6
$\begingroup$

Let $k$ be a field and $G$ a linear algebraic group over $k$. What is the group of $k$-rational points of $G$? By definition, $G$ is an algebraic variety. Suppose that $G$ is defined by polynomials $f_1, \ldots, f_n$. Then the set of $k$-rational points of $G$ is $A=\{a=(a_1, \ldots, a_n)\in k^n: f_1(a)=\ldots = f_m(a)=0\}$. Is $A = G$?

In $GL_n(k)$ case, what are the group of $k$-rational points of $GL_n(k)$? Thank you very much.

$\endgroup$
  • 1
    $\begingroup$ Yes, $A$ is the set of $k$-rational points of $G$. If $k$ is not algebraically closed, then $A$ and $G$ are fundamentally different (well at least in some cases). For instance if $A=\{ (0,0) \}$ and $k=\mathbb{R}$, then you can't tell if $G$ is defined by $x^2+y^2=0$ or by $(x=0,y=0)$ [other than the fact that only one has an algebraic group structure]. Similarly, if $k$ is finite, then $A$ doesn't tell you much about $G$. If $k$ is algebraically closed, then $A=G$ loses no information. $\endgroup$ – Jack Schmidt Jan 21 '14 at 18:05
7
$\begingroup$

A linear algebraic group $G$ is more than just a collection of points with an additional structure. So I think "does $A = G$" is the wrong question: one should think of these things as fundamentally different types of objects.

I am going to attempt to explain this without using too much algebraic geometry, and toward the end I am hiding some details. Please let me know if you want them filled in.

It is very common to think of a linear algebraic group $G$ defined over $k$ not as some definite object, but rather as a functor. This functor is from the category of fields over $k$ to the category of sets:

$$F \rightarrow G(F) \text{ (the F-rational points of } G).$$

So here $F$ is a field containing $k$. Actually, one allows any $k$-algebra $R$, but for this post I'm going to stick with fields. For many of the classical groups, you get what you'd expect. For example:

$$\text{GL}_n(F) = \text{collection of invertible matrices with coefficients in } F.$$

Note that $\text{GL}_{1,F}$ is defined by the equation $xy - 1$ inside $\mathbb{A}_F^2$.

However, some groups have weird behavior. Consider the circle group over $\mathbb{R}$, which cut out by the equation $x^2+y^2=1$ in $\mathbb{A}_\mathbb{R}^2$. The $\mathbb{R}$-rational points are the unit circle (whence the name), and hopefully it is clear that this is not the same as $\text{GL}_{1,\mathbb{R}}$. However when one considers $\mathbb{C}$-rational points, we have that

$$(x-iy)(x+iy) = x^2 + y^2 = 1$$

and the map sending $x \rightarrow x-iy = z$ and $y \rightarrow x+iy = w$ (with $zw = 1$) gives an isomorphism so that our group becomes $\text{GL}_{1,\mathbb{C}}$ (since we're over the complex numbers now).

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Very nice way to summarize the difficulties that can arise without getting too technical. $\endgroup$ – Tobias Kildetoft Jan 22 '14 at 14:47
  • $\begingroup$ @rghthndsd, thank you very much. Why we called these points rational? Are these points always given by rational functions? $\endgroup$ – LJR Jan 22 '14 at 16:20
  • $\begingroup$ @LJR: I do not know the etymology of "$k$-rational", but here is what I think is a good guess. It is a generalization of the situation over the integers: A point with integral coordinates is known as an integral point, and a point with rational coordinates (i.e. in $\mathbb{Q}$) is known as a rational point. Now when one wants to consider other fields $k$, a natural name is $k$-rational. Of course, one does not say "$\mathbb{Q}$-rational" (this would be silly), we still only say "rational". $\endgroup$ – RghtHndSd Jan 22 '14 at 16:35
  • $\begingroup$ @rghthndsd, thank you very much. $\endgroup$ – LJR Jan 22 '14 at 16:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.