# First and second fundamental forms

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I'm writing notes about the 3+1 formalism in general relativity, for myself. Inevitably I came across the notions of first and second fundamental forms. Mathematically, it is clear how these objects are defined: ($M$ is a 4-dim manifold with metric $g$, $\Sigma$ a hypersurface of $M$)

The first fundamental form is the induced metric on $\Sigma$, also given by the pullback of the spacetime metric $g$.

The second fundamental form $K: T_{p}(\Sigma)\times T_{p}(\Sigma)\rightarrow \mathbb{R}$ is given through the Weingarten map $\chi$, i.e. $(u,v)\mapsto -u\cdot \chi(v)$.

Now, I'm having difficulties with underlying physcial intuition for these two objects (especially the second fundamental form). Is there a way for a physicist to "visualise" them in a way? What kind of objects are these forms exactly?

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user ConciseAndClear

retagged Mar 25, 2014

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For the first fundamental form - if you've got two vectors tangent to $\Sigma$, and $\Sigma$ is embedded in $M$, and $M$ has a metric, just use the embedding to consider the vectors as living tangent to $M$ and use $M$'s metric to compute their inner product.

For the second fundamental form, basically, if you imagine a two surface $\Sigma$ embedded in $\mathbb{R}^3$, and you imagine the normals as arrows orthogonal to $\Sigma$ sticking out like a hedgehog's spines. Then if the surface is gently curved, the normals dont change much as you go from a point $x$ on $\Sigma$ to another point $x'$ on $\Sigma$ infinitesimally separated from $x$ by a vector tangent to $\Sigma$. Conversely if the embedding is highly curved, the normals change a lot when you do this small displacement.

The Weingarten map, your $\chi$, is just the map $$u\rightarrow \nabla_un$$ where $n$ is the normal to $\Sigma$ and $u$ is tangent to $\Sigma$, and this encodes how much the normals are changing when you nudge by $u$ along $\Sigma$.

The 2nd FF, or "extrinsic curvature" is just another way of representing the information in the Weingarten map. Explicitly, it's a bilinear form which maps a pair of vectors $u,v$ tangent to $\Sigma$ to a number $-u.\chi(v)$ ($\chi(v)$ is also tangent to $\Sigma$ so this makes sense).

BTW, since you're studying 3+1, this reference (which came up in one of Alex Nelson's posts), is really informative and steers you through the minefield of conflicting approaches.

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user twistor59
answered Apr 30, 2013 by (2,490 points)
thank you very much twistor59! that's exactly the kind of explanation I wanted!

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user ConciseAndClear
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To add a bit of fluff to twistor59's answer, let's take a bird's eye view of Riemannian geometry.

The Riemannian metric gives us the notion of lengths and angles as well as the concept of straight lines (geodesics).

Any submanifold inherits these notions from the ambient space, made explicit via the first fundamental form, which makes the submanifold is a Riemannian manifold in its own right.

As such, it comes with the notion of intrinsic curvature, eg manifest in the sum of the angles of a triagle formed by straight lines, which is independent of the embedding into any larger space.

However, there's a second notion of curvature, the extrinsic one, which does make use of this embedding, eg via normal vectors, osculating circles or approximation by paraboloids. The second fundamental form is of this type.

These different notions of curvature are of course related: You can get the intrinsic curvatue of a submanifold $N\hookrightarrow M$ (measured by the Riemann curvature tensor $R_N$) from its extrinsic curvature (measured by the second fundamental form $\mathrm{II}$) and the intrinsic curvature $R_M$ of the ambient space via $$\langle R_N(u,v)w,z\rangle = \langle R_M(u,v)w,z\rangle+\langle \mathrm I\!\mathrm I(u,z),\mathrm I\!\mathrm I(v,w)\rangle-\langle \mathrm I\!\mathrm I(u,w),\mathrm I\!\mathrm I(v,z)\rangle$$ (The formula was taken from this Wikipedia article).

This post imported from StackExchange Physics at 2014-03-22 17:27 (UCT), posted by SE-user Christoph
answered Apr 30, 2013 by (210 points)

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