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In Example 5. on this page (midway down the page) the author says that the definite integral

$$\int_a^b f(x) dx,$$

where $f: [a, b] \to \mathbb{R}$ can be interpreted as an integral of a connection. He says that if we set $X = [a, b]$ and let $(\mathbb{R})_{x \in X}$ be the trivial bundle over $X$ then the function $f$ induces a connection $\Gamma_f$ on this bundle by setting

$$\Gamma_f(x \to x + dx): y \to y + f(x)dx.$$ He says that the integral $\Gamma_f([a, b])$ of this connection along $[a, b]$ is the operation of translation by $\int_a^b f(x)$ in the real line.

  1. Is there a typo in the definition of the connection? Shouldn't the definition of $\Gamma_f$ map $y$ to $y + f'(x)dx$, not to $y + f(x)dx$?
  2. I don't understand why he is saying that the integral of this connection is equivalent to translation...surely the integral of the connection just outputs some arbitrary scalar, where is the notion of it representing translation coming from?

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In the diagram, vertical lines are fibres of the trivial bundle $[a, b] \times\ \mathbf{R}$, the horizontal line represents the "coordinate trivialization", the short segments (whose slopes at $(x, y)$ are $f(x)$) denote the horizontal spaces of the connection, and the dark curve is a horizontal section.

A connection on a trivial line bundle over an interval

  1. There's no typo in the definition, though it might have been friendlier to change notation $f \to f'$ so that the slope of the connection at $(x, y)$ was $f'(x)$ instead of $f(x)$.

  2. The integral of the connection is "translation by the integral" in the sense that if $\sigma_{1}$ is a "connection horizontal" section and $\sigma_{2}$ is the "coordinate horizontal" section satisfying $\sigma_{2}(a) = \sigma_{1}(a)$, then for each $t$ in $[a, b]$ the difference at $t$ is $$ \sigma_{1}(t) - \sigma_{2}(t) = \int_{a}^{t} f(x)\, dx. $$