25
$\begingroup$

Let $(M,g)$ be a Riemannian manifold of dimension $n$ with Riemannian connection $\nabla,$ and let $p \in M.$ Show that there exists a neighborhood $U \subset M$ of $p$ and $n$ (smooth) vector fields $E_1,...,E_n \in \chi (U),$ orthonormal at each point of $U,$ s.t. at $p,$ $\nabla_{E_i}E_j(p)=0.$

What I've found so far;

Let $V$ be a normal neighborhood at $p.$ Let $(W,x(x_1,...,x_n))$ be a local coordinate system at $p$ s.t. $x(W) \subset V.$ Let $\{X_i:=\frac{\partial}{\partial x_i}\}$ be the standard basis of $T_pM.$ By Gram-Schmidt, there exists an orthonormal basis $\{E_i\}$ of $T_pM.$ Consider the orthonormal frame $E_i(t),$ by parallel transporting $\{E_i(0)=E_i\}$ along a a geodesic $\gamma :I \to M$ starting at $\gamma(0)=p$ and ending in $\gamma(1)=q,$ where $q \in V.$ This gives us an orthonormal frame at each point of $V.$ To satisfy the second condition, let $E_i(t)=\sum_{l}a_{li}X_l$ and $E_j(t)=\sum_{s}b_{sj}(t)X_s.$ Then

$$\nabla_{E_i}E_j=\sum_{k}(\sum_{l,s} a_{li}(t)b_{sj}(t)\Gamma^k_{ls}+E_i(b_{kj}))X_k,$$

where $\Gamma^k_{is}$ are Christoffel symbols.

To have $\nabla_{E_i}E_j(p)=0,$ we must have $\sum_{l,s}a_{li}(0)b_{sj}(0)\Gamma^k_{ls}+E_i(b_{kj})(p)=0$ for each $k.$

How can I conclude the argument?

$\endgroup$
10
$\begingroup$

Let $(U,\phi)$ be a coordinate neighborhood of $p$ and let $g_{ij}$ and $\Gamma_{ij}^{k}$ denote the Riemannian metric tensors and the Christoffel symbols, respectively. If we recall that $g_{ij}=(E_i,E_j)$ and $\Gamma_{ij}^{k}=\nabla_{E_i}E_j(E_k)$ for all $1\leq i,j,k\leq n$ (where $M$ is a smooth $n$-manifold and $E_1,\dots,E_n$ are the coordinate frames on $(U,\phi)$), then we need only have $g_{ij}=\delta_{ij}$ on $U$ and $\Gamma_{ij}^{k}(p)=0$.

You are right in choosing a normal coordinate system at $p$, that is, choosing a normal coordinate neighborhood $(U,\phi)$ at $p$. Let me recall that this means choosing an orthonormal basis $F_1,\dots,F_n$ of the tangent space $T_p(M)$, choosing a star-shaped neighborhood of the origin in $T_p(M)$ which is diffeomorphically mapped onto $U$ under the exponential mapping $\text{exp}_p:D_p\to M$ (where $D_p$ is an open subset of $T_p(M)$ containing $0$) and defining $\phi=\exp_{p}^{-1}$ on $U$ (here we identify $T_p(M)$ with $\mathbb{R}^n$ by the linear isomorphism mapping $F_i$ onto $e_i$, $1\leq i\leq n$, where $e_1,\dots,e_n$ is the standard basis of $\mathbb{R}^n$).

In normal coordinates, the geodesics are of the form $y^i=a^it$ where $a_i$ is a constant for all $1\leq i\leq n$. If you prove this, then you should easily be able to see that $\Gamma_{ij}^{k}(p)=0$ for all $1\leq i,j,k\leq n$ by looking at the second-order ordinary differential equation of geodesics. You can also check that $g_{ij}(p)=\delta_{ij}$ for all $1\leq i,j\leq n$.

I hope this helps!

