Metric tensor of complex numbers & Hamiltonian Mechanics

Question

The Euclidean $\mathbb{R}^2$ geometric space can be mapped onto $\mathbb{C}$. In other words I see it like this $$\vec{v} = x\vec{x}+y\vec{y} = x\vec{1}+y\vec{i}= \begin{bmatrix}x \\y\end{bmatrix} $$ Where in the complex topology (loosely used here) $\vec{1},\vec{i}$ have a curvature given by: $$\eta=\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$$ as opposed to the normal Euclidean $\eta = \begin{bmatrix}1 & 0\\0 & +1\end{bmatrix}$. where the inner product is defined as $<v|v>=v \cdot\eta \cdot v$

This means that the gradient is $ \nabla =\begin{bmatrix}+ \frac{\partial }{\partial x }\\ -\frac{\partial }{\partial y }\end{bmatrix} $ which is very useful for Hamiltonian Mechanics. (1) Normally, the gradient gives the steepest descent, but what is its interpertation with this metric?

(2) I was hoping someone can explain to me what this metric tensor of complex numbers means, in terms of the curvature (or some geometric concept) and what its connection to complex algebra is. I know it somehow defines circles and makes vector algebra work well on them.

(3) Since Hamilton's equations of motion can be recast as $ \dot{\vec{v}} =\nabla H$, where H is the Hamiltonian, this also gives the phase space vector field. If you can provide some insight to the connection with Hamiltonian Mechanics, that would be wonderful.

[references are appreciated]

Answer 1

As I see it, passing to complex variables is just a simplification. Instead of having a metric tensor $\delta=$diag$(1,1)$, since $\mathrm{i}^2=-1$ you have to change it into $\eta=$diag$(1,-1)$ in order to obtain the same inner product.

If you consider a Hamiltonian system with $N$-degrees of freedom you might be interested in the simplification obtained by "complexifying" the phase space. Map $(q^i,p_i)$ to $z_i=q^i+\mathrm{i} p_i\in \mathbb{C}$. Then you can write the Hamilton equations of motion for that system in the more compact form

$$\mathrm{i}\frac{dz}{dt}= 2\nabla_\bar{z}H,$$ where $H=H(q^i(z),p_i(z))$ is the Hamiltonian and the components of $\nabla_\bar{z}\,f$ are $$\frac{\partial f}{\partial \bar{z}^i}:=\frac{1}{2}\left(\frac{\partial f}{\partial q^i}-\mathrm{i}\frac{\partial f}{\partial p_i}\right).$$