# Why is this implying a normal vector field?

Suppose $$\omega$$ is a $$n-1$$ form on a $$n-1$$ dimensional manifold and $$(a_1(x)dx_1 + ... + a_n(x)dx_n )\wedge \omega = c\Omega$$, with $$c \neq 0$$ and $$\Omega =dx_1\wedge...\wedge dx_n$$. Moreover $$(a_1(x),...,a_n(x))= a(x)$$ is a normal vector on the manifold, with a nonzero norm. Apparently with this you can say there exists a normal unit vector field on $$M$$, where you say $$n(x) = \frac{a(x)}{\| a(x)| }$$ if $$c >0$$ and $$n(x) = -\frac{a(x)}{\| a(x)| }$$ if $$c < 0$$. Why can you say this, how is from the wedge product seen that the normal vector fields is smooth, as in it "doesn't keep switching" .

• Is the manifold embedded in $\Bbb R^n$? – edm Dec 5 '18 at 4:03
• Yeeah it is in Rn – AkatsukiMaliki Dec 5 '18 at 7:06
• And is $a_1(x)dx_1 + \cdots + a_n(x)dx_n$ defined on the manifold or on the whole $\Bbb R^n$? – edm Dec 5 '18 at 11:07