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Suppose that $(U,\varphi)$ is a chart on manifold $M$, and $X,V$ are vector fields on manifold $M$, then we can write: $$X=\sum_{i=1}^{i=n}X^{i}\frac{\partial}{\partial x^{i}}$$ on $U$, and define a connection on $U$ by: $$D_{V}X=\sum_{i=1}^{i=n}(VX^{i})\frac{\partial}{\partial x^{i}}\cdots\tag{1}$$

Let $\{U_{j}\}_{j=1}^{\infty}$ be a locally finite covering of $M$, where each $U_{j}$ is coordinate neighborhood on $M$. Let $D^{j}$ be the connection on $U_{j}$ defined by (1) respectively . and let $\{f_{j}\}$ be the partitions of unity on $M$ that are subordinate to $\{U_{j}\}$.

Show that: $\sum_{j=1}^{\infty}f_{j}D^{j}$ is a connection on $M$.

Another question: suppose $X,V$ are two vcetor fields on manifold $M$, with the connection $\sum_{j=1}^{\infty}$ $f_{j}D^{j}$ defined above, how do we know that the new vector field $(\sum_{j=1}^{\infty}f_{j}D^{j})_{V}X$ is well defined on some intersections of the coordinate neighborhood, say $U_{j_{1}}$ and $U_{j_{2}}$ .

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Please take a look at how I've edited your MathJax code; I encourage you to emulate it. In particular, each "stretch" of mathematics should be in one piece of MathJax. For example, instead of writing $\frac{1}{2}$ $a$ $\otimes$ $b$, you should write $\frac{1}{2}a\otimes b$. – Zev Chonoles Mar 10 '13 at 15:29
You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. – Zev Chonoles Mar 10 '13 at 15:32
You should probably assume a little bit more - namely, that the covering $\{U_j\}$ of $M$ is locally finite. That is, for every $p\in M$, there is a neighborhood $V$ of $p$ for which $V\cap U_j = \emptyset$ except for finitely many values of $j$. Many definitions of "manifold" require the manifold to be paracompact. In this case, you are guaranteed that you can refine \{U_j\}$ to have this property. What this gains you is that at any point, your infinite sum (where you, a priori, have to worry about convergence, smoothness, etc.) is actually a finite sum, so none of these issues arise. – Jason DeVito Mar 10 '13 at 15:47
@Zev Chonoles :Thank you very much!I will improve my typing skill. – Wei Xia Mar 11 '13 at 2:14
@Jason DeVito :yes,I forgot that.Thanks for your reminding! – Wei Xia Mar 11 '13 at 2:25
up vote 2 down vote accepted

Let us write $D = \sum_{j=1}^{\infty}f_jD^j$. In order to check that $D$ is in fact a connection we need to verify three things:

1) Is it linear in $V$? That is, is it true that: \begin{equation} D_{g_1V_1+g_2V_2}X = g_1D_{V_1}X + g_2D_{V_2}X \end{equation} 2) Does it satisfy the Liebniz law in $X$: \begin{equation} D_{V}fX = (Vf)X + fD_VX \end{equation}

3) Is it linear over $\mathbb{R}$ in $X$?: \begin{equation} D_V(\alpha_1X_1+\alpha_2X_2) =\alpha_1D_VX_1 + \alpha_2D_VX_2 \end{equation} where $\alpha_1,\alpha_2\in \mathbb{R}$

It is easy to show that each `local' connection $D^j$ on $U_j$ satisfies all three of these. Now to show that $D$ satisfies $1$, $2$ and $3$, it will suffice to show that that it satisfies all three of these properties at each point $p$. So, choose a neighbourhood $W$ of $p$ such that $U_j\cap W = \emptyset$ for all but a finite number of $j$ (As per Jason De Vito's comment, we may do this if $\{U_j\}$ is a locally finite cover). Now if we want to calculate the value of $D_VX$ at $p$ it suffices to consider only the restrictions of $V$ and $X$ to $W$; $V|_{W}$ and $X|_{W}$. the hardest property to check is 2), so lets look at that: \begin{align} D_{V|_{W}}(gX_{W}) & = \sum_{j}f_jD^j_{V|_{W}}gX|_{W}\\ & = \sum_{j}(f_j(V_{W}g)X + gD^{j}_{V|_{W}}X) \\ & = (V_{W}g)X + gD_{V|_W}X \\ \end{align} where in the second line we used the fact that each $D^j$ satisfies the Liebniz law while in the third line we are using the fact that for any partition of unity, $\sum_{j}f_j = 1$, and because the sum restricted to $V$ is finite we avoid prickly issue of convergence. So our connection $D$ does indeed satisfy the Liebniz law on the set $W$ and in particular at $p\in W$. Showing that $D$ satisfies 1) and 3) is simpler.

As for your second question, let $W = U_{j_1}\cap U_{j_2}$. Again we can assume that $W$ makes intersection with only finitely many other $U_{j}$, call them $U_{j_3},\ldots U_{j_k}$. By simply writing out what $D_{V}X$ is on $W$: \begin{equation} D_{V}X = f_{j_1}D^{j_1}_{V}X + f_{j_2}D^{j_2}_{V}X+\ldots + f_{j_k}D^{j_k}_{V}X \end{equation} We see that it doesn't really depend on which coordinate neighbourhood we are considering (it involves information from all the coordinate neighbourhoods that intersect at $W$).

Hope this helps

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It help me a lot!Thank you! – Wei Xia Mar 11 '13 at 2:41

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