# Does the $(\mathbb Z/2)$-graded isomorphism $E_n \cong E_{n+2}$ have any nice properties?

+ 6 like - 0 dislike
302 views

This question assumes everything is dg. Let's decide to work over the "field" $\mathbb Q[\mu,\mu^{-1}]$ where $\mu$ has homological degree $+2$. Then chain complexes are just $\mathbb Z/2$-graded. In the sequel I will write "$X$" for $X\otimes_{\mathbb Q}\mathbb Q[\mu,\mu^{-1}]$ if the object $X$ was more standardly defined over $\mathbb Q$. If in fact the object $X$ was originally defined in terms of topological spaces, then I will write $X$ for $\mathrm C_\bullet(X)$, always with coefficients in $\mathbb Q[\mu,\mu^{-1}]$.

Let $n \geq 2$ be an integer. There are (zig-zags of quasi)isomorphisms of dg operads:

$$E_n \cong P_n \cong P_{n+2} \cong E_{n+2}.$$

The operad $E_n$ is the operad of little $n$-dimensional rectangles inside a large $n$-dimensional rectangle. The operad $P_n$ is the operad controlling $(n-1)$-shifted Poisson algebras, i.e. commutative algebras $A$ equipped with a Lie algebra structure on $A[n-1]$ which is a derivation in each variable.

The middle isomorphism $P_n \cong P_{n+2}$ is totally canonical: the data of a Poisson bracket $\{,\}$ is no different from the data of the Poisson bracket $\mu\{,\}$, since $\mu$ is invertible. The first and last isomorphisms are pretty horrible, and exist only because of the deep theorem called "Kontsevich formality". In particular, there is no canonical such isomorphism.

Is there anything that can be said about the composition $E_n \overset\sim\to E_{n+2}$? More precisely, is it possible to choose the two formality morphisms so that this composition has any nice property at all? Here are the types of "nice properties" that I could imagine:

• Perhaps this isomorphism is somewhat canonical? At least, perhaps a choice of it can be written down without going through all the machinery of Kontsevich formality?
• My (probably faulty) understanding is that you can get all formality morphisms uniformly once you choose a Drinfeld associator. Perhaps you need to use the same one?
• Both $E_n$ and $E_{n+2}$ are "Hopf operads", meaning there's a natural way to tensor together algebras for either one. Perhaps this isomorphism can be chosen to respect this structure? Actually, they are cocommutative Hopf algebras; perhaps that makes the question easier or harder?
• Perhaps the isomorphism (if both the above work out well) can be explained in terms of some topological construction?
• Kontsevich formality has an interpretation in terms of path integrals and quantum field theory (and gauge fixing and renormalization and counterterms and all that sort of stuff). Perhaps even if the isomorphism itself is not canonical, it can be interpreted as a "partial path integral" for a "relative" or "compactified" or "equivariant" or ... quantum field theory?
This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd
retagged Jan 6, 2015
A warm-up question: I don't really know what $\mathbb Q[\mu,\mu^{-1}]$ "means" in terms of derived geometry, but I'd love to. It's something like the functions on $\mathrm B \mathbb G_m$, or $\mathrm B^2 \mathbb G_m$, or something. I mean, $\mathrm{Spec}(\mathbb Q[\mu,\mu^{-1}])$ is a "(derived) abelian group scheme" that integrates the line in degree $2$, thought of as an abelian Lie algebra, and that line is something like $\mathrm B^2$(usual 1-dimensional Lie algebra). What is $\mathbb Q[\mu,\mu^{-1}]$ really?

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd
In physics, I've been told that $\mu$ is a "mass", or perhaps "mass squared", but I don't claim to understand why. I've also been told that $\mu = \hbar$, a claim that I understand even less.

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd
Here's a comment-question. The passage from a space $X$ to its $2$-periodic rational homology $H_{\bullet}(X) \otimes \mathbb{Q}[\mu, \mu^{-1}]$ is quite destructive. It factors through a less destructive thing, namely complex K-homology $KU_{\bullet}(X)$, and this in turn factors through an even less destructive thing, namely the free KU-module spectrum $KU \wedge \Sigma_{+}^{\infty} X$.

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Qiaochu Yuan
In particular, the passage to $2$-periodic rational homology that really lets us identify the $E_n$ and $E_{n+2}$ operads factors through a passage to free KU-module spectra, getting us a pair of operads in KU-module spectra which are isomorphic after passing to rational homotopy. But are they already isomorphic before doing that? (And can we write down an isomorphism having anything to do with complex Bott periodicity?)

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Qiaochu Yuan
@QiaochuYuan Unfortunately it seems very unlikely that those two operads in $KU$-modules are equivalent.

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Tyler Lawson
@QiaochuYuan: Interesting. As far as I know, formality requires dividing by arbitrary integers, so I don't expect an isomorphism between $E_n \wedge KU$ and $P_n \wedge KU$, at least not if my understanding of "$\wedge KU$" is correct. But either of the following would be awesome: (1) formality over $KU$; (2) interesting statements over $KU$, but no formality, hence interesting statements that don't require formality.

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd
Is there a single operad called E_n? Or is there an interesting connected space of operads called E_n with no distinguished base point? For the nonlinear version (operads in spaces), I hope that the "classifying space of E_n operads" is BO(n). The classifying space of E_n operads in your funny 2-periodic category must be a lot different. Maybe it is BO?

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user David Treumann
@DavidTreumann Is there a vector space call Q^n, or an interesting connected space of them? Often I feel like there's only one, since when I write Q^n, I usually mean that I have distinguished a basis. But then I don't know if I've distinguished a basis, or an ordered basis; for the former, perhaps there's BS_n of them. And of course often there's BGL(n) of them. Anyway, surely the space of E_n operads is a B(Aut E_n), and surely there's an action of O_n by changing the framing, but I would not expect it to be everything. Over Q, isn't Aut E_2 the Grothendieck-Teichmuller group?

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd
I didn't mean anything too metaphysical. Since the two-periodic E_n and E_{n+2} operads have actions of O(n) and O(n+2), something you could add to your list of hypothetical "nice properties" is that the isomorphism between them is equivariant for the inclusion O(n) --> O(n+2). Is that true?

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user David Treumann
@DavidTreumann Good question! Here's my understanding. There is a semidirect product $O(n) \ltimes E_n$ called "framed $E_n$", and its formality makes it isomorphic to what you get by adjoining to $P_n$ a generator that acts as primitive of the bracket in the Hochschild cohomology coming from the commutative multiplication. The isomorphism $P_n \cong P_{n+2}$ is compatible with this extra generator. So I think that there is a sense in which the isomorphism $E_n \cong E_{n+2}$ is equivariant for some sort of $O$ action, but I'd have to think a while to unpack the details.

This post imported from StackExchange MathOverflow at 2015-01-06 10:13 (UTC), posted by SE-user Theo Johnson-Freyd

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.