Lie subgroups of $GL_2(\mathbb C)$ which are not algebraic

Question

Why the Lie subgroups $\left\{ \exp\left( t \begin{bmatrix} 1 & 0\\ 0& i \end{bmatrix}\right); t\in \mathbb C\right\}$ and $\left\{ \exp \left(t\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix}\right); t\in \mathbb C\right\}$ are not algebraic subgroups of $GL_2(\mathbb C)$?

Answer 1

We need slightly more precision. The algebraic object is $\operatorname{GL}_2$, its $\Bbb C$-rational points are the "Lie group" $\operatorname{GL}_2(\Bbb C)$. We consider first the question:

Does there exist a subvariety $H\le \operatorname{GL}_2$, which inherits a group structure from $\operatorname{GL}_2$, such that $H(\Bbb C)$ is as a set the set $$ \left\{ \ \exp\left( t \begin{bmatrix} 1 & 0\\ 0& i \end{bmatrix}\right) \ :\ t\in \mathbb C\ \right\} \ ? $$ For this, let us compute the exponential, it is $$ \exp\left( t \begin{bmatrix} 1 & 0\\ 0& i \end{bmatrix}\right) = \begin{bmatrix} \exp(t) & 0\\ 0 & \exp(it) \end{bmatrix} \ , $$ and the projection on each of the diagonal entries, for $t$ running in $\Bbb C$, is $\Bbb C^\times$, but these diagonal entries do not satisfy any algebraic equation with complex coefficients. (Assume the converse, then restrict to $t\in \Bbb R$ and divide with the dominating term. The other answer is taking alternatively $t\in 2\pi \Bbb Z$.)

For the other set, we compute the exponential, it is $$ \exp \left(t\begin{bmatrix} 1 & 1\\ 0& 1 \end{bmatrix}\right) = e^t\begin{bmatrix} 1 & t\\ 0& 1 \end{bmatrix}\ , $$ and again, the entries $e^t$ and $te^t$ do not satisfy an algebraic equation. (We can again restrict to $t\in 2\pi i\Bbb Z$ to get a Zariski dense set for the component in position $(1,2)$, so this component is with full closure in the group, $\begin{bmatrix}1&*\\&1\end{bmatrix}(\Bbb C)\subset H(\Bbb C)$, but then how to realize "other elements" with the parameter $t$?

Answer 2

Assume otherwise. By imposing one additional polynomial identity, $d=1$, on the first group, you would then obtain a group isomorphic to $\Bbb Z$, but an algebraic group over $\Bbb C$ can only have finitely many or continuum-many points.

Answer 3

The matrices in the second group are of the form ($t\in\Bbb{C}$) $$ \left(\begin{array}{cc}e^t&te^t\\0&e^t\end{array}\right). $$ This group suffers from the same malady that Hagen diagnosed the first group with. If we restrict it by one more equation, $a_{22}=1$, then we get the matrices $$ \left\{\left(\begin{array}{cc}1&n 2\pi i\\0&1\end{array}\right)\bigg\vert\ n\in\Bbb{Z}\right\}. $$ But this group is countably infinite and cannot be algebraic.