1
$\begingroup$

In his paper where details for Tamarkin's proof of formality are given, Hinich considers a Koszul quadratic operad $P$, a graded $P$-algebra $H$, a $P_\infty$-algebra $X$ with $HX=H$ (as $P$-algebras I presume?) and then gives by HTT a $P_\infty$-algebra structure $Q$ on $H$ so it becomes equivalent to $X$. This is a map $Q : P^{\perp}(H)\to H$ or, what is the same, an element in the Lie algebra $\mathfrak g $ of coderivations of the free $P^{\perp}$-coalgebra $C = P^{\perp}(H)$. This is $\mathbb N$-graded by the arity of $P^{\perp}$.

He wants to prove (Theorem 4.1.3.) that if the inclusion $\mathfrak g_{\geqslant 1} \to \mathfrak g$ induces the zero map on $H^1$, then $H$ is intrinsically formal. In doing so, he uses the notation $C[\lambda]$. I am not sure if this means the 'degree wise' tensor product of $C$ with the polynomial algebra, i.e.

in each degree, this is $C(n)\otimes \lambda^{n-1}$

or the degree wise tensor product of $C$ with a polynomial algebra, i.e.

in each degree, this is $C(n)\otimes \mathbb k[\lambda]$.

Trying to understand the proof of the Theorem 4.1.3., it seems to me it should be the first. This makes sense since the inductive proof in (4.2.6) of page 9 seems to be extending the isotopy $\theta$ 'arity by arity', which is consistent with the first interpretation of the notation. My questions are as follows

  1. Could someone clarify this?
  2. In the proof in (4.2.6) Hinich also says ''where $Q'$ is some differential uniquely defined by the above formula'', but I can't figure out what formula he's referring to. Why does $Q'$ exist?
  3. In many lines of paragraph (4.2.6) (second display, fourth display and sixth display equations) Hinich writes $Q$ when I think he means $Q^0$, since after all he wants an isotopy from $C[\lambda]$ with $Q^0$ to $C[\lambda]$ with $Q^\lambda$. Is this the case?
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.