Subalgebras of matrices - MathOverflow

Subalgebras of matrices

2010-06-07T18:33:58Z

Given a matrix algebra over a field, can one describe all its subalgebras?

Answer by coudy for Subalgebras of matrices

2010-06-07T19:05:00Z

A starting point for a classification (up to conjugation) is the Burnside theorem:

there is no irreducible subalgebra in M_n(C) other than M_n(C) and {0}.

An elementary proof is given by Lomonosov Rosenthal (2003), I think it can be found online. There are also versions of the Burnside theorem for the field of real numbers R and the quaternions H.

With this theorem at hand, you can easily list all subalgebras of M_2(C). Restricting to the subalgebras containing id, we get the upper triangular matrices, upper triangular with the two diagonal terms being equal, diagonal matrices, diagonal matrices with the two diagonal terms being equal (and I think that's all, up to conjuguacy).

I would guess however, that there is no algorithm that can decide if two matrix algebras on some arbitrary field are isomorphic in general (but I may be wrong on that point).

Answer by Bugs Bunny for Subalgebras of matrices

2010-06-07T23:46:08Z

Probably, not in any useful way. You really want to restrict your class of subalgebras, say to irreducible ones, as in coudy's reply, in which case you will have Jacobson's density condition. I can imagine anice description of "projective" ones, i.e. the natural module is projective, or "completely reducible" ones.

But the general question is probably a wild problem...

Answer by Victor Protsak for Subalgebras of matrices

2010-06-08T00:18:09Z

I agree with Bugs Bunny. This is equivalent to classifying all algebras with a faithful $n$-dimensional representation. Of course, such algebras are finite-dimensional, but every finite-dimensional algebra, no matter how complicated, has a faithful representation (e.g. regular action), and representations can be arbitrarily complicated. So it is a wild problem.

Answer by Mariano Suárez-Alvarez for Subalgebras of matrices

2010-06-08T12:28:37Z

Fix $n$ and $d$ such that $1\leq d\leq n$, and consider $d$ generic matrices $x_1=(x_{1,i,j})_{1\leq i,j\leq n},\dots,x_{d}=(x_{d,i,j})_{1\leq i,j\leq n}$ (meaning: consider the polynomial ring $k[X]$ in $dn^2$ variables $x_{i,j,k}$ with $1\leq i\leq d$ and $1\leq j,k\leq n$, and construct the matrices there) There are $d^2$ polinomials $P_{i,j}\in k[X]$ whose vanishing express the fact that the matrix product $x_d\cdot x_d$ is a linear combination of the $x_1,\dots,x_n$, and there is a polynomial $U\in k[X]$ whose vanishing expresses the condition that the unit matrix $1\in M_n(k)$ is a linear combination of $x_1,\dots,x_d$, and there are polynomials whose non-vanishing express that the matrices $x_1,\dots,x_d$ are linearly independent over $k$.

It follows that the common zero set $\mathcal S$ of all these polynomials in $k^{dn^2}$ (well, to handle the non-vanishing ones, one needs to add a few more variables, and so on) can be identified with the set of subalgebras of $M_n(k)$ with a chosen basis. It is more or less clear that there is an action of $\mathrm{GL}(d,k)$ on $\mathcal S$, by «change of basis», whose orbits correspond to subalgebras of $M_n(k)$ of dimension $d$.

Now, it is not obvious that the quotient $\mathcal S/\mathrm{GL}(d,k)$ is a nice variety...

Maybe one can describe the quotient as the subvariety of the Grassmanian of subspaces of $M_n(k)$ of dimnsion $d$ satisfying appropriate conditions, but I cannot see off-hand how to express that a subspace is closed under matrix multiplication in term of its Plücker coordinates, though.

NB: Any parametrization of subalgebras of $M_n(k)$ is going to have parameters (ie, depend on a point in some 'variety', as the tautological parametrization with elements of $\mathcal S/\mathrm{GL}(d,k)$) as there are positive-dimensional families of subalgebras. The smallest example of a curve which does not generically repeat isomorphism types, I think, is the curve of 4-dimensional subalgebras of $M_4(k)$ of 'quantum exterior algebras'.