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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.

Christoph
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  • You can think of this as $O(n) / O(n-d)$, where $O(n-d)$ is embedding in $O(n)$ by putting the matrix in the lower right corner of a matrix and filling in the upper left with the $d \times d$ identity. This is called something like "the frame bundle" (perhaps "the orthogonal frame bundle"?), and comes up in discussions of Grassmannians, etc. (or so I recall from grad school). Not sure this helps, but might at least get you pointed to the right literature. – John Hughes Nov 19 '14 at 12:21
  • smooth and connected implies irreducible; homogeneous spaces of connected groups are smooth and connected (this remark is irresponsible and one should replace $\mathbb R$ with $\mathbb C$ to make it safer) – user8268 Nov 19 '14 at 14:42

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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).

Community
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Jesus RS
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  • I managed to check irreducibility in a different way: Cutting off the last $n-d$ rows of orthogonal $n\times n$ matrices, we get a surjection $SO(n)\to V$ for $n>d$, and $SO(n)$ is an irreducible real affine variety. Now I'm wondering if the complex Stiefel manifold is also an irreducible real affine variety. It isn't a complex variety since the equations involve complex conjugates… – Christoph Jan 14 '15 at 20:11
  • Yes, this works very well! Concerning the complex Stiefel manifold, it is connected, hence irreducible, as a complex algebraic variety. Now, it has an underlying real manifold structure. Examples of this: complex line is real plane, complex projective line is real sphere... Real dimension is topological dimension, twice the complex dimension. Thus, the complex Stiefel manifold is a real algebraic manifold, also connected, hence real irreducible. – Jesus RS Jan 14 '15 at 22:45