7
$\begingroup$

I am reading this paper Conjugate Gradient Algorithm for Optimization Under Unitary Matrix Constraint. It basically describes a variant of the conjugate gradient method when optimizing for unitary matrices which takes advantage of the Lie group structure of $U(n)$.

The question however is basic differential geometry and maybe Lie theory and this is what I am struggling with. Specifically, the following is stated:

Now, the parallel transport of a tangent vector $\tilde{X} = XW \in T_W U(n) , X \in \mathfrak{u}(n)$, w.r.t. the Riemannian connection along the geodesic $$ \mathcal{G}_W (t) = \exp(tS) W,\quad S\in \mathfrak{u}(n), \quad t\in \mathbb{R} \tag{1}$$ is given by $$\tau \tilde{X}(t) = \exp(tS/2) \; X \; \exp(−tS/2) \;G_W(t) \tag{2} $$ where $\tau$ denotes the parallel transport.

Here, $\mathfrak{u}(n) $ denotes the Lie algebra to $U(n)$ and the geodesic $\mathcal{G}_W(t)$ is such that it emanates from $W$ in the direction of the vector $(dR_W)_e(S)=SW \in T_WU(n)$. I should further note that $U(n)$ was equipped with a bi-invariant metric!

My questions:

  1. From the formula of the geodesic $\mathcal{G}_W(t)$, it seems to be that one can transport every vector on the Lie group just by left/right translation to every point. This seems much more natural than parallel transport. In the case of a bi-invariant metric, does it differ from parallel transport (which I understand depends on the geometry crucially)

  2. Could someone walk me trough how one obtains the formula for the parallel transport, namely (2)?

    In particular, this actually looks a like a right translation of the vector $\exp(tS/2) \; X \; \exp(−tS/2)$, too, i.e. $$ \tau \tilde{X}(t) = dR_{G_W(t)}\left(\exp(tS/2) \; X \; \exp(−tS/2)\right) \tag{3} $$ and the thing in the bracket is somehow the adjoint map but I still cannot puzzle these things together.

$\endgroup$

0

You must log in to answer this question.