I apologize in advance if my question is too elementary for MO.


It is a well known fact that the linear algebraic group $G = \mathsf{SO}\_n$ is connected, and there exist a few different proofs of this fact. One proof goes by showing that $G$ is generated by unipotent elements, and invoking the theorem that every linear algebraic group with this property is connected.


My question is about a different, more direct proof, involving the Cayley transform
$$ A \mapsto (I\_n + A)^{-1} (I\_n - A) , $$
which maps every matrix $A \in G(k)$ for which $I\_n + A$ is invertible (let's write $W$ for the set of such matrices), to a skew-symmetric matrix, and in fact, this map defines an isomorphism of varieties between the non-empty open subset $W$ of $G$, and an open subset of the irreducible variety $V$ of skew-symmetric matrices.


But how does one now conclude that $\mathsf{SO}\_n$ is connected? In particular, why is the closure of $W$ equal to $G$? Of course the complement of $W$ is given by the polynomial equation $\det(I\_n + A) = 0$, but how is this situation different from, for instance, the fact that $\mathsf{SO}\_n$ is the subvariety of $\mathsf{O}\_n$ defined by the polynomial equation $\det(A) - 1 = 0$ (or $\det(A) + 1 \neq 0$) whereas $\mathsf{O}\_n$ is not connected?


What subtlety am I missing?


**added:**


1. I'm assuming $\operatorname{char}(k) \neq 2$.
2. The underlying topology is the Zariski toplogy.
3. The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
4. It seems that the question is not too elementary...


**added:** I'll add a few more sentences to make clear what I mean by connected in the Zariski topology.


I'm considering $G$ as a group functor (i.e. a functor from the category of commutative $k$-algebras to the category of groups). Then $G$ is connected if and only if its coordinate algebra $k[G] = \mathcal{O}(G)$ has no non-trivial idempotents. In particular, in order to show that $G$ is connected, we may assume w.l.o.g. that $k$ is algebraically closed. [In fact, since $G$ is smooth (since $\operatorname{char}(k) \neq 2$), we may instead consider the group of $\bar{k}$-rational points $G(\bar{k})$, which brings us back to a more classical situation.]



Connectedness of the linear algebraic group SO_n
Each connected component of an algebraic group has the same dimension. Thus if it has a connected subvariety whose complement has a lower dimension, it is connected. (If it had two connected components, only one could contain the connected subvariety, and so the other would have a lower dimension.)


The subvariety where $I\_n+A$ is invertible is birational to $\mathbb A^{\frac{n(n-1)}{2}}$ and thus has dimension $\frac{n(n-1)}{2}$.


The complement is where $A$ has a $-1$ eigenvalue. But since every eigenvalue to an orthogonal matrix must have its inverse also an eigenvalue, the determinant is the product of all the eigenvalues which are their own inverse, which are just $-1$ and $1$. Thus, if the determinant is $1$, the number of $-1$ eigenvalues is even, so is at least $2$. We can split the matrix into a $2$-dimensional $-1$-eigenspace and a matrix in $SO(n-2)$. These things are paramaterized by a $SO\_{n-2}$-bundle on $G\_{2}^{n-2}$, whose dimension is $2(n-2)+\frac{(n-2)(n-3)}{2}=\frac{n(n-1)}{2}-1$. If the $-1$-eigenspace is more than two-dimensional there is more than one way to express a matrix in this way, but that can only decrease the dimension.


