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Editted to add Lemma 3.4.

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 $T_p(M)$.

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  • $\begingroup$ 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, $(\nabla_{X'}Y)_{p} = \sum b_{i}(p)(\nabla_{E_{i}}Y)_{p} = \sum a_{i}(p)(\nabla_{E_{i}}Y)_{p}=(\nabla_{X}Y)_{p}$. $\endgroup$ Apr 7, 2019 at 9:13
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    $\begingroup$ @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. $\endgroup$
    – Alex M.
    Apr 9, 2019 at 8:55
  • $\begingroup$ 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 $\tilde{X}$. $\endgroup$
    – user782220
    Apr 10, 2019 at 0:06

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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 $(\nabla_X Y)_p$).

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