Why is this point fixed?

Question

For an algebraic group $G$ that acts by $\sigma$ on a proper algebraic scheme $X/k$ and some closed point $x\in X$, we define the morphism $\psi_x: G\to X$ by $g\mapsto g\cdot x$. Say $\lambda$ is a $1$-parameter subgroup of $G$. We embed $\mathbb G_m$ into the affine line so that the map $\lambda_x:\psi_x\circ\lambda$ extends uniquely to a morphism $\widehat{\lambda_x}:\mathbb A_k^1\to X$ by the valuative criterion for properness.

I learned that the point $\widehat{\lambda_x}(0)$ is invariant under the action of $\lambda(\mathbb G_m)$, but I have no clue where this conclusion came from. Is it related to the uniqueness of the extension?

Thanks in advance. Any help is appreciated. Apologies for the very bad title.

Answer 1

I think that your question is really independent of $G$ so that we can work in the following (slightly adjusted) setting:

Let $k$ be a field and let $X$ be a proper algebraic $k$-scheme that is equipped with a $\mathbf{G}_m$-action. Let $x \in X(k)$ and denote by $\psi_x \colon \mathbf{G}_m \to X$ the action map. Then $\psi_x$ extends uniquely to a map $\widehat{\psi_x} \colon \mathbf{A}^1 \to X$ and we would like to see why $\widehat{\psi_x}(0) \in X^{\mathbf{G}_m}(k)$.

Now we have a natural action of $\mathbf{G}_m$ on $\mathbf{A}^1$ given by multiplication and this action of course extends the multiplication action of $\mathbf{G}_m$ on itself. The map $\psi_x$ is $\mathbf{G}_m$-equivariant by definition and we claim that this implies that also the extension $\widehat{\psi_x}$ is $\mathbf{G}_m$-equivariant. As $0 \in (\mathbf{A}^1)^{\mathbf{G}_m}(k)$ the claim then follows.

$\widehat{\psi_x}$ being $\mathbf{G}_m$-equivariant is equivalent to saying that the two maps $$ \mathbf{G}_m \times \mathbf{A}^1 \to X, \qquad (g, t) \mapsto \widehat{\psi_x}(g.t), \; g.\widehat{\psi_x}(t) $$ agree. As $\psi_x$ is $\mathbf{G}_m$-equivariant we already know that the two maps agree on $\mathbf{G}_m \times \mathbf{G}_m \subseteq \mathbf{G}_m \times \mathbf{A}^1$. As $\mathbf{G}_m \times \mathbf{G}_m$ is scheme-theoretically dense in $\mathbf{G}_m \times \mathbf{A}^1$ and $X$ is separated this implies that they already agree everywhere (see for example https://stacks.math.columbia.edu/tag/01RH).

Answer 2

Intuitively, $\lambda$ gives us a way to multiply elements of $X$ by nonzero numbers such that $(a b) x = a(b x)$, and we are extending this to something that lets us also multiply elements of $X$ by zero by taking

$$0 \cdot x = \lim_{a \to 0} \, a x$$

Of course if we multiply something by zero and then multiply it by something else, that ought to just be the same as multiplying by zero; i.e. we ought to be able to say something like

$$a(0 x) = a \lim_{b \to 0} \, b x = \lim_{b \to 0}\, a (b x) = \lim_{b \to 0}\, (a b) x = 0x.$$

If we're working over the complex numbers you can make this into a rigorous argument. In general, you'd use algebraic notions instead of limits.