Let $\lambda$ be an irrational number. Let $G \subset G_2(\mathbb{C})$ be defined as $G = \left\{ \mathrm{diag} (e^{ti}, e^{\lambda ti}) \mid t \in \mathbb{R} \right\}$. Prove that $G$ is not a manifold.
The definition I am using for a manifold of dimension $n$ is a set $X \subset \mathbb R^m$ such that, for all $x \in X$, there exists a neighborhood $V$ of $x$ in $X$ that is homeomorphic to an open set $U \subset \mathbb R^n$, where the homeomorphism $f\colon U \to V$ is smooth and has a smooth inverse (the latter meaning that $f^{-1}$ extends to a smooth map on an open subset of $\mathbb R^m$ containing $V$).
Thanks!
Hint: You can show that every suffciently small neighbourhood of $1\in G$ is disconnected