Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

For a given $n$, is there any characterization for the commutative subalgebras of $M_n(\Bbb{C})$? I would like to know how many commutative subalgebras there are for each possible dimension.

In view of Chapman's answer, I am refining my previous question:

Given $k\leq n$, is there any way of describing the commutative subalgebras of $M_n$ which are of dimension $k$.

share|improve this question
    
Some searching turned up eom.springer.de/m/m062790.htm. According to it there is a commutative subalgebra of dimension $r$ of the $n \times n$ matrices if and only if $$r< \left[ \frac{n^2}{4}\right] + 1$$ It mentions Schur's theorem, presumably the linear algebra version, but I don't immediately how it follows from that. The set of conjugacy classes of maximal commutative subalgebras is finite if $n<6$ and infinite if $n>6$. It does not explain the case $n=6$. –  Jan Jitse Venselaar Jun 22 '10 at 13:49
    
Of course, I meand $r \leq$... And Chapman's answer below has a much clearer explanation. –  Jan Jitse Venselaar Jun 22 '10 at 14:14
add comment

7 Answers 7

up vote 5 down vote accepted

If you are only concerned about commutative subalgebras of $M_n(\mathbb{C})$ then there is a fairly easy characterization. So any abelian algebra is generated by a single self adjoint element (spectral theorem). Call this element T. Then T is diagonalizable and so the algebra it form will be the algebra of polynomials over it. Since it is diagonalizable that is a unitarty $u$ with $uTu^*$ diagonal. And the algebra has dimension $k$ exactly when T has $k$ distinct non-zero eigenvalues.

Note: This is assuming that T is invertible. If T is not invertible then the polynomial algebra since it contains the constants will have dimension $k+1$.

So then we can view the algebra generated by T as an algebra of the form $u^*Au$ where A is an algebra of diagonal matrices.

share|improve this answer
6  
Owen, your remark only applies to self-adjoint subalgebras -- and I know we must all these days worship the god C-STAR ;) but some of us are reprobate in this regard. Try the algebra of 2 by 2 real-or complex matrices which satisfy the following constraints: top-left entry = bottom-right entry; bottom-left entry = 0. I'm guessing that you intended to restrict to self-adjoint subalgebras but forgot to specify this? –  Yemon Choi Jul 7 '10 at 3:29
    
(See also one of my favourite little facts: the Banach algebra C^1[0,1] embeds isomorphically as a closed subalgebra of a subhomogeneous von Neumann algebra...) –  Yemon Choi Jul 7 '10 at 3:30
    
Ah yes, I am most definitely restricting to self-adjoint algebras. –  Owen Sizemore Jul 7 '10 at 6:15
    
I actually needed a characterisation of self-adjoint commutative algebras, but I didn't realize it until I saw this answer. Thanks –  Carmen Jul 7 '10 at 11:04
add comment

See the related question at Dimension of subalgebras of a matrix algebra . In particular, I'd recommend the reference: M. Mirzakhani `A simple proof of a theorem of Schur' Amer. Math. Monthly 105 (1998), 260-262.

Schur's theorem states that the maximum dimension of a commutative subalgebra of $M_n(k)$ for a field $k$ is $1+[n^2/4]$. When $n=2m$ is even, such an algebra is given by matrices of the form $$\left(\begin{matrix} aI&B\\\\ O&aI \end{matrix}\right)$$ for $a\in k$ and $B$ an arbitrary $m$-by-$m$ matrix. Restricting $B$ to a linear subspace of $M_m(k)$ will give an commutative subalgebra of any dimension up to the maximum. These subalgebras have quite a simple structure up to isomorphism but their classification up to conjugacy by $\mathrm{GL}_n(k)$ looks intractable to me.

One can get commutative algebras not of this form, e.g., those of diagonal matrices. It may be possible to classify these up to isomorphism, but surely not up to conjugacy.

share|improve this answer
add comment

Although this is not an answer to the original question, readers might be interested to know of a cute open problem in the context of commutative subalgebras of $M_n(\mathbb{C})$. Let $A$ be such an algebra; we focus on the minimal number of generators of $A$ as a $\mathbb{C}$ algebra vis-a-vis the dimension of $A$ as a $\mathbb{C}$ space. If $A$ is generated by just one element, then by Cayley-Hamilton, the dimension of $A$ as a $\mathbb{C}$ space is bounded above by $n$. It is an old theorem of Gerstenhaber as also Motzkin and Taussky-Todd that if $A$ can be generated by two elements, then too, the dimension of $A$ is bounded above by $n$. OTOH, if $A$ needs at least four generators, there are examples to show that the dimension of $A$ can be greater than $n$. So here is the open question: are there commutative subalgebras $A$ that can be generated by three elements for which the dimension of $A$ as a $\mathbb{C}$ space is greater than $n$?

