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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?

Ted Shifrin
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Miles N.
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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…)).

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…?

peek-a-boo
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    Where did the preposition "over" come from? In my 50 years of studying, reading, and teaching this material it's always been "$f$ is transverse to $S$." (Guillemin and Pollack are fond of "transversal," but most mathematicians use that as a noun and use "transverse" as the adjective.) – Ted Shifrin Jul 29 '23 at 18:53
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    @TedShifrin from Dieudonne’s text (this was left as an exercise). But ok now that you mention it, “to” does indeed sound better (maybe “over” is a remnant from the original French?) – peek-a-boo Jul 29 '23 at 18:56
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    @peek-a-boo In French, I'd say "$f$ est transverse à $S$" or "$f$ est transversale à $S$". Over would translate as "au dessus de", which sounds very weird in my opinion in this context – Didier Jul 29 '23 at 19:01
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    or maybe it comes from the fact that when dealing with surjective maps or submersions or the like, we speak of “the fiber over…”, and so perhaps this was also adopted to transversality where we think of $f^{-1}(S)$ as “lying over $S$”. – peek-a-boo Jul 29 '23 at 19:02
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    @Didier yup, just checked the original French, and he does indeed say “…$f$ est transversale au-dessus de $Z$ en $x\in f^{-1}(Z)$…” (he happens to use $Z$ but i like $S$ more) – peek-a-boo Jul 29 '23 at 19:20
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    @peek-a-boo Looks like you were right, twice! Good to know, I somehow never heard it that way. Sounds weird to me, but maybe it was standard in ~50's mathematical French. – Didier Jul 29 '23 at 19:24
  • Thanks a lot for your answer and your useful reference! It seems that transversality is a core concept in differential topology but is seldom covered in a textbook in manifold optimization. For a correction, should the $\text{image}(Tf_x)$ actually be $\text{image}(D f_x)$, the image of the differential operator? – Miles N. Aug 04 '23 at 05:50
  • @MilesN. I use $Tf_x$ to mean the tangent map between tangent spaces $T_xX\to T_{f(x)}Y$ (also called the push-forward map of tangent spaces $f_{*,x}:T_xX\to T_{f(x)}Y$). Some people do indeed write $df_x$ or $Df_x$, but I rather reserve these for slightly different purposes. – peek-a-boo Aug 04 '23 at 10:46
  • @peek-a-boo, thanks very much. I am also curious about how to derive the expressions of the tangent space. It seems that the derivation for submanifolds like the Stiefel manifold is simple due to the fact that these manifolds are defined as level sets of a function (defined by an equation), and thereby the differentiation on both sides of the equation along with a use of the implicit function theorem can yield the result. However, how to analyze the manifold (defined by inequalities) in my question is still unclear to me. Could you please refer me to some related materials? – Miles N. Aug 06 '23 at 11:43
  • @MilesN. the theorem I quoted tells you how to compute the tangent space. The tangent space of the preimage is the preimage of tangent space. I suggest you write out explicitly what $S$ is, what $f$ is (in fact I already told you how to choose these for your particular situation), and pick an arbitrary point $x\in f^{-1}(S)$, and compute the tangent map $Tf_x$ etc. – peek-a-boo Aug 06 '23 at 12:48