I'm trying to get my head around the notion of a connection on a ''space'' $X$ over some field $k$. (I am mainly interested in connections in algebraic geometry as they can be used to define the de Rham realisation of the fundamental group, but please feel free to assume $X$ is a complex manifold or some other suitable space as this is about intuition.)

What I have heard about connections is that they ''connect'' the fibres of some bundle $\mathcal{E}$ over $X$ i.e. if $\mathcal{E}$ is a locally free sheaf on $X$ and there is a path $x\to y$ in $X$ between two points then there should be a morphism $\mathcal{E}_x\to\mathcal{E}_y$ of the stalks over these two points (as far as I understand this is parallel transport). This allows us to talk about monodromy representations of the fundamental group of $X$ etc.

But the definition I have read (e.g. in Szamuely's book Galois Groups and Fundamental Groups) is that a connection is a pair $(\mathcal{E},\nabla)$ where $\mathcal{E}$ is locally free over $X$ and $\nabla$ is a morphism of sheaves of $k$-vector spaces

$$\nabla: \mathcal{E}\to\mathcal{E}\otimes_{\mathcal{O}_X} \Omega^1_X$$

satisfying the Leibniz rule $\nabla(f m) = m\otimes df + f \nabla(m)$ for $f$ a section of $\mathcal{O}_X$ and $m$ a section of $\mathcal{E}$.

How are these two notions equivalent, and is there any further intuition about either one?

  • $\begingroup$ Connections also have a place in crystalline geometry where they can be seen as crystals on the infinitesimal site (restrict ourselves to char 0 geometry for this statement). If you want to hear more about this, I can write something :) $\endgroup$ – Alex Youcis Nov 7 '16 at 19:35
  • $\begingroup$ @AlexYoucis I'd be really happy to see connections appearing in any algebraic context, although at the moment I'm not really working with crystalline geometry, but if you don't mind writing something briefly about this I'd be very grateful to see it! $\endgroup$ – Alex Saad Nov 7 '16 at 19:58
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    $\begingroup$ Every single person involved in this post right now is named Alex $\endgroup$ – Alex Mathers Nov 7 '16 at 20:27
  • $\begingroup$ @AlexMathers this is becoming very uncanny! $\endgroup$ – Alex Saad Nov 7 '16 at 20:38
  • $\begingroup$ @AlexMathers I think it's about time we established a pecking order. Naturally I'm the head Alex :) $\endgroup$ – Alex Youcis Nov 22 '16 at 13:25

These two notions are equivalent. If you have connection $$ \nabla: \mathcal{E}\to\mathcal{E}\otimes_{\mathcal{O}_X} \Omega^1_X $$ and choose a path $\gamma(t): I=[0,1] \to X$ in $X$ you can define a parallel transport of a section $s$ of $\mathcal{E}$ as solution of the differential equation $$ \nabla_{\dot \gamma(t)} s (\gamma(t))=0, $$ satisfying initial condition $s(\gamma(0))=s_0$. Here $\nabla_{\dot \gamma(t)}$ is the substitution of the vector field $\dot \gamma$ in the 1-form $\nabla s$.

A 1-form is a section of the cotangent bundle $\Omega$, which is dual to the tangent bundle. You can always compute (or evaluate) section of the dual vector bundle on a section of a bundle and get a function. More generally, if you consider a 1-forms with values in a vector bundle (i.e. $\mathcal{E} \otimes \Omega$) you get a section of this bundle when substitute a vector field in this form.

In the other direction, if you have a notion of the parallel transport along a curve $\gamma$ in $X$ you can move section by some $\epsilon$ and take limit just as for ordinary derivatives, this will define for you $\nabla_{\dot \gamma(t)}$. Then, if you have any vector field $v$ on $X$ find an integral trajectory $\gamma$ of that vector field passing through a given point $x \in X$ $$ \frac{d}{dt} \gamma(t) = v(\gamma(t)) $$ and $\gamma(0)=x$. Use previous step to define $\nabla_{v}s=\nabla_{\dot \gamma(t)} s$. Now you have a map for any vector field on $X$, therefore you obtained a 1-form $\nabla s$ with value in $\mathcal{E}$.

  • $\begingroup$ Dear Alex, thank you for a great answer. Could you just elaborate on (for someone not well versed in differential geometry) what you mean by the substitution of a vector field into a 1-form? $\endgroup$ – Alex Saad Nov 7 '16 at 19:55
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    $\begingroup$ I've added an explanation for this. Do you need more details? $\endgroup$ – Alex Nov 7 '16 at 20:19
  • $\begingroup$ This is great, thank you! $\endgroup$ – Alex Saad Nov 7 '16 at 20:37

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