# The exponential map

## Geodesics

For a given point $p$ of a semi-Riemannian manifold $M$ with a connection $\nabla$, we can construct a family of geodesics with initial conditions

\begin{eqnarray} \gamma(0) &=& p\\ \dot{\gamma}(0) &=& v \end{eqnarray}for any $v \in T_p M$. Now if we pick a coordinate patch around $p$, such that for $U \ni p$, our chart is $\phi_U : U \to O \subset \mathbb{R}^n$, then our geodesic in terms of those components is

\begin{eqnarray} \phi_U \circ \gamma : I \subset \mathbb{R} &\to& O \subset \mathbb{R}^n\\ \lambda &\mapsto& \phi_U(\gamma(\lambda)) \end{eqnarray}This function, which we will generally write as $x^\mu(\lambda) = \phi_U(\gamma(\lambda))$, is simply a function between real spaces, with the usual definition of the derivative. For some simplicity first, we're going to define the coordinate basis. If we have the curves defined by $\phi_U(\gamma_\mu(\lambda)) = (\phi^0_U(p), \phi^1_U(p), \ldots, \phi^\mu_U(p) + \lambda, \ldots, \phi^n_U(p))$, the curve such that $\gamma_mu(0) = \phi_U(p)$ and that only varies in one coordinate, the vector defined by $\dot{\gamma}_\mu$ at $p$ will be called $\partial_\mu$. This means that we can describe our curves as

\begin{eqnarray} \phi_U(\gamma_\mu(\lambda)) = \phi_U(p) + \lambda e_\mu \end{eqnarray}with the derivative

\begin{eqnarray} \frac{d}{dt}\phi_U(\gamma_\mu(\lambda)) = e_\mu \end{eqnarray}If we now consider our curve $\phi_U \circ \gamma$, as it is $C^1$ at the point considered for us to differentiate it, we can use Taylor's theorem. Around our point $p = \gamma(\lambda_p)$, there exists a neighbourhood such that, for any $\lambda$ in that neighbourhood,

\begin{eqnarray} \phi_U \circ \gamma(\lambda) &=& \phi \circ \gamma(\lambda_p) + (\lambda - \lambda_p) \frac{d}{dt} (\phi_U \circ \gamma(\lambda)) |_{\lambda = \lambda_p} + (\lambda - \lambda_p)^2 h(\lambda_p)\\ &=& \phi(p) + (\lambda - \lambda_p) X(\lambda_p) + (\lambda - \lambda_p)^2 h(\lambda_p)\\ \end{eqnarray}$X$ is the function we got from deriving our curve's coordinates. Picking the canonical basis of $\mathbb{R}^n$, we get

\begin{eqnarray} \phi(p) + \lambda X(\lambda_p) &=& \phi(p) + \lambda \sum_\mu X^\mu(\lambda_p) e_\mu\\ &=& \phi(p) + \sum_\mu \lambda X^\mu(\lambda_p) e_\mu \end{eqnarray}Given the coordinate basis $\{ \partial_\mu \}$ of the tangent bundle at $\gamma(\lambda)$, we can express the derivative $\dot{\gamma}$ by

\begin{eqnarray} \dot{\gamma} &=& \frac{dx^\mu(\lambda)}{d\lambda} \end{eqnarray}...

Now let's consider our geodesics. At $p$, we have the family of curves $\gamma_{v}$, of curves with initial velocity $v$, and all must obey the geodesic equation. let's work them out in the coordinate basis :

\begin{eqnarray} \nabla_{\dot{\gamma}} \dot{\gamma} &=& \nabla_{u} u\\ &=& \nabla_{u^\nu \partial_\nu} u^\mu \partial_\mu\\ &=& u^\nu \left[ (\nabla_{\partial_\nu} u^\mu) \partial_\mu + u^\mu (\nabla_{\partial_\nu} \partial_\mu ) \right]\\ &=& u^\nu \left[ (\partial_\nu u^\mu) \partial_\mu + u^\mu {\Gamma^\sigma_{\mu\nu}} \partial_\sigma \right] \end{eqnarray}For a curve tangent, we have that $u^\nu \partial_\nu f = \dot{f}$. If we project everything on the coordinate basis, we obtain the set of equations

\begin{eqnarray} \dot{u}^\tau(\lambda) + u^\mu(\lambda) u^\nu(\lambda) {\Gamma^\tau_{\mu\nu}} = 0 \end{eqnarray}and, using the fact that our tangent is $u = \dot{x}$, we get the following set of equations

