The gauss map $ N$ is surjective if the surface is compact

Question

I'm having trouble in solving this (classical - I think) exercise. I have to adjust some detail.

Let M be a compact and orientable surface in $\mathbb{R}^3$ with genus $g \gt 0$ and Gaussian curvature K.

1) Let $M_{+} = \{p \in M | K(p) \geq 0\}$. Show that the Gauss map $N: M_{+} \to \mathbb{S}^2$ is surjective.

2) Show that $\int_{M} |K|dA \geq 8\pi$

My attempt is:

1)

Pick $\vec{v} \in \mathbb{S}^2$. Define the function $F: M \to \mathbb{R}$ with $F(p) = \vec{p} \cdot \vec{v}$. Since S is compact, F has a max $p_{0}$. Now take $\vec{v_{0}} \in T_{p_{0}}M$. So it exists a curve $\sigma: (-\epsilon, \epsilon) \to M$ with $\sigma(0) = p_{0}$ and $\sigma'(0) = \vec{v_{0}}$. Define the function $f: (-\epsilon, \epsilon) \to \mathbb{R}$ with $f(t) = \sigma(t) \cdot \vec{v}$. It has a max in 0, fo $f'(0) = 0$ and $f''(0) \leq 0$, Now, $f'(0) = \sigma'(0)\cdot\vec{v} = \vec{v_{0}}\cdot\vec{v} = 0$. So $\vec{v} \in (T_{p_{0}}M)^{\bot} = N_{p_{0}}$. So, $N_{p_{0}} = \pm\vec{v}$. (*) Then $f''(0) = \sigma''(0)\cdot\vec{v} = II_{p_{0}}(\vec{v_{0}})\cdot\pm N_{p_{0}}$, that is always negative or positive, so Gaussian curvature is positive.

My problem is (*). If I use Jordan-Brouwer theorem, I can choose a priori the normal vector pointing outside and I'm done, it will be positive. But I suppose I can conclude even without this strong theorem. My idea, but probably I'm wrong, is that if $N_{p_{0}} = -\vec{v}$ then $N_{p_{1}} = \vec{v}$ where $p_{1}$ is the minimum of F, but I don't know how to prove it.

2)

Here I have simply to prove that $\int_{M_{+}} |K|dA \geq 4\pi$, then it will follow from Gauss-Bonnet. The idea is clear: use the change of variable theorem.

$\int_{M_{+}} |K|dA = \int_{N^{-1}(\mathbb{S}^2)}|\det dN|dA \geq \int_{\mathbb{S}^2}dA = 4\pi$

My problem is: can I do that? N will not be in general injective nor a diffeomorphism, so I cannot use the "real" change of variable theorem. I put "$\geq$" because intuitively I restrict the integral on a domain that makes N bijective, but I don't know if this is correct.

Thanks in advance

Answer 1

For (2), the idea is to use the inverse function theorem. Take $v \in S^2$. Then, by part (1), there is some $w \in M_+$ such that $N(w) = v$. By the IFT, there are neighborhoods $U_w$ of $w$ and $U_v$ of $v$ such that $N\mid_{U_w} : U_w \to U_v$ is a diffeomorphism. Note $S^2 = \cup_v U_v$, so by compactness, $S^2 = \cup_{i=1}^n U_{v_i}$ for some $v_1,\dots,v_n \in S^2$. We can assume the sets $U_{v_1},\dots,U_{v_n}$ are disjoint (explanation below). Then, the sets $U_{w_1},\dots U_{w_n}$ are disjoint since $N \mid_{U_{w_i}}$ is a bijection for each $i$. Therefore, $4\pi = \int_{S^2} dA = \sum_{i=1}^n \int_{U_{v_i}} dA = \sum_{i=1}^n \int_{U_{w_i}} |\det dN| dA \le \int_{M_+} |K| dA$.

To see why we can assume $U_{v_1},\dots, U_{v_n}$ are disjoint, first note that replacing $U_{v_i}$ by two subsets whose union is $U_{v_i}$ is permissible, since the restriction of a diffeomorphism is a diffeomorphism. What we do is replace $U_{v_i}$ by $U_{v_i}$ intersected with all possible combinations of complements of $U_{v_1},\dots U_{v_{i-1}},U_{v_{i+1}},\dots,U_{v_n}$. It's hard to word this, so here's an example. Suppose we have three sets, $A,B,C$. We replace $A$ by $A \cap B \cap C, A \cap B^c \cap C, A \cap B \cap C^c, A \cap B^c \cap C^c$, and similarly with $B$ and $C$. It's easy to see that any pair of sets in this new collection (which has 12 elements) will be disjoint.

Answer 2

(1) For $v\in S^2$, if $t>0$ is sufficiently large, then $M$ and origin are in same side wrt $tv+v^\perp$. Then if we diminish $t$, we have some $t$ s.t. they meet. Hence $M$ has tangent plane whose out unit normal vector is $v$.

(2) For $v\in S^2$, define $h_v:\mathbb{E}^3\rightarrow \mathbb{R},\ h_v(x)=v\cdot x$. Then for a.e. $v$, $h_v$ is a morse function, i.e. the number of critical points is finite. If $M$ is a closed manifold not homeomorphic to a sphere, then $h_v$ has at least 4 critical points $p_i$. Here out unit normal at $p_i$ is $v$ or $-v$.

So recall $$ 16\pi \leq \int_{S^2}\ \sharp C(h_v) \ dv= \int_{S^2}\ \sharp n^{-1}(\pm v )\ dv =2\int_M\ |K| $$ where $\sharp$ is the number of elements in a set, $C$ is the set of critical points and $n$ is unit normal function on $M$.