Quotient map of Lie group structures for algebraic groups need not be quotient map of algebraic groups

Statement
It is possible to have an algebraic groups $$G_1,G_2$$ over a field admitting analytic structure (such as the field of real numbers or field of complex numbers) and a surjective function:

$$f:G_1 \to G_2$$

such that $$f$$ is a group homomorphism and can be viewed as a quotient map with respect to the induced Lie group structures on $$G_1$$ and $$G_2$$, but $$f$$ is not a morphism of algebraic varieties, i.e., it is not a regular morphism.

Related facts

 * Closed normal subgroup for Lie structure need not be closed normal subgroup of algebraic group

Over the real numbers
Let $$G_1 = \R$$ and $$H = \mathbb{Z}$$ be the subgroup of integers. Define $$G_2$$ to be the circle group in the usual way. $$G_2$$ is an algebraic group and also has an induced Lie group structure. Then:


 * $$H$$ is a discrete closed normal subgroup of $$G_1$$ with the Lie group topology, which is the usual Euclidean topology.
 * $$H$$ is not closed in $$G$$ under the Zariski topology because the only proper closed subsets of a one-dimensional space are the finite subsets.

The Lie quotient $$G_1/H$$ is isomorphic to $$G_2$$ as a Lie group. More explicitly, the mapping:

$$\! t \mapsto (\cos(2\pi t),\sin (2\pi t))$$

gives a surjective Lie group homomorphism from $$G_1$$ to $$G_2$$ with kernel $$H$$, and with $$G_2$$ acquiring the quotient topology.

However, this mapping is not a regular map, i.e., it is not a morphism of algebraic varieties (note that it uses the transcendental functions sine and cosine). In fact, there is no regular map that works.

Over the complex numbers
Let $$G_1 = \mathbb{C}$$ and $$H = \mathbb{Z}[i]$$ be the subgroup of Gaussian integers, i.e., numbers of the form $$\{a + bi \mid a,b \in \mathbb{Z} \}$$. Then:


 * $$H$$ is a discrete closed normal subgroup of $$G$$ with the Lie group topology, which is the usual Euclidean topology.
 * $$H$$ is not closed in $$G$$ under the Zariski topology because the only proper closed subsets of a one-dimensional space are the finite subsets.

The quotient $$G_1/H$$ can be realized as an elliptic curve group in multiple ways. The remainder of the explanation parallels the case of the reals.