Is the exterior algebra intrinsically formal?

Question

Following 4.6 and 4.7 of this paper by Seidel and Thomas, a graded algebra $A$ is called intrinsically formal if any two dgas with cohomology $A$ are quasi-isomorphic. There is a sufficient condition that says for an augmented graded algebra $A$, if the Hochschild cohomology $HH^q(A, A[2 − q]) = 0$ for all $q > 2$, then $A$ is intrinsically formal.

Is it possible to verify this condition for the exterior algebra on a vector space $V$? To be clear, it could be either $Sym(V[1])$ or $Sym(V[-1])$. The characteristic and the dimension of $V$ might be important. I'm focusing on char$=0$ or $2$ and dimensions $1$ and $2$.

We may also regard the exterior algebra as a superalgebra ($\mathbb{Z}/2$-graded algebra) so that $Sym(V[1])$ and $Sym(V[-1])$ are the same. Is there a similar sufficient condition for intrinsically formality (we also consider $\mathbb{Z}/2$-graded dgas)?

There is a related question on the computation, but I'm not sure if the answer is enough to verify the condition. There are also related papers 1 2 3, but all of them seem to deal with the exterior algebra as an ungraded algebra.

Answer 1

It's not intrinsically formal. Let $A=\Lambda(V)$ be the exterior algebra on an $n$-dimensional vector space $V$, it follows from the Hochschild-Konstant-Rosenberg isomorphism that

$\mathit{HH}^q(A,A[2-q])\cong\mathit{Sym}^q(V^\vee)\otimes\Lambda^2(V),$

which is large for $q>2$.

However, the following is true:

If $V\cong\mathbb{C}^2$, then a non-trivial $A_\infty$-deformation of $A=\Lambda(\mathbb{C}^2)$ cannot be a cyclic $A_\infty$-algebra. Roughly speaking, it means that the $A_\infty$-deformations do not preserve chain level Poincare duality. More general discussions can be found here https://arxiv.org/abs/2106.07692, Section 1.3.

Answer 2

Let me focus on the $\mathbb Z$-graded picture in characteristic $0$. (You can collapse the $\mathbb Z$-grading to a $\mathbb Z / 2$-grading if you like, but one should be a bit careful when generalizing this to characteristic $2$.)

The Hochschild cohomology groups $\mathrm{HH}^q (A, A [2-q])$ control higher multiplications $m_q$ which form part of an A$_\infty$ structure on $A$. The vanishing for $q > 2$ implies that there cannot be any cocycles defining higher multiplications which are not coboundaries, i.e. there are no "interesting" higher structures on $A$. This is one way of understanding intrinsic formality.

For the exterior algebra (or indeed any Koszul algebra) a useful trick can be to notice that the exterior algebra on $n$ generators of degree 1, say, is Koszul-dual to the symmetric algebra on $n$ generators of degree 0. But Koszul-dual algebras have "the same deformation theory". In fact, their (bigraded) Hochschild complexes are "isomorphic in the homotopy category of B$_\infty$ algebras", which implies in particular, that their DG Lie structures given by Hochschild differential plus Gerstenhaber bracket are L$_\infty$ quasi-isomorphic. This is a result of Keller [1]. (You can find a short summary of the general theory in §4 of [2].)

The Hochschild cohomology groups $\mathrm{HH}^q (A, A [2-q])$ comprise the Hochschild cohomology of $A$ in total degree 2. On the Koszul-dual side this just matches the usual Hochschild cohomology in degree $2$, i.e. $\mathrm{HH}^2 (A^!, A^!)$, whenever $A^!$ is trivially graded (which is the case for $A^! = \mathrm{Sym} (V^*)$). For the symmetric algebra, the latter is just $\mathrm H^0 (\Lambda^2 \mathcal T_{\mathbb A^n})$ via the HKR isomorphism, as noted in the answer by YHBKJ.

You can therefore understand A$_\infty$ deformations of the exterior algebra $\mathrm{Sym} (V [-1])$ in terms of associative deformations of the polynomial ring $\mathrm{Sym} (V^*)$, which are nothing but deformation quantizations of affine $n$-space. This link between A$_\infty$ deformations and quantizations was studied by Calaque, Felder and Rossi [3].

Under this isomorphism, the condition $q > 2$ just turns into polynomial degree $> 2$. That is, an A$_\infty$ deformation of the exterior algebra using cocycles defining classes in $\mathrm{HH}^q (A, A[2-q])$ for $q > 2$ correspond to deformation quantizations of Poisson structures of degree $\geq 3$.

Now, to come to your examples:

• If $\dim V = 1$, then $A = \mathrm{Sym} (V [-1]) = \Bbbk [X] / (X^2)$ with $|X| = 1$ and $A^! = \Bbbk [x]$ with $|x| = 0$. We have $\mathrm{HH}^q (A, A[2-q]) \simeq \mathrm{H}^0 (\Lambda^2 \mathcal T_{\mathbb A^1})_q = 0$ for all $q$ (where the subscript $_q$ means "bivector fields with homogeneous degree $q$ coefficients").
• If $\dim V = 2$, then $A \simeq \Bbbk \langle X, Y \rangle / (X Y + Y X, X^2, Y^2)$ and $\mathrm{HH}^q (A, A[2-q]) \simeq \mathrm{H}^0 (\Lambda^2 \mathcal T_{\mathbb A^2})_q \simeq \Bbbk^{q \choose 2}$, reflecting the fact that on $\mathbb A^2$ with coordinates $u, v$ any bivector is of the form $f \frac{\partial}{\partial u} \wedge \frac{\partial}{\partial v}$, and the space of homogeneous degree $q$ polynomials in $u, v$ has dimension $q \choose 2$.
  Note that in dimension $2$ there are no obstructions since $\mathrm{HH}^3 (A^!, A^!) \simeq \mathrm H^0 (\Lambda^3 \mathcal T_{\mathbb A^2}) = 0$, so you can essentially define higher multiplications $m_k$ on the generators freely, for example there is an A$_\infty$ structure determined by setting, say, $m_3 (X, Y, Y) = X Y$ (which is indeed of degree $2 - 3 = -1$), corresponding to the quantization of the Poisson structure determined by $\{ y, x \} = x y^2$, etc.

In higher dimensions, this will get more and more involved, as you can see from the Koszul-dual side, where deformation quantizations of (polynomial) Poisson structures is a nontrivial problem.

[1] Keller, Bernhard, Derived invariance of higher structures on the Hochschild complex
[2] Barmeier, Severin; Wang, Zhengfang, A$_\infty$ deformations of extended Khovanov arc algebras and Stroppel's Conjecture
[3] Calaque, Damien; Felder, Giovanni; Rossi, Carlo A., Deformation quantization with generators and relations, J. Algebra 337, No. 1, 1-12 (2011). ZBL1235.53095.