Orientation of submanifold

**Sogan** · February 7th 2011, 10:54 AM

Hello,

I try to solve this problem:

Let M, N be orientable Manifolds (of dimension m,n) and f:M->N a smooth submersion. => $G:=f^{-1}(p)$ 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 $G$ is an embedded submanifold. And therefore we have an canonical atlas for G. If $(U_i,f_i)$ is an atlas for M, then $(U_i \cap S, \pi \circ f_i)$ is an atlas for G, whereas $\pi$ is the projection on the first (m-n) coordinates.

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

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:
$g_j\circ f \circ (f_i)^{-1}$ whereas $g_j$ is a coordinate map of N and $f_i$ a coordinate maß of M.
Therefore we can put this forumula to the one above and get:
det $d( (g_j\circ f \circ (f_i)^{-1})\circ f_i) \circ((g_j\circ f \circ (f_i)^{-1}) \circ f_j)^{-1})>0$

Do you see, why this is greater than 0?

Regards

**xxp9** · February 9th 2011, 04:21 AM

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

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

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

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