---
tag: H8RN439272
kind: definition
name: Deformation of algebras
msc: [16S80]
defines: QP92G4ZGJE
uses: [04C5ZBG1WT, 3CTW1P4HBP, 033WGN00BF]
sources:
  - url: https://en.wikipedia.org/wiki/Deformation_(mathematics)
---

## Definition

Let $k$ be a [[field|04C5ZBG1WT]] and let $A$ be an [[associative algebra|3CTW1P4HBP]] over $k$ with product $(a, b) \mapsto ab$.
Let $A[\![t]\!]$ be the $k[\![t]\!]$-module of [[formal power series|033WGN00BF]] with coefficients in $A$.
A *formal deformation* of $A$ is a $k[\![t]\!]$-bilinear associative product $*$ on $A[\![t]\!]$ of the form
$$a * b = ab + t\,\mu_1(a, b) + t^2 \mu_2(a, b) + \cdots \quad (a, b \in A),$$
where each $\mu_i \colon A \times A \to A$ is $k$-bilinear.
Two formal deformations $*$ and $*'$ are *equivalent* if there is a $k[\![t]\!]$-linear map $T = \id + t T_1 + t^2 T_2 + \cdots$, with each $T_i \colon A \to A$ $k$-linear, such that $T(a * b) = T(a) *' T(b)$ for all $a, b$.

## 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.