# Connection does not depend on entire vector field

The book "An Introduction to Differentiable Manifolds and Riemannian Geometry" by William Boothby has the following definition of connection.

(3.1) Definition. A $$C^\infty$$ connection $$\nabla$$ on a manifold $$M$$ is a mapping $$\nabla: \mathfrak{X}(M) \times \mathfrak{X}(M) \to > \mathfrak{X}(M)$$ denoted by $$\nabla:(X,Y)\to \nabla_X Y$$ which has the linearity properties: For all $$f,g\in C^\infty(M)$$ and $$X,X',Y,Y'\in > \mathfrak{X}(M)$$, we have

(1) $$\nabla_{fX+gX'}Y = f(\nabla_x Y) + g(\nabla_{X'}Y)$$

(2) $$\nabla_X(fY+gY')=f\nabla_X Y + g\nabla_X Y' + (Xf)Y+(Xg)Y'$$

The book introduces a corollary which aims to show that $$(\nabla_X Y)_p=\nabla_{X_p} Y$$ because the definition of connection does not immediately state this. But I don't understand the proof of the below corollary. In particular its not clear to me what is $$\tilde{X}$$ doing and how its used to show dependence on $$X$$ at $$p$$ and not on the whole of $$X$$.

(3.4) Lemma. Let $$X,Y\in \mathfrak{X}(M)$$ and suppose that either $$X=0$$ or $$Y=0$$ on an open set $$U\subset M$$. If $$\nabla$$ is a connection [satisfying properties (1) and (2) of Definition 3.1], then the vector field $$\nabla_X Y = 0$$ on $$U$$.

(3.5) Corollary. Let $$p$$ be any point of $$M$$. If $$X,X'\in\mathfrak{X}(M)$$ such that $$X_p=X'_p$$, then for every vector field $$Y$$, $$(\nabla_X Y)_p=(\nabla_{X'} Y)_p$$. Denote this uniquely determined vector by $$\nabla_{X_p}Y$$. Then the mapping from $$T_p(M)\to > T_p(M)$$ defined by $$X_p\to \nabla_{X_p}Y$$ is linear.

Proof. Let $$U,\varphi$$ be a coordinate neighborhood of the point $$p$$. As in the proof of the lemma, there is a $$C^\infty$$ function on $$M$$ with $$\text{supp}(f)\subset U$$ and $$f\equiv1$$ on a neighborhood $$V$$ of $$p$$ (so $$\overline{V}\subset U$$). If $$X\in\mathfrak{X}(M)$$, then on $$U$$ we have $$X=\sum_{i=1}^n a_i E_i$$ with $$a_i\in C^{\infty}(U)$$ and $$E_1,\ldots,E_n$$ the vectors of the coordinate frames. We define $$\tilde{X}, \tilde{E_1},\ldots, \tilde{E_n}\in\mathfrak{X}(M)$$ and $$\tilde{a_1},\ldots,\tilde{a_m}\in C^{\infty}(M)$$ by $$\tilde{X}=f^2X,\tilde{E_i}=fE_i$$ and $$\tilde{a_i}=fa_i, i=1,\ldots,n$$, on $$U$$, and all to be zero (vectors and functions respectively) on the open set $$M-\text{supp}(f)$$. Then we have $$\tilde{X}=\tilde{a_1}\tilde{E_1}+\cdots+\tilde{a_n}\tilde{E_n}$$ on all of $$M$$; but on $$\overline{V}$$ this reduces to the equation above since $$\tilde{X}=X$$, $$\tilde{E_i}=E_i$$ and $$\tilde{a_i}=a_i$$ on this set. Applying Lemma 3.4 and property (1) of $$\nabla$$ gives $$\nabla_X Y=\nabla_{\tilde{X}} Y = \sum_{i=1}^n \tilde{a_i}\nabla_{\tilde{E_i}} Y \quad\text{on}\quad V$$. Hence $$(\nabla_X Y)_p = \sum \tilde{a_i}(p)(\nabla_{\tilde{E_i}} Y)_p = \sum a_i(p)(\nabla_{E_i}Y)_p$$, where the right-hand side depends only on the value $$X_p$$ of the vector field $$X$$ at $$p$$. This proves the first statement and the formula itself shows that the mapping $$X_p\to\nabla_{X_p}Y=(\nabla_X Y)_p$$ is a linear mapping of $$T_p(M)$$ into itself. For its value depends linearly on the components $$a_1(p),\ldots,a_n(p)$$ of $$X_p$$ relative to the basis $$E_{1p},\ldots,E_{np}$$ of $$T_p(M)$$.

• The role of $\tilde{X}$ and the like is to extend the formula $X=\sum_{i=1}^{n}a_{i}E_{i}$ on all of $M$ instead of just in the chart $(U, \phi)$. Then, for the part "$(\nabla_{X}Y)_{p}$ only depends on the value of $X$ in $p$",look at the last identity, $(\nabla_{X}Y)_{p} = \sum a_{i}(p)(\nabla_{E_{i}}Y)_{p}$. If $X'$ were another vector field with the same value in $p$ as $X$, say $X' = \sum b_{i}E_{i}$ in the same chart, but $b_{i}(p) = a_{i}(p)$, then, by the same reasoning, $(\nabla_{X'}Y)_{p} = \sum b_{i}(p)(\nabla_{E_{i}}Y)_{p} = \sum a_{i}(p)(\nabla_{E_{i}}Y)_{p}=(\nabla_{X}Y)_{p}$. – Matija Sreckovic Apr 7 at 9:13
• @user782220: I am afraid that this proof is as clear as it gets. No text on differential geometry does this kind of proof in full detail, it's a characteristic of this subject. The proof is indeed a bit sloppy: it is not clear where lemma 3.4 is used, and it is not clear why $(\nabla _{E_i} Y)_p$ does not depend on $f$, given that $E_i$ does. This last omission, in particular, makes the proof look a bit circular. You'll learn to live with these, though, that's everyday life in differential geometry. – Alex M. Apr 9 at 8:55
• So am I understanding this correctly that the subtle point is that linearity (property 1) cannot be applied to directly show that $\nabla_X Y= \sum_{i=1}^n a_i \nabla_{E_i} Y$ on $V$ because the $E_i$ are not defined on all of $M$. And that is the reason for going through $\tilde{X}$. – user782220 Apr 10 at 0:06

Well, look at the final formula $$(\nabla_X Y)_p = \sum \tilde{a_i}(p)(\nabla_{\tilde{E_i}} Y)_p = \sum a_i(p)(\nabla_{E_i}Y)_p.$$
The $$E_i$$ do not depend on what $$X$$ are: they are just the coordinate vector fields on our chosen coordinate chart at $$p$$. The coefficients $$a_i(p)$$ depend only on $$X_p$$, since they are defined as the coefficients of $$X_p$$ with respect to the basis $$(E_1)_p,\dots,(E_n)_p$$ for $$T_p(M)$$. So if we had any other vector field $$X'$$ with $$X'_p=X_p$$, then we would have $$a_i'(p)=a_i(p)$$ for each $$i$$ where $$a_i'$$ is defined correspondingly for $$X'$$. From the formula above, we conclude that $$(\nabla_X Y)_p=(\nabla_{X'}Y)_p$$.
The role of $$\tilde{X}$$ here is just to be able to write $$X$$ as a linear combination of the $$E_i$$: since the $$E_i$$ are only defined locally near $$p$$, we must multiply everything by the bump function $$f$$ so we can ignore everything outside a neighborhood of $$p$$ (and Lemma 3.4 tells us this won't change $$(\nabla_X Y)_p$$).