To any space $X$ you can associate its de Rham space $X_{dR}$. Vector bundles on $X_{dR}$ are the same thing as vector bundles on $X$ with a flat connection.

Can anything like this be said for meromorphic connections?

For instance, a naive idea is that there might literally be a space $X_{mdR}$ whose vector bundles are vector bundles on $X$ with a flat meromorphic connection. Then there would be maps $\eta_{dR}\to X_{mdR}\to X_{dR}$, where $\eta$ is the generic point of $X$ (on which all meromorphic bundles are holomorphic), suggesting that maybe $X_{mdR}$ could be constructed as some sort of intermediate object between $\eta_{dR}$ and $X_{dR}$.

I'm aware that Deligne has a structure theorem for regular meromorphic connections, so maybe it's more reasonable to restrict to the regular meromorphic case.

  • $\begingroup$ Did you mean to draw your arrows the other way? I see the map $\eta_{dR} \to X_{dR}$ but not vice versa. $\endgroup$ – Will Sawin Jan 13 at 19:26
  • $\begingroup$ @WillSawin Whoops, thanks for pointing that out! $\endgroup$ – Meow Jan 13 at 20:47
  • $\begingroup$ Have you tried to interpret $X_{dR}$ as a groupoid? $\endgroup$ – Leo Alonso Jan 14 at 12:06
  • $\begingroup$ For the Riemann surface case, you should have a look at the papers/work of Philip Boalch. He studies both the regular and irregular case, and develops a pretty extensive theory. $\endgroup$ – Andy Sanders Jan 15 at 22:45

Here is one way to construct an object $X_{mdR}$ for $X$ say, a complex variety. Recall that $X_{dR}$ can be constructed as follows. Let $\hat{X}$ be the formal completion of the diagonal inside $X\times X$. Then $X_{dR}$ is the quotient of $X$ by the equivalence relation defined by $\hat{X}$.

Let $\eta$ denote the generic point of $X$. We can define an affine scheme $\hat{\eta}'$ as the inverse limit of the $\hat{U},$ as $U$ ranges over all affine opens in $X$. (I believe, though I didn't carefully check, that this is not the formal completion of $\eta$ inside $\eta\times\eta$.) It comes with two natural maps $p_1,p_2:\eta'\rightarrow\eta$ making $\hat{\eta}'$ an equivalence relation. Then we can define $X_{mdR}$ to be the quotient of $\eta$ by $\hat{\eta}'$.

A vector bundle on $X_{mdR}$ is a vector bundle $V$ on $\eta$ with an isomorphism $p_1^*V\cong p_2^*V.$ We can find an open $U_1$ so that $V$ can be defined over $U_1$, and we can find a smaller open $U_2\subseteq U_1$ so that the isomorphism $p_1^*V\cong p_2^*V$ is defined over $\hat{U_2}$. (This follows from the general framework of "Noetherian approximation", but can also be seen explicitly.) Thus $V$ comes from a vector bundle with connection on $U_2$.

That being said, I'm not convinced that this is a good way of thinking about meromorphic connections. My experience is that in non-finite type settings, usually trying to work with $D$-modules via constructions like the de Rham stacks runs into issues fairly quickly. For starters, I'm not sure if this works well in families, i.e., if there is any nice interpretation of vector bundles on $A\times X_{mdR}$.

  • $\begingroup$ Are you claiming that vector bundles on your prestack are the same thing as vector bundles with connection on an open subset of $X$? The concern I would then have is that these do not have a unique extension to a meromorphic connection on all of $X$ (e.g. by Deligne's theorem). $\endgroup$ – Meow Jan 16 at 13:31
  • 1
    $\begingroup$ Ah OK, I misunderstood. I assumed by meromorphic connection you simply meant a $\mathcal{D}$-module which is intermediate extended, but you actually want that plus the data of a lattice. Let me think and see if I can say anything about that case. I am tempted to just take the coproduct of $X_{mdR}$ and $X$ over $\eta$ - let me think if that works... $\endgroup$ – dhy Jan 16 at 13:37
  • $\begingroup$ What would the maps $X_{mdR}\to X,\eta$ be? I agree it feels like something like this should work. $\endgroup$ – Meow Jan 17 at 13:36
  • $\begingroup$ @Meow I wanted to take the pushout along the maps $\eta\rightarrow X,X_{mdR}$. Sorry, I got distracted; I'll try to make some time to think about this tomorrow. $\endgroup$ – dhy Jan 20 at 0:49
  • 1
    $\begingroup$ @Meow I am now very confused. Are you working in the algebraic or the analytic world? If you are working in the analytic world then I agree with what you are saying, but then I don't know how to make sense of even the normal de Rham stack $X_{dR}$. If you are working in the algebraic world then I see no difference between meromorphic connections and $D$-modules on $U$. (In particular, $\mathcal{O}_X(D)$ and $\mathcal{O}(U)$ coincide in the algebraic world no?) $\endgroup$ – dhy 22 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.