\begin{eqnarray} \dot{x}^\mu(\lambda) &=& u^\mu(\lambda)\\ \dot{u}^\tau(\lambda) &=& - u^\mu(\lambda) u^\nu(\lambda) {\Gamma^\tau_{\mu\nu}} \end{eqnarray}We have $2n$ set of equations, for the function $y = (x, u) : \mathbb{R} \to \mathbb{R}^n \times \mathbb{R}^n$. We can rework it a bit as

\begin{eqnarray} \dot{y}(\lambda) &=& F(\lambda, y(\lambda))\\ y(0) &=& (x_0, v) \end{eqnarray}with $F(\lambda, y(\lambda)) = (u, - u^\mu(\lambda) u^\nu(\lambda) {\Gamma_{\mu\nu}})$. This is simply a first order initial value problem. If $\Gamma$ is uniformly Lipschitz continuous (this is true if $g$ is $C^2$), then by the Picardâ€“LindelĂ¶f theorem, there exists some $\varepsilon > 0$ such that there exists a unique solution $y(\lambda)$ for $\lambda \in [\lambda_0 - \varepsilon, \lambda_0 + \varepsilon]$. Our value of $\varepsilon$ depends obviously on both the point considered $x_0$ and the initial velocity $v$, so that we require

\begin{equation} |\lambda| < \varepsilon(x_0, v) \end{equation}## The normal neighbourhood

From this, we can build a set of coordinates around $p$. For initial values $p$ and $v$, we have a mapping from $[-\varepsilon, \varepsilon]$ to some coordinates of $M$ :

\begin{equation} x^\mu = f^\mu(\lambda, x_0, v) \end{equation}for $f$ a solution of the geodesic equation. From the properties of the geodesic equation, we can see that a rescaling of $v$ leads to a rescaling of $\lambda$. Consider the geodesic with initial velocity $v / k$, $k \in \mathbb{R}$. If we take the same solution $x^\mu(k \lambda)$, a factor of $k^2$ drop out of the geodesic equation, but does not change the solution. On the other hand, we have $\dot{x}(k\lambda)\_{\lambda = 0} = kv$. Therefore, rescaling both $\lambda$ and $a$ will lead to

\begin{equation} f^\mu(\lambda, x_0, v) = f^\mu(k \lambda, x_0, v / k) \end{equation}From this, we can define the exponential map thusly : the exponential map is the mapping of the initial conditions $(x_0, v)$ to the further point of this geodesic for $\lambda = 1$. This is true for any point on this geodesic within the range of uniqueness, since we can always rescale the velocity for this. We can of course define it in a coordinate-independent way :

\begin{equation} \exp_{p}(v) = \gamma_{p,v}(1) \end{equation}which we can map back to our coordinate via

\begin{equation} f(1, x_0, v^\mu) = \exp_{\phi^{-1}(x_0)}(v^\mu \partial_\mu) \end{equation}Let's consider the domain $O \subset T_pM$ on which the exponential map is defined, ie

\begin{equation} O = \{ v \in T_pM | \text{There us a unique geodesic $\gamma$ such that $\gamma(0) = p$, $\dot{\gamma}(0) = v$, and $\gamma(1)$ is defined.} \} \end{equation} If the exponential map is defined for $p, v$ (ie $v \in O$), it is also defined for $p, v / k$, $k > 0$. The domain of the exponential map is therefore star-shaped. We will now show that $O$ is an open set.Take a coordinate neighbourhood $U$ of $p$, with a chart $\phi$. Take the local trivialization of $\pi^{-1}(U)$, with local coordinates $(x^\mu, v^\mu)$. Consider the vector field defined by

\begin{equation} G(x, v) = v^\sigma \frac{\partial }{\partial x_\sigma} - v^\mu v^\nu {\Gamma^\sigma}_{\mu\nu}(x) \frac{\partial }{\partial v_\sigma} \end{equation}The integral curves of $G$ are exactly the first-order system of ODEs we saw previously, so that every integral curve of this vector field is a geodesic, and reciprocally, any geodesic $\gamma$ lifts to an integral curve by taking $\gamma \to (\gamma(\lambda), \dot{\gamma}(\lambda))$.

[...]### Totally normal neighbourhood

An even more desirable property we could want is to have a neighbourhood for which every two points can be linked by a geodesic.

## Fermi normal coordinates

## Bibliography

Last updated :

*2020-03-05 15:00:40*