Finding covariant derivative of a riemanian submanifold

Question

Hi, I have a question about properties which are common to a manifold and its submanifolds. I start with the metric. $ M \subset N, dim(M) = m, dim (N) = m+1 $ let $ g^N $ be the metric of N, so that $ (N,g^N) $ is a riemanian manifold and N is a submanifold. Now, I'm looking at N and I'm trying to understand what does $ g^M $ looks like. WLOG I assume that in every point $ p \in M $ there exists $ \phi $ a homemorphism of a neighbourhood of p to $ U \subset R^{m+1} $ $ p = \phi(U^1,...,U^m,U^{m+1} = 0) $ I call the reduced $ \phi, \psi $. Now, I can see that $ \partial \psi / \partial u^j = \partial \phi / \partial u^j $ for $ 1 \leq j \leq n $ and that, $ \\ \partial \psi / \partial u^{m+1} = 0 $ (by definition) so I conclude that in U coordinates, $ g^N $ has the form $ \left(\begin{array}{cc}A_{m \times m}&*\\***&B_{1 \times 1}\end{array}\right) $ This must be this way, of the inner product will not be induce correctly from N to M. A is exactly $ g^M $ Now, I'm trying to check the Cristoffel symbols (so I could know what the covariant derivative is). I use the formula $ \Gamma^k_{i j} = 1/2 * g^{k l} ( \partial g_{l j} / \partial u^i + ...)$ And here is my problem. the factors in the brackets are identical for M and N, but I cant say the same about $ g^{k l} $. If I could determine that * from above is zero (?) then I could say that the inverse of $ g^N $ is $ \left(\begin{array}{cc}A^{-1}&0\\C&D\end{array}\right) $ but unfortunately, I dont know if I can choose coordinates, so that this property holds. Can I somehow make it happen? or is there another way to compute $ \Gamma^M $ from $ \Gamma^N $?

thanks

Answer 1

If I understand your notation correctly, then your question is a bit confused, because $g^N$ has to be a symmetric matrix, so that "$***$" = "$*$". The condition that $g^N$ is block diagonal does not have to hold; it says that the tangent vector of the last coordinate, $\partial/\partial u^{m+1}$, is perpendicular to the surface $M$. On the other hand, there always exist local coordinates with this property. If you take any local coordinates for $M$, you can evolve them for a short time with the normal surface flow. You can even get the condition $B = 1$ in a local chart.

Also, there certainly is another way to get the covariant derivative on $M$ and its Christoffel symbol. Namely, if you apply the covariant derivative $\nabla^N$ to a tangent vector field $v$ on $M$ in some tangent direction $w$, you get a vector field $\nabla^N_w(v)$ on $M$ that does not have to be tangent. You should then just project this derivative $\nabla^N_w(v)$ orthogonally onto the tangent bundle $TM$. The orthogonal projection is a useful tensor field $P$ defined on the tangent bundle $TN$ restricted to $M$, and you can write an explicit expression for the covariant derivative $\nabla^M$, or the Christoffel symbol or even the curvature tensor, in terms of $\nabla^N$ and this tensor field $P$. Actually, I am not entirely sure that this method is algebraically all that different, but it is at least conceptually different.

Answer 2

@Tamir: what book are you learning from? It seems that you are missing/confounding some elementary concepts.

This really belongs as a clarification of my earlier comment, but it doesn't look like it will fit in the comment box. Hence now it lives as an answer.

To start with we set some notations.

$(N,g^N)$ is a Riemannian manifold. This means that $N$ is a smooth manifold (locally diffeomorphic to domains in $\mathbb{R}^{m+1}$) and $g^N|_p$ at some point $p\in N$ is a positive definite (and hence non-degenerate) symmetric bilinear form on $T_pN$. So given an arbitrary vector $v\in T_pN$, $g^N(v)$ is a co-vector, or an element of $T^*_pN$. $T_pN$ and $T_p^*N$ are not the same space; the metric however induces a canonical isomorphism between the two (since the metric can be understood as a non-degenerate map between the two vector spaces).

