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Let $F=\{F_p; p\in M\}\subseteq TM$ be a rank $k$ smooth distribution. Can anyone explain-me what is the set $$\displaystyle\nu(F)=\frac{T_pM}{F_p}.$$

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1 Answer 1

I'm not sure exactly what you are looking for. Given $p$, we have $F_p$ is a subspace of $T_pM$. Thus $\displaystyle \frac{T_pM}{F_p}$ is just the quotient vector space.

If there is an integral submanifold for the distribution, then $F_p$ will be the tangent space to that submanifold at $p$. When you quotient, you kill this part off, so $\nu(F)$ will be $N_p$, the normal bundle for the integral submanifold at $p$.

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I was looking for a geometrical interpretation.. –  PtF Apr 24 '13 at 17:56
    
@PtF Excellent. Good thing I gave one. –  Matt Apr 24 '13 at 18:00
    
Yeah, thanks =D –  PtF Apr 24 '13 at 18:02
    
I found one too: the elements of the set $T_pM/F_p$ are "lines" which are parallel to the subspace $F_p$.. –  PtF Apr 24 '13 at 18:03
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