# Are exterior algebras intrinsically formal as associative dg algebras?

(Cross-posted from mathematics stackexchange.)

Fix a finite dimensional vector space $$V$$ over a field of characteristic zero, and let $$R=Sym(V[1])$$ be the free graded commutative algebra generated by $$V$$ in cohomological degree $$-1$$, but thought of as a formal associative dg algebra (we forget that it is commutative).

Then it is easy to see that (edit: any $$A_{\infty}$$-deformation of) $$R$$ is almost formal, meaning that in a minimal $$A_{\infty}$$-model, the higher products eventually vanish. Indeed, $$R$$ is concentrated in non-positive cohomological degrees, so all products of an $$A_{\infty}$$-structure $$\mu_{n} : R^{\otimes n} \rightarrow R[2-n]$$ have to vanish for $$n>>0$$.

Question 1: Is $$R$$ intrinsically formal, so that for any deformation, $$\mu_{n}$$ can be assumed to be zero for $$n>2$$?

Question 2: What if $$R=Sym(V[-1])$$, so that $$R$$ is concentrated cohomologically in non-negative degrees? (I'm thinking of the cohomology ring of a torus, but as an associative algebra, rather than a commutative algebra.) Can there be non-trivial $$A_{\infty}$$-structures on $$R$$, or is $$R$$ intrinsically formal?

• You say that you consider $R$ as a formal associative DG-algebra. This means that you endow it with the trivial differential, hence it is itself a minimal model. There are no higher $\mu_n$'s. Commented Jan 31, 2020 at 14:41
• Thanks. I mean to ask whether an exterior algebra has non-trivial $A_{\infty}$-deformations, or whether it is `intrinsic formality'. I'll clarify.
– TBRS
Commented Jan 31, 2020 at 14:57
• It seems that the answer to Q2 is positive, at least in characteristic zero. If $R$ has cohomology $Sym(V[-1])$, then I think you can show it has an augmentation $R \rightarrow k$, and that the Koszul dual $R^{!}=RHom_{R}(k,k)$ is quasi-isomorphic to $Sym(V^{*})$. Taking the Koszul dual again, we get a quasi-isomorphism $R \simeq Sym(V[-1])$, as desired. Is there a more elementary argument? I still don't know about Q1, or about positive characteristic.
– TBRS
Commented Feb 1, 2020 at 9:40
• The Koszul dual $R^{!}=RHom_{R}(k,k)$ is not quite $Sym(V^{*})$, but rather the completion thereof. So more naturally to work with the dual coalgebra $Sym(V)$. But I still think the argument is valid.
– TBRS
Commented Feb 2, 2020 at 19:22
• I deleted my answer to the question - as pointed out by Bertram Arnold the theorem I quoted doesn't apply to the situation here. Commented Feb 3, 2020 at 14:00

Here are examples of nontrivial $$A_\infty$$-structures extending the product on $$\operatorname{Sym}(\mathbb R^3[-1])$$ and $$\operatorname{Sym}(\mathbb R^5[-1])$$, respectively. Below the fold, I have kept my original answer, which got the main ideas right but almost all degrees wrong.

Fix coordinates $$\xi_1,\xi_2,\xi_3$$ on $$\mathbb R^3[-1]$$ and set $$\mu_3(f_1,f_2,f_3) = \xi_2\xi_3 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ .$$ This defines a $$A_\infty$$-structure: Every term in $$[\mu_3,\mu_3](f_1,\dots,f_5) = \sum \mu_3(\dots,\mu_3(\dots),\dots)$$ contains $$\xi_1^2 = 0$$ and hence vanishes; similarly, $$[\mu_3,\mu_2](f_1,\dots,f_4)$$ vanishes if one of the $$f_i$$ is a multiple of $$\xi_2$$ or $$\xi_3$$, and it also vanishes if one of them is the identity since $$[-,-]$$ preserves normalized Hochschild cochains. Thus the only expression we have to check is \begin{align*} [\mu_2,\mu_3](\xi_1,\xi_1,\xi_1,\xi_1) &= \color{red}{-}\xi_1\mu_3(\xi_1,\xi_1,\xi_1) -\mu_3(\xi_1^2,\xi_1,\xi_1) \\ &\ + \mu_3(\xi_1,\xi_1^2,\xi_1) - \mu_3(\xi_1,\xi_1,\xi_1^2) \\ &\ + \mu_3(\xi_1,\xi_1,\xi_1)\xi_1\\ &= -\xi_1\xi_2\xi_3 + \xi_2\xi_3\xi_1 = 0 \end{align*} (The red sign comes from the Koszul sign rule.)

