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My teacher defined the pullback bundle and gave me an example:

Let $\gamma:(-1,1) \to \mathbb{R}^2$ be a smooth curve. Then for $w \in \Gamma(\gamma^*T_p{\mathbb{R}^2})$, the pullback connection is $$\nabla_{\frac{\partial}{\partial x}}w = \text{directional derivative along curve}$$

Can someone explain this? So this is the directional derivative of $w$ along $\frac{\partial}{\partial x}$? What would such a $w$ look like?

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up vote 6 down vote accepted

The term 'directional derivative' needs to be replaced by the term 'covariant derivative'. Directional derivatives take functions as second inputs, but in this discussion, we are taking vector fields, such as the vector field $ w $ mentioned by the OP when stating the problem, as second inputs (I am using the convention $ {\nabla_{\text{First input}}}(\text{Second input}) $). We thus need an appropriate generalization of the concept of a directional derivative, and that leads one to the concept of a covariant derivative (an affine connection).

According to the definition of a pullback connection, the $ \gamma $-pullback to $ (-1,1) $ of an affine connection $ \nabla $ on $ \mathbb{R}^{2} $, hereafter denoted by $ {\gamma^{*}}(\nabla) $, satisfies the following relation: $$ (\star) \quad {[{\gamma^{*}}(\nabla)]_{\frac{\partial}{\partial x}}}({\gamma^{*}}(X)) = {\gamma^{*}} \left( {\nabla_{{d \gamma} \left( \frac{\partial}{\partial x} \right)}}(X) \right) = \gamma^{*} \left( {\nabla_{\gamma'}}(X) \right). $$ The expression at the right end is the $ \gamma $-pullback of the covariant derivative of $ X $ with respect to $ \gamma' $ (the vector field along $ \gamma $), and I have used the fact that $ \left[ {d \gamma} \left( \dfrac{\partial}{\partial x} \right) \right]_{x} = \gamma'(x) $ for all $ x \in (-1,1) $. Hence, your statement could be more precisely phrased as $$ {[{\gamma^{*}}(\nabla)]_{\frac{\partial}{\partial x}}}({\gamma^{*}}(X)) = \text{The $ \gamma $-pullback of the covariant derivative of $ X $ along $ \gamma $}. $$

The most important thing to observe is that in ($ \star $), the pullback connection $ {\gamma^{*}}(\nabla) $ is not defined for all vector fields on $ (-1,1) $, but only for vector fields that have been pulled back from those on $ \mathbb{R}^{2} $. In other words, $ {\gamma^{*}}(\nabla) $, as it currently stands, is defined only for elements of $ {\gamma^{*}}(\Gamma(T \mathbb{R}^{2})) $. However, $ {\gamma^{*}}(\nabla) $ can be extended so that it is defined for all sections of the pullback tangent bundle (i.e., defined for all $ s \in \Gamma({\gamma^{*}}(T \mathbb{R}^{2})) $), and such an extension can be shown to be unique. There is a need to extend because we only have $ {\gamma^{*}}(\Gamma(T \mathbb{R}^{2})) \subseteq \Gamma({\gamma^{*}}(T \mathbb{R}^{2})) $ but would like the pullback connection to be defined for all sections of the pullback tangent bundle. In the end, we require $ w $ to be a section of $ {\gamma^{*}}(T \mathbb{R}^{2}) $ and not an arbitrary vector field on $ (-1,1) $.

Now, one may be wondering why $ \nabla $ accepts $ \gamma' $ as first input although it is a vector field that is defined only on the image of $ \gamma $. Do we not need the first input of $ \nabla $ to be a vector field defined on all of $ \mathbb{R}^{2} $? How then can $ \gamma^{*} \left( {\nabla_{\gamma'}}(X) \right) $ be a well-defined object? The reason is as follows: If we extend $ \gamma' $ to two different vector fields $ Y_{1} $ and $ Y_{2} $ on $ \mathbb{R}^{2} $, then the point derivations $ [{\nabla_{Y_{1}}}(X)]_{p} $ and $ [{\nabla_{Y_{2}}}(X)]_{p} $ at any point $ p \in \text{Im}(\gamma) $ must be identical (this is a basic fact about affine connections in differential geometry). It thus follows that for all $ x \in (-1,1) $, \begin{align} \left[ \gamma^{*} \left( {\nabla_{Y_{1}}}(X) \right) \right]_{x} &= \left[ {\nabla_{Y_{1}}}(X) \right]_{\gamma(x)} \quad (\text{By the definition of $ \gamma^{*} $.}) \\ &= \left[ {\nabla_{Y_{2}}}(X) \right]_{\gamma(x)} \quad (\text{By the preceding comment.}) \\ &= \left[ \gamma^{*} \left( {\nabla_{Y_{2}}}(X) \right) \right]_{x} \quad (\text{By the definition of $ \gamma^{*} $ again.}) \end{align} Therefore, $ \gamma^{*} \left( {\nabla_{\gamma'}}(X) \right) $ is indeed a well-defined object.

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Thank you very much. great answer. – maximumtag Dec 4 '12 at 17:56

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