# 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.

$(N,g^N)$ 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.