# existence of affine connection on manifold

I am studying Riemannian Geometry following my professor's notes. On the proof of existence of affine connection on a $C^\infty$ manifold, the notes states:

By partition of unity, a connection can be locally induced from standard connection on $\mathbb{R}^n$.

I understand the standard connection on $\mathbb{R}^n$, which is a generalization of directional derivative at a point. But how can it induce a connection on the manifold?

• The easiest solution is to construct an immersion into $R^{2n}$, $n=dim(M)$, and then use the induced affine connection from $R^{2n}$. Jan 25, 2015 at 0:58

The key point is that a finite convex combination of connections on $M$ is itself a connection. (It's easy to check this by hand -- that the combination is convex is what gives you the Leibniz rule.)

So, we cover $M$ by a countable set of charts $\lbrace (\psi_i, U_i) \rbrace$ such that each point $p \in M$ has a neighborhood intersecting only finitely many of the $U_i$. We let $\lbrace \varphi_i \rbrace$ be a partition of unity subordinate to $\lbrace U_i \rbrace$.

On each $U_i$ there exists a connection $\nabla_i$ obtained from the standard connection on $\mathbb{R}^n$ and the chart map $\psi_i$. (This is the connection determined by $(\nabla_i)_{\partial_j} \partial_k = 0$ whenever $\partial_j, \partial_k$ are coordinate vector fields coming from $\psi_i$.)

It follows that the map $$\nabla = \sum_i \varphi_i \nabla_i,$$ which is a finite convex combination of connections in a neighborhood of any point of $M$, is itself a connection on $M$.

• May you explain what do you mean by "On each $U_i$ there exists a connection obtained from the standard connection on $\mathbb{R}^n$ and the chart map." I know in $\mathbb{R}^n$ there exist such connection, but how to induce a connection on manifold?
– John
Jan 25, 2015 at 3:31
• @JohnZHANG You have coordinate functions $x^1, \dots, x^n$ on $U_i$ coming from the chart, and these give vector fields $\partial_1, \dots, \partial_n$ on $M$. (This is standard differential geometry: by $\partial_j$ I mean $\partial/\partial x^j$.) The connection $\nabla_i$ is determined by $(\nabla_i)_{\partial_j} \partial_k = 0$. Jan 25, 2015 at 3:48
• What do you mean by "the connection $\nabla _i$ is determined by $(\nabla_i)_{\partial _j}\partial _k=0$"? Do you mean you set these values=0, and extend it to a connection based on the requirement of definition? If so, what it's called "obtained from the standard connection on $\mathbb{R}^n$"( originally I thought I need to define a map on $M$ based on the directional derivative on $\mathbb{R}^n$ and prove it's a connection).
– John
Jan 25, 2015 at 7:45
• Yes, we extend to a connection; in particular, any vector field $X$ on $U_i$ can be written $X = \sum f_j \partial_j$ for some smooth functions $f_j$ on $U_i$, and then the properties of a connection determines $$(\nabla_i)_{(\Sigma f_j \partial_j)} (\Sigma g_k \partial_k).$$ This is the standard connection on $\mathbb{R}^n$, or on $\psi_i(U_i) \subseteq \mathbb{R}^n$ if you like, moved over in the obvious way to $U_i$ via the diffeomorphism $\psi_i$. (If two manifolds are diffeomorphic, obviously a connection on one induces a connection on the other.) Jan 25, 2015 at 7:53
• Thanks for your explanation. Let me clarify it again. Given a chart $(U_i,\phi_i)$ around $p$, and a section $Y\in\Gamma(TM)$, then $\forall x\in U_i$, $Y(x)$ is only a vector in $T_xM$, which can be represented as a vector in the same $\mathbb{R}^n$ (since $(U_i,\phi_i)$ is fixed). Hence the standard connection on $\mathbb{R}^n$ can be applied here (this is why it's called "induced by $\mathbb{R}^n$"). Is my understanding correct?
– John
Jan 25, 2015 at 12:37