Relationship between decompositions of a $G$-variety $V$

Question

Let $V$ be a variety over a field $k$, and let $G$ be an algebraic group over $k$ which acts morphically on $V$. $V$ has three canonical decompositions, and I'm interested in the relationships between them. We can write

$$V=\bigcup_{\substack{\text{irr.}\\[.03in]\text{comp.}}} X=\bigsqcup_{\substack{\text{conn.}\\[.03in]\text{comp.}}}U=\bigsqcup_{G-\text{orbits}} \mathcal{O}$$

where each decomposition union of some basic building block of the variety $V$ (disjoint in the case of connected components and $G$-orbits). Are any of these decompositions 'finer' than another? Specifically, here are three related questions:

1. Is each connected component of $V$ actually a disjoint union of $G$-orbits? (For this do we require the connectedness of $G$?)
2. Is each irreducible component also a disjoint union of $G$-orbits?
3. Are there any similar relationships between irreducible and connected components?

If $G$ is connected, and some orbit of $G$ meets two connected components, then the orbit is not connected. If all orbits of a connected group $G$ are themselves connected (is this true?), this would answer my first question affirmatively.

In the case the answer to any of these questions is no, counterexamples would be greatly appreciated. Thanks in advance.

Answer 1

I believe if $G$ irreducible, then we have the relationships

$$|C_V|\le |I_V|\le |O_V|$$

where $C_V$, $I_V$, and $O_V$ are the sets of connected components, irreducible components, and $G$-orbits, respectively.

First, notice that irreducible components are connected (all non-empty open sets intersect), so that each irreducible component is contained in a unique connected component. For each connected component, choose some irreducible component inside. This gives an injection $C_V\hookrightarrow I_V$.

Next, choose an $x\in V$. This gives a continuous map $\varphi_x:G\to V$ which sends $g$ to $g\cdot x$. The image of this map is the orbit of $x$, $\mathcal{O}_x$, and since $G$ is irreducible, $\mathcal{O}_x$ is irreducible (continuous images of irreducible sets are irreducible). Now, for any irreducible component, $X$, choose some $x\in X$ not contained in any other irreducible component. The orbit of $x$ will now be contained in $X$, and not in any other irreducible component. This gives us the injection $I_V\hookrightarrow O_V$.

Does this look right?