# An exercise in the Riemannian geometry book

If $M$ is a smooth closed $n$-dimensional Riemannian manifold which is Riemannian embedded in $\mathbb R^{n+1}$, then there exists a point $p \in M$ such that the sectional curvatures at $p$ are all positive.

Can any one give me a hint for this problem? I was considering the maximum $p$ of function $|x|^2$ on $M$, then near $p$, $M$ is "wrapped" by some $S^n$ and has the same tangent space as $S^n$. But I am stuck there.

We consider the functions $L_q(x)=|x-q|^2$. Then we have a maximum $p$ of $L_q$ and we fix the unit vector $v=\frac{p-q}{|p-q|}$ throughout, so $v$ is the normal vector at $p$. Now if we set $q(t)=p+tv$, then when $t\leq-|p-q|$, $p$ is always the maximum of function $L_{q(t)}$ (this is true if we draw a ball at $q(t)$ with radius $|p-q(t)|$, then all $M$ is contained in this ball).

Therefore when $t$ sufficiently tends to $-\infty$, the Hessian $L_{q(t)}$ is always semi-positive definite. Now if we fix a coordinate neighborhood aroud $p$, then Hessian matrix $H$ of $L_{q(t)}$ at $p$ is given by $H=2(F-tS)$ where $F,S$ are the first and second fundamental forms of $M$ at $p$. So we conclude that $S$ has to be semi-positive definite.

But how can we move further to say $S$ is positive definite?

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There is a trivial counter example, namely, $S^1\subseteq \mathbb{R}^2$. Perhaps you're assuming $n>1$? Also, for the $n=2$ case, see math.stackexchange.com/questions/89061/…. –  Jason DeVito Oct 20 '12 at 5:16
@Hezudao: concerning your edit (partial progress): my answer solves the problem along these lines, except that it's simpler (using linear functions rather than distance squared) - and complete :) –  user8268 Oct 24 '12 at 21:04

There exists a linear function $L:\mathbb R^{n+1}\to\mathbb R$ such that $L|_M$ has non-degenerate critical points (by Sard's theorem). If you choose a maximum of $L$ on $M$ then the second quadratic form at that point is positively definite, hence all the sectional curvatures at that point are positive.

edit (why positive 2nd fund. form implies positive curvature):

if $K$ is the 2nd fundamental form then the curvature tensor is given by $$(R(u,v)u,v)=K(u,u)K(v,v)-K(u,v)^2$$ and if $u,v$ are linearly independent and $K$ is positively definite then the RHS is positive. Sectional curvatures are the LHS if $(u,u)=(v,v)=1$ and $(u,v)=0$.

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I know the Hessian of $L$ at the maximum is positive-definite, but how can you say the sectional curvatures at that point are positive? –  Hezudao Oct 20 '12 at 17:48
@Hezudao: see the new part of my answer –  user8268 Oct 20 '12 at 20:05
I know that if the second fundamental form is positive definite, then all sectional curvatures are positive. But can you clarify why you can deduce that second fundamental form is positive definite by the positive definiteness of $L$ at the point? –  Hezudao Oct 20 '12 at 23:03
@Hezudao: the 2nd fund. form at $P\in M$ is (by definition) the Hessian at $P$ of $L|_M$ (provided $L$ has unit gradient and that $P$ is a critical point of $L|_M$) –  user8268 Oct 21 '12 at 14:43

One way to obtain the Levi-Civita connection for embedded submanifolds in $\mathbb{R}^{n+1}$ is to take the covariant derivative of a vector field with respect to the standard flat connection in $\mathbb{R}^{n+1}$, and then project the resulting vector into the tangent space. Thus the covariant derivative of a curve through the point of tangency on the manifold matches that of the sphere. Consequently, so does the Riemann curvature. (I'm a little fuzzy on this step: do I need the vector fields to be equal in a neighborhood, instead of just at a point?). So at the point of tangency, the sectional curvatures of the manifold match those of the sphere.

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I think this is not right. Since $(\nabla_XY)_p$ depends not only on $X_p,Y_p$, but also depends on the value of $Y$ at least in a curve whose tangent vector at $p$ is $X_p$ . –  Hezudao Oct 20 '12 at 4:37
For the sectional curvatures of the submanifold to match the sphere, you need the sphere to osculate the submanifold. In general, you could have a sphere of very large radius, and consequently very low curvature, tangent to a point of sharp curvature. But you don't need the curvatures to match, you just need to estimate the submanifold's curvature below by the sphere's curvature. –  Neal Oct 20 '12 at 5:22