This $$A_\infty$$-algebra has a nontrivial Massey product $$0\notin \langle\xi_1,\xi_1,\xi_1\rangle = \xi_2\xi_3 + \xi_1A + A\xi_1$$ and hence is not formal.

Essentially the same argument applies to the $$A_\infty$$-structure on $$\mathbb R^5[-1]$$ defined by the only nontrivial bracket $$\mu_3(f_1,f_2,f_3) = \xi_2\xi_3\xi_4\xi_5 (\partial_{\xi_1}f_1)(\partial_{\xi_1}f_2)(\partial_{\xi_1}f_3)\ .$$

Recall that we can associate to any (cohomologically) graded algebra $$(A,\mu_A)$$ the Hochschild cochains $$C^n(A) = \prod_{p+q = n}\operatorname{Hom}^p(A^{\otimes q},A)$$ which come equipped with a differential defined by precomposing with the multiplication of cyclic neighbours in the tensor product. As the totalization of a double complex, it also comes with a canonical filtration, namely $$C^n_{\ge m}(A) = \prod_{{\substack{p+q=n\\q\ge m}}} \operatorname{Hom}^p(A^{\otimes q},A)$$. The skew-symmetrization of the pre-Lie structure $$f\circ g(c_1,\dots,x_{m+n-1}) = \sum_{i=1}^m \pm f(x_1,\dots,g(x_i,\dots,x_{i+n-1}),\dots,x_{m+n-1})$$ defines a filtered shifted dgla structure on $$C^*(A)$$, and $$A_\infty$$-structures with trivial differential $$(A,0,\mu_A,\mu_3,\dots)$$ correspond to Maurer-Cartan elements $$(\mu_3,\dots)$$ in $$C^2_{\ge 3}(A)$$.

Now consider any such element $$\mu$$; its component of lowest filtration defines a degree $$2$$ cohomology class in the associated graded, since $$\mathrm d\mu = -\frac{1}{2}[\mu,\mu]$$ lies in higher filtration (as $$[C^p_{\ge m}(A),C^q_{\ge n}(A)]\subset C^{p+q-1}_{\ge m+n-1}(A)$$). If this cohomology class is $$0$$, i.e. $$\mu = \mathrm d\eta$$ up to terms of higher filtration, we can act by the gauge symmetry $$\mu\mapsto e^{-\eta}\circ\mu = \mu - \mathrm d\eta\, + (\text{higher filtration})$$ to obtain an equivalent Maurer-Cartan element which lies in higher filtration. Since the Hochschild cochains are complete with respect to the canonical filtration, the limit of this process is well-defined, i.e. if all obstructions vanish we obtain that our Maurer-Cartan element is gauge equivalent to $$0$$.

If $$A = \operatorname{Sym}(V)$$ is the commutative algebra on a graded vector space $$V$$, we can compute $$C^*(A)$$ with its filtration explicitly: We have $$C^*(A) \simeq \operatorname{Ext}_{A\otimes A^{op}}(A,A)$$, and since the multiplication map $$A\otimes A^{op}\to A$$ is $$\operatorname{Sym}$$ of the codiagonal $$V\oplus V\to V$$, we obtain a Koszul resolution $$A\simeq (\operatorname{Sym}(V\oplus V\oplus V[1]),\mathrm d)$$. Explicitly, we choose a basis $$x_i$$ of $$V$$; then the resolution is generated by elements $$x_i,x_i'$$ in degree $$|x_i|$$ and $$y_i$$ in degree $$|x_i|-1$$, with $$\mathrm dy_i = x_i' - x_i''$$. (This is where we use characteristic $$0$$.) It follows that $$C^*(\operatorname{Sym}(V))\simeq \widehat{\operatorname{Sym}}(V\oplus V^*[-1])$$, where the filtration is given by the degree in the $$V[-1]$$-variables, and the hat indicates that we take the completion with respect to this filtration.

