# Show that $F^*\omega$ is an orientation form of $M$.

$F:M\rightarrow N$ local diffeomorphism between oriented manifolds of positively dimension. Suppose that with respect to positively oriented charts of $M$ and $N$, the Jacobian of $F$ has positive determinant. Let $\omega$ be an orientation form of $N$. Show that $F^*\omega$ is a form of orientation for $M$.

My idea is to show that the set $\{dF_pv_1,...,dF_pv_n\}$ is a positively oriented of $T_qN$, $q=F(p)$ and $\{v_1,...,v_n\}$ is an arbitrary basis of $T_pM$ positively oriented. But I do not know how.

Remark: If it is possible, I wanted it to be proved by the definition of orientation form, i.e., that would show that $F^*\omega$ is a form of orientation of $M$ directly from the definition of orientation form.

• Where you assume what about $\{v_1,\dots,v_n\}$? Do you have to do this for an arbitrary collection of vectors, or can you choose them thoughtfully so that you can use the hypothesis about the Jacobian? – Ted Shifrin Jun 7 '18 at 23:19
• @TedShifrin Max edited – Takashi Jun 7 '18 at 23:41
• No, you want $\{v_1,\dots,v_n\}$ to be a positively oriented basis! – Ted Shifrin Jun 8 '18 at 0:16
• @TedShifrin Yes, I corrected now. – Takashi Jun 8 '18 at 0:37

## 1 Answer

$\textbf{Hint}$ :
For coordinate basis, it goes like this. Let $\omega$ is a positively oriented orientation form for $N$. For any $p\in M$ and oriented smooth charts $(U,x^i)$ and $(V,y^i)$ we have \begin{align*} (F^*\omega)_p\bigg( \frac{\partial}{\partial x^1}\bigg|_p,\dots,\frac{\partial}{\partial x^n}\bigg|_p \bigg) &= \omega_{F(p)} \bigg(dF_p\bigg(\frac{\partial}{\partial x^1}\bigg|_p\bigg), \dots, dF_p\bigg(\frac{\partial}{\partial x^n}\bigg|_p\bigg)\bigg) \\ &= \det \bigg(\frac{\partial \widehat{F}^j}{\partial x^i}(p) \bigg)\, \omega_{F(p)} \bigg( \frac{\partial}{\partial y^1}\bigg|_p,\dots,\frac{\partial}{\partial y^k}\bigg|_p \bigg). \end{align*} By hypothesis it has positive value. So $F^*\omega$ is positively oriented.

For arbitrary oriented basis $\{v_1,\dots,v_n\}$, you have to show that $(F^*\omega)_p(v_1,\dots,v_n)>0$. The arguments is similar, only this time write $v_i = v_i^j \partial_{x^j}$, and see what happens to $(F^*\omega)_p(v_1,\dots,v_n)$.

• If it is possible, I wanted it to be proved by the definition of orientation form, ie, that would show that $F^*\omega$ is a form of orientation of $M$ directly from the definition of orientation form. – Takashi Jun 8 '18 at 0:53
• I understood, thanks! – Takashi Jun 8 '18 at 21:43