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Let $F:M^n\to(N^{n+1},\bar{g})$ be a smooth immersion of a hypersurface in a Riemannian manifold. For simplicity, we assume both $M$ and $N$ are oriented closed manifolds. If $\nu$ denotes a unit normal vector field along the hypersurface, I would like to derive the Gauss-Weingarten equations in local coordinates $(\phi;x^i)$ around $p\in M$: $$\begin{align} \frac{\partial^2 F^\alpha}{\partial x^i\partial x^j}-\Gamma_{ij}^k\frac{\partial F^\alpha}{\partial x^k}+\overline{\Gamma}_{\beta\delta}^\alpha\frac{\partial F^\beta}{\partial x^i}\frac{\partial F^\delta}{\partial x^j}&=-h_{ij}\nu^\alpha,\tag{1}\\ \frac{\partial\nu^\alpha}{\partial x^i}+\overline{\Gamma}_{\beta\delta}^\alpha\frac{\partial F^\beta}{\partial x^i}\nu^\delta&=h_{ij}g^{j\ell}\frac{\partial F^\alpha}{\partial x^\ell}.\tag{2} \end{align}$$ $g$ is the induced metric on the hypersurface, and $h_{ij}$ gives the coordinate representation of the scalar second fundamental form $A$ in $(x^i)$. To derive the equations, we must have a solid knowledge of how the definition of $A$ works, but unfortunately I'm not quite aware of it. What the research paper told me is that $A(p)$ is a bilinear form $A(p):T_p M\times T_p M\to\Bbb{R}$ given by $$h_{ij}=\langle\overline{\nabla}_{e_i}\nu,e_j\rangle=-\langle\nu,\overline{\nabla}_{e_i}e_j\rangle\tag{3}$$ I was told $(e_i)$ is an orthonormal frame, but what do I do with it if I am to work with local coordinates $(x^i)$? Also, why isn't there any $F$-related terms in (3)? Can I possibly interpret $\overline{\nabla}_{e_i}e_j$ as $\frac{\partial^2 F}{\partial x^i\partial x^j}$, please?

My idea would be that because coordinate vector fields $\frac{\partial}{\partial x^i}$ form a local frame for $M$ and the differential $dF$ is injective, we can see that $\nu$ and $\frac{\partial F}{\partial x^i}\overset{\color{red}?}{:=}dF(\frac{\partial}{\partial x^i})$ form a basis pointwise on $N$. And that $F^\alpha$ should be the coordinate expression of $F$; to be precise, we should have $$\frac{\partial F}{\partial x^i}=\frac{\partial F^j}{\partial x^i}\frac{\partial}{\partial y^j},$$ where $y^j$'s are coordinates $(\psi;y^j)$ around $F(p)$ and $F^j$ is the $j$-th component of $\psi\circ F\circ\phi^{-1}$.

The meaning of the Gauss-Weingarten equations has a clear picture in differential geometry of curves and surfaces, but how do I generalize the derivation there to obtain the equations in question? Thank you.

Edit. The following are what equations (1) and (2) would be like in classical differential geometry (do Carmo, 2/e): these are nothing but linear combinations of the basis vectors with $N$ denoting a unit normal vector field along $S$. And I don't know why they become so ugly in an abstract setting.

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    $\begingroup$ Doing differential geometry with moving frames and differential forms is the cleanest, most conceptual approach. Anything with coordinates is going to be horrendously messy. Of course, $\partial F/\partial x^k$ is the analogue of $\mathbf x_{u^k}$ in your surface situation. In the surface equations you gave, these equations define the Christoffel symbols and the second fundamental form, and the last two equations tell you nothing other than $N_u\cdot N = 0$, etc. So you need to edit your post to include the relevant definitions. $\endgroup$ Dec 14, 2023 at 17:06
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    $\begingroup$ Oh, the equations are not much uglier in the general setting. What you're forgetting is that $N$ is no longer flat Euclidean space, so its geometry has to enter in. But, as I said, I recommend learning moving frames if you want clear and intuitive statements about submanifold geometry. $\endgroup$ Dec 14, 2023 at 17:09
  • $\begingroup$ Yeah, the metric on $N$ deviates $N$ from the flat manifold $\Bbb{R}^{n+1}$, so a nontrivial $\overline{\Gamma}$ term is sure to come in, so a straightforward expansion in basis vectors simply doesn't work in this situation. $\endgroup$
    – Boar
    Dec 14, 2023 at 23:25
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    $\begingroup$ The main point is that you need a connection to differentiate $F$. It does not take its values in a fixed vector space, but instead on a curve manifold. $\endgroup$ Dec 14, 2023 at 23:30

