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

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Why do you want to know? –  François G. Dorais Jun 7 '10 at 18:41

4 Answers 4

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).

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I am not sure I know why Burnside theorem implies the list of subalgebras of M_2(C) you suggested. (I am interested in subalgebras with 1, so your list is ok to me). Also, what if instead of $C$ you take $k(X)$ where $k$ is any field? –  chana Jun 9 '10 at 15:34
@chana. Burnside theorem shows that a proper subalgebra of M_2(C) is reducible. That means that there is a non-zero vector in $C^2$ which is an eigenvector for all matrices in the subalgebra. Using that vector as the first element of a new basis, you can check that, in that new basis, all elements of the subalgebra are upper triangular matrices. Then it is not very difficult to list all the subalgebras of 2x2 upper triangular matrices (First case: all elements are diagonal matrices. Second case: there is an element which is not diagonal). –  user6129 Jun 10 '10 at 7:55

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.

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The question is asking for the subalgebas of one matrix algebra, as far as I can tell, which is a different problem than classifying all f.d. algebras with a faithful representation (ie, all algebras) –  Mariano Suárez-Alvarez Jun 8 '10 at 12:10
Yes, you can do it for a few small values of $n$, but there will be (a) moduli of associative algebras of a given dimension $k$; and (b) moduli of individual wild algebras present even for a fixed $n$ once it is large enough. –  Victor Protsak Jun 8 '10 at 15:34
Heh. Of course there will be module as in (a)! I was just observing that the OP's question was slightly different than the one you answered. What do you mean by "moduli of individual wild algebras"? –  Mariano Suárez-Alvarez Jun 8 '10 at 20:34
I am referring to "wild representation type": classification of its modules contains as a subproblem classifying all pairs of matrices up to conjugation. By Drozd's theorem, a finite-dimensional algebra $A$ is either of finite, tame, or wild type. There are parameters in isomorphism classes of $A$-modules $\implies$ $A$ is not finite or tame $\implies$ $A$ is wild. –  Victor Protsak Jun 9 '10 at 0:07

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...

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Even projective is probably too much to ask for: the regular module of an $n$-dimensional algebra is an $n$-dimensional projective module. So in this case, an apparent subproblem is to classify all $n$-dimensional associative algebras. Good luck with that... –  Victor Protsak Jun 9 '10 at 6:29
I disagree, Victor. The problem is "to describe" and not "to classify". If $V$ is a regular representations then choosing a generic element identifies $V$ with the algebra. Thus, it is exactly describing associative algebra structures on $V$. This can be described in a useful way by writing multiplication tensor!! –  Bugs Bunny Jun 9 '10 at 18:15
And, off course, classifying them will require quotients and basically hopeless unless you restrict a certain class of algebras, like semisimple or with square of Jacobson rad zero or 3-nilpotent. –  Bugs Bunny Jun 9 '10 at 18:17

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'.

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This variety has been studied; see Remark 1.2 in front.math.ucdavis.edu/0608.5491 for a survey of the literature. –  David Speyer Jun 8 '10 at 14:09
I think that at least part of the problem is that the geometric invariant theory quotient only captures generic orbits, whereas for the true classification, you need to know $\textit{all}$ orbits. So the quotient in the sense of classifying all isomorphism types will be a horrible non-separated mess with pieces of different dimensions. This may conceivably happen even if the generic orbits have a manageable description (e.g. think of $AB=BA$ where $A$ and $B$ are semisimple). –  Victor Protsak Jun 8 '10 at 15:44
Ah! Thanks for the reference, David. –  Mariano Suárez-Alvarez Jun 8 '10 at 20:32
Mariano, this quotient space will bring you no sunshine. Think, for example, of n-tuples of matrices: we all know and love the basic invariants (traces of products) but the problem of classifying them is as wild as a hare. –  Bugs Bunny Jun 9 '10 at 18:04
@BB: I did not expect sunshine, but was only describing the obvious approach to obtaining a set of moduli :) –  Mariano Suárez-Alvarez Jun 9 '10 at 20:38

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