Now given $M$ a $m$-dimensional smooth submanifold of $N$. The identity map $\iota:M\to N$ is an embedding. The tangent space $T_pM$ for $p\in M$ naturally embeds into $T_{p}N$ by the tangent bundle map $d\iota: TM\to TN$. (Note that, however, without the Riemannian structure there's no natural embedding of the dual space $T^*_pM$ into $T_p^*N$; but don't worry about that now, since we won't need it.)

Now I give an explicit construction of a coordinate system in a neighborhood $U$ of $p$ such that the metric $g^N$ is block diagonal along points $q \in U\cap M$.

First fix a neighborhood $V\subset M$ of $p$, and an arbitrary coordinate system $u^1,\ldots,u^m$ on $V$, with $p$ at the origin. At every point $q\in V$ the vectors $\partial_i = \partial/\partial u^i$ span the tangent space $T_qM$. Now consider the image of $\partial_i$ under the map $d\iota$, call them $e_i = d\iota \partial_i$. By elementary linear algebra since $T_qM\subset T_qN$ has co-dimension 1, there exists a unique vector $n_q$, up to scaling, such that $g(n_q,e_i) = 0$ for all $i$. (The reason I used $e_i$ instead of $\partial_i$ on $T_qN$ is because without a full set of coordinates [we're still missing one] it doesn't make sense to speak of the "coordinate derivative", which is obtained by varying one coordinate value while holding the remainder fixed.)

In any case, so in $T_qN$ for any point $q\in M$, we now have a set of basis vectors $e_1,\ldots,e_m,n_q$. Normalize $n_q$ so that $g(n_q,n_q) = 1$ and $\eta(e_1,e_2,\ldots,e_m,n_q) > 0$ where $\eta$ is the volume form on $N$. (So now we fix the size and orientation of the field $n_q$.) Now we extend $n_q$ somehow: pick your favourite way. One possible way is to extend $n_q$ by the geodesic flow for some short period of time. Let's call $\gamma(t,q)$ the geodesic starting from the point $q$ with initial speed $n_q$ at the time $t$. By possibly shrinking the neighborhood $V$, there exists some $\epsilon > 0$ such that $\gamma: (-\epsilon,\epsilon)\times V \to N$ is injective.

Now let the neighborhood $U = \gamma( (-\epsilon,\epsilon)\times V )$. Define the coordinate on $U$ thus: for any point $x\in U$, let $\tilde{u}^i(x) = u^i\circ \pi_2\circ\gamma^{-1}(x)$, where $\pi_2: (-\epsilon,\epsilon)\times V \to V$ is projection onto the second component. In other words, at a point $x$, find the unique geodesic $\gamma(t,q)$ that passes through $x$, and set the value $\tilde{u}^i(x) = u^i(q)$. To complete the coordinate system you let $\tilde{u}^{m+1}(x) = \pi_1\circ \gamma^{-1}(x)$, where $\pi_1$ is projection onto the first component, or that if $\gamma(t,q)$ is the geodesic, set $\tilde{u}^{m+1}(x) = t$.

Now you simply check that by definition, the surface $\{t = 0\} \cap U = V$. And that along this surface $\partial/\partial \tilde{u}^{m+1} = n_q$, and $\partial/\partial \tilde{u}^i = e_i$ for the other $i$'s. So that for any $q\in V$ the metric $g^N$ is block diagonal. Furthermore, observe that $$ \partial_{m+1} g(\partial_{m+1},\partial_{m+1}) = 2 g(\partial_{m+1},\nabla_{\partial_{m+1}}\partial_{m+1}) = 0$$ by the geodesic equation, you have that $g(\partial_{m+1},\partial_{m+1}) = 1$ on $U$. Also, use that $$ \partial_{m+1} g(\partial_i,\partial_{m+1}) = g(\nabla_{\partial_{m+1}}\partial_i,\partial_{m+1}) = g(\nabla_{\partial_i}\partial_{m+1},\partial_{m+1}) = \frac12 \partial_i g(\partial_{m+1},\partial_{m+1}) = 0$$ (first equality uses the geodesic equation again, the second one uses that Levi-Civita connection is torsion free, and the Lie bracket of coordinate vector fields vanish). We see that $g(\partial_i,\partial_{m+1}) = 0$ on $U$. So in the whole neighborhood $U$, we have that $g^N$ is block diagonal, with the block $B$ exactly 1.