# (Non-)algebraic groups: regularity of multiplication does not imply regularity of Inversion?

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 $$\GL_n$$.

• If the variety is normal, this follows from Zariski’s Main Theorem. You can probably prove normality using homogeneity (without inversion). Nov 17 at 12:45
• 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. Nov 24 at 14:50
• 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. Nov 24 at 14:59
• 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. Nov 24 at 16:42