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In algebraic geometry I learned that $GL(n, \mathbb R)$ is Zariski-closed (every linear algebraic group is a closed subgroup of it, and it is itself linear algebraic), and as the Zariski topology is coarser than the usual topology it must be closed in the standard topology, but by the defining condition $\det(A) \ne 0$ it is open as well, which is not possible in the standard topology, as this space is connected.

So, something is wrong here, but I do not see it? Could anyone please explain this to me?

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    $\begingroup$ Closed in $\mathbf{R}^{n^2}$? Surely that's false. $\endgroup$ – Eoin Sep 1 '17 at 19:48
  • $\begingroup$ Proof: let $z$ be a variable, $1$ the identity matrix. Then $z\cdot 1$ is in $GL_n$ for all $z\neq 0$. So $\lim_{z\rightarrow 0} z\cdot 1$ is a limit point not contained in $GL_n$. $\endgroup$ – Eoin Sep 1 '17 at 19:51
  • $\begingroup$ @Eoin Okay, so where does my reasoning with the Zariski topology breaks down? It seems valid to go that way, so paradoxical to me... $\endgroup$ – StefanH Sep 1 '17 at 19:53
  • $\begingroup$ The Zariski topology is horrible so don't be discouraged. $\endgroup$ – Randall Sep 1 '17 at 19:54
  • $\begingroup$ @StefanH You didn't give a reason for GL_n to be closed in the Zariski topology. The condition every linear algebraic group is a subgroup of GL_n doesn't imply anything topologically. That GL_n is determined by the nonvanishing of the determinant does imply that it is open in $\mathbf{R}^{n^2}$ however. $\endgroup$ – Eoin Sep 1 '17 at 19:55
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$GL(n)$ is an algebraic group, that is, it is a variety (described by polynomials) and the group multiplication and inverse are also polynomial. For example, $GL(2)$ is the set of all $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with $ad-bc\ne 0$, which can also be described as the set of all $(a,b,c,d,D)$ that are zeroes of the single polynomial $(ad-bc)D-1$. Note that we are seeing five, not four variables here! Hence we are taling about a (Zariski) closed subset of $F^{n^2+1}$, not of $F^{n^2}$.

Also note that adding the $D$ variable is not just a trick to convert an inequality into an equality. Instead, it is essential to make inversion polynomial, the inverse of $(a,b,c,d,D)$ being $(dD,-bD,-cD,dD,ad-bc)$.

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$GL(n, \mathbb R)$ is Zariski-closed (every linear algebraic group is a closed subgroup of it, and it is itself linear algebraic)

Every linear algebraic group is a closed subgroup of $GL(n,\mathbb{R})$ (that is, closed as a subset of the space $GL(n,\mathbb{R})$ with the Zariski topology). This tells you $GL(n,\mathbb{R})$ is closed as a subset of itself, but that is obvious. What it does not tell you is that $GL(n,\mathbb{R})$ is closed as a subset of $\mathbb{R}^{n^2}$, and in fact your argument shows it is not closed as a subset of $\mathbb{R}^{n^2}$.

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Let $F$ be a field, and $\mathbb{A}_F^m$ the affine $m$-space. By definition, $\mathbb{A}_F^m$ is the affine scheme $\textrm{Spec}(F[T_1, ... , T_n])$, whose underlying topological space is the set of prime ideals of $F[T_1, ... , T_n]$ in the Zariski topology. The closed irreducible subsets of $\mathbb{A}_F^m$ are in one to one correspondence with the prime ideals of $F[T_1, ... , T_n]$.

Consider the determinant $\textrm{det} = \textrm{det}(T_{ij})$, which is a polynomial in $n^2$ variables. If $R = F[T_1, ... , T_n]$, then the open set

$$\{ \mathfrak p \in \mathbb{A}^{n^2} : \textrm{det} \not\in \mathfrak p\}$$

inherits the structure of a affine scheme, which is called $\textrm{GL}_n$. Its coordinate ring is $R_{\textrm{det}}$, the localization of $R$ at $\textrm{det}$. It is not a closed subset of $\mathbb{A}^{n^2}$.

But there is another scheme, isomorphic to $\textrm{GL}_n$, whose underlying space is a closed set in affine space. Namely, the closed subscheme of $\mathbb{A}^{n^2+1} = \textrm{Spec}(R[Y])$ corresponding to the prime ideal $I = Y \cdot \textrm{det}(T_{ij}) - 1$ of $R[Y] = F[T_{ij},Y]$.

To give an isomorphism of these schemes is the same as giving an isomorphism of their coordinate rings

$$F[T_{ij},Y]/I \rightarrow R_{\textrm{det}}$$

which is easy.

When $F$ is a topological field, e.g. $\mathbb{R}$, then the $F$-rational points $$\textrm{GL}_n(F) := \textrm{Hom}_{\textrm{$F$-sch}}(\textrm{Spec}(F),\textrm{GL}_n)$$

inherit a group structure as well as a topology, namely the subspace topology from $\mathbb{A}_F^{n^2}(F) = F^{n^2}$.

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