2 Answers 2

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$\newcommand\bp{\bar\partial}\newcommand\bn{\overline\nabla}\newcommand\bG{\overline\Gamma}$

ANOTHER WAY:

First, verify that each side of (1) and (2) transform under change of coordinates on both $M$ and $N$ like components of a tensor. It then suffices to verify the equations under a single choice of coordiinates. The trick now is to find a set of coordinates in a neighborhood of $p \in M$ and one in a neighborhood of $F(p) \in N$ that simplify the equations at $p$. I believe the right choice is to use geodesic normal coordinates centered at $p$ on $M$ and extend them to coordinates on $N$ using geodesics normal to $F(M)$. It's worth working this out carefully, too.

DIRECT APPROACH:

The second fundamental form is a symmetric $2$-tensor $h$. So if $$ h_{ij} = A(e_i,e_j) = \langle\overline\nabla_{e_i}\nu,e_j\rangle, $$ then given any two vectors $v = v^ie_i, w=w^je_j \in T_pM$, $$ A(v,w) = A(v^ie_i,w^je_j) = v^iw^jA(e_i,e_j) = v^iw^j\langle \overline\nabla_{e_i}\nu,e_j\rangle = \langle \overline\nabla_v\nu,w\rangle.$$

ADDED: But before we go any further, we should be more precise about this and also clarify what we mean by $\overline\nabla_v\nu$, since $v \in T_pM$ and $\nu(p) \in T_{F(p)}N$.

The connection acting on $\nu$ is the Levi-Civita connection on $N$ pulled back to the pullback bundle $F^*T_*N$, which is a vector bundle over $M$. In particular, given a section $\nu: M \rightarrow T_*N$ where $\nu(p) \in T_{F(p)}N$ and $v \in T_pM$, the covariant derivative of $\nu$ is defined to be $$\bn_v\nu(p) = \bn_{F_*v}\bar\nu(F(p)), $$ where $\bar\nu$ is the section of $T_*N$ restricted to $F(M)$ such that $$\nu = \bar\nu\circ F. $$ Also, the correct definition of $A$ is $$ A(v,w) = \langle \bn_v\nu,F_*w\rangle. $$

The calculations in coordinates start like this:

Let $\partial_1, \dots, \partial_n$ be the coordinate vector fields on $M$ and $\bp_1, \dots, \bp_{n+1}$ be the coordinate vector fields on $N$. Since $$ F_*\partial_i = \partial_iF^\beta\bp_\beta, $$ it follows that $$ \bn_{\partial_i}\bp_\beta = \bn_{\partial_iF^\alpha\bp_\alpha}\bp_\beta = \partial_iF^\alpha\bn_{\bp_\alpha}\bp_\beta = \partial_iF^\alpha\bG_{\alpha\beta}^\delta\bp_\delta. $$ Therefore, \begin{align*} h_{ij} &= A(\partial_i,\partial_j)\\ &= \langle\bn_{\partial_i}\nu,F_*\partial_j\rangle\\ &= \langle\bn_{\partial_i}(\nu^\beta\partial_\beta),\partial_jF^\delta\partial_\delta\rangle\\ &= \partial_jF^\delta\langle \partial_i\nu^\beta \partial_\beta+\nu^\gamma\bn_{\partial_i}\partial_\gamma,\partial_\delta\rangle\\ &= \partial_jF^\delta(\partial_i\nu^\beta + \nu^\gamma\partial_iF^\alpha\bG_{\alpha\gamma}^\beta)\langle\partial_\beta,\partial_\delta\rangle\\ &= \bar g_{\beta\delta}\partial_jF^\delta(\partial_i\nu^\beta + \nu^\gamma\partial_iF^\alpha\bG_{\alpha\gamma}^\beta). \end{align*} It follows that \begin{align*} g^{j\ell}\partial_\ell F^\alpha h_{ij} &=\partial_\ell F^\alpha g^{j\ell}\partial_jF^\delta \bar g_{\beta\delta}(\partial_i\nu^\beta + \nu^\gamma\partial_iF^\mu\bG_{\mu\gamma}^\beta). \end{align*} I beliieve the way to finish the calculation is to use the fact that $\bn_i\nu \in F_*T_pM$ and verify that the linear map \begin{align*} T_{F(p)}N &\rightarrow F_*T_pM\\ V^\beta\bp_\beta &\mapsto V^\beta\partial_\ell F^\alpha g^{j\ell}\partial_jF^\delta \bar g_{\beta\delta}\bp_\alpha \end{align*} is orthogonal projection.

