18
$\begingroup$

If $n$ is given and $A$ is a subalgebra of $M_n(\mathbb C)$, the algebra of $n \times n$ matrices with entries in the field of complex numbers, then what are the possible values of dimension of $A$ as a vector space over $\mathbb C$?

$\endgroup$
5
  • $\begingroup$ Ok, I just found out that there is an interesting result due to Schur which gives a partial answer to my question. Here it is for those who are interested: If F is a field, then there exists a "commutative" subalgebra A of M_n(F) with dim_F A = k if and only if k \leq [n^2/4] + 1, where [ ] is the floor function. I'm starting to think that there exists a subalgebra of M_n(F) of any dimension! $\endgroup$
    – abcba
    Mar 28, 2010 at 8:02
  • 1
    $\begingroup$ @abcba: here's a hint for constructing that commutative subalgebra: write an n x n matrix as (A B;C D) with A,B,C,D n/2 x n/2 matrices, and then consider the space with A=C=D=0. To get further start eating into B. Add scalar multiples of the identity if you're the sort of person whose algebras have to contain 1. $\endgroup$ Mar 28, 2010 at 8:05
  • 6
    $\begingroup$ Is there an 8 dimensional subalgebra of M_3? $\endgroup$ Mar 28, 2010 at 8:35
  • 5
    $\begingroup$ One can get all dimensions up to $n(n+1)/2$ by using subalgebras of upper triangular matrices. We can alo get some larger examples by the construction $(A\ B;0\ D)$ where $A$ and $D$ run through given subalgebras of $M_k$ and $M_{n-k}$ and $B$ is arbitrary. Some dimensions are not accessible by these constructions, e.g., dimension $8$ when $n=3$. Are there any subalgebras with these dimensions? $\endgroup$ Mar 28, 2010 at 8:49
  • 3
    $\begingroup$ A nice proof of Schur's theorem is at M. Mirzakhani `A simple proof of a theorem of Schur' Amer. Math. Monthly 105 (1998), 260-262. $\endgroup$ Mar 28, 2010 at 8:51

3 Answers 3

12
$\begingroup$

Rough answer : almost all small dims can appear, there are some restrictions to large dims.

For example, considering 1 matrix all dims between 1 and n appear. Taking centralizers of these all numbers of the form sum a_i^2 where a is a partition of n appear.

In general, consider k-tuples of positive integers a and b such that their scalar product a.b=n (a should be thought of as the Morita setting, b as the matrix-sizes of the semi-simple part of the subalgebra), then any number of the form

sum b_i^2 + subsum b_ib_j

is possible (here 'subsum' means that one takes all terms b_xb_y for all x,y in a substring

1 <= i_1 < i_2 < ... < i_l <=k for any 0<=l<=k)

Edit : the subsum gives the dimension of the Jacobson radical. This answer cannot be the final one, as it only detects the subalgebras of global dimension 1. For example any n-diml algebra can be embedded in nxn matrices.

There are some obvious restriction wrt large dimensions. For example, there cannot be an 8-dml subalgebra of 3x3 matrices as its semi-simple part can be at most C x M_2(C) and so its dimension must be smaller or equal to 7.

For general n there cannot be subalgebras with dimensions between the dim of the largest parabolic subgroup of GL(n) and n^2.

Edit : a closely related question can be found here : problems concerning subspaces of mxm matrices.

$\endgroup$
5
$\begingroup$

Soit $E$ un $\mathbb C$-espace vectoriel de dimension $n$. J'ai démontré entre autres les deux résultats suivants dans un article à paraître dans la revue française Quadrature :

  • On suppose que $k$ vérifie les inégalités $k \ge 2$ et $k^{2}\le n$. Soit $\mathcal{A}$ une sous-algèbre de $\mathcal{L}(E)$ qui vérifie la relation $n^{2}-kn+k^{2}-k+1 < \dim \mathcal{A} < n^{2}-kn+n.$ Alors, $\mathcal{A}$ vérifie la relation $\dim \mathcal{A}=n^{2}-kn+k^{2}.$

  • Soient $n$ un entier naturel et $p$ un entier de l'intervalle $[0,n^{2}].$ On suppose $p$ écrit sous la forme $p=n(n-k)+t,\ 0\le t \le n-1$. Alors il existe une sous-algèbre de dimension $p$ dans $\mathcal M_n (\mathbb C )$ si et seulement s'il existe une sous-algèbre de dimension $t$ dans $\mathcal M_k(\mathbb C)$.

$\endgroup$
2
3
$\begingroup$

I think that the fact that every proper subalgebra is contained in am maximal parabollic follows immediately from Jacobson's density theorem because if a subalgebra does not preserve any subspace, then $C^n$ is a simple module for it. This is of course true over any field.

In the case of Lie algebras rather than associative algebras, then a classification of maximal subalgebras of finite dimensional simple Lie algebras over the complex numbers was obtained by Dynkin. In the positive characteristic case a classiifcation can probably be obtained using arguments which were used for the classifcation of maximal subgroups of finite simple groups. This is at least what I understood talking to Liebeck and Seitz, but I am not an expert on these matters.

However, in the Lie case an elementary argument that the maximal dimension of a proper subalgebra of $sl_n(F)$ is $n^2-n$, assuming $F$ has characteristic different than 2 can be found in Y. Barnea and A. Shalev, Hausdorff dimension, pro-p groups, and Kac-Moody algebras, Trans. Amer. Math. Soc. 349 (1997), 5073-5091 (Theorem 1.7). Other related stuff (related to possible dimensions) but more on the group theoretic side can be found in the same paper. A generalization of this to other classical Lie algebras can be found in Abért, Miklós; Nikolov, Nikolay; Szegedy, Balázs Congruence subgroup growth of arithmetic groups in positive characteristic. Duke Math. J. 117 (2003), no. 2, 367--383 (Theorem 4).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.