If \(\mathcal{M}\) is an Euclidean submanifold, many intrinsic quantities can be expressed using information from the ambient Euclidean space. Here are two examples where the Weingarten map plays an important role: first is the Riemannian hessian of a smooth map, and second is the Ricci curvature tensor. Assume that \(f:\mathcal{M}\to\mathbb{R}\) is a smooth function and \(\widehat{f}:\mathbb{R}^d\to\mathbb{R}\) is a smooth extension of \(f\). In data science, it is much more convenient to compute the gradient or hessian of \(\widehat{f}\) than those of \(f\). Absil etc. presented the relations and applied them in Riemannian optimization. On the other hand, Ricci curvature is also a hot topic in manifold learning. An old paper of Hicks proves that there is a simple relation between Ricci map and Weingarten map if the manifold is a hypersurface. Many problems concerning Ricci curvature thus descend to the estimation of the Weingarten map.
Riemannian Hessian
Let \(\mathcal{M}\) be an \(m\)-dimensional submanifold in the \(d\)-dimensional Euclidean space \(\mathbb{R}^d\). Let \(\pi_x:\mathbb{R}^d\to T_x\mathcal{M}\) be the orthogonal projection to the tangent space \(T_x\mathcal{M}\). It turns out that the gradient of \(f\) is nothing but the projection of gradient of \(\widehat{f}\). Denote \(\partial\widehat{f}\) to be the usual gradient for \(\widehat{f}\) and \(\text{grad}(f)\) to be the Riemannian gradient for \(f\). Then we have
\(\text{grad}(f)_x=\pi_x(\partial\widehat{f}_x)\)
To check this equality, note that for any tangent vector \(v\in T_x\mathcal{M}\) and smooth curve \(\gamma\) in \(\mathcal{M}\) with \(\gamma(0)=x\) and \(\gamma'(0)=v\) we have \[ \langle v,\text{grad}(f)_x\rangle=v(f)=\frac{\rm d}{\rm dt}|_0f(\gamma(t))=\frac{\rm d}{\rm dt}|_0\widehat{f}(\gamma(t))=\langle v,\pi_x(\partial\widehat{f}_x)\rangle \] Let \(\nabla\) be the Riemannian connection on \(\mathcal{M}\). For any vector fields \(X\) and \(Y\), by definition, \[ \nabla_XY=\pi(\partial_XY) \] Define the Riemannian hessian of \(f\) at \(x\) by \[ \text{Hess}f_x(v)=\nabla_v\text{grad}(f) \] It is easy to check that \[ \langle\text{Hess}f_x(v),w\rangle=v(w(f))-\langle\nabla_vw,\text{grad}(f)\rangle=\langle\text{Hess}f_x(w),v\rangle \] Thus, \(\text{Hess}f_x:T_x\mathcal{M}\to T_x\mathcal{M}\) is a self-adjoint operator. However, the original definition is difficult to use in practice. Further computation shows that \[ \text{Hess}f_x(v)=\pi_x\partial_v(\pi\partial\widehat{f})=\pi_x((\partial_v\pi)\partial\widehat{f}_x+\partial^2\widehat{f}_x(v)) \] where we have used the product rule. If we understand \(\pi\) as matrices then \(\partial_v\pi\) contains elements which are derivatives of those in \(\pi\). From the result we see that Riemannian hessian of \(f\) differs from the projection of usual hessian of \(\widehat{f}\) by a term involving the covariant derivative of \(\pi\). Let \(\pi^\perp:\mathbb{R}^d\to T^\perp_x\mathcal{M}\) be the projection operator to the normal space at \(x\). Using the identity \(\text{id}=\pi+\pi^\perp\), we have \[ \pi_x\partial_v\pi=-\pi_x\partial_v\pi^\perp \] In addition, since \(\pi\pi^\perp=0\), we have \[ 0=\pi_x\partial_v(\pi\pi^\perp)=\pi_x(\partial_v\pi^\perp)_x+\pi_x(\partial_v\pi)_x\pi^\perp_x \] which implies \(\pi_x(\partial_v\pi^\perp)_x\pi_x=0\) and \(\pi_x(\partial_v\pi)\partial\widehat{f}_x=-\pi_x(\partial_v\pi^\perp)_x\pi_x^\perp\partial\widehat{f}_x\).
