1. ## Orientation of submanifold

Hello,

I try to solve this problem:

Let M, N be orientable Manifolds (of dimension m,n) and f:M->N a smooth submersion. =>$\displaystyle G:=f^{-1}(p)$ is orientable.

Ok i have a proof, but it can't be correct, since the same "proof" would also work for any embedded submanifold.:

We know that $\displaystyle G$ is an embedded submanifold. And therefore we have an canonical atlas for G. If $\displaystyle (U_i,f_i)$ is an atlas for M, then$\displaystyle (U_i \cap S, \pi \circ f_i)$is an atlas for G, whereas $\displaystyle \pi$is the projection on the first (m-n) coordinates.

Now we have to show that det $\displaystyle d(\pi \circ f_i) \circ(\pi \circ f_j)^{-1})>0$

And this follows, since the f_i are orientable.(?) This can't be. (because not every submanifold is orient.)
But why is it >0?? We know also that the projection is in some nbh. equal to the map:
$\displaystyle g_j\circ f \circ (f_i)^{-1}$ whereas $\displaystyle g_j$ is a coordinate map of N and $\displaystyle f_i$ a coordinate maß of M.
Therefore we can put this forumula to the one above and get:
det $\displaystyle d( (g_j\circ f \circ (f_i)^{-1})\circ f_i) \circ((g_j\circ f \circ (f_i)^{-1}) \circ f_j)^{-1})>0$

Do you see, why this is greater than 0?

Regards

2. Use the local normal form of a submersion, f can be written as
$\displaystyle f( x^1, ..., x^m) = (x^1, ..., x^n)$. And $\displaystyle (x^{n+1}, ..., x^m)$ are the coordinates of the sub-manifold G.

Suppose $\displaystyle (u^1, ..., u^m)$ is another local chart of M and $\displaystyle (u^1, ..., u^n)$ the chart of N with non-empty intersection with the above charts respectively. Since $\displaystyle (u^1, ..., u^n)$ and $\displaystyle (x^1, ..., x^n)$ both are coordinates of N, $\displaystyle (u^1, ..., u^n)$ is only the function of $\displaystyle (x^1, ..., x^n)$, i.e. it does not depend on $\displaystyle (x^{n+1}, ..., x^m)$. Also $\displaystyle (u^{n+1}, ..., u^m)$ is only the function of $\displaystyle (x^{n+1}, ..., x^m)$.

Since M and N are both orientable we can choose the sign of the coordinates so that the Jocobian $\displaystyle \frac{du^1 \wedge ... \wedge du^m}{dx^1 \wedge ... \wedge dx^m} > 0$ and $\displaystyle \frac{du^1 \wedge ... \wedge du^n}{dx^1 \wedge ... \wedge dx^n} > 0$, which implies that $\displaystyle \frac{du^{n+1} \wedge ... \wedge du^m}{dx^{n+1} \wedge ... \wedge dx^m} > 0$