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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.)

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

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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 $n=1$).

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  • $\begingroup$ That's excellent - thank you very much. $\endgroup$
    – aceincc
    Commented Jun 2, 2023 at 8:51
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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.

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