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Consider a pair of matrices $(c_1, c_2)$. The words "it generates the 2-dimensional Lie algebra", means that there exists a pair of scalars $k_1$, $k_2$, such that $$[c_1, c_2] = k_1 c_1 + k_2 c_2,$$ where $[a,b]$ is the "commutator" $ab-ba$.

Not any pair generates 2-d Lie algebra.

Question: what is the dimension of the subset of matrices $(c_1,c_2)$ which generate 2-d Lie algebra? at least for $2\times 2$ matrices ?

It is clearly greater than $n^2+n$ since I can take - $c_1$ - arbitrary (so $n^2$) and $c_2$ - commuting with $c_1$, which gives ($+n$ for generic matrices). But in this way I get only $[c_1, c_2] =0$.

The motivation for the question comes from this question on MathOverflow.

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2 Answers 2

It seems the dimension is n^2+n. The subtlety is that this manifold is highly reducible, there are about n^2+1 components, and n^2 of them of small dimensions, the only one component of dimension n^2+n is [A,B]=0.


Consider [A,B]=aA+bB. Take A - diagonal matrix with different values on the diagonal. Consider a, b non-zeros.

Remark 1. [A, * ] - has always a trivivial diagonal.

Corollary diag(B) = - aA/b.

Remark 2. [A, *] on the space of matrices with trivial diagonal has eigen matrices E_{ij} with eigenvalues (a_i - a_j).

Corollary off-diag of B is equal to kE_{ij} for some i,j and b=(a_i-a_j).

Hence if we fix A - diagonal, and a,b (non-zeros), then matrix B is determined uniquely as B= -aA/b + E_{ij}/(a_i-a_j)

Hence the dimension of the of the variety for fixed a,b which are non-zero is n^2.

While for a,b=0 we have n^2+n .

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Alexander, if you want the dimensions of the components of the variety $V=\{(A,B)\in M_n(\mathbb{C})|[A,B]\in span(A,B)\}$, then you cannot fix the coefficients except if they are $0$ , that is a special case. Let $V_A=\{B|(A,B)\in V\}$ where $A$ is a generic matrix. Clearly, the component of $V_A$ with maximal dimension is the commutant $C(A)$ that has dimension $n$. Now consider $W_A=V_A\setminus C(A)$ ; then, for every $n$, the component of maximal dimension has dimension $2$ and not $0$. Indeed such matrices $A,B$ are said quasi commutative and they are simultaneously triangularizable.

Then the case $n=2$ is easy. Choose randomly $A=\begin{pmatrix}a_1&a_2\\0&a_3\end{pmatrix}$ and seek $B$ in the form of a upper triangular matrix s.t. $[A,B]=uA+vB$ with $(u,v)\not=(0,0)$.

More generally, for every $n$, you randomly choose $u,v$ (there are exceptional values that do not work !) and you obtain a sole solution in $B$.

EDIT: in fact we obtain $2$ types of components that have dimension $2$. Let $(a_i)_i$ be the diagonal of the triangular matrix $A$.

Type 1: the previous one that is locally parametrized par $(u,v)$ where $v\not=a_i-a_j$ with $i<j$.

Type 2: Let $v=a_i-a_j$ where $i<j$. There is a component parametrized by $u$ and an entry of $B$. There are $\binom{n}{2}$ such components.

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