# Non-reductiveness of upper trianguler matrices $\mathfrak{t}_{n}$ as a Lie algebra

I am having difficulty with a rather simple proposition.

Let $$\mathfrak{t}_{n}$$ constitute the $$n \times n$$ upper triangular matrices over $$\mathbb{C}$$ with the usual matrix commutator, and $$\mathfrak{n}_{n}$$ those that are strictly upper triangular. A Lie algebra is reductive if it is the direct sum of its center and its simple ideals.

I would like to show that $$\mathfrak{t}_{n}$$ is not reductive for all $$n>1$$.

Intuitively, I would expect that because $$\mathfrak{t}_{n}$$ contains $$\mathfrak{n}_{n}$$ as an ideal, and $$\mathfrak{t}_{n} / \mathfrak{n}_{n} \cong \{ n \times n$$ diagonal matrices$$\}$$, which is not an ideal in $$\mathfrak{t}_{n}$$ (by considering the $$n=2$$ case and computing a commutator), that $$\mathfrak{t}_{n}$$ is not reductive; as the ideal $$\mathfrak{n}_{n}$$ has no complement. However, I am not convinced that this is the proper reasoning, nor rigorous.

In particular, how would this show that the direct sum of the center and simple ideals is not $$\mathfrak{t}_{n}$$ ($$\mathfrak{n}_{n}$$ is notably not semisimple by the characterization of finite dimensional semisimple Lie algebras)? Can one show that there are no non-Abelian simple proper ideals? Even for the $$n=2$$ case, I can think of no more elegant solution to show non-reductiveness other than brute ideal-checking with the three vector subspaces.

I would appreciate any direction towards the general case, or a nice way to do the $$n=2$$ case that is indicative towards the prior. (Preferably with as little machinery as possible, as the problem itself seems quite simple.)

• Show that any simple ideal in that algebra has dimension 1 (so technically is not simple…) Commented Jun 2, 2023 at 8:40

## 2 Answers

This can be solved, for example, by using some properties of reductive Lie algebras.

Suppose $$\mathfrak g$$ is reductive. Then, as you mentioned, $$\mathfrak g = Z_{\mathfrak g} \oplus \mathfrak g_{ss}$$, where $$Z_{\mathfrak g}$$ is the center and $$\mathfrak g_{ss}$$ is the semisimple part, i.e. the part which splits as direct sum of the simple ideals: $$\mathfrak g_{ss} = \mathfrak g_1 \oplus \dots \oplus \mathfrak g_k$$.

Now, $$[\mathfrak g_i, \mathfrak g_j]$$ is zero if $$i \neq j$$ (this is what it means to have a direct sum) and is an ideal of $$\mathfrak g_i$$ if $$i=j$$. Because $$\mathfrak g_i$$ is simple it must be $$[\mathfrak g_i, \mathfrak g_i] = \mathfrak g_i$$.

Thus, $$[\mathfrak g_{ss},\mathfrak g_{ss}] = \mathfrak g_{ss}$$ and, since $$Z_g$$ is abelian, $$[\mathfrak g,\mathfrak g] = \mathfrak g_{ss}$$ so that $$\mathfrak g = [\mathfrak g,\mathfrak g] \oplus Z_{\mathfrak g}$$.

In your case, $$Z_{\mathfrak t_n}$$ is the scalar multiples of the identity and $$[\mathfrak t_n,\mathfrak t_n] = \mathfrak n_n$$, and the direct sum of those does not give the whole $$\mathfrak t_n$$ (unless $$n=1$$).

• That's excellent - thank you very much. Commented Jun 2, 2023 at 8:51

The Lie algebra $$\mathfrak{t}_n$$ is solvable, because the commutator series terminates with zero. A nonzero solvable Lie algebra is reductive if and only if it is abelian, because the solvable radical of a reductive Lie algebra is the center, which is abelian. However, obviously $$\mathfrak{t}_n$$ is not abelian for $$n>1$$, and hence not reductive.