Timeline for Is there a "spectral exterior algebra" construction in higher algebra?
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6 events
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| Sep 6, 2021 at 22:52 | comment | added | Emily | @ARockandaHardPlace Thanks! I'm really shocked at how sneaky a situation this is: if you replace $\mathsf{ModSp}_R$ by $\mathrm{N}_{\bullet}(\mathsf{Mod}_R)$ and apply this construction to $R$, then I think this gives you $R[t]\cong\mathrm{Sym}_R(R)$ for $k=0$ and then $\bigwedge^{\bullet}_RR$ for $k\geq 1$, at which point it stabilises. Similarly, the $n$-categorical version of this stabilises at $k=n$. But when you pass from abelian groups to spectra this gives something totally different, and we get the symmetric algebra only for $k=\infty$! | |
| Sep 6, 2021 at 8:46 | comment | added | A Rock and a Hard Place | An interesting thing to ponder is to combine the two constructions: consider the (possibly $n$-times sheared) free $\tau_{\le k}(S)$-graded $\mathbb E_\infty$-$R$-algebra generated by the $n$-fold suspension $\Sigma^n(R)$. No idea what you could say about or do with this $(k, n)$-bigraded objects, but it looks fun! :) | |
| Sep 6, 2021 at 8:44 | comment | added | A Rock and a Hard Place | So it's not about "higher exteriority", which is to me maybe about changing the commutativity constraint, i.e. symmetric group actions and things like that - as discussed above, that has to do with "free graded algebras" (in whatever sense we're thinking about) on suspensions. Instead, the commutativity constraint is the same, we're just imposing it with varying levels of homotopy coherence. | |
| Sep 6, 2021 at 8:41 | comment | added | A Rock and a Hard Place | This is a very interesting idea too, though I believe it's doing something slightly different. Namely, for $k = 0$, we get $\tau_{\le 0}(S) = \mathbf Z$, and so we are looking at the free $\mathbf Z$-graded $\mathbb E_\infty$-$R$-algebra, which is the polynomial $\mathbb E_\infty$-algebra $R[t]$. For $k=\infty$, we obviously get the free $\mathbb E_\infty$-ring $S\{t\}$ back, so what the construction for $0 < k <\infty$ seems to be doing is building in ever more of the higher grading-coherences that are there in an $S$-grading, other than just the underlying $\mathbf Z$-grading.. | |
| S Sep 6, 2021 at 4:20 | history | answered | Emily | CC BY-SA 4.0 | |
| S Sep 6, 2021 at 4:20 | history | made wiki | Post Made Community Wiki by Emily |