2
$\begingroup$

Let $\Sigma$ be a smooth hypersurface of a $d$ dimensional smooth Riemannian manifold $(\mathcal M, G)$; we may see $G_x$ as a mapping from $T_x(\mathcal M)$ into $T_x^*(\mathcal M)$ so that $$ \langle G_x X, X\rangle_{T^*_x(\mathcal M), T_x(\mathcal M)}=G_x(X). $$ Working in a neighborhood of a point of $\Sigma$ we may consider a parametrization from a chart $U$ of $\Sigma$ into a chart $W$ of $\mathcal M$, $$ \mathbb R^{d-1}\supset U\ni u\mapsto x(u)\in W\subset\mathbb R^d, \quad \text{rank $x'(u)=d-1.$} $$ We define the first fundamental form $g_u$ as $$ \langle g_u T, T\rangle_{E^*, E}=\langle G_{x(u)} x'(u)T, x'(u)T\rangle_{F^*, F}, \quad T\in \mathbb R^{d-1}=E, \quad F=T_{x(u)}(\mathcal M), $$ and $g_u$ is obviously a positive definite quadratic form on $E$. We introduce $N(u)$ as a normal vector to $\Sigma$ at $x(u)$ and that vector $N(u)$ belongs to $F$ and is such that $$ \langle G_{x(u)} x'(u)T, N(u)\rangle_{F^*, F}\equiv 0. $$ We define the covector $\nu(u)=G_{x(u)} N(u)$ and we have $ \langle \nu(u),x'(u)T\rangle_{F^*, F}\equiv 0, $ so that $$ \langle \nu'(u) T,x'(u)T\rangle_{F^*, F}+\langle \nu(u),x''(u)T^2\rangle_{F^*, F}=0. $$ I want to define the second fundamental form in the direction $N$ as $$ \langle\omega_u T, T\rangle_{E^*, E}=\langle \nu(u),x''(u)T^2\rangle_{F^*, F}, $$ and the Gauss curvature as $$ K=\frac{\det(\omega_u)}{\det (g_u)}. $$ The above description is standard when $\mathcal M$ is the Euclidean $\mathbb R^d$. Is it correct in this more general case ?

$\endgroup$

1 Answer 1

3
$\begingroup$

To be honest, I find your formulas hard to follow. I hope the following is useful:

I define the second fundamental form for a submanifold $\Sigma \subset \mathcal{M}$ of any codimension as follows: Let $X \in T_x\Sigma$ and $Y$ be a tangent vector field along $T_*\Sigma$. There are two different connections, the Levi-Civita connection $\nabla^\Sigma$ induced by the metric on $\Sigma$ and the connection $\nabla^{\mathcal{M}}$ induced by the metric on $\mathcal{M}$. It is straightforward to check that $\nabla^{\mathcal{M}}_XY$ is well defined, even though $Y$ is defined only on $\Sigma$. Then $\nabla^{\mathcal{M}}_XY$ can be decomposed into two components, one tangent to $\Sigma$ and the other normal. It turns out that the tangential component is $\nabla^{\Sigma}_XY$ and the normal component is the second fundamental form (viewed as a normal-vector-valued symmetric tensor on $T_*\Sigma$), which I'll denote by $H$: $$ \nabla^{\mathcal{M}}_XY = \nabla^{\Sigma}_XY + H(X,Y). $$ It is also straightforward to show that $H(X,Y) = H(Y,X)$. Since the definition of $H$ shows that $H$ depends only on the value of $X$ at the point $x \in \Sigma$, the symmetry of $H$ implies it is a tensor.

In codimension $1$, the standard second fundamental is simply $\nu\cdot H$, where $\nu$ is a unit normal vector$. It is also easy to see that the definition above is the same as the standard one using the Gauss map.

Also, as Oliver Nash points out, the second fundamental form can be viewed as a bundle map $H: S^2T_*\Sigma \rightarrow N_*\Sigma$ (this is the adjoint of Oliver's ap) or, equivalently, a section of the bundle $N_*\otimes S^2T^*$.

$\endgroup$
5
  • 1
    $\begingroup$ Maybe worth adding that $H$ is often usefully viewed as a family of quadratic forms on the tangent space, parameterised by the conormal bundle. I.e., we can regard $$H : N^* \to S^2T^*\Sigma$$ over $\Sigma$ (using the metric to identify $T^*$ and $T$). Also, it can also be obtained as the derivative of the Gauss map. $\endgroup$ Commented Jan 10, 2018 at 11:53
  • $\begingroup$ Oliver, thanks for pointing that out. Also, is there an easy way to use the Gauss map when the codimension is greater than $1$? I've been wondering about that. In principle, the Gauss map can be defined using Grassmannians, but it gets complicated. I learned how to handle this using moving frames from a paper by Griffiths, but is there an easier way? $\endgroup$
    – Deane Yang
    Commented Jan 10, 2018 at 15:40
  • $\begingroup$ IIUC, the way I think about this is if $f : X \to Y$ is an immersion of smooth manifolds, $\dim X = k$ and we have a fixed parallelization $TY \simeq Y \times V$ (for some vector space $V$). Then we can regard $df \in \Omega^1(X, V)$ as a map: $$\mathcal{G} : X \to Gr(k, V).$$ Since the tangent space to the Grassmannian at $U \in Gr(k, V)$ is naturally $Hom(U, V/U)$, we have a natural decomposition of the pull back: $$\mathcal{G}^*TGr(k, V) \simeq T^*X\otimes N$$ where $N$ is the normal bundle of $f$. $\endgroup$ Commented Jan 10, 2018 at 22:23
  • $\begingroup$ Thus we can regard $d\mathcal{G} \in \Omega^1(X, T^*X\otimes N) = \Omega^0(T^*X\otimes T^*X\otimes N)$ but since this is a second derivative and partial derivatives commute, it is symmetric and so in fact we obtain an element of $$\Omega^0(S^2T^*X\otimes N).$$ $\endgroup$ Commented Jan 10, 2018 at 22:26
  • $\begingroup$ Maybe worth emphasising that what I describe above obviously only works when $Y$ is flat. Presumably your question was what about when it isn't, so this might not be of interest to you after all. This POV also works for projective varieties but that may be as far as it goes. (It's a long time since I've read that lovely Griffiths and Harris paper.) ¯\_(ツ)_/¯ $\endgroup$ Commented Jan 11, 2018 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.