# Formulating a Method of Steepest Ascent on Lie Groups

1. Jan 30, 2014

### Mandelbroth

Suppose we have a compact Lie group $G$, and a differentiable function $f:G_0\to\mathbb{R}$ from the identity component of $G$ to the real numbers. I'm looking to maximize the value of this function.

Being something of a neophyte at optimization, especially of this kind, I decided to stick with something I thought I knew well: the method of steepest ascent. Long story short, I'm having trouble with generalizing the concept to Lie groups.

My idea was fairly simple. We start with some point $p$ on the identity component of $G$, and then we take the logarithm of this point (that is, we take the inverse of the exponential map) because we want to add something to it. As a note, I justified this by saying that, if the point had more than one value for the logarithm, I could just pick one (every point on $G_0$ has at least one logarithm). Then, I would add some multiple of $\nabla f_p$ to $\log(p)$, and finish by exponentiating to get $p'=\exp(c\nabla f_p + \log(p))=\exp(c\nabla f_p)p$. Repeat.

The problem is, I can't figure out what I would use for the analogue of the gradient. Maybe I'm just not seeing something? I don't know. Any nudge in the right direction would be greatly appreciated. Thank you.

Last edited: Jan 30, 2014
2. Jan 30, 2014

### jgens

Not sure if this helps (the question is way outside my area of study), but you can always fix a (bi-invariant) Riemannian metric and define gradients using that.

3. Jan 30, 2014

### Mandelbroth

I'm unfamiliar with this concept, but Wikipedia claims to know something about this. To fact-check, you're suggesting that I could introduce a Riemannian metric $g$ and define $\nabla f_p$ by $g_p(\nabla f_p, X_p)=X_p(f)$?

Wouldn't that put $\nabla f_p$ on $T_p G$ and not $T_e G$ (where $e$ is the identity on $G$), though?

4. Jan 30, 2014

### jgens

Correct definition. You could also use pushforwards on the relevant left-multiplication map to map vectors TGp into vectors TGe. Invariance of the metric should ensure this is pretty well-behaved with respect to the gradient too.

Last edited: Jan 30, 2014