Now let us specialize to the two cases in your question, namely $$V$$ is concentrated in degree $$-1$$ or $$+1$$. In the second case, $$C^*(A)$$ is concentrated in nonnegative degrees, and $$HH^2(A) := H^2(C^*(A))\cong \operatorname{Sym}^* V^*\otimes \Lambda^2 V$$, with the relevant part in $$\operatorname{Sym}$$-filtration $$\ge 3$$. In the first case, $$V\oplus V^*[-1]$$ lives in degrees $$-1$$ and $$2$$, and the deformation group is given by $$HH^2_{\ge 3}(A) = \prod_{p\ge 3} \Lambda^{2p-2}V\otimes \operatorname{Sym}^p V^*$$ and thus also does not vanish. Indeed, Kontsevich's formality theorem shows that Maurer-Cartan elements of $$\hbar C^*_{\ge 3}(A)[[\hbar]]$$ are in bijection with those of its cohomology, i.e. an infinitesimal deformation can be extended to a formal deformation iff it satisfies the Maurer-Cartan equation in in the cohomology. In some cases, the formal infinite series involve only finitely many terms and therefore give rise to actual deformations, as is the case in the two examples above.

• Thanks! I like this argument. I haven't seen before the extra filtration on Hochschild cochains, and its relation to formality. Could you provide some pointers to the literature where I can see other examples of this kind of argument?
– TBRS
Commented Feb 3, 2020 at 10:15
• I suppose the filtration restriction on $HH^{*}(A)$, to get minimal $A_{\infty}$-structures rather than non-minimal and curved gadgets, should be related to the Quillen sequence $A \rightarrow {\rm Der}^{\ast}(A) \rightarrow HH^{*}(A)[1]$. In Ultimately, it should be expressible in terms of ${\rm Der}^{\ast}(A)$ rather than cochains, no? And in this graded commutative case, I think the map $A \rightarrow Der^{*}(A)$ is zero, so that probably $Der^{*}(A)$ is just embedding to contain the filtered piece you are describing.
– TBRS
Commented Feb 3, 2020 at 10:24
• Note that there seems to be a dual $*$ missing somewhere. I think actually $HH^{*}(Sym(V))=Sym(V) \otimes Sym(V^{*}[-1])$. Note that is certainly true if $V$ is in degree $0$, by HKR. If correct, then when $V$ is of pure degree $1$, $V^{*}[-1]$ is of pure degree $0$, and so I would get $HH^{2}(A)=\bigwedge^{2}V$. Or did I mess up degrees? But it kind of makes sense. I guess there should be non-homogeneous deformations of $A$ by "super Clifford algebras".
– TBRS
Commented Feb 3, 2020 at 13:10
• I agree with this answer. But let me add that there is a subtlety involved in making sense of this type of argument. To be precise in the second paragraph we are taking the transfinite composition of infinitely many gauge transformations and for this to be meaningful we also need to be sure that the sequence of gauges can be made to converge to zero with respect to the filtration. So we should prove in general that obstructions vanish in the cohomology of $C^{2}_{\geq n}$, rather than that their images in $HH^2$ vanish. Commented Feb 3, 2020 at 13:39
• An aside: deformations of $Sym(V[-1])$ to a non-minimal $A_{\infty}$-algebra arise in nature. Interpret $Sym(V[-1])$ as Lie algebra cochains for the abelian Lie algebra $V^{*}$, and now deform the bracket on $V^{*}$ to something non-trivial. That's dual to deforming the differential in $Sym(V[-1])$.
– TBRS
Commented Feb 4, 2020 at 8:28