Recall that the Weingarten map at \(x\) for \(\xi\in T^\perp_x\mathcal{M}\) is defined by \(A_{x,\xi}(v)=-\pi_x\partial_v\tilde{\xi}\) where \(v\in T_x\mathcal{M}\) and \(\tilde{\xi}\) is any extension of \(\xi\). The identity implies that \[ A_{x,\xi}(v)=-\pi_x\partial_v(\pi^\perp\tilde{\xi})=-\pi_x(\partial_v\pi^\perp)_x\xi-\pi_x\pi_x^\perp\partial_v\tilde{\xi}=-\pi_x(\partial_v\pi^\perp)_x\xi \] Utilizing the above results we have \[ \text{Hess}f_x(v)=\pi_x\partial^2\widehat{f}_x(v)+A_{x,\pi_x^\perp\partial\widehat{f}_x}(v) \]
Ricci Map
Suppose \(\mathcal{M}\) is a hypersurface in Euclidean space. i.e. \(\mathcal{M}\) is of codimension \(1\). Let \(\xi\) be a local unit normal vector field. The second fundamental form can be simplified as \[ h(X,Y)=\partial_XY-\nabla_XY=\langle\partial_XY,\xi\rangle\xi=\langle Y,A_\xi(X)\rangle\xi \] where we have used the identity \(X\langle Y,\xi\rangle=\langle\partial_XY,\xi\rangle+\langle Y,\partial_X\xi\rangle\) and the definition of Weingarten map. Recall that for Euclidean submanifolds Gauss equation reads \[ 0=\tilde{R}(X,Y)Z=R(X,Y)Z-A_{h(Y,Z)}X+A_{h(X,Z)}Y \] Substituting the expression of the second fundamental form into above, we have \[ R(X,Y)Z=\langle A_\xi(Y),Z\rangle A_\xi(X)-\langle A_\xi(X),Z\rangle A_\xi(Y) \] Let \(Z_1,\cdots,Z_m\) be an orthonormal base of principal vectors of \(A_\xi\). i.e. \(A_\xi(Z_i)=k_iZ_i\) where \(k_i\)'s are principal curvatures. Let \(H=\sum k_i\) be the mean curvature and define \(Ric:T_x\mathcal{M}\to T_x\mathcal{M}\) by \(Ric(X)=\sum_{i=1}^m R(X,Z_i)Z_i\), called Ricci map. Then we have \[ \begin{aligned} Ric(X)&=\sum_{i=1}^mR(X,Z_i)Z_i\\ &=\sum_{i=1}^m\langle A_\xi(Z_i),Z_i\rangle A_\xi(X)-\langle A_\xi(X),Z_i\rangle A_\xi(Z_i)\\ &=\sum_{i=1}^m\langle k_iZ_i,Z_i\rangle A_\xi(X)-\langle A_\xi(X),Z_i\rangle k_iZ_i\\ &=H A_\xi(X)-\sum_{i=1}^m\langle A_\xi(X),A_\xi(Z_i)\rangle Z_i\\ &=H A_\xi(X)-\sum_{i=1}^m\langle A_\xi^2(X),Z_i\rangle Z_i\\ &=H A_\xi(X)-A_\xi^2(X) \end{aligned} \] Therefore we have proved the following identity relating the Ricci map with Weingarten map \[ A_\xi^2-HA_\xi+Ric=0 \] An obvious corollary is that every principal vector is also a Ricci vector. i.e. \(Ric\) has the same eigenvectors as \(A_\xi\).
Example. Let \(\mathcal{M}\) be a surface in the three space. The characteristic equation for Weingarten map is \(A^2-HA+KI=0\) where \(K\) is Gauss curvature. Comparing both equations we have \(Ric=KI\).