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Fact/motivation: $\DeclareMathOperator{\inv}{inv}\DeclareMathOperator{\GL}{GL}$ If $G$ is a smooth manifold of dim.$n$ and a group, s.t. the multiplication $$ m \colon G \times G \to G $$ is smooth, then it implies that the inversion $$ \inv \colon G \to G $$ is smooth too. This is e.g. due to the 'Implicit Function Theorem' and can be found for example in https://services.math.duke.edu/~bryant/ParkCityLectures.pdf (The author does not require for a Lie group to have smooth inversion).

But I don't know of any similar statement for regular maps in algebraic geometry. So my question is:

Question:

Does a similar statement (not) hold for algebraic (affine) groups? That is, is there any group, for instance of the form $G = V(I) \subset \mathbb{A}^n(k)$ over some algebraically closed field $k$ ($I \subset k[X_1,\ldots, X_n]$ ideal), such that the multiplication is regular but the inversion is not?

Non-example:

All I could think of was for example $\mathbb{A}^1(k)^{\times}$, but this is in general not closed inside $\mathbb{A}^1(k)$ (unless $k$ finite but then every map is polynomial), and if one 'closifies' it, i.e. understands it rather as $V(XY-1) \subset \mathbb{A}^2(k)$, this forces the inversion automatically to be polynomial (so this is just $\mathbb{G}_m(k)$). Same of course applies for $\GL_n$.

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    $\begingroup$ If the variety is normal, this follows from Zariski’s Main Theorem. You can probably prove normality using homogeneity (without inversion). $\endgroup$ Nov 17 at 12:45
  • $\begingroup$ I'm not sure I follow your definitions: Is this a counter-example for you? Let $I$ be the ideal of $\mathbb{Q}[x]$ generated by $x(x^2-2)$. We have a regular map $\mu: V(I) \times V(I) \to V(I)$ given by $\mu(x,y) = 0$. This equips the points of $V(I)(\mathbb{C})$ with the structure of a monoid which is not a group. However, if we restrict our attention to $V(I)(\mathbb{Q})$, we get a (trivial) group. $\endgroup$ Nov 24 at 14:50
  • $\begingroup$ Hi David, no, it isn't since the inversion on $V(XY - 1)$ is given by $(x,y) \mapsto (y,x)$, which is polynomial, although if I would look at the open subset $k^{\times} \subset k = \mathbb{A}^1(k)$, then $k^{\times}$ would have a regular multiplication, but the inversion would not necessarily be regular. For the sake of simplicity I shall assume $k$ to be algebraically closed. $\endgroup$ Nov 24 at 14:59
  • $\begingroup$ Re, if you want the assumption that $k$ is algebraically closed to be part of the question (I can't tell if that is what you mean), then it is best to edit it into the question itself. $\endgroup$
    – LSpice
    Nov 24 at 16:42

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