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Prove that the Möbius band is not orientable.

I know that in the Möbius band the central circle is orientable. If I let $Y$ be the Möbius band and $Z$ be a compact submanifold of $Y$ with $\dim(Z)=\frac 12 \dim(Y)$. From one of my exercise I have show that if $Z$ is globally defined by independent function then $I_2(Z,Z)=0$. The central circle is orientable but it is not definable by an independent function, so there is no smooth choice of orientation for all the tangent space $T_y(Y)$. If $Y$ can't be given an orientation, it's not orientable.

I'm not sure if my explaining is understandable. Please help me improve it.

Brian M. Scott
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Diane Vanderwaif
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    If it's orientable, you can have arrows sticking out of each point (continuously). But you can't assign them like that on the central circle for obvious reasons because it starts looping on itself in the opposite direction. – user2345215 Nov 15 '14 at 17:20
  • thanks, your explaination is much simplier and easier to understand than mine. – Diane Vanderwaif Nov 15 '14 at 17:57

1 Answers1

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The key fact here is that a hypersurface $Z$ in an orientable manifold $Y$ is orientable if and only if its normal bundle $N(Z,Y)$ is trivial. Here we have $Z$ the central circle in the Möbius strip $M$, and $I_2(Z,Z)=1$, so, by the Tubular Neighborhood Theorem, $N(Z,Y)$ cannot be trivial.

Ted Shifrin
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