Deformation of algebras

Remarks

Comparing the coefficients of $t$ in $(a * b) * c = a * (b * c)$ gives $$a \mu_1(b, c) - \mu_1(ab, c) + \mu_1(a, bc) - \mu_1(a, b) c = 0,$$ so $\mu_1$ is a Hochschild $2$-cocycle, and equivalent deformations have cohomologous $\mu_1$. Example: for any $T = \id + t T_1 + \cdots$, the product $a * b = T^{-1}(T(a) T(b))$ is a formal deformation, since it is the original product carried over by $T$; it is called trivial. Non-example: on $A = k[x]$, let $\mu_1(f, g) = f(0) g(0)$ and $a * b = ab + t \mu_1(a, b)$. For $a = x$, $b = c = 1$ the left side of the cocycle identity is $x \cdot 1 - 0 + 0 - 0 = x \ne 0$, so $*$ is not associative. If the Hochschild cohomology $HH^2(A, A)$ is zero, then every formal deformation of $A$ is trivial, and $A$ is called rigid.