Dimension of subalgebras of a matrix algebra - MathOverflow

Dimension of subalgebras of a matrix algebra

2010-03-28T05:46:51Z

If n is given and A is a subalgebra of M_n(C), the algebra of n-by-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 C?

Answer by lieven lebruyn for Dimension of subalgebras of a matrix algebra

2010-03-28T09:27:19Z

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.

Answer by Yiftach Barnea for Dimension of subalgebras of a matrix algebra

2010-03-30T20:36:04Z

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).

Answer by abou for Dimension of subalgebras of a matrix algebra

2011-07-03T22:18:25Z

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 $\mathcal M_k(\mathbb C)$.