# Existence of diffeomorphism so that $\int (\phi^* f) \omega = 0$

Can someone help me with this question? Is a qual exam question and I have no idea how to tackle it.

Prove or disprove: Let $f\in \mathcal{C}^\infty(S^n)$ be a smooth function and $x_1, x_2\in S^n$ be two points such that $f(x_1)<0<f(x_2)$. Let $\omega$ be a smooth nonvanishing $n$-form. Then there is a diffeomorphism $\phi\colon S^n \rightarrow S^n$ such that $\int_{S^n} (\phi^* f)\omega =0$.

• Hint: find a one parameter family if diffeomorphisms and use intermediate value theorem. – Moishe Kohan Jun 3 '16 at 20:35
• Is there any way you could make the title less vague? I don't know differential geometry so I can't, but surely you can give a better overview than "geometry question" – pjs36 Jun 3 '16 at 22:48

If $\int f \omega = 0$, set $\phi =\text{id}$ and we are done. Assume that $\int f \omega <0$ (If not, consider $-f$). Since $\omega$ is nonvanishing, We also assume that $\omega$ is positive, so $$\mu(A) := \int_A \omega >0$$ for all nonempty open sets $A$.

Let $x_0 \in \mathbb S^n$ such that $f$ is positive. Then by continuity there is an open ball $U$ centered at $x_0$ so that $f \ge c>0$ on $U$. Let $\{\phi_r\}_{r >0}$ be the one parameter family of diffeomorphisms so that $\phi_1 = \text{id}$, $\phi_s U \subset \phi_r U$ if $s<r$ and $$\bigcup_r \phi^{-1}_r U = \mathbb S^n\setminus \{-x_0\}.$$ (see the construction below) Note that $$g(r)= \int_{\mathbb S^n} (\phi_r^* f)\omega$$ is continuous and $g(1) <0$. Since $f$ is uniformly bounded below by $-m$, we have

$$\begin{split} \int_{\mathbb S^n} (\phi_r^* f)\omega &= \int_{\phi_r^{-1}\ U} (\phi_r^*f) \omega + \int_{\mathbb S^n\ \setminus \phi^{-1}_r\ U} (\phi_r^*f)\omega\\ &\ge c \mu(\phi^{-1}_r U) - m \mu( \mathbb S^n \setminus \phi^{-1}_r U) \\ &\to c \mu(\mathbb S^n) >0 \end{split}$$

as $r\to 0$, where $\mu(A) = \int_A \omega$. Thus there is $r_0$ so that

$$\int_{\mathbb S^n} (\phi_{r_0}^* f)\omega =0.$$

To construct the family of diffeomorphisms, consider the stereographic projection at $\psi : \mathbb S^n \setminus \{-x_0\} \to \mathbb R^n$ at $x_0$. Let $U = \psi^{-1}(B_1)$, where $B_1$ is the unit ball in $\mathbb R^n$. Then define $\phi_r : \mathbb S^n \to \mathbb S^n$ by

$$\phi_r(x) =\begin{cases} \psi^{-1} (r\psi(x)) & \text{if }x\neq -x_0 \\ x_0 &\text{if }x = -x_0 .\end{cases}$$

Note that $\phi_r$ is smooth even at $-x_0$, since if you use the stereographic projection $\overline \psi$ at $-x_0$, you get (Since $\overline \psi \circ \psi (x) =\frac{x}{|x|^2}$)

$$\phi_r(x) = \begin{cases} \overline\psi^{-1} (\frac 1r \overline \psi(x)) &\text{if } x\neq x_0 \\ x_0 & \text{if }x = x_0.\end{cases}$$

Note that you can also construct a one parameter family of diffeomorphisms using the gradient vector (with the standard metric on $\mathbb S^n$) of the function $$F(x) = x\cdot x_0,$$ where the dot product is the standard one on $\mathbb R^{n+1}$.

• +1 well done. I interpreted the hint incorrectly; I thought about this idea (using the IVT on the integral), but I couldn't figure out how to do so. – Moya Jun 5 '16 at 4:48

Based on the hint:

Since $S^n$ is path connected, there is path $\gamma:[0,1]\to S^n$ such that $\gamma(0)=x_1,\gamma(1)=x_2$. We can take this path to be injective (think about $S^2$ using angular coordinates). Push the vector field $\dfrac{\partial}{\partial t}$ forward using $\gamma$ to a vector field $V$ on $\gamma([0,1])$. Since this is an embedded submanifold which is closed in $S^n$, we can extend this to a compactly supported vector field (also called $V$) defined on $S^n$ (see Lee, chapter 8).

Because $S^n$ is compact, $V$ admits a complete flow, i.e. a smooth 1-parameter family of diffeomorphisms $\Phi(t,\cdot)=\phi_t:S^n\to S^n$. Consider the map $$F=f\circ \Phi(\cdot,x_1):\mathbb{R}\to\mathbb{R}$$Then $F(0)=f(\Phi(0,x_1))=f(x_1)<0$ and $F(1)=f(\Phi(1,x_1))=f(x_2)>0$ (check that $\Phi(1,x_1)=x_2$). By the intermediate value theorem, there is a $t_0\in(0,1)$ with $F(t_0)=0$. Now consider the form $(\phi_{t_0}^{\ast}f)\omega$....

Still working on it.