The calculation for (1) is: \begin{align*} h_{ij} &= A(\partial_i,\partial_j)\\ &= \langle\nabla_{\partial_i}\nu,F_*\partial_j\rangle\\ &= \partial_i\langle\nu,F_*\partial_j\rangle - \langle \nu, \bn_i(F_*\partial_j)\rangle\\ &= -\langle\nu, \bn_{\partial_i}(\partial_jF^\alpha\bp_\alpha)\rangle\\ &= -\langle\nu, (\partial^2_{ij}F^\alpha +\partial_iF^\delta\partial_jF^\beta\bG_{\delta\beta}^\alpha)\bp_\alpha\rangle\\ &= -\langle\nu, (\partial^2_{ij}F^\alpha - \Gamma_{ij}^k\partial_kF^\alpha+\partial_iF^\delta\partial_jF^\beta\bG_{\delta\beta}^\alpha)\bp_\alpha\rangle\\ &= -\langle \nu, \nabla^2_{ij}F\rangle. \end{align*} Therefore, $$ \nabla^2F = A\nu. $$ In any codimension, the Hessian of $F$ is actually a bundle map $$\nabla^2F: S^2T_pM \rightarrow T^\perp_pM, $$ where $T^\perp_pM$ is the orthogonal complement of $F_*T_pM \subset T_{F(p)}N$. In codimension greater than $1$, there is no unique normal, so the second fundamental form is defined to be the bundle map $\nabla^2F$.

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    $\begingroup$ So that's a good question. The differential of $F$, also known as the pushforward map, is an injective linear map $F_*: T_pM \rightarrow T_{F(p)}$, and $F_*(\partial_i) = \partial_iF^\alpha\partial_\alpha$. So it kind of doesn't matter which one you use to define $h_{ij}$. The tricky part is understanding the connections used in the calculation and which Christoffel symbols should appear. $\endgroup$
    – Deane
    Dec 17, 2023 at 3:00
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    $\begingroup$ I've rewritten my answer to be more precise with the notation. I'm sure there are typos. $\endgroup$
    – Deane
    Dec 17, 2023 at 4:58
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    $\begingroup$ I made some revisions. $\endgroup$
    – Deane
    Dec 17, 2023 at 15:37
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    $\begingroup$ Excuse me, how do you explain $$ -\langle\nu, (\partial^2_{ij}F^\alpha +\partial_i F^\delta\partial_jF^\beta\overline{\Gamma}_{\delta\beta}^\alpha)\partial_\alpha\rangle= -\langle\nu, (\partial^2_{ij}F^\alpha - \Gamma_{ij}^k\partial_k F^\alpha+\partial_iF^\delta\partial_j F^\beta\overline{\Gamma}_{\delta\beta}^\alpha)\partial_\alpha\rangle?$$ The unbarred Christoffel symbols just sprang up from nowhere. $\endgroup$
    – Boar
    Dec 19, 2023 at 8:36
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    $\begingroup$ Good question! I had the opposite problem. I knew the term had to be there, but I couldn't figure out how to justify it. It definitely does not need to be there. However, if you put it in, something nice happens. The right factor becomes a tensor, so it is a coordinate-independent local invariant of the map (with respect to the Riemannian metrics). It is well defined for any smooth map, even if it is not isometric, and is called the Hessian for obvious reasons. For example, a smooth map is harmonic if $g^{ij}\nabla^2_{ij}F = 0$. $\endgroup$
    – Deane
    Dec 19, 2023 at 15:19
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Due to my poorly-grounded knowledge, I'm NOT sure if what I'm going to write is mathematically rigorous. Therefore, if anyone is willing to offer justification or even defy it with sensible reasons, I would be very grateful to have you help me rest in peace.

