Why is $GL(n, \Bbb R)$ open in the Zariski topology?

Question

The general linear group $GL(n,\mathbb{R})$ is said to have Zariski topology, but I do not understand how. Can anyone explain this to me?

Answer 1

The Zariski topology is a topology well-suited to the study of algebraic varieties, the principal objects of study in algebraic geometry:

For any field $\Bbb F$, a set $X \subseteq \Bbb F^n$ is defined to be closed in the Zariski topology on iff it has the form $$V(S) := \{(x_1, \ldots, x_n) \in \Bbb F^n : f(x_1, \ldots, x_n) = 0 \,\forall f \in S\}$$ for some set $S$ of polynomials $f_a(x_1, \ldots, x_n) \in \Bbb F[x_1, \ldots, x_n]$.

The general linear group $GL(n, \Bbb F)$ is usually defined to be set of invertible $n \times n$ matrices over $\Bbb F$, that is, the $n \times n$ matrices with nonzero determinant. On the other hand, the determinant map $\det: M(n, \Bbb F) \to \Bbb F$ is a polynomial in the entries $x_{11}, \ldots, x_{nn}$ of the argument matrix $X$, where $M(n, \Bbb F)$ is the space of $n \times n$ matrices over $\Bbb F$, which we can identify with $\Bbb F^{n^2}$ in the obvious way. Thus, the set $$V(\{\det\}) := \{X \in M(n, \Bbb F) : \det X = 0\} \subset M(n, \Bbb R) \cong \Bbb F^{n^2}$$ is closed in the Zariski topology, and hence its complement, namely, $$V(\{\det\})^c = \{X \in M(n, \Bbb F) : \det X \neq 0\} = GL(n, \Bbb F),$$ is open in the topology.

Answer 2

Yes, the space of all n-by-n matrices with entries in field F itself admits the structure of an affine space and hence also admits the Zariski topology. The general linear group is, as the previous poster said, a Zariski-open subset of the affine space. The general linear group may itself be given the subspace topology induced from the Zariski topology on the affine space, and in this way, the general linear group may be said to "have" the Zariski topology.