Is the space of orthogonal vectors a manifold?

Question

It is known that the subspace formed by $q$ orthonormal vectors in the $p$-dimensional Euclidean space, $$\mathcal{M}=\{ X\in\mathbb{R}^{p\times q} : X^\top X = I_q \}, $$ is a manifold. Specifically, it is called a Stiefel manifold and denoted as $\mathrm{St}(p,q)$.

Now the question is, whether $$\mathcal{M}=\{ X\in\mathbb{R}^{p\times q} : X^\top X = \text{diag}(\lambda_1,\dots,\lambda_q),\ \lambda_1,\dots,\lambda_q > 0 \}, $$ is still a manifold? If the answer is yes, how to derive its normal space, tangent space and the retraction operation onto the manifold?

Answer 1

Here, you’ll find the notion of transversality helpful (ok here you don’t exactly need the full generality of transversality, but this is still a good concept to know).

Definition. (Transversality of a map over a submanifold)

Let $X,Y$ be smooth manifolds, $S\subset Y$ an embedded submanifold and $f:X\to Y$ a smooth map. We say $f$ is transverse to $S$ at a point $x\in f^{-1}(S)$ if \begin{align} T_{f(x)}Y=\text{image}(Tf_x)+T_{f(x)}S. \end{align} If this condition holds for all points $x\in f^{-1}(S)$, we say that $f$ is transverse to $S$.

So, transverality of a map means that the tangent linear map $Tf_x:T_xX\to T_{f(x)}Y$ “has a big enough image” so that it together with $T_{f(x)}S$ make up the entire tangent space $T_{f(x)}Y$. As a special case, note that if $f$ is a submersion, then it is transverse to every submanifold of $Y$, simply because the image of $Tf_x$ is already $T_{f(x)}Y$ (due to it being a surjective map). Another case to note is that if $S=\{s\}$ is a singleton then $f$ being transverse to $S$ means exactly that $s$ is a regular value of $f$ (i.e at every point of the level set $f^{-1}(\{s\})$, the tangent map must be surjective).

The regular-value theorem is a special case of the following theorem in which $S$ is taken to be a singleton:

Theorem.

Let $X,Y$ be smooth manifolds, $S\subset Y$ an embedded submanifold and $f:X\to Y$ a smooth map which is transverse to $S$. Then, $f^{-1}(S)$ is an embedded submanifold of $X$ with same codimension as $S$, i.e $\dim X-\dim f^{-1}(S)=\dim Y-\dim S$. Furthermore, for each $x\in f^{-1}(S)$ we have that $T_x[f^{-1}(S)]=(T_xf)^{-1}[T_{f(x)}S]$, i.e the tangent space to the preimage of $S$ (under $f$) is the preimage (under the tangent map of $f$) of the tangent space of $S$.

Once you know what the tangent spaces are, you can of course figure out what the orthogonal complements are.

As for the proof, you can easily google it; you can also find its proof in Guillemin and Pollack (and see here for the motivation of this theorem (which is essentially what’s in Guillemmin and Pollack…)).

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As in the case of the Stiefel manifold, apply this with $X=M_{p\times q}(\Bbb{R})$ and $Y=\text{Sym}_{q\times q}(\Bbb{R})$ and $S$ being the positive-diagonal matrices (why are they a submanifold?). What is the dimension of $S$? So what is its codimension? Hence the dimension of your $\mathcal{M}$ is…?