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Dear @ehsanmo, thank you for the correction (I have fixed it). The equalities $\Gamma_{ij}^{k}=\nabla_{E_i}E_j(E_k)$ are the definitions of the Christoffel symbols (for all $1\leq i,j,k\leq n$). For example, please see the relevant Wikipedia article: Christoffel symbols. $\endgroup$ – Amitesh Datta Feb 16 '12 at 8:33
  • 1
    $\begingroup$ @ehsanmo No, you are right. However, the question is to find some coordinate neighborhood of $p$ in which the given equalities (concerning the Riemannian metric tensor and the Christoffel symbols) are valid; of course, these equalities will not hold in all coordinate neighborhoods (as you observed, however, they hold in a suitable normal coordinate neighborhood). (Moreover, although the Levi-Civita connection on a Riemannian manifold is uniquely determined by the Riemannian metric, its values on coordinate frames depend on the coordinate frames in question.) $\endgroup$ – Amitesh Datta Feb 16 '12 at 11:28
  • 9
    $\begingroup$ I have two issues with the construction and the answer: 1) Why does the parallel transport necessarily produce smooth(!) vector fields (the value at each point is the result a different differential equations...) 2) Using the Christoffel symbols implies that the frame {Ei} is the standard frame associated with the coordinates. This, if true, is not clear from the construction of the frame using parallel transport. Am I not seeing something? $\endgroup$ – user38075 Aug 16 '12 at 21:58
  • 1
    $\begingroup$ Just for future readers, and in response to user38075's comment: this answer shows that the Christoffel symbols corresponding to the normal coordinates vanish and that the corresponding coordinate fields are orthonormal at $p$. These coordinate fields are not however guaranteed to be orthonormal in a neighborhood of $p$. To get the orthonormal frame sought by the question one can apply Gram-Schmidt to the normal coordinate fields. You then have to prove though that the resulting frame satisfies $\nabla_{E_i}E_j(p) = 0$. $\endgroup$ – Nick Strehlke Jan 12 '15 at 21:57
  • 1
    $\begingroup$ @ZachBlumenstein all geodesics through $p$ are in the form of $y^i=a^i t$ but this doesn't hold for any other points in $U$, you should be careful dealing with base points $\endgroup$ – user360777 May 4 '17 at 9:09
9
$\begingroup$

Let $U=B_r(p)\subset M^n$ be a normal neighborhood. For each $q\in U$, there is a normalized geodesic $\gamma_q$ joining $p$ with $q$ (radial geodesic). Let $\{v_1,\ldots,v_n\}$ be an orthonormal basis of $T_pM$ and let $\{V_1,\ldots,V_n\}$ be their respective parallel transports along $\gamma_q$. For each $j=1,\ldots,n$, define the field $E_j$ by $$E_j(q)=V_j(d(p,q)),$$ where $d$ is the Riemannian distance. One has that $E_j$ is a $C^{\infty}$ field, because the curves $\gamma_q$ vary $C^{\infty}$ with $q$, in the sense that the EDO's of the geodesics $\gamma_q$ have their coefficients depending $C^{\infty}$ on $q$.

Now, consider $\sigma_i(s)$ the normalized geodesic such that $\sigma_i(0)=p$ and $$\sigma_i'(0)=v_i=V_i(0)=E_i(p).$$ One has, $$\nabla_{E_i}E_j(p)=\nabla_{E_i(p)}E_j=\nabla_{\sigma_i'(0)}E_j=\frac{D(E_j\circ\sigma_i)}{ds}(s)\Big|_{s=0}$$ Since $(E_j\circ\sigma_i)(s)=V_j(d(p,\sigma_i(s)))=V_j(s)$ is a parallel field along $\gamma_{\sigma_i(s)}=\sigma_i\big|_{[0,s]}$, we have that $$\nabla_{E_i}E_j(p)=\frac{D(E_j\circ\sigma_i)}{ds}(s)\Big|_{s=0}=\frac{DV_j}{ds}(0)=0.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ The accepted answer to this question is really a non-answer, and more of a refresher on what normal coordinates are. This answer should IMO be the accepted answer since it shows how to construct the sought-after frame, and proves it is $C^\infty$, and satisfies $\nabla_{E_i}E_j(p) = 0$. $\endgroup$ – Alex Ortiz Aug 17 at 19:01
2
$\begingroup$

Choose a normal neighborhood $U\subset M$ containing $p$. Let $\varphi:U\to\mathbb{R}^n$ be a diffeomorphism and $\mathbb{i}:T_pM\to\mathbb{R}^n$ be an isomorphism. Suppose $\{e_i\}_{i=1}^n$ is an orthonormal base in $\mathbb{R}^n$. Then $\widetilde E_i:=\mathbb{i}^{-1}(e_i)$ is a base in $T_pM$. By using the Gram-Schmidt method, we can find an orthonormal base $\{F_i\}_{i=1}^n$ in $T_pM$, such that $g(F_i,F_j)=\delta_{ij},$ and $F_1=\widetilde E_1$.

