Why is $\mathbb R^n$ under the Zariski topology not a topological group?

Question

Reasons that $(\mathbb R^n, +, \mathcal Z)$ is not a topological group: Given any two distinct points $\vec{p},\vec q \in \mathbb R ^n$ let $P$ be the unique hyperplane through $\vec p$ which is perpendicular to the vector $\vec p- \vec q$. $P$ is closed with respect to $\mathcal Z$ and hence its complement is an open set containing $\vec q$ but not $\vec p$. Therefore the space is $T_0$. If it were a topological group, this would imply it were Hausdorff, which is not the case.

Reasons that $(\mathbb R^n, +, \mathcal Z)$ should be a topological group. It is both a group an a topological space. Moreover all polynomials $(\mathbb R^n, \mathcal Z) \to (\mathbb R, \mathcal Z)$ are continuous. So any map of the form $\vec x \mapsto (p_1(\vec x), p_2(\vec x), \ldots, p_n(\vec x)) $ or $(\vec x, \vec y ) \mapsto (p_1(\vec x, \vec y), p_2(\vec x, \vec y), \ldots, p_n(\vec x, \vec y) $ where $p_i$ are polynomials, is continuous, and the operations of addition and negation are both of this form.

Could someone point out where my misapprehension?

Answer 1

The problem is that the product topology is not the same as the Zariski topology. So, while maps $\mathbb{R}^{2n}\to\mathbb{R}$ are continuous, polynomial ones, this is in the Zariski topology, not the product topology.

In fact, every $T_0$ topological group is already Hausdorff. Every variety over $\mathbb{R}$ is $T_0$ but is not Hausdorff.

EDIT: Let me emphasize precisely what goes wrong. In your argument, you claim that the map $\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}^n$ given by $(x,y)\mapsto x-y$ is continuous, since it has polynomial coefficients, where $\mathbb{R}^n\times\mathbb{R}^n$ is given the product topology. If this were true, then the preimage of $0$ would be closed, since $0$ is closed in $\mathbb{R}^n$ with the Zariski topology. But, the preimage is precisely the diagonal $\Delta_{\mathbb{R}^n}\subseteq\mathbb{R}^n\times\mathbb{R}^n$. But, the diagional being closed is equivalent to being Hausdorff, and since $\mathbb{R}^n$ is not Hausdorff with the Zariski topology, this is a contradiction. Thus, the map $(x,y)\mapsto x-y$ is not continuous.

Answer 2

Let me mention that the correct "replacement" for the usual euclidean space in algebraic geometry is the affine space. In particular, this means that you don't have the product topology of the underlying topological spaces (which wouldn't make much sense anyway since this doesn't incorporate the algebraic structure). Also, group objects can be defined in any category with products (actually you don't need products). Then $(\mathbb{R}^n,+)$ is a group object in $\mathrm{Top}$ i.e. a topological group, whereas $(\mathbb{A}^n,+)$ is a group object in $\mathrm{Var}$, i.e. an algebraic group. You can write this group as $(\mathbb{G}_a)^n$ (in order to distinguish it from $\mathbb{A}^n$ which usually has no group structure attached to it). It is also known as the $n$-dimensional torus.