I have often heard the slogan that "a matrix algebra has no deformations," and I am trying to understand precisely what that means. While I would be happy with more general statements about finite-dimensional semisimple algebras over non-necessarily algebraically closed fields, I am mainly interested in the case when the base field is $\mathbb{C}$, so the question is phrased in terms of matrix algebras over $\mathbb{C}$. I will state the question in three different contexts, in increasing order of interest to me, and say what I know about each one.


Formal Deformations
-------------------


A formal deformation of an associative $k$-algebra $A$ is a $k[[h]]$-bilinear multiplication on $A[[h]]$ of the form
$$ a \cdot b = ab + m\_1(a,b)h + m\_2(a,b)h^2 + \dots $$
where the first term is given by the original multiplication on $A$. This type of deformation is connected with the Hochschild cohomology $HH^\bullet(A)$. As I understand it, for a semisimple algebra $A$, the Hochschild cohomology vanishes in degree $\ge 1$, and this implies that a matrix algebra has no deformations in the formal sense. So I guess there isn't much of a question so far.


Deformation of Structure Constants
----------------------------------


In this context, we fix a natural number $n$ and a vector space $V$ with a fixed basis $a\_1, \dots, a\_n$, and we consider associative algebra structures on $V$. The *structure constants* of an algebra $A$ with underlying vector space $V$ with respect to this basis are the complex numbers $c\_{ij}^k$ such that
$$ a\_i a\_j = \sum\_{k}c\_{ij}^k a\_k, $$
and these numbers clearly determine the algebra. This gives us a point in $\mathbb{C}^{n^3}$. Associativity gives us some polynomial constraints on the structure constants, so we can think of the collection of $n$-dimensional associative algebras as some kind of algebraic variety in $\mathbb{C}^{n^3}$. (I know very little algebraic geometry, so please excuse/correct me if am using the wrong terminology.) Since one can take many different bases for an algebra, this variety overcounts associative algebras.


I don't know how this is proved, but as I understand things, semisimplicity is a Zariski-open condition in the variety of associative algebra structures, so semisimplicity is preserved under small deformations, and so a small deformation of $M\_n(\mathbb{C})$ is still semisimple.


### Question 1



> 
> Is the property of being a matrix algebra preserved under deformation of structure constants?
> 
> 
> 


Deformation of Relations
------------------------


This is the most flexible setting, and the one in which I am most interested. Here we consider algebras with a fixed collection of generators and a fixed number of relations, and we allow the relations to vary smoothly. This allows the dimension of the algebra to vary as well. However, this is also the hardest notion to formalize.


An example will illustrate what I mean. Consider the family of algebras
$$ A\_t = \mathbb{C}[x]/(tx^2 -x),$$
where $t$ varies in $\mathbb{C}$. Clearly there is a drop in dimension at $t=0$, although generically the dimension is constant. A slightly more complicated example is given by the family $B\_t$, where 
$$ B\_t = \mathbb{C} \langle x,y \mid x^2 = y^2 = 0, \quad xy + yx = 2t \rangle.$$
In this example the dimension is constant, but at $t=0$ one has the exterior algebra $\Lambda(\mathbb{C}^2)$, while for $t \neq 0$ this is the Clifford algebra of $\mathbb{C}^2$ with respect to a nondegenerate bilinear form, and hence $B\_t \simeq M\_2(\mathbb{C})$ for $t \neq 0$.


### Question 2



> 
> What is a good way to formalize this notion of deformation?
> 
> 
> 


A first guess would be something like this: fix a space of generators $V$, a number of relations $k$, and a base manifold $M$ (in my above examples $M=\mathbb{C}$). To limit things a bit, I guess I would like to consider only quadratic-linear relations. Allowing these relations to vary means giving a smooth function 
$$F : M \to \mathrm{Gr}\_k(\mathbb{C} \oplus V \oplus V^{\otimes 2})$$
and considering the "bundle" (sheaf?) of algebras 
$$ A\_p = T(V)/ \langle F(p)\rangle, \quad p \in M $$
over $M$.


However, I'm not sure if that completely captures what I want. I might also insist that the projection of each $F(p)$ onto $V^{\otimes 2}$ has no kernel (which is not the case for the first example I described, but is the case for the second).


### Question 3



> 
> Given the setup just described (or an appropriately modified version of your choosing), what are some conditions on $M$, $F$ such that the property "$A\_p$ is a matrix algebra" is local on $M$?
> 
> 
> 


For this question I am mainly interested in the cases when $M = \mathbb{R}$ or $M = \mathbb{C}$, although more general base spaces are of course interesting as well.


I have heard about something called the *Azumaya locus* which may be related to this, but I don't really know anything beyond the name.


I didn't realize this would be so long! If you're still with me, thanks for reading.



A matrix algebra has no deformations?
***Deformation of relations***


Answer to **question 2** is the following: a deformation of an algebra $A\_0$ parametrized by a pointed affine scheme $\*\to X=Spec(B\to k)$ is the data of a $B$-algebra $A$ such that $A\_0\cong A\otimes\_B k$. 


Observe that the kind of deformations your are looking at in question 1 are those that are called "flat". 




---


***Deforming structure constants of matrix algebras***


In order to find the answer to **question 1** (which is yes) I would suggest you to use [Artin–Wedderburn theorem](http://en.wikipedia.org/wiki/Artin-Wedderburn_theorem). 


EDIT 1: more precisely, over $\mathbb{C}$ any finite-dimensional (semi-)simple algebra is isomorphic to (a Cartesian product of) matrix algebras.


EDIT 2: Anyway, it seems that there is a more general phenomenon (see the fondational paper of Gerstenhaber, *On the deformation of rings and algebras*, Ann. of Math. 79 (1), (1964),
59–104): 
$$
\textrm{absolute rigidity}\Rightarrow \textrm{analytic rigidity}\Rightarrow \textrm{geometric rigidity}
$$
where *absolute rigidity* means that $HH^2(A)=0$, *analytic rigidity* means that there are no non-trivial formal deformations of $A$, and *geometric rigidity* means that the $GL(V)$-orbit of the point defined by $A=(V,c\_{ij}^k)$ in the variety of structure constants on $V$ is open (this tells you in particular that sufficiently small deformations are trivial). You can conclude by using that 
$$
\textrm{semi-simplicity}\Rightarrow\textrm{absolute rigidity}
$$




---


***Azumaya algebras***


Finally, if I understand well your **question 3**, your guess seems correct: the right notion to look at is precisely the one of [Azumaya algebra](http://en.wikipedia.org/wiki/Azumaya_algebra) (if you are interested by the relation to Cherednick algebras mentionned in the answer of Daniel Larsson, I would suggest you to take a look at the last two or three chapters of [this book](http://arxiv.org/abs/math/0606233) by Pavel Etingof). 




---


*Final remark*


Let me also mention that there is a lot of work about deformation theory of quadratic algebras by sub-quadratic ones (see references given in [this MO answer](https://mathoverflow.net/questions/80627/koszul-duality-between-weyl-and-clifford-algebras/80666#80666) - EDIT: sorry, you already know this since you're the one who actually asked the question!)


