In the book, optimization on matrix manifolds by Absil, Chapter 3 pg 59, it is given that retraction on the orthogonal manifold $O_n$, using the Cayley's transform, is given by $$ R_{X}(X \Omega)=X\left(I-\frac{1}{2} \Omega\right)^{-1}\left(I+\frac{1}{2} \Omega\right) $$ I am trying to derive to this from first principles but stuck mid-way.

My attempt:

Let $Q \in O_n$ and $\Omega$ be a skew symmetric matrix. By Cayley's transform, for each $\Omega$ we can find a $Q$ such that $$Q = \left( I - \Omega \right) \left( I + \Omega \right)^{-1}$$ Let $\phi$ be the mapping from $[Q^{-1},I - \Omega] \rightarrow I+\Omega$. Therefore, we can write $\pi_{1} \circ \phi^{-1}$ as matrix that takes in $I+\Omega$ and outputs $Q^{-1}$.

\begin{align} R_{X}(X+X\Omega) &:=\pi_{1}\left(\phi^{-1}(X+X\Omega)\right) \\ & = X \pi_{1}\left(\phi^{-1}(I+\Omega)\right) \\ & = X Q^{-1} \\ & = X \left( I + \Omega \right) \left( I - \Omega \right)^{-1} \\ & = X \left( I - \Omega \right)^{-1} \left( I + \Omega \right) \qquad \quad \text{as the matrices commute}\\ \end{align}

I am not sure how to get the the $\frac{1}{2}$ as mentioned in the textbook. Can you please help. Thank you.

Update: One of the answers says that this is a typo in the book. Can others please confirm?

Below is a picture of the relevant text and symbols. enter image description here

  • 1
    $\begingroup$ If the following papers are helpful: "A Riemannian conjugate gradient method for optimization on the Stiefel manifold", "A Feasible Method for Optimization with Orthogonality Constraints". $\endgroup$ – River Li 8 hours ago

Hint: Let $\Omega$ be skew-symmetric of order $n$. Then $\frac{1}{2} \Omega$ is also skew-symmetric. Hence, by Cayley's transform there exists a $Q \in O_n$ such that $$Q = (I - \tfrac{1}{2} \Omega) (I + \tfrac{1}{2} \Omega)^{-1}$$.

  • $\begingroup$ I am not sure how it helps because our input to retraction mapping is of the form $(X + X\Omega)$, not $(X + 0.5 X\Omega)$ $\endgroup$ – kasa Nov 12 at 6:45
  • $\begingroup$ You are right, but I don't see a problem if you redefine $R_X$ to be $$R_X(\xi) := \pi_1 (\phi^{-1}(X + \tfrac{1}{2} \xi)).$$ With that modification the proof of Proposition 4.1.2 would still work, hence it would still define a retraction. $\endgroup$ – Math Tourist 9000 Nov 12 at 19:45
  • $\begingroup$ You cannot redefine retraction as $$R_X(\xi) := \pi_1 (\phi^{-1}(X + \tfrac{1}{2} \xi))$$. The proof is for $R_X(\xi) := \pi_1 (\phi^{-1}(X + \xi))$, introducing $\frac{1}{2}$ will contradict the local rigidity property of the retraction. $\endgroup$ – kasa Nov 13 at 7:33
  • $\begingroup$ Ok, nevermind.I believe there's a typo in the book.From Proposition 4.1.2 we know that $R_X(\xi) := \pi_1(\phi^{-1}(X + \xi))$ is a retraction for any $\xi$ in the neighbourhood of the origin of $T_X M$. Let's take $\xi=\tfrac{1}{2} X\Omega$.Then, by what we just said $R_X(\frac{1}{2} X\Omega)$ is a retraction and specially, let's take $\phi: (Q^{-1}, I - \frac{1}{2} \Omega) \mapsto I + \frac{1}{2} \Omega$ (obtained from the Cayley transform).Now, repeating your steps with this, you get the formula from the book except $R_X(\frac{1}{2} X\Omega)$ is on the left hand side and not $R_X(X\Omega)$. $\endgroup$ – Math Tourist 9000 Nov 13 at 16:50

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