Let us begin to derive identity (1). As the scalar second fundamental form is named $A$, let us choose the vector-valued second fundamental form to be $\mathbf{A}=-A\nu$, where $\nu$ is the unit normal. Then we begin our derivation with what is called the Gauss formula in John Lee's Introduction to Riemannian Manifolds. Please turn to its 228th page to find $$\overline{\nabla}_X Y=\nabla_X Y+\mathbf{A}(X,Y).\tag{4}$$ In terms of our notation, Lee derives this formula by working with the inclusion map $\iota:M^n\hookrightarrow(N^{n+1},\bar{g})$. I admit this is not the circumstance under which the original question is discussed, but I think there might be some kind of identification to make (4) applicable to the question. Anyway, I need covariant derivatives on $M$ to help bring in the terms with $\color{red}{\text{unbarred}}$ Christoffel symbols. Now we feed $X,Y$ in (4) with coordinate vector fields, and the right-hand side becomes $$\begin{align} \nabla_{\partial_i}\partial_j+\mathbf{A}(\partial_i,\partial_j)&=\Gamma_{ij}^k\partial_k-h_{ij}\nu^\alpha\partial_\alpha\\ &=\Gamma_{ij}\partial_k F^\alpha\partial_\alpha-h_{ij}\nu^\alpha\partial_\alpha, \end{align}$$ adopting @Deane's convenient notation. Notice that the unbarred Christoffel symbols have been ushered in. Now let us look at the other side of (4), which is likewise fed with coordinate vector fields to get $$\begin{align} \overline{\nabla}_{\partial_i}\partial_j&=\overline{\nabla}_{\partial_i}(\partial_j F^\alpha\partial_\alpha)=\partial_{ij}^2 F^\alpha\partial_\alpha+\partial_j F^\delta\overline{\nabla}_{\partial_i}\partial_\delta\\ &=\partial_{ij}^2 F^\alpha\partial_\alpha+\partial_j F^\delta\overline{\nabla}_{\partial_i F^\beta\partial_\beta}\partial_\delta\\ &=\partial_{ij}^2 F^\alpha\partial_\alpha+\partial_j F^\delta\partial_i F^\beta\overline{\nabla}_{\partial_\beta}\partial_\delta\\ &=\partial_{ij}^2 F^\alpha\partial_\alpha+\partial_j F^\delta\partial_i F^\beta\overline{\Gamma}_{\beta\delta}^\alpha\partial_\alpha. \end{align}$$ Finally, we equate these two results to obtain (1).

To attempt identity (2), let us begin with an intermediate result which we can obtain from @Deane's answer: $$h_{ij}=\partial_j F^\delta(\partial_i\nu^\beta+\nu^\gamma\partial_i F^\alpha\overline{\Gamma}_{\alpha\gamma}^\beta)\bar{g}_{\beta\delta}$$ With this equation, we go on to multiply both sides by $\partial_k F^\beta$ to see $g_{ij}=\partial_i F^\alpha\partial_j F^\delta\bar{g}_{\alpha\delta}$ in action. Then we get $$h_{ij}\partial_k F^\beta=(\partial_i\nu^\beta+\nu^\gamma\partial_i F^\alpha\overline{\Gamma}_{\alpha\gamma}^\beta)g_{kj}.$$ Now here's the biggie: I'm not sure if it is legitimate to multiply both sides by the inverse $g^{ik}$. But if you were convinced to do so, you'd be getting $$g^{ik}h_{ij}\partial_k F^\beta=\partial_j\nu^\beta+\nu^\gamma\partial_j F^\alpha\overline{\Gamma}_{\alpha\gamma}^\beta,$$ which is literally (2) since you can exercise symmetry of $A$.

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