# induced connection on open sets

Let $M$ be a smooth manifold. If $\nabla$ is a linear connection on $M$, I would like to induce a unique linear connection on an open subset $U\subseteq M$. I know that for all $p\in U$ there is a natural isomorphism $T_pU\cong T_pM$, so I can restrict global vector fields to local vector fields on $U$. Unfortunately there are some local vector fields on $U$ that don't came from a restriction of global vector fields.

For this reason I can't find a reasonable linear connection $\nabla^U$ over $U$ induced by $\nabla$. I need help.

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The connection $\nabla$ on a manifold $M$ is a local operator. The value of $\nabla_X(Y)$ at a point $p \in M$ depends only on $X_p$ and the value of $Y$ in an arbitrary small neighborhood around $p$. This is enough to define the connection on $TU$ when $U \subset M$ is an open subset, without extending the vector fields involved to the whole of $M$. More generally, you may want to read about the pullback of a connection which allows you to restrict a connection to more general submanifolds and even more.

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The proposition about the local behavior of the connection says: If Y and Z are GLOBAL vector fields that coincide on an open neighborhood of $p$ then $\nabla_XY=\nabla_XZ$ (The same thing is true for $X$). I don't understand in which way I can apply this statement to the problem. – Dubious Dec 8 '12 at 17:58

Take an open cover $\{U_j\}$ of $U$ given by relatively compact open sets in $U$ and a partition of unity $\{\phi_j\}$ subordinated to $\{U_j\}$ (i.e. $\mathrm{supp}\phi_j\subset U_j$).

For any vector-field $X$ on $U$, $X=\sum \phi_j\cdot X$ and $\phi_j X$ is a vector-field on $M$. Therefore we can define $$\nabla^U_XY=\sum\nabla_{\phi_j X}Y$$ and $$\nabla^U_YX=\sum\nabla_Y\phi_j X$$ The definition is locally meaningful, because the covering is locally finite, and so are the sums.

Hope it helps.

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why $\phi_jX$ is a vector field on $M$? For every $p\in U$ we have that $(\phi_jX)_p=\phi(p)X_p$, but if $p$ is in $X\setminus U$ what is the sense? – Dubious Dec 8 '12 at 17:42
The point is that on $X\setminus U$, we have $\phi(p) = 0$, so it doesn't matter what you choose for $X_p$ - even if you choose some horribly discontinuous thing for $X_p$, $\phi(p) X_p = 0_p$. So $\phi X$ is still a smooth vector field on all of $M$. – Jason DeVito Dec 8 '12 at 18:01
@Galoisfan: Well, I'm thinking more about $\phi X$ than $\sum \phi_j X$. Note that $\phi X$ is not an extension of $X$ to all of $M$ because there are points $p\in U$ with $\phi(p)X_p \neq X_p$. It's more like: Shrink $X$ to an even smaller open subset and then extend the smaller thing to all of $M$. – Jason DeVito Dec 8 '12 at 18:16
No, $\sum\phi_j X$ is the vector-field $X$, which is a section of $TM$ on $U$; but the things you add up, $\phi_j X$ for each $j$, can be extended to global sections, just define them to be zero outside $U$. A locally finite series of global sections is by no means bound to be a global section, unless you have some kind of uniform estimates. – wisefool Dec 8 '12 at 18:18
Actually, you can combine what I'm saying with what levap said... with the terminology of Jason DeVito, you shrink down $X$ considering it only around a point $p$ and cutting it off with a smooth function, then you extend this things to be 0 outside, obtaining a global object. The germ of this vector-field in $p$ and the germ of the original one coincide, therefore you can define the connection on germs and, by the locality property, you know it's well defined and that everything works. – wisefool Dec 8 '12 at 18:21