# Metric tensor of complex numbers & Hamiltonian Mechanics

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]

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ps. My background is in physics so I apologize for the surely sloppy mathematical notation and nomenclature, it demonstrates a lack of understanding on my part, apologies :) –  AimForClarity Jan 11 '13 at 20:49
The complex plane equipped with that (1,-1) bilinear form has a hyperbolic geometry (as opposed to the Euclidean geometry you get with the (1,1) metric.) That's about all I can think of saying... :/ –  rschwieb Jan 11 '13 at 21:47
@rschwieb Thank you, how do you see the hyperbolic nature mathematically is it just the $\nabla \vec{v}$, where $\vec{v}=(x,y)$? –  AimForClarity Jan 11 '13 at 22:07
@AimForClarity I'm not sure I understand why you write "curvature" for $\eta$. I do not understand that part of the question. –  c.p. Jan 11 '13 at 22:28
Well I'm not sure I do wither. I wrote that because It seems to me that this is like the Riemann curvature tensor used in general relativity, although this is more of a hunch than any hard proof. I am not really familiar with differential geometry or topology. Maybe someone else can shed light on this. –  AimForClarity Jan 11 '13 at 22:32

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 $$\frac{\partial f}{\partial \bar{z}^i}:=\frac{1}{2}\left(\frac{\partial f}{\partial q^i}-\mathrm{i}\frac{\partial f}{\partial p_i}\right).$$
the $2$ factor comes from the definition of the partial with respect to $\bar{z}$. // These equations are $N$ complex equations, so $2N$ real eqs. I think I saw that once on one of Marsden's books. Let me search... –  c.p. Jan 11 '13 at 22:47