# Formality of Ext algebras and direct sums

Does taking direct summands/sums preserve formality of ext-algebras? More precisely:

Given an abelian category, say linear over a field and with enough injectives, one gets an $A_\infty$-srutcture on the $ext$-algebras of its objects. Let $X,Y$ be objects of our category. What is the relation between the following assertions (with additional assumptions if necessary)?

1) $Ext^\bullet(X,X)$ is formal and $Ext^\bullet(Y,Y)$ is formal

2) $Ext^\bullet(X\oplus Y,X \oplus Y)$ is formal

3) Two out of $Ext^\bullet(X,X)$, $Ext^\bullet(Y,Y)$ and $Ext^\bullet(X\oplus Y,X \oplus Y)$ are formal

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My understanding is that formality of the DGA $Ext^\bullet(X\oplus Y,X\oplus Y)$ implies 1), but also formality of $Ext^\bullet(X,Y)$ and $Ext^\bullet(Y,X)$ as bimodules over $Ext^\bullet(X,X)$ and $Ext^\bullet(Y,Y)$, and that this is a much stronger condition.

For instance, let $(E,p)$ be an elliptic curve over a field, work in the abelian category of coherent sheaves, let $X=\mathcal{O}$ and let $Y=\mathcal{O}_p$ be the skyscraper at $p$.

Then $Ext^\bullet(X,X)$ and $Ext^\bullet(Y,Y)$ are both (intrinsically) formal, but $Ext^\bullet(X\oplus Y,X\oplus Y)$ knows the affine coordinate ring of $E\setminus\{ p\}$ for the cubic embedding into $\mathbb{P}^2$. That's because one can iteratively build $\mathcal{O}(np)$ for $n>0$ as a twisted complex in $X$ and $Y$ (namely, $\mathcal{O}((n+1)p)$ is the twist of $\mathcal{O}(np)$ along the spherical object $Y$). Over an algebraically closed field, this gives a $j$-line of quasi-isomorphism classes of $A_\infty$-algebras $Ext^\bullet(X\oplus Y,X\oplus Y)$.

As requested a bit more detail on why 2) implies 1), probably by too clunky an argument. Let $A=Ext^\bullet(X\oplus Y, X\oplus Y)$. We can regard this as an ordinary graded $K$-algebra, in which case non-formality of the $A_\infty$-structure is detected by the primary deformation class in $HH^\bullet_K(A,A)$. That is: after transferring the DG structure to a minimal $A_\infty$-structure on $A$ using homological perturbation theory, the composition $\mu^3$ defines a Hochschild cocycle. If it is a coboundary then we can kill $\mu^3$ by a gauge transformation which leaves $\mu^1$ and $\mu^2$ untouched, whereupon $\mu^4$ is a cocycle; and so on. If the structure is not formal, one will eventually obtain a non-trivial Hochschild class, called the primary deformation class.

We can alternatively regard $A$ as a 2-object graded-linear category, i.e., an algebra over $R=K\oplus K$, in which case non-formality is detected by a primary class in $HH^\bullet_R(A,A)$, defined similarly. But one checks using the bar resolution that $HH^\bullet_R(A,A)\cong HH^\bullet_K(A,A)$ as $K$-modules. Hence, if the algebra is formal, then so is the category; the restriction of the categorical primary deformation class to endomorphisms of $X$ is then trivial.

The references I tend to use for this sort of thing are the first chapter of Seidel's book "Fukaya categories and Picard-Lefschetz theory", and also his paper "Homological mirror symmetry for the quartic surface", but there are certainly other possibilities.

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Thanks for this fast answer! Can you explain how formality of $Ext^\bullet(X\oplus Y, X\oplus Y)$ implies formality of $Ext^\bullet(X,X)$? Also I must admit, that I don't know what formality of $Ext^\bullet(X,Y)$ means. –  Jan Weidner May 7 '12 at 19:59
In fact, in Tim Perutz's example, the $A_{\infty}$-algebra $Ext^{\bullet}(X\oplus Y, X \oplus Y)$ knows $D^{b}_{Coh}(E)$ and hence $E$, since once you can build powers of an ample globally generated line bundle, you can build the whole derived category. –  Chris Brav May 7 '12 at 20:06
The formality of the cross-terms was carelessly phrased, but it means formality as bimodules. I've added something about why 2) implies 1). –  Tim Perutz May 7 '12 at 21:38
Chris: that's true, but it sounds a bit back-to-front to me: the derived category isn't a complete invariant for projective varieties in general, but once you have those powers you have the homogeneous coordinate ring. –  Tim Perutz May 7 '12 at 21:55
Tim: yes I agree. The choice of a generator for the derived category gives more information. But if $X$ is a smooth projective variety and I give you $D^{b}_{Coh}(X)$ as an abstract $A_{\infty}$-category (with no t-stucture or symmetric monoidal structure) together with a generating set of the form $\mathcal{O}_{X}, \mathcal{O}_{X}(1), \cdots, \mathcal{O}_{X}({\rm dim}\; X)$, then you have no way of knowing that the generator is of this form (using only the $A_{\infty}$-structure, and so I think no way of constructing the homogeneous coordinate ring. –  Chris Brav May 8 '12 at 6:29