Faithful representation of the Heisenberg group I have been trying to solve a problem concerning the Heisenberg Lie group $H$. Show that there does not exist a faithful representation $\rho:H\to\text{GL}(2,\mathbb{R})$.
Any ideas about how to solve it would be very appreciated.
 A: Let $\mathfrak{h}_m$ be a $2m+1$-dimensional Heisenberg Lie algebra, with basis $e_1,\ldots e_m,f_1,\cdots ,f_m,z$ and Lie brackets $[e_i,f_i]=z$ for all $i$. One can show that the minimal dimension of a faithful representation of $\mathfrak{h}_m$ is $\mu(\mathfrak{h}_m)=m+2$ - see here, Lemma $1$. For the $3$-dimensional Heisenberg Lie algebra this says $\mu(\mathfrak{h}_1)=3$. Hence there is no $2$-dimensional faithful representation $\rho \colon\mathfrak{h}_1\rightarrow \mathfrak{gl}_2(\mathbb{R})$.
A: The Lie group representation $\rho$ induces a Lie algebra representation $$\pi:\mathfrak{h}\to\mathfrak{gl}(2,\mathbb{R})$$
Now, $\mathfrak{h}$ is a nilpotent Lie algebra, so by Engel's theorem, there is a basis of $\mathbb{R}^2$ in which all matrices of $\pi(\mathfrak{h})$ are strictly upper-triangular. But the space of strictly upper-triangular matrices in $\mathfrak{gl}(2,\mathbb{R})$ is $1$-dimensional, while $\mathfrak{h}$ is $3$-dimensional. So $\pi$ is not injective, and hence $\rho$ cannot be injective either.
A: Here's a proof (I won't even assume $\rho$ continuous). Assume by contradiction that there exists a homomorphism $H\to\mathrm{GL}_2(\mathbf{C})$ that is nontrivial on $[H,H]$. Then it maps $[H,H]$ to a nontrivial subgroup of $\mathrm{SL}_2(\mathbf{C})$. Since the squaring map is surjective on $[H,H]$, it's also true on $\rho([H,H])$, and hence the latter cannot be equal to $\{I_2,-I_2\}$. So $\rho([H,H])$ contains a matrix which is not a scalar multiple of the identity. Since $[H,H]$ is central, $\rho(H)$ is contained in the centralizer of $\rho([H,H])$. But any non-scalar matrix in $\mathrm{GL}_2(\mathbf{C})$ has an abelian centralizer. It follows that $\rho(H)$ is abelian and hence $\rho([H,H])$ is trivial, a contradiction.
Corollary: there is no injective homomorphism $H\to\mathrm{GL}_2(\mathbf{R})$.
