Finding covariant derivative of a riemanian submanifold

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

• You don't need * to be all zero. You just need it to be zero along $M$. So just take an arbitrary local coordinate on $M$ to start. The metric defines along $M$ the normal direction. Choose a field of normal vectors, extend the field arbitrarily in a thickened slab around $M$. Flow $M$ along the vector field. Then the flow $t$ gives the "vertical coordinate". The lie transport of the local coordinates on $M$ gives the coordinate on the slab. And along $M$ the total metric $g^N$ is block diagonal. – Willie Wong Jul 26 '10 at 0:19
• (Of course when you do this you have to be careful when you compute the curvature; beware of taking additional vertical derivatives!) – Willie Wong Jul 26 '10 at 0:22
• OK I understand this, but I'm still not sure how do I demonstrate that $g^N$ would be diagonal. Intuitively - Suppose I have a "vertical vector" $a \in M \subset N$ then in the local coordinates you showed me $g^N * a$ as a matrix operating on a vector, should give me a vector which is in $T_a N$ but not in $T_a M$ And this shows that the matrix elements "*" would have to be zero. But again, this is intuitive, from linear algebra. Can you please help me to understand this delicate point? thanks for the time, Tamir – tamir Jul 26 '10 at 17:37

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.

• Hi Greg, I'm still not sure about two points here, so let me see if I get it correctly. – tamir Jul 26 '10 at 18:10
• In the first part, given a metric tensor, is it OK just to choose new local coordinates and preserve the tensor? I think that as long as I choose a smooth coordinate change, than maybe g takes another form, but it still operates the same on members of $TN$ . It's like doing $w (C^TgC) v = (Cw)^T * g * (Cv)$ right? (if it's not too much trouble - if it is so, since $g$ is symmetric and strictly positive, so it can always be diagonalized and therefore we have sort of "principal" directions where the arc length is just $ds^2 = Udu^2 + Vdv^2 + W*dw^2 + ...$ ?) – tamir Jul 26 '10 at 18:10
• About the second approach you showed me, I see that $P$ is doing the "magic" there. But does $P$ have an "analytical" definition of some sort that I can work with? My problem started with the fact that when I tried to project $\nabla^N$ to $\nabla^M$ I took $\nabla^N_{\partial i}\partial j$ and I wrote it explicitly with $\partial k$ and "threw away" the term with $k = m+1$. This is the projection as I understand it. But the Christoffels in the other $\partial k, k \neq m+1$ stuck me. Because then, I didn't know what how to "project" them. Can you give me anopther word on that, please? – tamir Jul 26 '10 at 18:10
• PS (no place left up there) thanks for the time and effort, Tamir. – tamir Jul 26 '10 at 18:11

@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.