algebras of functions.

5. • CommentRowNumber5.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 31st 2015
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   Author: DavidRoberts
   Format: MarkdownItexAh, ok, thanks.

   Ah, ok, thanks.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeApr 1st 2015
   • (edited Apr 1st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItex[ wrote something here, but will need to think more first ]

   [ wrote something here, but will need to think more first ]

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeApr 1st 2015
   • (edited Apr 1st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexsecond attempt: I am thinking now that the obvious axiom to add to narrow in on supergeometry proper is simply that the operator $\rightrightarrows$ that projects onto even-degrees is an $\mathbb{A}^1$-localization, for an object $\mathbb{A}^1$ to be denoted $\mathbb{R}^{0|1}$: $$ \rightrightarrows \;\simeq\; loc_{\mathbb{R}^{0|1}} $$ If that is true, then the condition $\stackrel{\rightrightarrows}{\mathbb{R}^{0|1}} \simeq \ast$, which in the supergeometric model reflects the sign rule in the form $\theta^2 = 0$, is automatic. Moreover, this should actually be true in the supergeometric model: localization at $\ast \to \mathbb{R}^{0|1}$ means to pass to objects $X$ for which $$ Hom(\mathbb{R}^{0|1},X) \longrightarrow Hom(\ast,X) $$ is an equivalence. But on the left we get a point in $X$ with an odd extension, and on the right just the point, so this says that there is no way to make an odd extension in $X$, hence that $X$ is even-graded.

   second attempt:

   I am thinking now that the obvious axiom to add to narrow in on supergeometry proper is simply that the operator $\rightrightarrows$ that projects onto even-degrees is an $\mathbb{A}^1$-localization, for an object $\mathbb{A}^1$ to be denoted $\mathbb{R}^{0|1}$:

   $$\rightrightarrows \;\simeq\; loc_{\mathbb{R}^{0|1}}$$

   If that is true, then the condition $\stackrel{\rightrightarrows}{\mathbb{R}^{0|1}} \simeq \ast$, which in the supergeometric model reflects the sign rule in the form $\theta^2 = 0$, is automatic.

   Moreover, this should actually be true in the supergeometric model: localization at $\ast \to \mathbb{R}^{0|1}$ means to pass to objects $X$ for which

   $$Hom(\mathbb{R}^{0|1},X) \longrightarrow Hom(\ast,X)$$

   is an equivalence. But on the left we get a point in $X$ with an odd extension, and on the right just the point, so this says that there is no way to make an odd extension in $X$, hence that $X$ is even-graded.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeApr 1st 2015
   • (edited Apr 1st 2015)
   • PermaLink
   Author: Urs
   Format: MarkdownItexI have written that out now in notes [here](http://ncatlab.org/schreiber/show/Modern+Physics+formalized+in+Modal+Homotopy+Type+Theory). Comes out quite nicely, I would think. (And I should have seen this long before, but it's a good sign if the formalism is more clever than I am.) So we go through three levels of oppositions, starting from that of the ground topos, and on the first and third stage we may ask for objects exhibiting the leftmost adjoint by localization. This way we find two singled-out objects, and they have the interpretation of $\mathbb{R}$ and $\mathbb{R}^{0|1}$. That's nice, because in other threads I was looking for general abstract reasons to consider specifically cocycles of the form $\mathbf{B}\mathbb{R}^{0|p} \to \mathbf{B}^{2} \mathbb{R}^d$. This is getting closer now.

   I have written that out now in notes here. Comes out quite nicely, I would think. (And I should have seen this long before, but it’s a good sign if the formalism is more clever than I am.)

   So we go through three levels of oppositions, starting from that of the ground topos, and on the first and third stage we may ask for objects exhibiting the leftmost adjoint by localization. This way we find two singled-out objects, and they have the interpretation of $\mathbb{R}$ and $\mathbb{R}^{0|1}$.

   That’s nice, because in other threads I was looking for general abstract reasons to consider specifically cocycles of the form $\mathbf{B}\mathbb{R}^{0|p} \to \mathbf{B}^{2} \mathbb{R}^d$. This is getting closer now.

9. • CommentRowNumber9.
   • CommentAuthorDavid_Corfield
   • CommentTimeApr 2nd 2015
   • (edited Apr 2nd 2015)
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexIf we have _relative_ cohesion, as in $$ Sh_\infty\left(SmthMfd, Sh_\infty\left(Sch_{\mathbb{Z}}\right) \right) $$ over the base $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$, are we likely to see example of relative 'solid' cohesion? Maybe $$ Sh_\infty\left(SmthSuperMfd, Sh_\infty\left(Sch_{\mathbb{Z}}\right) \right)? $$

   If we have relative cohesion, as in

   $$Sh_\infty\left(SmthMfd, Sh_\infty\left(Sch_{\mathbb{Z}}\right) \right)$$

   over the base $Sh_\infty\left(Sch_{\mathbb{Z}}\right)$, are we likely to see example of relative ’solid’ cohesion? Maybe

   $$Sh_\infty\left(SmthSuperMfd, Sh_\infty\left(Sch_{\mathbb{Z}}\right) \right)?$$
10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeApr 2nd 2015
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    Author: Urs
    Format: MarkdownItexYes, that would all go through unmodified.

    Yes, that would all go through unmodified.

11. • CommentRowNumber11.
    • CommentAuthorDavid_Corfield
    • CommentTimeApr 2nd 2015
    • PermaLink
    Author: David_Corfield
    Format: MarkdownItexAre there any signs of this already being glimpsed in that specific case? What would they be called, smooth $E-\infty$-supergroupoids relevant for differential superalgebraic K-theory?

    Are there any signs of this already being glimpsed in that specific case? What would they be called, smooth $E-\infty$-supergroupoids relevant for differential superalgebraic K-theory?

12. • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeApr 2nd 2015
    • PermaLink
    Author: Urs
    Format: MarkdownItexMaybe. It would seem to be suggestive that the super-grading should somehow be unifiedwith that in $E_\infty$-rings. But I still don't really see through this yet.

    Maybe. It would seem to be suggestive that the super-grading should somehow be unifiedwith that in $E_\infty$-rings. But I still don’t really see through this yet.