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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 3rd 2017
   • (edited Mar 3rd 2017)
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   Author: Urs
   Format: MarkdownItexI finally gave _[[spectral super-scheme]]_ an entry, briefly stating the idea. This goes back to the observation highlighted in [Rezk 09, section 2](https://ncatlab.org/nlab/show/spectral+super-scheme#Rezk09). There is some further support for the idea that a good definition of supergeometry in the spectrally derived/$E_\infty$ context is nothing but $E_\infty$-geometry over even periodic ring spectra. I might add some of them later. Thanks to Charles Rezk for discussion (already a while back).

   I finally gave spectral super-scheme an entry, briefly stating the idea.

   This goes back to the observation highlighted in Rezk 09, section 2. There is some further support for the idea that a good definition of supergeometry in the spectrally derived/$E_\infty$ context is nothing but $E_\infty$-geometry over even periodic ring spectra. I might add some of them later.

   Thanks to Charles Rezk for discussion (already a while back).

2. • CommentRowNumber2.
   • CommentAuthorDavidRoberts
   • CommentTimeMar 3rd 2017
   • PermaLink
   Author: DavidRoberts
   Format: MarkdownItexDoes this relate to any grading by $\mathbb{S}$, or augmentation or any other such structure?

   Does this relate to any grading by $\mathbb{S}$, or augmentation or any other such structure?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMar 3rd 2017
   • (edited Mar 3rd 2017)
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   Author: Urs
   Format: MarkdownItexThat $\mathbb{Z}$-grading on $\pi_\bullet(E)$, that's $\pi_0(\mathbb{S})$-grading. Indeed, by [Sagave-Schlichtkrull, theorem 1.7-1.8](https://ncatlab.org/nlab/show/infinity-group%20of%20units#SagaveSchlichtkrull11) every connective $E_\infty$-ring is canonically $\mathbb{S}$-graded. So $E_\infty$-geometry already is the higher analog of super-geometry, but of $\mathbb{Z}$-graded supergeometry, not of the proper $\mathbb{Z}/2$-graded supergeometry. The remaining problem is to turn this categorified/homotopified $\mathbb{Z}$-graded supergeometry into genuine $\mathbb{Z}/2$-graded supergeometry. This is achieved by requiring an even periodiv ground $\infty$-ring

   That $\mathbb{Z}$-grading on $\pi_\bullet(E)$, that’s $\pi_0(\mathbb{S})$-grading.

   Indeed, by Sagave-Schlichtkrull, theorem 1.7-1.8 every connective $E_\infty$-ring is canonically $\mathbb{S}$-graded.

   So $E_\infty$-geometry already is the higher analog of super-geometry, but of $\mathbb{Z}$-graded supergeometry, not of the proper $\mathbb{Z}/2$-graded supergeometry.

   The remaining problem is to turn this categorified/homotopified $\mathbb{Z}$-graded supergeometry into genuine $\mathbb{Z}/2$-graded supergeometry. This is achieved by requiring an even periodiv ground $\infty$-ring

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 16th 2017
   • (edited Mar 16th 2017)
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   Author: Urs
   Format: MarkdownItexI have added a brief mentioning of the corresponding concept of the _[spectral superpoint](https://ncatlab.org/nlab/show/spectral+super-scheme#SpectralSuperpoint)_.

   I have added a brief mentioning of the corresponding concept of the spectral superpoint.