I'm interested in the affine variety $$ V = \left\{ \, A\in \mathbb R^{d\,\times\, n} \, \middle| \, A\,A^T = I \, \right\} \subseteq \mathbb R^{d\, \times\, n}, $$ where $n\ge d$ and $I$ is the $d\times d$ unit matrix. This is the set of real $d\times n$ matrices with orthonormal rows, so another way to look at this is the configuration space of $d$ orthonormal vectors in $\mathbb R^n$.
For $n=d$ we have $V=\operatorname{O}(n,\mathbb R)$, so $V$ is not irreducible since it splits in two components where $\det(A)=1$ and $\det(A)=-1$, respectively. I'm guessing that this decomposition is already the decomposition of $V$ into irreducible components for $d=n$, but I'm not sure how to show that or where to look for a reference.
For $n>d$ I would guess that $V$ is irreducible, but I don't know how to figure out whether that is true.
I'd appreciate it, if anybody could provide insight or references on this problem.
This variety $V$ is called Stiefel manifold. See
http://en.wikipedia.org/wiki/Stiefel_manifold
There you find a lot of properties, in particular that $V$ is connected for $d<n$ as you guess. The essential point here is that while you cannot move continuously an ortonormal basis into another if they give opposite orientations, when $d<n$ you can make the move jumping outside of the $d$-plane they generate. Consider a $d$-frame $\{u_1,\dots,u_d\}$. Since $d<n$ there is some line ortogonal to $L=L[u_1,\dots,u_d]$, generated by a unitary vector $w$, and you can rotate $u_d$ in the plane $\Pi=L[u_d,w]$ to get $-u_d$, say you have a continuous path of unitary vectors $\gamma(t)\in\Pi$ with $\gamma(0)=u_d,\gamma(1)=-u_d$ (here we jump off $L$). In the end, $$ \varGamma(t)=\{u_1,\dots,u_{d-1},\gamma(t)\} $$ is a path in $V$ connecting $\{u_1,\dots,u_d\}$ to $\{u_1,\dots,-u_d\}$, two frames with opposite orientation. Thus all frames generating the same $d$-plane are connected by a path. Then one must connect different $d$-planes, which is just to say that the Grassmanian $G$ of $d$-planes in $\mathbb R^n$ is connected. This is also classical. For instance
Fundamental groups of Grassmann and Stiefel manifolds
Next for irreducibility, yes, smooth and connected implies irreducibility in the real case. This is just the identity principle: if a polynomial $f$ does not vanish on $V$, smooth and connected, $\{f=0\}\cap V$ has empty interior. It follows that if neither $f$ nor $g$ vanish, we have $V\setminus\{fg=0\}=(V\setminus\{f=0\})\cap (V\setminus\{g=0\})$, and an intersection of two dense open sets is open dense, hence not empty (we don't need Baire here). It is good to remark that the problem is the converse: irreducible smooth does not implies connectedness (non-singular real cubics provide examples).