Suppose $G\in GL_{n_1}(\mathbb C),H\in GL_{n_2}(\mathbb C)$ are matrix Lie groups such that $\theta:G\to H$ is a Lie homomorphism . Then is the image of G under the map necessarily a matrix Lie group?
I thought the answer would be positive.
Definition of matrix Lie group: a subgroup of $GL_n(\mathbb C)$ such that any sequence converges to an element in A or to an element which is not invertible; i.e., closed under $GL_n(\mathbb C)$
I attempted like the following:
$A_n\in G$ such that $A_n \to A$ as $n\to \infty$
As $\theta $ is a continuous map it must converge and limit is in range .
So its image must be a closed subgroup of $H$.
Hence a matrix Lie group.
Is this correct? Or there exists some counterexample to claim?
Any help will be appreciated.
Let $G$ be the group of reals under addition, identified with the matrix Lie group of 1x1 matrices with positive determinant via the exponential map $x\mapsto e^x$; let $a$ be an irrational number and consider $f:G\rightarrow GL(2,\mathbb C)$ given by
$$ f(x)= diag(e^{2i\pi x},e^{2i\pi ax}) . $$
You can prove that the image of $f$ is not closed in $GL(2,\mathbb C)$.
Remark. This is basically the same example provided by @Jason DeVito but rewritten it to make matrices appear explicitly.