Gauss Equation

We take two traces of the Gauss equation to get a scalar relationship between intrinsic and extrinsic curvature quantities of an embedding.

10 February 2025

The most fundamental equation of Riemannian geometry is the Gauss equation, which relates intrinsic and extrinsic curvatures of an embedding Σ⊂M\Sigma\subset M. Here's the full beast.

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Theorem [Gauss]. Given a embedding of a hypersurface Σn−1\Sigma^{n-1} into a Riemannian manifold MnM^n, for vector fields X,Y,Z,W∈X(Σ)X,Y,Z,W\in \mathfrak{X}(\Sigma),

Rm(X,Y,Z,W)=RmΣ(X,Y,Z,W)−A(X,W)A(Y,Z)+A(X,Z)A(Y,W).\begin{equation} \begin{split} Rm(X,Y,Z,W)&= Rm_\Sigma(X,Y,Z,W)\\ &- A(X,W)A(Y,Z)\\ &+ A(X,Z)A(Y,W). \end{split} \end{equation}

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For notation, a quantity with a subscript of Σ\Sigma means that it's associated with the hypersurface Σ\Sigma, while no subscript means that it's associated with the ambient manifold MM. In the above, RmRm denotes the (0,4)(0,4) Riemannian curvature tensor, A(X,Y):=⟨∇XN,Y⟩A(X,Y):=\langle \nabla_X N, Y\rangle denotes the second fundamental form, and NN denotes a unit normal to Σ\Sigma. Later, we use RicRic to denote Ricci curvature, RR to denote scalar curvature, and H:=∑i=1n−1A(ei,ei)H:=\sum_{i=1}^{n-1} A(e_i,e_i) to denote mean curvature.

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Geometers typically find RmRm too difficult to study and historically they've reduced complexity by taking traces. We claim that after taking two traces, Gauss equation reduces to the following scalar formula.

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Theorem [Gauss]. Under the same assumptions as above,

Ric(N,N)=12(R−RΣ+H2−∣A∣2).\begin{equation} \begin{split} Ric(N,N)=\frac{1}{2}(R-R_\Sigma +H^2-|A|^2). \end{split} \end{equation}

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Proof: To this end, we will take traces wrt an orthonormal frame E:={e1,...,en−1}E:=\{e_1,...,e_{n-1}\} along Σ\Sigma. We first take a single trace in the X,WX,W slots to get

Ric(Y,Z)−Ric(N,N)=RicΣ(Y,Z)−(∑i=1n−1A(ei,ei))⏟=HA(Y,Z)+∑i=1n−1A(ei,Z)A(Y,ei).\begin{equation} \begin{split} Ric(Y,Z) - Ric(N,N)&= Ric_\Sigma(Y,Z)\\ &- \underbrace{\bigg(\sum_{i=1}^{n-1} A(e_i,e_i)\bigg)}_{=H}A(Y,Z)\\ &+\sum_{i=1}^{n-1} A(e_i,Z)A(Y,e_i). \end{split} \end{equation}

The extra negative term in the top line comes from the fact that we took trace over only n−1n-1 directions, and E∪NE\cup N is an orthonormal frame for MM. Second, we take a trace in the Y,ZY,Z slots to get

R−2Ric(N,N)=RΣ−H(∑j=1n−1A(ej,ej))⏟=H+∑i,j=1n−1A(ei,ej)A(ej,ei)⏟=A(ei,ej).=RΣ−H2+∣A∣2,\begin{equation} \begin{split} R - 2Ric(N,N)&= R_\Sigma- H\underbrace{\bigg(\sum_{j=1}^{n-1} A(e_j,e_j)\bigg)}_{=H}\\ &+\sum_{i,j=1}^{n-1} A(e_i,e_j) \underbrace{A(e_j,e_i)}_{=A(e_i,e_j)}.\\ &=R_\Sigma-H^2+|A|^2, \end{split} \end{equation}

which gives the result. Again, an extra negative term comes in during this round of trace. The last line follows from the definition of norm for a (0,2)(0,2)-tensor. In applications, particularly those pertaining to rigidity, we usually only care that it is a non-negative function. Finally, we show that AA is symmetric. We compute

A(X,Y)−A(Y,X)=⟨∇XN,Y⟩−⟨∇YN,X⟩=X⟨N,Y⟩⏟=0−⟨N,∇XY⟩−Y⟨N,X⟩⏟=0+⟨N,∇YX⟩=⟨N,[X,Y]⟩=0.\begin{equation} \begin{split} A(X,Y)-A(Y,X)&=\langle \nabla_X N,Y \rangle - \langle \nabla_Y N,X \rangle\\ &=\underbrace{X\langle N,Y\rangle}_{=0} - \langle N,\nabla_X Y\rangle\\ &- \underbrace{Y\langle N, X \rangle}_{=0}+ \langle N,\nabla_Y X\rangle\\ &=\langle N, [X,Y] \rangle\\ &=0. \end{split} \end{equation}

The second and third equalities come from the defining properties of the Levi-Civita connection. The last equality follows since [X,Y]∈X(Σ)[X,Y]\in \mathfrak{X}(\Sigma), as X,YX,Y both are.