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From the discussion at Hochschild cohomology and A-infinity deformations, it seems that general Hochschild cohomology classes correspond to deformations where the deformation parameter can have nonzero degree.

So I have some naive and maybe stupid questions:

How can I interpret this geometrically? What is the "base space" of the deformation? What kind of object is it?

In other words, what is the "Spec" of a graded ring or a graded algebra (e.g. $k[t]$ or $k[[t]]$ or $k[t]/(t^n)$ with the variable $t$ having some nonzero degree)?

(… maybe what I'm really asking is: Is there a theory of "schemes" where the "affine schemes" correspond to graded commutative rings rather than commutative rings?)

LSpice
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Kevin H. Lin
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    This looks like derived algebraic geometry (for which there seems to be more than one school of thought), but where the differentials vanish. In this setting I think one usually views Spec of a graded ring as ringed space whose underlying topological space is Spec of the degree zero subring, but with a graded sheaf of functions. – S. Carnahan Jul 14 '10 at 23:58
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    To me, it looks like $G_m$-equivariant scheme theory... though maybe this is not useful for the more specific questions above. – Marty Jul 15 '10 at 00:01
  • @Scott: I was going to guess that it had something to do with "derived algebraic geometry" -- whatever that means... But in for instance Lurie's stuff (which is the only thing under the heading of "DAG" that I've read), it seems that he deals with $E_\infty$ ring spectra, rather than dg rings or dg algebras... – Kevin H. Lin Jul 15 '10 at 00:12
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    It's all the same stuff rationally. – Aaron Bergman Jul 15 '10 at 00:16
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    To amplify Aaron's comment, in characteristic zero $E_\infty$ ring spectra are an equivalent notion to commutative DGAs. One needs to be slightly careful tossing around the derived-algebraic-geometry or $\mathbb{G}_m$-equivariant monikers, though, because Hochschild cohomology (or something like it) really classifies deformations as an associative DGA, not a commutative one. – Tyler Lawson Jul 15 '10 at 01:50
  • ...but even having said that, it really does sound like what you're looking for is derived algebraic geometry. – Tyler Lawson Jul 15 '10 at 02:12
  • Sometimes Lurie deals with $E_\infty$ rings, sometimes simplicial commutative rings. He never deals with commutative dgas, because they work badly (except in characteristic zero, where they're all the same). – Ben Wieland Jul 15 '10 at 02:19
  • For what it's worth, I am personally only interested in characteristic zero things (for the most part). – Kevin H. Lin Jul 15 '10 at 03:11
  • @BenWieland I've never heard about this badness of dga in positive characteristic. Could you elaborate on this (references)? – user40276 Jul 15 '15 at 22:49
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    @user40276 In positive characteristic, (graded) commutative DGAs are not the same as $E_{\infty}$-DGAs. Dyer-Lashof/Steenrod operations are obstructions to strictification and they do not generically vanish unless you are in characteristic $0$. If you want a model structure, you have to consider these weaker notions as they are homotopical and the strict notion is not. Does this help? – Sean Tilson Jul 16 '15 at 12:00
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    @user40276 a good exercise is to see that the free functor from chain complexes to CDGAs does not preserve weak equivalences. I think that it fails even with the simplest acyclic complex. Tyler Lawson uses a different failure of a Quillen adjunction to show that a particular model structure does not exist. – Ben Wieland Jul 16 '15 at 22:08
  • @BenWieland Thanks for the response. – user40276 Jul 25 '15 at 09:16
  • @SeanTilson Thanks, this helped me. But let me ask for a clarification. Do you mean that the category of DGAs and $E_{\infty}$-DGAs are not even weak equivalent (by some suitable equivalence, for instance, the usual one on simplicial categories)? Anyway, do you know have a reference for this condition for the strictification using Dyer-Lashof operations? – user40276 Jul 25 '15 at 09:26
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    The definition of the DL operations has built in to it the choices of homotopies that make things commutative. Therefore, if the the algebra were already commutative you could take the constant homotopy and it would be evident that the operation vanished. The operations are also homotopy invariant in a certain sense (a weak equivalence between suitably bifibrant algebras etc.). So they non-vanishing of the operations tells you that it is not equivalent. This does not mean that vanishing gives you a strictification a priori. I would be happy for a reference for that as well. If you are ... – Sean Tilson Jul 25 '15 at 09:52
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    ...asking about the vanishing of operations in characteristic 0, this is a simple computation in group homology of symmetric groups with coefficients in $\mathbb{Q}$, there isn't any outside degree 0. Mike Mandell has a paper on comparisons of TAQ in many different settings that might be relevant. In particular, he spells out when things are equivalent to $E_{\infty}$-dgas. – Sean Tilson Jul 25 '15 at 09:54
  • @SeanTilson Ok!Thanks!I will check this reference. – user40276 Jul 27 '15 at 00:27
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    @user40276 I did my exercise and I must restrict the claim: The free CDGA on an acyclic complex supported in degrees $2n$ and $2n+1$ is equivalent to the ground ring, as expected, but the the free CDGA on an acyclic complex supported in degrees $2n-1$ and $2n$ has nontrivial homology. I am using a homological differential that lowers degree; with the cohomological convention, the two cases switch. – Ben Wieland Jul 28 '15 at 18:28

2 Answers2

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One possible answer is in Toën-Vezzozi paper From HAG to DAG, who were themselves inspired by Ciocan-Fontanine and Kapranov (Derived Quot schemes and Derived Hilbert schemes).

This approach works well in characteristic zero (otherwise one has to deal with simplicial commutative rings or $E_\infty$-ring spectra, like in Lurie's work).

Martin Sleziak
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DamienC
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maybe these notes by Vezzosi can be helpful for some

http://www.dma.unifi.it/~vezzosi/papers/derivedintctgtcplx.pdf

pro
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