Why is this point fixed? 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.
 A: 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).
A: 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.
