Geometric intuition behind the second fundamental form and curvature

Question

I am closely following Lee's book on Riemannian Manifolds.

Let $(M, g)$ and $(\tilde{M}, \tilde{g})$ be a Riemannian manifolds such that $\iota: M \rightarrow \tilde{M}$ is an immersion and $g$ is the pullback of $\tilde{g}$ under $\iota$. Let $\tilde{\nabla}$ be a Riemannian connection on $\tilde{M}$ and $X, Y$ two arbitrary vector fields on $M$. Extend $X, Y$ arbitrarily to $\tilde{M}$so that the covariant derivative $\tilde{\nabla}_XY$ is well defined, which we can decompose as $$\tilde{\nabla}_XY = (\tilde{\nabla}_XY)^T + (\tilde{\nabla}_XY)^\perp$$ where $(\tilde{\nabla}_XY)^T$ is the projection of $\tilde{\nabla}_XY$ onto the tangent bundle $TM$ and $(\tilde{\nabla}_XY)^\perp$ is the image of the projection of $\tilde{\nabla}_XY$ onto the normal bundle $NM$. We then define the second fundamental form as $$\mathrm{I\!I}(X,Y) := (\tilde{\nabla}_XY)^\perp.$$

It can be shown that if $\nabla$ is the Riemannian connection on $M$ then $(\tilde{\nabla}_XY)^T$ is nothing but $\nabla_XY$, and so $$\tilde{\nabla}_XY = \nabla_XY + \mathrm{I\!I}(X,Y).$$

I understand the above pretty well, the second fundamental form tells you how much the two covariant derivatives differ. Where I begin to get lost is what happens after this. Everything is still clear algebraically, but less so geometrically and thinking in terms of curvature.

Lee proves two nice equations relating the second fundamental form and curvature:

The Weingarten Equation: $$\langle \tilde{\nabla}_XN, Y\rangle = -\langle N, \mathrm{I\!I}(X,Y)\rangle$$ and the Gauss Equation: $$\tilde{RM}(X,Y,Z,W) = RM(X,Y,Z,W) - \langle \mathrm{I\!I}(X, W), \mathrm{I\!I}(Y,Z) \rangle + \langle \mathrm{I\!I}(X, Z), \mathrm{I\!I}(Y,W) \rangle.$$

Lee also says:

For any vector $V \in T_pM$, $\mathrm{I\!I}(V, V)$ is the $\tilde{g}$-acceleration at $p$ of the $g$-geodesic $\gamma_V$. If $V$ is a unit vector, $|\mathrm{I\!I}(V, V)|$ is the $\tilde{g}$-curvature of $\gamma_V$ at $p$.

and

In the special case in which $\tilde{M}$ is $\mathbb{R}^m$ with the Euclidean metric, we can make this geometric interpretation even more concrete: $\mathrm{I\!I}(V, V)$ is the ordinary Euclidean acceleration of the geodesic in $M$ with initial velocity $V$.

I read in some other sources that the motivation for the second fundamental form comes from looking at a hypersurface in Euclidean space where there is a well defined normal vector and this is what requires us to consider an ambient manifold $\tilde{M}$ in the construction. But other than this I am lost on what the normal component of the covariant derivative has to do with curvature. I think this is also why I don't fully appreciate the two quotes passages above. Furthermore, I don't think I have a good intuition for "intrinsic" vs "extrinsic" curvature of a curve in $M$.

How can I see all this geometrically?

Answer 1

Start from the equation that for all vector fields $X,Y$ on $M$, \begin{align} \widetilde{\nabla}_XY&=\nabla_XY+\mathrm{I\!I}(X,Y). \end{align} Now, ‘cheat’ slightly (really we’re just using basic facts about covariant derivatives not requiring the full strength of knowing the entire vector field… but this is completely trivial if you adopt the Ehresmann definition of a connection) to take $X=Y=\gamma’$ for a smooth curve $\gamma:I\subset \Bbb{R}\to M$ to get \begin{align} \widetilde{\nabla}_{\gamma’}\gamma’&=\nabla_{\gamma’}\gamma’+\mathrm{I\!I}(\gamma’,\gamma’). \end{align} So, rearranging, \begin{align} \mathrm{I\!I}(\gamma’,\gamma’)&=\widetilde{\nabla}_{\gamma’}\gamma’-\nabla_{\gamma’}\gamma’\\ &=\text{acceleration of $\gamma:I\to M\subset \widetilde{M}$ measured using $\widetilde{\nabla}$}\\ &\,\,\,\,\,-\text{acceleration of $\gamma:I\to M$ measured using $\nabla$}. \end{align} So, (the quadratic form associated to) the second fundamental form tells us how much the acceleration of a curve $\gamma$ differs when you compute it in the ambient space vs in the submanifold. This explains both the quotes:

