5
$\begingroup$

I am currently working in a Riemannian manifold optimization problem, with very little formal background on differential geometry.

The optimization variable lives in a generalized complex Stiefel Manifold $$ST(n,p,B)=\{X\in\mathbb{C}^{n\times p}: X^*BX=I_p\}$$ where $B\succ0$ is positive definite, with the usual metric $\rho(X,Y)=\mathrm{Re}(\mathrm{trace}(X^*Y))$. I have obtained the geometric characterizations of the manifold (tangent space, projection, retraction, gradient and Hessian) that allows me to run optimization procedures such as trust-regions or others.

Now, the cost function has rotational unimodal invariance on every column of $X$, i.e., the cost function presents invariance over the group $\mathcal{T}(p)$ of diagonal unitary matrices of size $p$: $$f(X)=f(XT)\qquad\text{with }T\in\mathcal{T}(p)$$ where $$\mathcal{T}(p)=\{T\in U(p):T\text{ diagonal}\}=\{T\in U(p):T=\mathrm{diag}(e^{j\theta_1},\ldots,e^{i\theta_p})\}=U(1)^{\times p},$$ is a group, as is the maximal torus of $U(p)$ and is free, proper, etc. This fact calls for the definition of a quotient manifold with equivalence class $$X\sim Y\Rightarrow Y=XT,\,T\in\mathcal{T}(p)$$ or, equivalently, $$[X]=\{XT:T\in\mathcal{T}(p)\}.$$

I have been struggling with deriving the quotient nature of the manifold $ST(n,p,B)/\!\!\sim$, i.e. $ST(n,p,B)/\mathcal{T}(p)$. I did this successfully in the case of $p=1$: complex vectors with rotational invariance, or $\mathbb{C}^n/U(1)$, but the more general setting is a little more confusing.

If you have any ideas on how to represent the tangent space (in particular, horizontal and vertical spaces), projection and retraction, I would really appreciate the help. I can also add more information and additional derivations. In particular, a derivation of the quotient geometry from the canonical complex Stiefel manifold, i.e.

$$\mathcal{M}=St(n,p)/U(1)^{\times p},\quad St(n, p)=\{X\in\mathbb{C}^{n\times p}:X^*X = I_p\}$$ would help greatly.

Thanks!

$\endgroup$
3
  • 1
    $\begingroup$ Computationally, you can "cheat" by working with selecting particular representative matrix in the Stielfel manifold itself and just make sure to take orthogonal complement of the $U(1)^{\times n}$-action whenever needed. $\endgroup$ Jul 19, 2020 at 1:53
  • $\begingroup$ Thanks for your comment. I do understand the idea, but I also wanted to go through the actual derivation to enhance comprehension :) $\endgroup$
    – cjferes
    Jul 19, 2020 at 2:28
  • 1
    $\begingroup$ Also, I understand the idea behind the group action, the orthogonal complement, but I still struggle when defining those explicitly. Again, I'm still very new to the topic. $\endgroup$
    – cjferes
    Jul 19, 2020 at 5:03

1 Answer 1

0
$\begingroup$

Ok, so I still don't have a complete answer but it's the best I have been able to do during the week. This is good enough for the purposes of starting this conversation and having references for anyone interested, and seeing that the question had a significant amount of views and no answer yet, I'll post my work so far and accept that as an answer. I will also post any further progress.

Edit: I found an expression for the projection to horizontal space. Added for completeness.


After studying a few resources I found the Ph.D. thesis from T.A. Palka where they introduce the so-called "Basis manifold", which corresponds to a quotient manifold of the complex Steifel manifold under the group action of unitary diagonal matrices. i.e. $St(\mathbb{C}^{n\times p})/\mathcal{U}(1)^{\times p}$. This is equivalent to setting $B=I_p$ in my own formulation, which is straightforward to adapt so it still fulfills the purpose (although is something I have yet to do for my own purposes).

Consider the Steifel manifold as a quotient manifold of the set of unitary matrices $\mathcal{U}(n)$ over the action of unitary matrices $\mathcal{U}(n-p)$, i.e. $St(\mathbb{C}^{n\times p})=\mathcal{U}(n)/\mathcal{U}(n-p)$. The equivalence class is described in the ambient space as $$[\,Z\,]=Z\begin{bmatrix}I_n&0\\0&Q\end{bmatrix}\,\qquad Q\in\mathcal{U}(n-p),$$ that is, the first $p$ columns of $Z$ remain unchanged. This is obviously well known, but it helps to understand the method used to describe normal and tangent spaces that is then used for the manifold of interest. This is related to the direct representation of matrices in the Stiefel manifold $X\in St(\mathbb{C}^{n\times p})$ as $$Z=\begin{bmatrix}X&X_{\perp}\end{bmatrix}\quad\Rightarrow\quad [\,Z\,]=\begin{bmatrix}X&X_{\perp}\end{bmatrix}\begin{bmatrix}I_p&0\\0&Q\end{bmatrix}$$ where $X^*X_{\perp}=0$ by definition. In particular, under the direct representation, is useful to know that the tangent space of the complex Stiefel manifold is $$\mathcal{T}_XSt=\{X\Omega+X_{\perp}B,\quad \Omega=-\Omega^*\in\mathbb{C}^{p\times p}\,\text{anti-Hermitian}, B\in\mathbb{C}^{(n-p)\times p}\,\text{arbitrary}\},$$ with the projection operator to tangent space given by $$ \begin{align} \mathrm{Proj}_{{X}}^{\mathrm{T}}({W})&={W}-{X}\mathrm{herm}\big({X}^*{W}\big), \end{align}$$ where $\mathrm{herm}(A)=0.5(A+A^*)$ is the Hermitian part of matrix $A$.

