The update makes it much clearer what is wanted for the metric.
Let $d\Theta^2$ denote the spherical metric
(induced by the Euclidean) on $S^{n-1}=\{ x\in {\Bbb R}^n: \|x\|=1\}$ and
write $r=\sqrt{x_1^2+\cdots + x_n^2}$.
The standard Euclidean metric on ${\Bbb R}^n$ may then be written as follows:
$ \sum_i dx_i^2 = dr^2 + r^2 d\Theta^2 $.
The central idea in the current post
is to change the sign on the angular part, i.e. we
consider the punctured Euclidean space
$X={\Bbb R}^n\setminus\{0\}$ equiped
with the following $(1,n-1)$ pseudo-Riemannian metric:
$$ g= dr^2 - r^2 d\Theta^2 =
2 dr^2 - \sum_{i=1}^n dx_i^2$$
We wish to describe geodesics in $(X,g)$. In relativistic terminology,
a tangent vector with $g(v,v)>0$ is time-like,
$g(v,v)<0$ is space-like, while $g(v,v)=0$ corresponds to a
ligth-cone vector.
It is of interest to note that the metric
is invariant under the orthogonal group which implies that there is
angular momentum conservation: A geodesic starting at some
given position in space and in a given direction will always stay in the
span of those two directions, i.e. it suffices to restrict
our attention
to those two dimensions. So let us write:
$$ g= dr^2 - r^2 d\phi^2$$
with $(r,\phi)$ being standard polar coordinates in the plane.
Geodesics in normal Riemannian geometry are paths between
points that are extremal for the length
$\int \sqrt{g(\dot{x},\dot{x})} dt$.
Normalizing to constant speed it
is equivalent to be extremal for the action functional:
$$ S = \int g(\dot{x},\dot{x}) dt =
\int {\cal L} (r,\phi,\dot{r},\dot{\phi}) dt,$$
with the Lagrangian ${\cal L} =
\dot{r}^2 - r^2 \dot{\phi}^2 $.
So we take extremality of this action to define geodesics in the
present context.
An extremal path verifies Lagrange's equations:
$$
2 \ddot{r} =
\frac{d}{dt} \frac{\partial L}{\partial \dot{r}} =
\frac{\partial L}{\partial r} = - 2 r \dot{\phi}^2
\;\; \mbox{and} \; \;
\frac{d}{dt} \left( r^2 \dot{\phi} \right) =
\frac{d}{dt} \frac{\partial L}{\partial \dot{r}} =
\frac{\partial L}{\partial r} = 0.
$$
The last implies the above-mentioned angular conservation: $r^2 \dot{\phi}=A$
for some constant $A$. Similarly, there is conservation of
energy (Hamiltonian):
$$ E =
\dot{r}\frac{\partial L}{\partial \dot{r}} +
\dot{\phi} \frac{\partial L}{\partial \dot{\phi}} -
L = 2 L - L = L $$
So
$\dot{r}^2 - r^2 \dot{\phi}^2 = E$ for some constant $E$.
First case: If $\dot{\phi}=0$ at some instant of time then $A=0$ and
$\phi$ is a constant of motion. We may solve to get: $\dot{r}=\pm\sqrt{E}$
which is just a linear motion in time, $r(t) = \pm\sqrt{E}(t-t_0)$
(the geodesic ceases to exist when $r(t_0)=0$).
Second case: When $\dot{\phi}\neq 0$, it has a constant sign
(same as the sign of $A$). We may then by the implicit function theorem
write $r=r(\phi)$ so that $\dot{r} = r'(\phi) \dot{\phi}$.
Then $(r'^2-r^2) \dot{\phi}^2 = E$ and inserting
the angular momentum conservation we deduce
the following equation for the trajectories:
$$ r'^2 - r^2 = \frac{E}{A^2} r^4$$
Subcases:
a) $E=0$: We get $r'=\pm r$ or $r(\phi) = \exp (\pm (\phi-\phi_0))$.
b) $E<0$ (space-like trajectories): Write $r=1/u$ and solve the resulting ode for $u$.
