TITLE: Connectedness of the linear algebraic group SO_n
QUESTION [15 upvotes]: 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:

I'm assuming $\operatorname{char}(k) \neq 2$.
The underlying topology is the Zariski toplogy.
The proof I'm mentioning in the second paragraph is only valid in characteristic $0$.
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.]

REPLY [5 votes]: This is a more elementary solution, which would make sense in the context of Springer's book.  We consider
$SO(V,\langle,\rangle)$, or for short $SO(V)$, where $V$ is $n$-dimensional and $\langle,\rangle$ a nondegenerate
symmetric bilinear form on $V$. (Recall that any two such are equivalent.)
We'll show by induction on $n$ that $SO(V)$ is connected.
Springer's exercise shows that any $g \in SO(V)$ that doesn't have eigenvalue $-1$ is in the identity component of $SO(V)$.
When $n = 1$, $SO(V) = 1$ and when $n = 2$ we see directly that $SO(V)$ is connected. (In fact,
$SO_2 = \{\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) : a^2 + b^2 = 1\}$ and
$SO_2 \cong \mathbb G_m$ as algebraic groups via $\left(\begin{matrix} a & b\\ -b & a\end{matrix}\right) \mapsto a+bi$
with $i^2 = -1$.)
In general, fix $g \in SO(V)$ and we'll show that $g$ is contained in the identity component.
Note that $V = \oplus V_\lambda$ is a direct sum of generalised eigenspaces and that $V_\lambda \perp V_\mu$ if
$\lambda \ne \mu^{-1}$. Thus we get an orthogonal decomposition $V = V_{-1} \perp V_1 \perp \perp_{\{\lambda \ne \lambda^{-1}\}}
(V_\lambda \oplus V_{\lambda^{-1}})$. In particular, $V_{-1}$ is a nondegenerate subspace and we have the
orthogonal decomposition of $g$-invariant subspaces $V = V_{-1} \perp V_{-1}^\perp$. Also, we get that $\dim V_\lambda = \dim V_{\lambda^{-1}}$
if $\lambda \ne \lambda^{-1}$, so by considering the determinant we see that $\dim V_{-1}$ even (as remarked by Will), say it equals $2d$.
It follows that $g$ is contained in the subgroup $SO(V_{-1}) \times SO(V_{-1}^\perp)$ of $SO(V)$. By induction
we are thus reduced to the two extreme cases: $V = V_{-1}$ or $V = V_{-1}^\perp$. The second case is treated by the
Springer exercise, so we may assume that all eigenvalues of $g$ are $-1$ and $n = 2d$.
Now $-g$ is unipotent, hence by the Springer exercise we know it is in the identity component of $SO(V)$.
Multiplying by $-1 \in SO(V)$ we deduce that $g$ is in the same component as $-1$. Finally, after choosing an
appropriate basis $-1$ lies in the subgroup $SO_2 \times \dots \times SO_2$ of $SO_{2d}$, so we are done by the $n = 2$ case.