no classification of nilpotent lie groups

Question

there is no classification of (simply connected) nilpotent lie groups, but I am tempted to try to generalize the construction of the Heisenberg group. For an upper triangular matrix:

$$ \left( \begin{array}{cccc} 1 & \boxed{\cdot}& \boxed{\cdot}& \boxed{\cdot} \\ 0 & 1 & \boxed{\cdot} & \boxed{\cdot} \\ 0 & 0 & 1 & \boxed{\cdot}\\ 0 & 0 & 0 & 1 \end{array} \right) $$

here I use a $4 \times 4$ matrix, but could be $n \times n$. Each $\boxed{\cdot} \in \big\{ \mathbb{R}, \{ 0\} \big\} $ and since Lie groups are closed under multiplication (or lie algebras are closed under commutators if we populate one square with an $\mathbb{R}$) we might initially put a $\{0\}$ but have to complete to an $\mathbb{R}$.

If my combinatorial construction could generate make all simply connected nilpotent Lie groups, I would have found a basic classification theorem. Therefore, are there any such groups that live outside this classification?

The idea is... they are all matrix groups, but since there is no classification, I should be able to make a fairly general construction and there will be groups falling outside of this class.

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What if I changed the Euler number of the circle bundle over the torus? Does this have a matrix representation?

$$ (x,y,z) (x,y,z) = (\,x+x \,,\, y+y+ {\color{#9FB825}n}\,xz \,,\, z+z\,) $$

There are lots of examples of these simply connected nilmanifolds coming from matrices, but it's not hard to find examples without an obvious matrix representation (e.g. changing the structure constants), or a reasonable-looking classifiction scheme (that will eventually fail).

Answer 1

Here is a simple example of an uncountable family of nilpotent Lie groups. Fix real numbers $\lambda_1, \dots, \lambda_n$. Fix integers $k_1, \dots, k_n$. Let $V$ be the set of all functions of the form $f(x)=\sum_i e^{\lambda_i x} p_j(x)$ for $p_j(x)$ a polynomial function of degree at most $k_j$. Let $G=\mathbb{R} \times V$ with multiplication $(x_0,f_0(x))(x_1,f_1(x))=(x_0+x_1,f_0(x)+f_1(x-x_0))$. I proved in my paper https://arxiv.org/abs/1406.2487 that the values of $\lambda_j$ and $k_j$ are invariants under isomorphism. The proof is a long computation, expanding out the condition of being a Lie group morphism. It is easy to see that $G$ is nilpotent.

Edit: It is not so easy to see, because it isn't true. Actually this is solvable, but not nilpotent unless all $\lambda_j$ vanish. The paper mentioned above, Chong-Yun Chao, "Uncountably many nonisomorphic nilpotent Lie algebras", Proc. AMS. has a very simple construction. Take a vector space with basis $x_1,\dots,x_{n-4},y_1,\dots,y_4$ with brackets $[x_i,x_j]=\sum_{k=1}^4 c^k_{ij} y_k$ and all other brackets zero. He proves that if we take algebraically independent $c^k_{ij}$ over the rationals, then the various algebras we obtain for different choices of these $c^k_{ij}$ are not isomorphic over the real numbers. The paper is only 3 pages long and the argument is elementary.