# Classical deformation of algebras

Given a complex manifold (or a smooth scheme) $X$, the classical (infinitesimal) deformation theory is parametrized by the first cohomology with coefficients in the tangent sheaf $H^1 (X, T_X)$.

Modern deformation theory uses the second Hochschild cohomology $HH^2 (X)$, which parametrizes deformations of the algebra of functions on $X$. By the HKR isomorphism $HH^2$ splits into a direct sum $$HH^2 (X) \cong H^2 (X, \mathcal O_X) \oplus H^1 (X, T_X) \oplus H^0 (X, \Lambda^2 T_X)$$

A Hochschild 2-cocycle corresponding to a classical deformation (sometimes called "commutative deformation") should give a commutative algebra, which may (or may not, depending on integrability of the new complex structure) be a (equally) commutative algebra of functions on a (homeomorphic) surface with a different complex structure.

The term $H^0 (X, \Lambda^2 T_X)$ corresponds to (almost-)Poisson structures on $X$. On the algebra level, Poisson structures are those corresponding to making the product non-commutative.

I would like to know what the term $H^1$ corresponds to on the level of algebras.

For example, given an algebra of functions on a complex surface, e.g. $A = \mathbb C [a, b, c] / \langle ab-c^2 \rangle$, how can I construct a new commutative algebra (a "commutative" deformation of $A$), starting from the generators $a, b, c$ and relation $ab = c^2$.

• I would say $HH^2(X)$ parametrizes deformations of the category of quasi-coherent sheaves on $X$, not the algebra of functions. If $X$ is affine, there is certainly no $H^1(X, T_X)$ term. However, you seem to be looking not at global functions but at the homogeneous coordinate ring which depends on an ample line bundle. I wouldn't expect general deformations in $H^1(X, T_X)$ to preserve this line bundle. The deformations that do are parametrized by normal vectors to the projective embedding of $X$. Dec 28, 2014 at 18:53
• @PavelSafronov Thank you for your comment! Maybe I've been thinking about $HH^2$ in the wrong way and the terms $H^1$ and $H^0$ correspond to classical deformations of the space and non-commutative deformations of the algebra of functions, respectively, so that $H^1$ wouldn't necessarily be a commutative deformation of the algebra of functions... Dec 28, 2014 at 19:32