Let $(\widetilde{\mathcal{M}},\widetilde{g})$ be some pseudo-Riemannian manifold and $(\mathcal{M},g)$ a Riemannian submanifold of co-dimension $1$, i.e. a hypersurface. Then, the fundamental form is the symmetric $2$-tensor field defined by

$$k(X,Y):=\langle N,\widetilde{\nabla}_{X}Y\rangle$$

for all $X,Y\in\mathfrak{X}(\mathcal{M})$, where $\widetilde{\nabla}$ is the Levi-Civita connection of the ambient manifold $\widetilde{\mathcal{M}}$, $N$ the unit normal vector and $X,Y$ are arbitrarely extended to whole $\widetilde{\mathcal{M}}$ (the choice of extension does not matter). Since $N$ is a normal vector, this can also be written as

$$k(X,Y):=-\langle \widetilde{\nabla}_{X}N,Y\rangle$$

In physics texts, I have seen that in coordinates $k_{\mu\nu}=-N_{\mu;\nu}=-\widetilde{\nabla}_{\nu}N_{\mu}$ and I would like to derive this equality.

My attempt: Let us use that $k$ is symmetric. Then:


which is seems to be different from


Is there something wrong in my calculation? Or are the two expressions somehow equivalent?

  • $\begingroup$ In the last equality of the first line of your equation, you made a mistake using the definition of the Christoffel symbol: should be $\widetilde{\Gamma}^{\mu}_{\alpha\nu}$ instead of $\widetilde{\Gamma}^{\beta}_{\alpha\nu}$. $\endgroup$
    – Jotabeta
    Nov 7, 2022 at 11:16
  • $\begingroup$ Sorry, there was a typo. Is fixed now. $\widetilde{\nabla}_{\partial_{\nu}}\partial_{\alpha}=\widetilde{\Gamma}^{\beta}_{\alpha\nu}\partial_{\beta}$ is correct. I wrote $\partial_{\alpha}$ instaed, that was a typo. $\endgroup$ Nov 7, 2022 at 11:29
  • $\begingroup$ @G.Blaickner Hello, may I have your definition of submanifolds? Did you use a general immersion $F:\mathcal{M}\to\widetilde{\mathcal M}$ (not necessarily the inclusion map)? Thank you. $\endgroup$
    – Boar
    Oct 28, 2023 at 16:27
  • 1
    $\begingroup$ @Wombat I took the definition of Lee's book on smooth manifolds: $\mathcal{M}\subset\widetilde{\mathcal{M}}$ equipped with a topology (not necessarily subspace topology) and smooth structure such that the inclusion is an immersion. But, there is no other definition afaik. On wikipedia, they define immersed submanifolds as the image of an injective immersion, but this definition really is equivalent. $\endgroup$ Oct 30, 2023 at 9:56
  • 1
    $\begingroup$ I mean, take an abitrary injective immersion $F:\mathcal{M}\to\widetilde{\mathcal{M}}$ (note that injectivity is necessary, since otherwise $F(\mathcal{M})$ does not necessarily has the structure of a manifold). Then $F(\mathcal{M})$ is an immersed submanifold w.r.t. the inclusion, as in the definition of Lee's book. $\endgroup$ Oct 30, 2023 at 10:00

1 Answer 1


The mistake is $\partial_\nu N^\alpha g_{\alpha\mu} = \partial_\nu N_\mu$. This does not hold, because $\partial_\nu N^\alpha$ is not a tensor (i.e., it does not transform like a tensor when you change coordinates). Instead, you have to do the following: \begin{align*} \partial_\nu N^\alpha g_{\alpha\mu} &= \partial_\nu(N^\alpha g_{\alpha\mu}) - N^\alpha\partial_\nu g_{\alpha\mu}\\ &= \partial_\nu N_{\mu} - N^\alpha(\Gamma_{\nu\alpha}^\beta g_{\beta\mu} +\Gamma_{\nu\mu}^\beta g_{\beta\alpha}) \end{align*}

It's also worth noting that the identity can also be proved using what I like to call "differentiation by parts" as follows: \begin{align*} k(X,Y) &= \langle N,\widetilde\nabla_XY\rangle\\ &= \partial_X\langle N,Y\rangle - \langle \widetilde\nabla_XN,Y\rangle\\ &= - \langle \widetilde\nabla_XN,Y\rangle, \end{align*} because $\langle N,Y\rangle = 0$.

  • $\begingroup$ To make this clearer for me, I want to evaluate your equation for a the case of a sphere, where the normal is given by $N = (\sin(\theta)cos(\phi),\sin(\theta)\sin(\phi),\cos(\theta))$. But I got so very confused: The indices $\nu, \mu, \alpha, \beta$ take values $1,\,2,\,3$, because the normal has three components? But then, I should derive respect to $(x,y,z)$ and not $(\theta,\phi)$? (Disclaimer: very new to differential geometry and self-taught) $\endgroup$
    – Ale
    May 9, 2023 at 15:03
  • $\begingroup$ @Ale, Which book are you studying from? Recall that when you study a curve in 3-space, you have to parameterize the curve, which means that you have a map from a single variable $t$ to the coordinates $x,y,z$. In other words, the coordinates all become functions of the parameter $t$. When you need to differentiate something along a curve, you differentiate with respect to $t$ and not with respect to $x, y, z$. The same is for a surface. You have to parameterize it using 2 parameters, so $x,y,z$ are functions of the 2 parameters (sometimes called $u,v$). Here, your parameters are $\theta,\phi$. $\endgroup$
    – Deane
    May 10, 2023 at 2:37
  • $\begingroup$ I'm using Lee's "Introduction to Riemannian manifolds" and Kreyszig's "Differential geometry". $\endgroup$
    – Ale
    May 10, 2023 at 7:55
  • $\begingroup$ I state my question more precisely: I parametrize the sphere by $X(\theta,\phi) = (\sin(\theta)\cos(\phi),\sin(\theta)\sin(\phi),\cos(\theta)) = N$. Then I would have the following terms of $\partial_{\nu}N^{\alpha}g_{\alpha \mu}$: $\partial_{\theta}N^{\theta}g_{\theta \mu}, \partial_{\theta}N^{\phi}g_{\phi \mu}, \partial_{\phi}N^{\theta}g_{\theta \mu}, \partial_{\phi}N^{\phi}g_{\phi \mu}$. But this means that there should be only two components of N but I see 3. Which are the $N^{\theta}$ and $N^{\phi}$ components? $\endgroup$
    – Ale
    May 10, 2023 at 8:06
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    $\begingroup$ I don't understand your notation. $N$ is a $3$-dimensional vector normal to the sphere and $\partial_\theta X, \partial_\phi X$ are $3$-dimensional vectors tangent to $S$. The second fundamental form is given by the $2$-by-$2$ matrix $$ \begin{bmatrix} -\partial_\phi X\cdot \partial_\phi N & -\partial_\phi X\cdot \partial_\theta N \\ -\partial_\theta X\cdot \partial_\phi N & -\partial_\theta X\cdot \partial_\theta N \end{bmatrix} = \begin{bmatrix} N\cdot\partial^2_{\phi\phi}X & N\cdot\partial^2_{\phi\theta}X \\ N\cdot\partial^2_{\theta\phi}X & N\cdot\partial^2_{\theta\theta}X \end{bmatrix}. $$ $\endgroup$
    – Deane
    May 10, 2023 at 17:30

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