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Let $k$ be a field of characteristic zero, $X=\mathbb{G}_{m, k}=\mathrm{Spec}\ k[t, t^{-1}]$ the multiplicative group over k and $E=\mathcal{O}_X$ the trivial line bundle.

Consider the connection $\nabla: E \to E \otimes_{\mathcal{O}_X} \Omega^1_X$ defined by $\nabla(1)=\alpha \frac{dt}{t}$ for some $\alpha \in \mathbb{C}$, that is:

$\nabla=d+\alpha \frac{dt}{t}\wedge$

My question is: has $(E, \nabla)$ regular singularities?

Thanks for you help

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  • $\begingroup$ p^4: I didn't downvote you and yes it does. $\endgroup$ Commented Feb 5, 2013 at 16:19
  • $\begingroup$ Dear Donu, could you explain why ? It seemed pretty obvious to me but then I found the paper mi.fu-berlin.de/users/esnault/preprints/helene/… page 2 and I was confused about the statement on sections $\endgroup$
    – pppp
    Commented Feb 5, 2013 at 16:39
  • $\begingroup$ It's obvious if you can use the standard Fuchs criterion that the singularities of the connection are logarithmic (cf Deligne's book). I get $t^s$ as a basis of horizontal sections in the example in the linked preprint. I'm sure they mean something else, but I don't have the time to read it carefully right now. $\endgroup$ Commented Feb 5, 2013 at 16:57
  • $\begingroup$ That's the point. I don't see what exp(-t) has to do with horizontal sections $\endgroup$
    – pppp
    Commented Feb 5, 2013 at 17:43

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Dear $p^4$,

I had a chance to glance further at the 4 author paper linked in your comment. Their connection form is $-dt + sdt/t$. This has double pole at $\infty$, so it isn't regular*. However, the connection in your question has logarithmic singularities at $0$ and $\infty$, so it is regular, as I said earlier. There is no contradiction.

*(Afterthought) This is a bit sloppy, since the Fuchs criterion is only a sufficient condition for regularity. But to see the non regularity, observe that the solution they write down $exp(t)t^{-s}$, which is correct (!), has bad singularities at $\infty$.

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