On exercise 2.6.4 (page 32) of Andreas Gathmann's Algebraic Geometry class notes, you can read the following:

"Let $X$ be an affine variety, and let $G$ be a finite group. Assume that $G$ acts on $X$, i.e that for every $g \in G$ we are given a morphism $g:X\rightarrow X$ (denoted by the same letter for simplicity of notation), such that $(g \circ h)(P)=g(h(P))$ for all $g,h \in G$ and $P \in X$"

I'm having doubts here because I can´t seem to grasp the usual action concept. More specifically, the only thing I read from this definition is that Gathmann's saying that $G$ acts on $X$ if we're simply given a collection of morphisms on $X$... Also, the condition $(g\circ h)(P)=g(h(P))$ seems to be just how one defines composition of set-theoretical maps, making it irrelevant.

What would seem to make sense to me is that we start with a map $G \times X \mapsto X\;,\;(g,P)\mapsto g.P $ satisfying the usual conditions:

  1. $e.P=P\;,\; \forall P \in X$ (where $e \in G$ is the identity)
  2. $(gh).P=g.(h.P),\; \forall g,h\in G,\;\forall P \in X$

i.e "the usual action concept" which I mentioned before, such that for all $g\in G$, the map $g:X\rightarrow X,\;P\mapsto g.P$ (also denoted with the same letter for simplicity of notation) is an affine variety morphism. But in this case we know that these maps are actually automorphisms of $X$ (the inverse of $g$ being given by $g^{-1}$ which is also a morphism), so why would Gathmann only refer to these maps as morphisms?

I am a bit confused with what is Gathmann trying to say here...Am I overlooking something or doing something wrong?

Thank you in advice.

  • 2
    $\begingroup$ $g\circ h$ is an element of $G$ (on which, presumably, $\circ$ is the operation), not a composition of functions. This is exactly what you are calling $gh$. $\endgroup$ – Slade Mar 15 '15 at 0:38
  • $\begingroup$ @Slade: Hmm, that woud make more sense. It doesn´t seem to follow from this that the identity element fixes every point in $X$, and consequently that the morphisms $g:X\rightarrow X$ are not necessarily bijective (and even less automorphisms of $X$). But this seems to agree with Gathmann only refering to them as morphisms. Can we say that here we're defining a somewhat weaker notion of action? (i.e by excluding condition 1. I mentioned on the post, we don´t even have a natural representation of $G$ as a group of permutations on $X$) $\endgroup$ – Pedro A. Matos Mar 15 '15 at 0:56
  • $\begingroup$ One usually defines a group action as a map $G\to\operatorname{Aut}(X)$. Gathmann seems to be (maybe accidentally) defining it as a map $G\to\operatorname{End}(X)$. Unfortunately, the latter is a monoid, not a group, and in particular the axiom for a monoid homomorphism does not yield a group homomorphism (as it does for a map of groups). As such, this is probably a good definition of a monoid action, and a bad definition of a group action. $\endgroup$ – Slade Mar 15 '15 at 18:07
  • $\begingroup$ @Slade: a monoid homomorphism does yield a group homomorphism; what Gathmann writes is even weaker, namely a semigroup homomorphism. $\endgroup$ – Qiaochu Yuan Mar 16 '15 at 4:33

You're correct; Gathmann's definition is incomplete. It does not imply that the map $\rho$ sending a group element $g \in G$ to a map $X \to X$ sends the identity to the identity and inverses to inverses. It does imply that $\rho(e^2) = \rho(e) = \rho(e)^2$, so the image of the identity $e \in G$ must be an idempotent, but it may be a nontrivial idempotent. You should assume that he meant the correct thing.

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  • $\begingroup$ (For example, pick a point $x \in X$. Then Gathmann's definition allows every $\rho(g)$ to be the map $X \to X$ which sends every point of $X$ to $x$.) $\endgroup$ – Qiaochu Yuan Mar 15 '15 at 3:32

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