Let H be the Heisenberg group defined as $ H=\left\{\begin{pmatrix} 1&a&c\\ 0&1&b\\ 0&0&1\\ \end{pmatrix}: a,b,c \in \mathbb{R}\right\}$. The lie algebra $ \mathfrak{h} $ is spanned by the matrices $ \left\{h_i=\begin{pmatrix} 0&1&0\\ 0&0&0\\ 0&0&0\\ \end{pmatrix}, \begin{pmatrix} 0&0&0\\ 0&0&1\\ 0&0&0\\ \end{pmatrix}, \begin{pmatrix} 0&0&1\\ 0&0&0\\ 0&0&0\\ \end{pmatrix}\right\} $. The adjoint rep of H is given by $ X\mapsto X^{-1}hX $, where $ X \in H, h \in \mathfrak{h}$. \
Is the adjoint rep of H decomposible? $ h_3 $ forms an invariant subspace and $ h_2 \mapsto h_1+bh_3, \text{ } h_2\mapsto h_2+ah_3 $ under the adjoint rep, if $ X=\begin{pmatrix} 1&a&c\\ 0&1&b\\ 0&0&1\\ \end{pmatrix} $. My answer would be no, since the subspace spanned hy $ {h_1,h_2} $ is "shifted" in the direction of $ h_3 $. How can i make this argument formal?
We will use standard linear algebra to show that any non-zero $H$-invariant subspace of $\mathfrak{h}$ contains $h_3$, and hence the intersection of any two non-zero invariant subspaces is non-zero (in particular, $\mathfrak{h}$ is indecomposable). The idea is simply to use the fact that a linear operator on a finite dimensional vector space has at least one eigenvector, combined with the observation that the generators of $H$ have very few eigenvectors.
Let $M \subseteq \mathfrak{h}$ be a non-zero $H$-invariant subspace and consider the conjugation action of $$g_1=\left(\begin{matrix} 1 & 1& 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)$$ on $M$. By direct calculation the only eigenvalue for the $g_1$ action on $\mathfrak{h}$ is $1$, and the corresponding eigenspace is the span of $h_1$ and $h_3$ (with your notation above). Thus $M$ contains a non-zero linear combination of $h_1$ and $h_3$. Since each of $M$ and the span of $h_1$ and $h_3$ is stable by $$g_2=\left(\begin{matrix} 1 & 0& 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} \right)$$ so is their intersection, which is non-zero by the above reasoning. Thus the intersection must contain a non-zero eigenvector for the conjugation action of $g_2$. These are precisely the non-zero linear combinations of $h_2$ and $h_3$, so it follows that $h_3 \in M$.