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discrete G-set

Last updated January 27, 2022

Definition

Examples

Galois actions See: Galois group

Since $G$ is compact, we can see Property 2, that all orbits are finite, directly:

In fact, for each $\alpha\in F^s$ the orbit $G\alpha$ is precisely the set of roots of the minimal polynomial for $\alpha$. A minimal polynomial is seperable, irreducible, and monic. Conversely, any polynomial satisfying those conditions is the minimal polynomial of some element in $F^s$. So there is a bijection

\begin{gather} {G\alpha}{\alpha\in F^s} \quad \leftrightarrow \quad {\text{monic, sep, irred, polys over }F}, \
G
\alpha = {\alpha_1,\dots,\alpha_n}\mapsto \prod_{i=1}^n(x-\alpha_i) \ {\text{roots of }f\text{ in }F^s}\leftarrow f \end{gather}

where $F^s$ denotes the separable closure.

One aspect of Grothendieck’s Galois theory is the weaking of the left side (i.e. the set of orbits) to the set of finite $G$-sets (i.e. the set being acted on in finite). This is a weakening because we have just said that every orbit is finite, implying it is a finite $G$-set.

Properties

  1. $X$ is a discrete $G$-set iff the orbit $G_x\subset G$ is open for all $x\in X$.
    1. Recall the orbit is $Gx={gx\mid g\in G}$.
    2. Recall the stabilizer $G_x={g\in G\mid gx=x}$.
  2. By the orbit-stabilizer theorem, if $G$ is compact and $X$ is a discrete $G$-set, then all orbits $G_x$ are finite.
    1. The orbit-stabilizer theorem says there is a bijection $G/Gx = G_x$.
  3. Forms a category, $G_F\text{Set}$.
    1. objects are discrete $G_F$-sets.
    2. morphisms are $G_F$-equivariant maps
      1. i.e. $f:S\to T\in \text{Set}$ such that $f(\sigma s)=\sigma f(s)$ for all $\sigma\in G$.

Proofs

2 Continuous image of compact set is bounded, so that $Gx\subset X$ is bounded…


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