• if $\gamma$ is a geodesic in $M$, then the second term drops out, and we get $\mathrm{I\!I}(\gamma’,\gamma’)=\widetilde{\nabla}_{\gamma’}\gamma’$, i.e it equals simply the ambient acceleration.
• by taking the norm on both sides of the equation above, we see that $\left\|\mathrm{I\!I}(\gamma’,\gamma’)\right\|_{\widetilde{g}}=\left\|\widetilde{\nabla}_{\gamma’}\gamma’\right\|_{\widetilde{g}}$, and if $\gamma’$ has constant norm 1 (for a geodesic, this is equivalent to requiring $\|\gamma’(0)\|=1$) then by definition, the RHS is the curvature of $\gamma$ in $(\widetilde{M},\widetilde{g})$.

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Now, let’s think about the following.

1. In the first bullet point, $\gamma$ is a geodesic in $M$, meaning it is “the straightest possible curve in $(M,g)$”.
2. But our computation shows that $\widetilde{\nabla}_{\gamma’}\gamma’$ is not necessarily $0$, i.e $\gamma$ is not necessarily a straight curve in the ambient $(\widetilde{M},\widetilde{g})$. So, this discrepancy must come because $M$ itself is sitting inside of $\widetilde{M}$ in some wonky manner.
3. the actual value of the ambient acceleration is given by $\mathrm{I\!I}(\gamma’,\gamma’)$, which means the amount of wonkiness is measured precisely by $\mathrm{I\!I}$.

As a very concrete example, think about $S^1$ sitting inside of $\Bbb{R}^2$ (with standard structures), and consider the curve $\gamma(t)=(\cos t,\sin t)$. This is the unit speed parametrization of $S^1$, so it is a geodesic in $S^1$, i.e as far as the circle is concerned, $\gamma$ describes a ‘straight line’, but clearly we would not say $\gamma$ is a straight line in $\Bbb{R}^2$.

Another way to see quite directly what the second fundamental form measures, is via Weingarten’s equation. The quantity $\widetilde{\nabla}_XN$, roughly speaking, is measuring how much the normal $N$ is changing as you move it along the submanifold $M$ (in the direction $X$). If this vanishes then it should mean that the submanifold is not ‘bending’ in the ambient space. In other words, we’re looking at how the normal direction changes in order to detect how the submanifold is embedded.

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Much more conceptually and abstractly, you should think of having a big vector bundle $(E,\pi,M)$ with a linear connection $D$. Then, imagine having a direct sum subbundle decomposition $E=L\oplus L’$. The connection $D$ induces connections $\nabla$ in $L$ and $\nabla’$ in $L’$, which can then be put back together to give a connection $\nabla^{\oplus}=\nabla\oplus\nabla’$ in $L\oplus L’=E$. So, we have two connections in $E$, namely the original $D$ and the disassembled-then-reassembled $\nabla^{\oplus}$. The difference $A=D-\nabla^{\oplus}$ (which recall by abstract theory is an $\text{End}(E)$-valued $1$-form on $M$, or equivalently a bilinear bundle morphism $TM\oplus E\to E$) is the full shape tensor (also called the second fundamental form, though I only use this term in a more special case). It tells us how exactly the bundles $L,L’$ sit inside of $E$, and how to detect wonkiness in the manner they sit inside the bigger $E$. You can then start asking about how curvatures of $D$ is related to that of $\nabla^{\oplus}$; the answer is given by the Gauss-Codazzi equations. Once you know a bit about manipulating vector-bundle-valued forms, these equations (which are a great historic achievement) pop out almost immediately; see here.