Under the same premise, the Basis manifold $\mathcal{B}(\mathbb{C}^{n\times p})=St(\mathbb{C}^{n\times p})/\mathcal{U}(1)^{\times p}=\mathcal{U}(n)/\left(\mathcal{U}(1)^{\times p}\times\mathcal{U}(n-p)\right)$, and the equivalence class in $\mathcal{U}(n)$ ambient space is defined as $$[\,Z\,]=Z\begin{bmatrix}D&0\\0&Q\end{bmatrix}\,\qquad D\in\mathcal{U}(1)^{\times p},\quad Q\in\mathcal{U}(n-p),$$ or in Stiefel ambient space as $$[\,X\,]=XD\,\qquad D\in\mathcal{U}(1)^{\times p}.$$

The thesis then goes into describing the vertical and horizontal spaces (whose direct sum is equal to tangent space), although is a little obtuse: the thesis focuses on estimation bounds and not on optimization per se. In particular, the thesis itself does not present the projector operators to the horizontal space, which is desired, or a general version of the components of the horizontal space. Nevertheless, by going through the new book from N. Boumal I think I've been able to find the conditions that describe the horizontal space in a general manner.

For a quotient manifold, we separate the tangent space in two subspaces: the vertical space $\mathcal{V}_X$ and the horizontal space $\mathcal{H}_X$. The former contains the vectors tangent to the equivalence class for $X$. The argument in Boumal's book, Proposition 9.3, is that the vertical space corresponds to the tangent space to the fiber of the equivalence class, which can be obtained as follows. For a particular point $X$, we define its fiber as $\mathcal{F}=\{XD:D\in\mathcal{U}(1)^{\times p}\}$. All vectors of the fiber are of the form $c'(0)$ for some smooth path $c:\mathbb{R}\to \mathcal{F}$ such that $c(0)=X$, which actually is of the form $c(t)=XD(t)$ for $D:\mathbb{R}\to\mathcal{U}(1)^{\times p}$ with $D(t)=I_p$. Thus, all tangent vectors in the vertical space are of the form $XD'(0)$. Now, the tangent space of $\mathcal{U}(1)^{\times p}$ is the set of diagonal matrices with imaginary elements, which corresponds to the Lie algebra of unitary diagonal matrices $\mathfrak{t}(p)$ (argument which comes from the fact that the set $\mathcal{U}(1)^{\times p}$ is the maximal torus of $\mathcal{U}(p)$). Thus, $$\mathcal{V}_X=\{XT,\,\,T\in\mathfrak{t}(p)\}.$$

On the other hand, $\mathcal{H}_X$ is the orthogonal complement of $\mathcal{V}_X$, that is, it contains all vectors in the tangent space that are orthogonal to the equivalence class: $$\begin{align}\mathcal{H}_X=(\mathcal{V}_X)^{\perp}&=\{H\in \mathcal{T}_XSt:\langle H,V\rangle=0\,\,\,\forall\,V\in\mathcal{V}_X\}\\ &=\{H\in \mathcal{T}_XSt:\langle H,XT\rangle=0\,\,\,\forall\,T\in\mathfrak{t}(p)\}\\ &=\{H\in \mathcal{T}_XSt:\langle X^*H,T\rangle=0\,\,\,\forall\,T\in\mathfrak{t}(p)\}, \end{align}$$ that is, all vectors $X^*H$ have to be orthogonal to imaginary diagonal matrices.

Moreover, as $H\in\mathcal{T}_XSt$, we know that $H=X\Omega+X_{\perp}B$ for some $\Omega$ anti-Hermitian and some arbitrary $B$. Then we have that $$\begin{align}X^*H&=X^*\left(X\Omega+X_{\perp}B\right)=I_p\Omega+0B=\Omega \end{align}$$ that is, $X^*H$ is anti-Hermitian.

These two conditions on $X^*H$ (orthogonality to $\mathcal{U}(1)^{\times p}$ and anti-Hermitian) define the horizontal space. It is clear then that the horizontal space is composed of matrices $H$ such that $X^*H$ are anti-Hermitian with a zero diagonal.

Now we find the projection to horizontal space. Let $\mathrm{ddiag}(M)$ the operator that yields a diagonal matrix whose elements are the same diagonal elements of $M$. Then, the operation that projects into the horizontal space is then $$ \begin{align} \mathrm{Proj}_{{X}}^{\mathrm{H}}({W})&=\mathrm{Proj}_{{X}}^{\mathrm{T}}({W})-{X}\mathrm{ddiag}\big({X}^*\mathrm{Proj}_{{X}}^{\mathrm{T}}({W})\big)\\ &={W}-{X}\mathrm{herm}\big({X}^*{W}\big)-{X}\mathrm{ddiag}\big({X}^*W-\mathrm{herm}({X}^*W)\big). \end{align}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .