Timeline for Interpolating between the flat and smooth affine lines in spectral algebraic geometry
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2021 at 15:08 | comment | added | Emily | I'd really love a copy of the draft version! I've sent you an email :) | |
| Sep 8, 2021 at 15:07 | comment | added | Emily | In any case, I think something that would be quite cool to understand is how the shearing and $\mathsf{Free}^k$ constructions are related: they both encode the Koszul rule in some form, but they do so in very different and most likely completely inequivalent ways! | |
| Sep 8, 2021 at 15:07 | comment | added | Emily | In particular, this construction also equips the category $\mathsf{Fun}(\tau_{\leq1}\mathbb{S},\mathsf{Mod}_R)$ with a different monoidal structure (Day convolution), whereas the usual one for naive-commutativity is the pointwise monoidal structure (...is it?) | |
| Sep 8, 2021 at 15:07 | comment | added | Emily | and otherwise $ab=\sigma_{\deg(a)+\deg(b)}(ba)$ if $\deg(a)\deg(b)$ is odd. So when these automorphisms are given by $\sigma_k(a)=-a$, this rule reduces to the usual $\mathbb{Z}$-graded Koszul rule, and in general lax symmetric monoidal functors $\tau_{\leq k}\mathbb{S}\to\mathsf{Mod}_R$ will already encode a generalised form of the Koszul rule for $k\geq1$! | |
| Sep 8, 2021 at 15:06 | comment | added | Emily | The $1$-truncation has $\mathbb{Z}$ as its set of objects, but it also has information about $\pi_1(\mathbb{S})\cong\mathbb{Z}_2$ encoded in the morphisms, and when you consider lax monoidal functors from it, you get a $\mathbb{Z}$-graded $R$-algebra $M_\bullet$ together with a collection of order $2$ automorphisms $\sigma_k\colon M_k\to M_k$. If you require additionally this functor to be symmetric, then (at least in the $1$-categorical setting), this gives an encoding of the Koszul rule: the braided monoidal condition becomes $ab=ba$ if $\deg(a)\deg(b)$ is even,... | |
| Sep 8, 2021 at 15:06 | comment | added | Emily | There's a point here which I think might be confusing in the discussion (I feel like we are talking past each other a small bit, though I apologise if this is a misunderstanding! As I'm sure you've noticed, I'm quite confused!): the zero-truncation of the sphere spectrum as a monoidal category is $\mathbb{Z}_{\mathsf{disc}}$, and lax monoidal functors from it to $\mathsf{Mod}_R$ are naive-graded-commutative $R$-algebras. | |
| Sep 8, 2021 at 13:19 | comment | added | A Rock and a Hard Place | Regarding my master's thesis though: I'm glad you enjoyed it, but you should know that it has a fair amount of mistakes (not the least of which is a too-optimistic correspondence between two notions of $S$-grading, based on a misunderstanding of a result by Hess-Shipley). I plan to post a revised version of it .. eventually, maybe :) If you're interested, I'd be happy to share a draft version upon request (myb shoot me an email?) | |
| Sep 8, 2021 at 13:10 | comment | added | A Rock and a Hard Place | Yes, the correspondence to lax symmetric monoidal functors $\mathbf Z\to\mathrm{Mod}_R$, and commutative graded $R$-algebra wants you to consider the naive commutativity. To get the Koszul sign rule, you can instead use the shearing trick mentioned a few questions back: that really amounts to equipping $\mathrm{Fun}(\mathbf Z, \mathrm{Mod}_R)$ with a different symmetric monoidal structure. The $1$-shearing not being symmetric monoidal is good, because Koszul and naive sign rules are different. $2$-shearing not being symmetric monoidal in char $p$ or spectrally is less intuitive though ... :) | |
| Sep 8, 2021 at 12:49 | comment | added | Emily | *"its underlying monoid in $\mathsf{Mod}_R$": this makes no sense; it should instead read something like "underlying $R$-module equipped with this weird Koszul-rule-commutative kind of algebraic structure" :P | |
