Does a one-parameter subgroup of a reductive group 'act reductively'?

96 Views Asked by At

For the present question I am only interested in the case where the underlying field is $\mathbb{C}$.

Hilbert fourteenth can be stated as follows:

Let $V$ be an algebraic variety and $G$ a linear algebraic group acting algebraically on it. Is the rings of invariants $\mathbb{C}[V]^G$ finitely generated?

It is thanks to Nagata that we know this is not the case, yet it has been proved that if $G$ is reductive then $\mathbb{C}[V]^G$ is finitely generated. I came across a problem concerning subgroups of reductive groups and I tryed to verify this naive statement:

Let $G$ be a linear algebraic reductive group acting on an algebraic variety $V$ and $H \subset G$ be an algebraic subgroup. Then $\mathbb{C}[V]^H$ is finitely generated.

It is again by the counter-example of Nagata that the above is false, nevertheless I can't prove that the following is false:

Let $G$ be a linear algebraic reductive group acting on an algebraic variety $V$ and $h: \mathbb{C} \to G$ be an injective algebraic homomorphism. Then $\mathbb{C}[V]^{h(\mathbb{C})}$ is finitely generated.

The main example for me would be $G=SL_2(\mathbb{C})$ and $h(\mathbb{C})$ consist of upper triangular unipotent matrices.

Do you know whether the last statement is true or false? Any reference that I can look at would also be very welcome. Thank you!