The work on this question has led naturally to study the variety of commuting triples of matrices (a subvariety of affine 3n^2 space over $\mathbb{C}$). This variety has been proved to be reducible (by Guralnick) for $n\ge 32$ (since sharpened to $n\ge 29$ by others); one set of enquiries focuses on determining irreducibility of this variety for smaller $n$. I believe it is known now that for $n\le 7$, this variety is irreducible. This translates to the result that for $n\le 7$, such an $A$ as above must indeed have dimension bounded by $n$.

Given the reducibility for general $n$ of this variety, another set of enquiries focuses on studying specific subvarieties of this commuting triples variety. For instance, studying the varieties of commuting triples $(M_1, M_2, M_3)$ where $M_3$ is fixed to be a matrix in which each eigenvalue appears in at most two blocks (aka "$2$-regular") leads naturally to generalized tangent varieties over determinantal varieties. On the other hand, fixing $M_3$ to be the nilpotent Jordan block of size $k$ repeated $m$ times along the diagonal (so $n = mk$) leads to generalized tangent varieties over the variety of commuting pairs of $m\times m$ matrices.

Interesting stuff, this!

share|improve this answer
add comment

The interesting part of this is the nilpotent part. You could ask for (but in vain, it seems) a classification of all nonunital commutative subalgebras of End(V) such that every element is nilpotent, or equivalently all (unital) commutative subalgebras of End(V) such that every element is scalar+nilpotent. I believe that choosing a subalgebra $A\subset End(V)$ is the same as (1) choosing a splitting $V=V_1\oplus\dots\oplus V_k$ into nontrivial subspaces, (2) choosing a nilpotent subalgebra $A\subset End(V_i)$ for each $i$, and (3) letting $A$ be $\prod_iA_i$. That is, if you define the $V_i$ in terms of $A$ as the largest subspaces preserved by $A$ on which $A$ acts by scalar+nilpotent then that is a splitting of $V$ and $A$ is the full product of its images in the $End(V_i)$.

share|improve this answer
add comment

This is a reply to Tom's reply. Let's stick to commutative subalgebras of $M_n(k)$ where $k$ is algebraically closed. Let $A$ be a unital commutative subalgebra of $M_n(k)$. Then $N=k^n$ is a faithful $A$-module. Let $J$ be the Jacobson radical of $A$. Then $A/J$ is a commutative semisimple algebra and so is isomorphic to $k^r$ for some $r$. By Wedderburn's principal theorem, $J$ splits off from $A$, that is there is a subalgebra $B$ of $A$ such that $B$ is isomorphic to $A/J$ and $A=B\oplus J$. In the algebra isomorphism $B\cong k^r$, the projection onto each factor defines an idempotent in each. Let them be $e_1,\ldots,e_r$. Then the $e_i$ are pairwise orthogonal and add to $1$. Then $N=e_1N\oplus\cdots\oplus e_rN$, so let $V_i$ equal $e_i N$. Similarly, as $A$ is commutative, $A=e_1 A\oplus\cdots \oplus e_r A$ and $A_i=e_iA$ is an algebra with identity $e_i$. Again, as $A$ is commutative $A_i$ acts faithfully on $V_i$ but annihilates $V_j$ for $j\ne i$. Also each element of $A_i$ is scalar plus nilpotent. This confirms Tom's observation (unless I have made a blunder somehwere (quite possible)).

share|improve this answer
    
I should add that Tom's observation essentially reduces the isomorphism classification of commutative subalgebras of $M_n(\mathbb{C})$ to that of nilpotent commutative subalgebras. –  Robin Chapman Jun 22 '10 at 16:13
add comment

I think the general problem that given $k \leq n$, does there exist a maximal commutative subalgebras of dimension $k$ is still open. See for example Laffey's paper, where he proves that $k > (2n)^{2/3} -1 $.

share|improve this answer
add comment

Regarding the question: ``Given $k \le n$, is there any way of describing the commutative subalgebras of $M_n$ which are of dimension $k$:'' It is easy to see that the set of commutative subalgebras of $M_n$ of any dimension $d$ form a subvariety of the Grassmanian $G(d,n^2)$ of $d$-dimensional subspaces of $n^2$ dimensional space (all over the given base field $\mathbb{C}$, of course). It is easy to write down the equations locally, in any patch of the Grassmanian where a given determinant of size $d$ doesn't vanish: In such a patch, it is standard that one can extract basis vectors for a given subspace $W$ as $\pm$ its Plucker coordinates. So, if $b_1$, $\dots$, $b_d$ are basis vectors, our equations read $b_ib_j$ (the matrix product written out as a vector) wedged with $b_1 \wedge\dots\wedge b_d = 0$, $b_ib_j - b_jb_i = 0$, and $I_n$ (the indentity matrix written out as a vector) wedged with $b_1 \wedge\dots\wedge b_d = 0$.

It is known for instance that the variety of commuting subalgebras of $M_n$ of dimension $n$ is not irreducible for large $n$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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