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Let $\pi:E\longrightarrow M$ a smooth fibre bundle. A connection is a linear bundle homomorphism $\Phi:TE\longrightarrow TE$ such that $\Phi$ is a projection to the vertical bundle $VE\subset TE$.

I read that a connection in $E$ is equivalent to a section $\Gamma:E \longrightarrow J^1E$, and the space of connection is an affine space.

What is this relationship, explicitly? and why it's an affine space?

Thanks

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    $\begingroup$ It is clear affine linear combinations $(1-t)\Phi_0+t\Phi_1$ of connections form a connection, following your initial definition, because the projection condition holds when you make such a combination. Hence an affine space. $\endgroup$
    – Ben McKay
    Commented Jun 22, 2020 at 19:32
  • $\begingroup$ "A connection is a linear bundle homomorphism $\Phi:TE\longrightarrow TE$" - This can't be, cause otherwise it would be $\mathcal{C}^\infty$-linear. $\endgroup$
    – Qfwfq
    Commented Jun 22, 2020 at 21:41
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    $\begingroup$ Dear Qfwfq, of course $\Phi$ is not the covariant derivative of a section, but you can easily obtain the covariant derivative by defining $\nabla s=\psi\circ \Phi\circ Ds$, where $Ds$ is the differential and $\psi\colon s^*VE\to E$ is the natural identification of the vertical bundle along a section. In fact, the two notions of a (linear connection) and a projection operator $\Phi$ satisfying certain linearity conditions are equivalent. $\endgroup$
    – Sebastian
    Commented Jun 23, 2020 at 5:56
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    $\begingroup$ Dear alexpglez98, a projection to the vertical bundle is equivalent to the projection to a complementary bundle, usually called the horizontal bundle. If you specify a section of the 1-jet bundle you get a horizontal bundle (as the image). This horizontal bundle is complementary to the vertical bundle and hence yields projections to the two summands. $\endgroup$
    – Sebastian
    Commented Jun 23, 2020 at 5:59
  • $\begingroup$ Thanks Ben and Sebastian. I understand now more or less what you say about the image. If $\Gamma:E \longrightarrow J^1E$ is a section, then $\Gamma(\xi)=j^1_p\phi$ for some section $\phi$. The image is $d\phi_p(T_pM)\subset T_\xi E$, it's well-defined because depends of the germ of $\phi$ at $p$ not on the section. But I don't see why this collection forms a horizontal subspace. $\endgroup$
    – alexpglez98
    Commented Jun 24, 2020 at 15:45

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