You end up with
(modulo mistakes in my calculations):
$$ r(\phi) = \frac{A/\sqrt{-E}}{\cosh(\phi-\phi_0)} $$
c) $E>0$ (time-like trajectories):
$$ r(\phi) = \frac{A/\sqrt{E}}{\sinh(\phi-\phi_0)} $$
Symmetries: A part from the rotational symmetry I don't think that
there are any other. The fact that $(r,\phi)$ is identified
with $(r,\phi+ 2\pi)$ gives a topological constraint which
prevents us from doing Lorentz-like transformations.
My answer to the original post:
A suggestion: In ${\Bbb R}^n$, write $r\cdot r'$ and $|r|$ for the Euclidean scalar product and length, respectively.
Define two infinitesimal (Riemannian) pseudo-metrics between infinitesimal close vectors $r$ and $r+dr$ (with $r\neq 0$):
$$ ds_1 = \left| \frac{r}{|r|} \cdot dr \right|
\; \; {\rm and} \; \;
ds_2 = \left|dr - \frac{r}{|r|}\left( \frac{r}{|r|} \cdot dr\right) \right|$$
$ds_1$ measures the radial distance, $ds_2$ the rotational. If you want to represent them as complex numbers you may set $dz=ds_1+i ds_2$. Then, $|dz|$ (modulus of complex number) corresponds to the infinitesimal Euclidean distance.
You may then measure "complex" path lengths: If $r(t)$, $t\in [0,1]$ is a $C^1$ curve then
$$ L={\rm len}_{\Bbb C} (r,[0,1]) = \int_0^1 \left| \frac{r}{|r|} \cdot \dot{r} \right| dt + i \int_0^1 \left|\dot{r} - \frac{r}{|r|}\left( \frac{r}{|r|} \cdot \dot{r}\right) \right|dt$$
separates the usual length of the curve into the radial part (real) and the rotational part (imaginary). The modulus of $L$ is equivalent (though not necessarily equal) to the usual Euclidean length of the path.
Depending on your purpose with defining a complex distance, there might be an ambiguity as to the definition of the distance between two finite vectors $r_1$ and $r_2$ as you would have to specify what a geodesic is in this picture. A perhaps natural choice is to define a geodesic as a path that minimizes Euclidean distances. Then geodesics are straight line segments and the complex distance may be (partially) calculated as follows:
Let $r'$ be the point on the line segment $[r_1;r_2]$ closest to the origin (could be one of the end-points) and let $a\geq 0$ be the distance from the line through $r_1$, $r_2$ and the origin.
For $0< u\leq v$ write
$ \Theta(u,v) = u\ln \frac{v+\sqrt{v^2-u^2}}{u} $. Then if $r'$ is not one of the end-points:
$$ d_{\Bbb C}(r_1,r_2) = \left(|r_1|+|r_2|-2|r'| \right)
+
i \left(\Theta(a,|r_1|+ \Theta(a,|r_2|)-2\Theta(a,|r'|)\right)$$
while if $r'$ is one of the end-points you get the simpler:
$$ d_{\Bbb C}(r_1,r_2) = \left||r_2|-|r_1|\right|
+ i \; a \left| \ln \frac{|r_2|+\sqrt{|r_2|^2-a^2}}{|r_1|+\sqrt{|r_1|^2-a^2}}\right| $$
separating into how much you move radially and rotationally along the geodesic.
In the limit $a\rightarrow 0$ the imaginary part vanishes as wanted since $r_1$ and $r_2$ are proportional in that case.
The world lines that extremize the proper time between A and B are those which satisfy the Lagrange equation:
\begin{equation}
-\frac{d}{d \sigma} \left(\frac{\partial L}{\partial({dx^{\alpha} / d \sigma})} \right)+\frac{\partial L}{\partial x^{\alpha}}=0
\end{equation}
Where the Lagrangian is:
\begin{equation}
L= \left(-g_{\alpha \beta} \frac{d x^{\alpha}}{d \sigma} \frac{d x^{\beta}}{d \sigma} \right)^{\frac{1}{2}}
\end{equation}
The Schwarzschild metric in geometrical units is
\begin{equation}
\left(
\begin{array}{cccc}
-(1-\frac{2 m}{r}) & 0 & 0 & 0 \\
0 & (1-\frac{2 m}{r})^{-1} & 0 & 0 \\
0 & 0 & r^2 & 0 \\
0 & 0 & 0 & r^2 \sin ^2(\theta ) \\
\end{array}
\right)
\end{equation}
The parameter t is therefore called world time. If m = 0 the Schwarzschild metric reduces to the Minkowski metric; if $m \neq 0$ it is singular for $r = 0$. For r large with respect to m the Schwarzschild metric coincides with the Newtonian approximation, since $2m/r$ is the Newtonian potential outside a spherical body of mass m centered at $r = 0$. In addition, when
$1−(2m/r)$ vanishes for $r = 2m$, the metric appears to be singular there. This singularity can be removed by choosing a different coordinate system.