Now for any $q\in U_p:=U\setminus\{p\}$, there exists a unique $v_q\in T_pM$, such that $\exp_p(v_q)=q$. Let $\gamma_{p,q}(t) :=\exp_p(tv_q)$ and $\widetilde F_i$ be the $g$-parallel vector field along $\gamma_{p,q}$ such that $\widetilde F_i(0)=F_i$. Let $E_i(q):=\widetilde F_i(1)$ and $E_i(p):=F_i$. Then, $\{E_i\}_{i=1}^n$ forms an orthonormal base at each $q\in U$. It is sufficient to prove that $\nabla_{E_i}E_j(p)=0$.

For any $v=\sum_{i=1}^nv_iE_i(p)\in T_pM$. Consider $\gamma_v(t):=\exp_p(tv)$, which is a geodesic with $\dot\gamma_v(t)=\sum_{i=1}^nv_iE_i(\gamma_{p,q}(t))$ since $E_i$ is parallel along each geodesic. Then, we have $\ddot\gamma_v(t)=0$ for any $t\in[0,1]$ by that $v_i$ is a constant and the continuity of $\ddot\gamma_v$ at the endpoints. Then, by the geodesic equation, we have \begin{align*} \sum_{i,j=1}^n\Gamma_{ij}^k(p)v_iv_j= -(\ddot\gamma_v)_k(0)=0,\quad\text{ for any }k\in\{1,\ldots,n\}. \end{align*} Due to the arbitrary of $v\in T_pM$, we can deduce that $\Gamma_{ij}^k(p)=0$ for any $i,j,k\in\{1,\ldots,n\}$, which follows, by the definition of $\Gamma_{ij}^k$, that $g(\nabla_{E_i}E_j(p),E_k(p))=0$ for any $i,j,k\in\{1,\ldots,n\}$, which further implies that $\nabla_{E_i}E_j(p)=0$ for any $i,j\in\{1,\ldots,n\}$.$\square$

In your proof, it is not necessary to use $\{X_i\}$ again in the last paragraph, since $\{X_i\}$ will not give any more information on the Christoffel coefficients.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ If $E_1=\tilde F_1$ how can you guarantee that $\{F_i\}_{i=1}^n$ is orthonormal? It seems to me you can only tell orthogonality $\endgroup$ – miraunpajaro May 9 at 10:58
  • $\begingroup$ @miraunpajaro Are your asking for the normality of $\{E_i\}_{i=1}^n$? The normality of $\{F_i\}_{i=1}^n$ is trivial because you can always normalize vectors obtained from the Gram-Schmidt method. For $\{E_i\}_{i=1}^n$, at each point $q$, every $E_i(q)$ is a vector in a $g$-parallel vector field $\widetilde F_i$, which is a normal vector field since $F_i$ is normal and the length should be maintained along a geodesic. $\endgroup$ – Yufeng Lu May 16 at 6:44
0
$\begingroup$

Recall parallel transport is a isometry, i. e. if $P=P_{\gamma, t_0, t}: T_{c(t_0)}M\rightarrow T_{c(t)}M$ is parallel transport then

$<u, v>_{c(t_0)} = <P(u), P(v)>_{P(c(t_0))}$

then, because you get ${E_i}$ is orthonormal, $<E_i, E_j> = \delta_{i j}$ is the kronecker delta. Therefore $g_{ij} = <E_i, E_j> = \delta_{i j}$ thus $\Gamma_{ij}^k(p) = 0$.

Thereupon $\nabla_{E_i}E_j(p) = \sum_k \Gamma_{ij}^k(p)E_k(p) = 0$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.