The two terms you wrote down are roughly the horizontal and vertical parts of the metric. Roughly speaking, the first part gives the metric for the first factor $T_pM$ of $T_{p,v}TM$. The second part gives the metric for the second factor $T_vT_pM$.
By the definition of the projection operator $\pi$ and the definition of the covariant derivative on $(M,\langle\cdot\rangle)$, it is clear that the expression you wrote down is coordinate independent. There are several things to check to make sure that it is a metric
- It is positive definite
- It is bilinear
- It is tensorial
Since we already have coordinate independence, it is most convenient to work over a fixed coordinate system.
Let $\{x^1,\ldots,x^n\}$ be a system of coordinates for $M$; this can be extended to a system of local coordinates $\{x^1,\ldots,x^n;y^1,\ldots,y^n\}$ for $TM$ where $(x,y)$ corresponds to the point $(p,v)\in TM$ with $p$ the point in $M$ specified by $x$, and $v\in T_pM$ given by $\sum y^i\partial/\partial x^i$.
At a fixed point $(p,v)$, an element of $T_{p,v}TM$ can be then described by
$$ V = \sum \xi^i \frac{\partial}{\partial x^i} + \sum \zeta^i \frac{\partial}{\partial y^i}$$
with the projection
$$ d\pi(V) = \sum \xi^i \frac{\partial}{\partial x^i} $$
Express the curve $\alpha(t) = (p,v)(t)$ in this coordinates we have that the condition $\alpha'(0) = V$ is simply the statement that $\frac{d}{dt}p^i(0) = \xi^i$ and $\frac{d}{dt}v^i(0) = \zeta^i$.
With this we can compute
$$ \frac{D}{dt}v^i(0) = \frac{d}{dt} v^i(0) + \Gamma^i_{jk}\left(\frac{d}{dt}p^j(0)\right)\left(v^k(0)\right) $$
using the definition, and where $\Gamma$ is the Christoffel symbol of the Riemannian metric on $M$. This we immediately see to be
$$ \frac{D}{dt}v^i(0) = \zeta^i + \Gamma^i_{jk}v^k(0)\xi^j $$
which in fact is a linear map from $T_{p,v}TM$ to $T_pM$.
Now the three properties are easily checked:
- It is tensorial because the expressions are completely independent of which curve $\alpha$ is chosen, as long as $\alpha'(0)= V$.
- It is bilinear because $T_{p,v}TM\ni V\mapsto (d\pi(V),\frac{D}{dt}v(0))\in T_pM \oplus T_pM$ is linear, and the Riemannian metric on $M$ is bilinear
- Since the Riemannian metric induced on $T_pM\oplus T_pM$ is positive definite, to prove that the new object is also positive definite, it suffices to check that the map $V\mapsto (d\pi(V),\frac{D}{dt}v(0))$ is injective. But this is true by direct inspection.
Best Answer
The exponential mapping creates a metric of the form
\begin{align*} ds^2=dr^2+\psi(r,\theta)^2d\theta^2, \end{align*}
locally in a sufficiently small neighbourhood of any point $p$ on a two-dimensional manifold. Your condition expresses a local symmetry: on a sufficiently small neighbourhood of $p$ the metric is invariant with respect to a translation in the $\theta$ coordinate (we could call it a 'local rotation').
It makes a lot of difference if you want that kind of metric in one point or in every point, and in the latter case if the function $\psi$ has to be the same for every point (it isn't in your examples with $\psi^2(r)=r,r^3\ldots$).
If the same function $\psi(r)$ is valid around every point $p$ then we are in the realm of surfaces of constant curvature; to see this, note that the curvature for a general $\psi(r)$ is given by
$$K=-\frac1{\sqrt G}\frac{\partial^2\sqrt G}{\partial r^2}$$
where $G=1.\psi^2(r)$ is the determinant of the matrix expressing the components of the metric tensor. This shows that the function $\psi(r)$ uniquely determines the curvature at the origin of the coordinate system. If all origins have the same function $\psi(r)$ then all origins have the same scalar curvature.
Thus constant curvature implies $\psi''/\psi$ must be constant. Depending on the sign of the constant (the opposite of the sign of the curvature), the solutions of that differential equation are: first-degree polynomials (your example of a Euclidean plane), a scaled, shifted version of the hyperbolic sine (your example of a hyperbolic sphere) and a scaled, shifted version of the ordinary sine (a sphere). The shift can be eliminated by requiring $\psi(0)=0.$
The sphere of radius $R$ has $\psi(r)=\sin(r/R).$