The geodesic equations based on the Schwarzschild metric above are
\begin{equation}
\left(
\begin{array}{ccc}
\frac{d u_0}{d \tau} = \frac{2 m u_0 u_1}{2 m r-r^2} \\
\frac{d u_1}{d \tau} = \frac{m (2 m-r) u_0^2}{r^3}-\frac{m u_1^2}{2 m r-r^2}-(2 m-r) u_2^2-(2 m-r) \sin ^2(\theta) u_3^2 \\
\frac{d u_2}{d \tau} = \cos (\theta) \sin (\theta ) u_3^2-\frac{2 u_1 u_2}{r} \\
\frac{d u_3}{d \tau} = -\frac{2 (u_1+r \cot (\theta ) u_2) u_3}{r} \\
\end{array}
\right)
\end{equation}
where $u_{\alpha}$ are the component of the four proper velocity vector while $\tau$ is the proper time. When looking at the metric you can see that is independent of $t$, thus there is a Killing vector associated with this symmetry under displacement of the time coordinate t with components $\psi=(1,0,0,0)$. The first equation can be written as:
\begin{equation}
\frac{d}{d \tau} \left((1-\frac{2 m}{r}) u_0(\tau) \right) = 0
\end{equation}
which is integrated to give
\begin{equation}
\left((1-\frac{2 m}{r}) u_0(\tau) \right) =- g_{00} \frac{dt}{d \tau}= E
\end{equation}
with E constant.
Now you have to remember that the dot product is obtained by computing the first fundamental form.The first fundamental form is the restriction of the usual dot product in $\mathbb{R}^3$ to the tangent plane $T_p S$. Given two vector a and b
the first fundamental form $Ip (a, b)$ at the point $p$ can be expressed as the bilinear form $a^T g b$, where g is the matrix. In the case of the xy-plane or a cylinder $g$ is a diagonal matrix with diag(g) = (1,1). In this case the dot product of $a$ and $b$ is
\begin{equation}
(a_1,a_2) \left(
\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}
\right) (b_1,b_2)^T= a_1 b_1+ a_2 b_2
\end{equation}
However is the matric is not flat we have to account for the terms $g_{\alpha, \beta}$. So yes, $U^aV_a$ is a dot product, with the index a running from 1 to the 4, since you are in a 4 dimensional space. This dot product can be written as
\begin{equation}
g_{00} u_0 v_0-g_{00}^{-1} u_1 v_1+r^2 u_2 v_2+ r^2 \sin^2(\theta)u_3 v_3
\end{equation}
However, The 4-velocity of a stationary observer is $(u_0,0,0,0)$, so for a stationary observer we have only the term $g_{00} u_0 v_0$.
Consider now the lapse of proper time τ between two events at a fixed spatial point (differentiation with respect the position is zero) in Schwarzschild space-time:
\begin{equation}
ds^2=-d \tau^2=-g_{00} dt^2
\end{equation}
hence
\begin{equation}
d \tau=\sqrt{g_{00}} dt
\end{equation}
The element of proper time dτ is measured by a clock at the particular point, while the element of world time dt is fixed for the whole manifold. Since $g_{00}<1$, so $d \tau < dt$ i.e. clocks go slower in a gravitational field. In fact time itself goes slower in a gravitational field. In flat spacetime $m$ turn down to zero so that $dt=d \tau$, and a clock records the coordinate time $t$.
Best Answer
Given your update, the $r=const$ slices are uniform in $t$:
$$ \mathrm{d}\tau = \sqrt{\left(1- \frac{r_s}{r}\right)^{-1}} \,\mathrm{d}t = (\mathrm{const}) \cdot \mathrm{d}t$$
So you get an ordinary Euclidean line.