| Sep 8, 2021 at 3:31 | comment | added | Emily | (P.S. When you mentioned Proj in algebraic geometry, you reminded me that I really need to read your thesis. I had a memory of it as being really good reading, and looking into it again made me notice that this is actually an understatement, it is absolutely wonderful! :) | |
| Sep 8, 2021 at 3:31 | comment | added | Emily | I think I'm certainly a bit confused with these: is it right to say that the free "commutative $R$-algebra with a $\mathbb{Z}$-gradation" on $R$ is $R[t^{\pm1}]$, and that the "free Koszul-rule-$\mathbb{Z}$-graded-commutative" $R$-algebra on $R$ is $\bigwedge^\bullet_R(R)$? I've seen the later statement in the nLab, and this is why I came up with this construction in the other question. | |
| Sep 8, 2021 at 3:31 | comment | added | Emily | ...that its underlying monoid in $\mathsf{Mod}_R$ (i.e. its left Kan extension along the terminal functor $\tau_{\leq k}\mathbb{S}\to\Delta^0$) is not necessarily an $\mathbb{E}_{\infty}$-ring (and probably isn't one for $k>1$). When you wrote $\mathsf{Free}^\infty(\mathbb{S})\cong\mathbb{S}\{t^\pm\}$, did you mean the usual $\mathbb{E}_{\infty}$-commutativity? | |
| Sep 8, 2021 at 3:31 | comment | added | Emily | When I wrote this idea up on my other question, I was thinking about this Koszul kind of graded-commutativity (which agrees with the naive one for $k=0$): $\mathsf{Free}^{k}(R)$ is the free $\mathbb{E}_{\infty}$-monoid on $\Delta_R$ with respect to Day convolution on $\mathsf{Fun}(\tau_{\leq k}\mathbb{S},\mathsf{Mod}_R)$, and is hence not necessarily $\mathbb{E}_{\infty}$ in the naive sense, meaning... | |
| Sep 8, 2021 at 1:28 | comment | added | A Rock and a Hard Place | About the first question though, no, a lax symmetric monoidal functor from $\Omega^\infty(S)$ is not the same thing as an invertible object. You're right, my discussion of symmetric monoidal functors above isn't perfect, as we really should be talking about lax symmetric monoidal ones instead. | |
| Sep 8, 2021 at 1:22 | comment | added | A Rock and a Hard Place | Perhaps you are confused about the kind of graded commutativity we are considering here. We are talking about 'naive' graded commutativity, e.g. $ab = ba$ regardless of the degrees of $a$ and $b$, as opposed to the graded-commutativity of the Koszul sign rule $ab = (-1)^{|a||b|}ba$. The notion under discussion here is relevant for instance for doing Proj in algebraic geometry. In this sense, the exterior algebra isn't actually commutative, so it surely can't be a free commutative graded algebra. Instead, that role is played by the symmetric algebra. | |
| Sep 7, 2021 at 16:30 | comment | added | Emily | (Or rather $\bigwedge^\bullet_R(M)$ is the free $\mathbb{Z}_{\geq0}$-graded-commutative $R$-algebra on $M$, I think?) | |
| Sep 7, 2021 at 16:26 | comment | added | Emily | ...which is the free $\mathbb{Z}$-graded-commutative $R$-algebra (though of course everything is different in the $\infty$-setting!). [2/2] | |
| Sep 7, 2021 at 16:26 | comment | added | Emily | The reason I was thinking we would get exterior algebras in the $1$-categorical case is because a lax symmetric monoidal functor $\Omega^{\infty}(S)\to\mathrm{N}_{\bullet}(\mathcal{C})$ is the same thing as a lax symmetric monoidal functor $\mathsf{Ho}(\Omega^{\infty}(S))\to\mathcal{C}$, and the latter has the explicit description given here, which includes $\mathbb{Z}$-graded commutative $R$-algebras as a special case. So I thought the free such lax symmetric monoidal functor would be the exterior algebra too, [1/2] | |
| Sep 7, 2021 at 16:26 | comment | added | Emily | Thanks! Does the correspondence $$\left\{\begin{gathered}\text{symmetric strong monoidal}\\\text{functors $\Omega^{\infty}(S)\to\mathcal{C}$}\end{gathered}\right\}\cong\left\{\text{invertible objects of $\mathcal{C}$}\right\}$$ work even for symmetric lax monoidal functors $\Omega^{\infty}(S)\to\mathcal{C}$? | |
| Sep 7, 2021 at 12:29 | history | answered | A Rock and a Hard Place | CC BY-SA 4.0 |