Grassmanians are homogeneous spaces.
In the real case, you have the oriented Grassmanian $G^0(k,\mathbb{R}^n)$ of oriented $k$-planes in $\mathbb{R}^n$ is diffeomorphic to $SO(n)/\left(SO(k)\times SO(n-k)\right)$ where $SO(n)$ is the collection of $n\times n$ special orthogonal matrices.
Likewise, the nonoriented Grassmanian $G(k,\mathbb{R}^n)$ of nonoriented $k$-planes in $\mathbb{R}^n$ is diffeomorphic to $SO(n)/S(O(k)\times O(n-k))$.
Finally, the complex Grassmanian, $G(k,\mathbb{C}^n)$, is diffeomorphic to $SU(n)/(SU(k)\times SU(n-k))$ where $SU(n)$ denotes the $n\times n$ special unitary matrices.
(Some slight modifications may be necessary when $k=0$ or $k=n$).
Once you have written them like this, you have a general theorem that given compact Lie groups $G$ and $H$, then $H\rightarrow G\rightarrow G/H$ is a fiber bundle. In particular, we can use the long exact homotopy sequence.
It follows immediately that the complex Grassmanian is simply connected because $SU(n)$ is both connected and simply connected.
In the real case, a bit more work needs to be done. For the oriented Grassmanian, it's enough to note that the canonical map $SO(k)\rightarrow SO(n)$ is a surjection on $\pi_1$ as soon as both $n$ and $k$ are bigger than 1, (isomorphism when $n,k>2$) and is always an isomorphism on $\pi_0$. Thus, the real oriented Grassmanian is simply connected.
This also gives the answer for the unoriented real Grassmanian because there is a natural double covering $G^0(k,\mathbb{R}^n)\rightarrow G(k,\mathbb{R}^n)$ given by forgetting the orientation. Hence, the real unoriented Grassmanian has $\pi_1=\mathbb{Z}/2\mathbb{Z}$.
Alternatively, note that the induced map from $S(O(k)\rightarrow O(n-k))$ to $SO(n)$ is an isomorphism on $\pi_1$, but that $S(O(k)\times O(n-k))$ has more than one component.
It is enough to check that the preimage $A = \pi^{-1}(B)$ in $X_{n,k}$ is dense ($\pi(\bar A) \subset \overline{\pi(A)}$ for continuous $\pi$). Now reduce to the case $W = \{0\}_k \times \mathbb{R}^{n-k}$. Then $\pi^{-1}(B)$ are the $n\times k$ matrices with the leading $k\times k$ minor $\ne 0$. This is clearly dense
in $X_{n,k}$, since it is dense in the set of all $n\times k$ matrices.
It is interesting to consider Plücker coordinates.
The condiition $V\cap W = 0$ is a linear condition $\ne 0$ in the Plücker coordinates of $V$.
Best Answer
To understand how to choose between Grassmann and Stiefel manifolds, it helps to understand better the difference. In the following I will only look at the case of real vector space $\mathbb{R}^n$. Much information can be found at https://en.wikipedia.org/wiki/Stiefel_manifold and https://en.wikipedia.org/wiki/Grassmannian.
The Grassmannian $Gr(k,n)$ is the (compact) manifold of all $k$-dimensional linear subspaces in $\mathbb{R}^n$. So the one-dimensional case $k=1$ is real projective space. The Stiefel manifold $V_k(n)$ is (compact) manifold of all (orthogonal) $k$-frames (hereafter we leave out "orthogonal" which is always implied). A $k$-frame is a set of $k$ orthonormal vectors in $\mathbb{R}^n$, so can alternatively be described as the manifold of $n\times k$ column orthogonal matrices.
Let us compare the definitions for a few low-dim cases:
- $Gr(1,n)$ and $V_1(n)$
$Gr(1,n)$ is the set of all line through the origin, while $V_1(n)$ is the set of all unit vectors, so is the unit sphere. To each point in $Gr(1,n)$ there corresponds two (antipodal) unit vectors in $V_1(n)$.
- $Gr(2,n)$ and $V_2(n)$
$Gr(2,n)$ is the set of all 2-planes (through the origin) while $V_2(n)$ is all 2-frames in $\mathbb{R}^n$. To each point in $Gr(2,n)$ (that is, plane) there corresponds in $V_2(n)$ the set of all orthogonal bases for that plane.
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General case
There is a natural projection $p\colon V_r(n) \mapsto Gr(r,n)$ which to each frame in $V_r(n)$ sends it to the plane in $Gr(r,n)$ of which it is a basis. In this way, we can define $Gr(r,n)$ as a quotient space of $V_r(n)$.
What does all this mean for your optimization problem? If you are only interested in linear subspaces, use $Gr(r,n)$, but if how you parametrize the space with an orthogonal basis is important, use the Stiefel manifold.
But, algorithmically, it could be easier to represent a Stiefel manifold, so you can use that but build into the step-finding algorithm that you avoid walking to a new frame which is a basis for the same subspace. See Optimization Algorithms on Matrix Manifolds