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Let $G$ be either a compact, simple, simply-connected Lie group, or a simply-connected complex reductive group (so either $SU(n)$ or $SL(n,\mathbb{C})$, for instance), or even complex affine. I don't know the correct setting for this question, so assume the case that makes it work. Please bear with the tentativeness of this question.

Consider the space $\mathcal{A}$ of polynomial or Laurent connections (again, take the option that works or makes sense) on the trivial $G$-bundle on $\mathbb{C}^\times$. This is something like an ind-scheme (perhaps only in the reductive case, and possibly only after picking a faithful representation). In the case of smooth connections on $G\times S^1$ (and probably for analytic too) we have a holonomy map $\mathcal{A}^{sm} \to G$, and this is a surjective submersion. I think there is also such a map for the case of $\mathcal{A}$, going around $S^1\subset \mathbb{C}^\times$.

In the event this exists, how algebraic is this map? Is it surjective?

My thought is that this may be a cover in some generalised sense. Maps from ind-schemes to schemes aren't so common, though that I know how to find how they work or what they do. There are some notes by Gaitsgory that don't help me at all.

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    $\begingroup$ I don't know the general answer, but would guess no. Think about the connection on $\mathbb{C}^\times$ with covariant derivative $\frac{\partial}{\partial x}-a/x$. This has holonomy around the circle given by $e^a$, which is about as transcendental as it gets. In general, anything that involves solving a differential equation (as holonomy does) is really bad news for staying algebraic. $\endgroup$
    – Ben Webster ♦
    Commented Aug 2, 2015 at 15:44
  • $\begingroup$ Hmm, I should throw log-algebraic into the mix as well, now you mention it. Not sure that fixes the example you mention, though... (which I don't quite understand: is G=C^×?) $\endgroup$
    – David Roberts ♦
    Commented Aug 2, 2015 at 17:08
  • $\begingroup$ @BenWebster that's doesn't look like a polynomial connection, though... :-) (perhaps it is: this is pushing at the boundaries of where I can make sense of the geometry.) $\endgroup$
    – David Roberts ♦
    Commented Aug 2, 2015 at 22:54
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    $\begingroup$ It's a Laurent polynomial connection! You're right that it doesn't completely kill things, but it was meant as an illustration of how "solve a differential equation" is an extremely non-polynomial operation. $\endgroup$
    – Ben Webster ♦
    Commented Aug 2, 2015 at 23:50
  • $\begingroup$ @BenWebster Note that a map of ind-schemes $\mathcal{A} \to G$ lives in the set $\lim_{i\in I}Sch(\mathcal{A}_i,G)$, where $\mathcal{A} = colim_{i\in I}\mathcal{A}_i$ in the category of ind-schemes and each $\mathcal{A}_i$ is a scheme. I may be wrong, but the exponential function is quite possibly an element of this set (I think one can take $I=\mathbb{N}$, and the $n^{th}$ coordinate as the $n^{th}$ Taylor polynomial of exp) $\endgroup$
    – David Roberts ♦
    Commented Aug 3, 2015 at 2:13

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Connections on the trivial $G$-bundle can be identified with maps from the base to $\mathrm{Lie}(G)$ via $x\frac{d}{dx} + f(x) \leftrightarrow f(x)$. One way to present this as an ind-scheme is by choosing generators of $\mathcal{O}_{\mathrm{Lie}(G)}$, and considering the finite dimensional affine space of ring maps $\mathcal{O}_{\mathrm{Lie}(G)} \to \mathbb{C}[x,x^{-1}]$ for which the generators land in $\bigoplus_{n=-N}^N \mathbb{C} x^n$.

Following Ben Webster's example in the comments, consider the tangent field $x \frac{d}{dx} - a$ for $a \in \mathrm{Lie}(G)$. This connection lies in the very first affine space of our sequence, with $N=0$. If $a$ is semisimple, then solutions to the equation $x\frac{df}{dx} = af$ look like "$cx^a$". The holonomy with respect to the basepoint $1$ is exponential in $a$ - in the one-dimensional case, identifying $\mathrm{Lie}(G)$ with $\mathbb{C}$, we get $e^{2 \pi i a}$.

In general, this example gives us an exponential map from $\mathrm{Lie}(G)$ to $G$. This (and more generally, the holonomy map) exists in the category of (ind) complex analytic spaces, but outside the case of unipotent groups it is not algebraic.

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  • $\begingroup$ Thanks for finally disabusing me of a vague hope! I guess, then, one can see that (for Lie groups of the sort I mention) the holonomy map is surjective. I would hope that this is some sort of submersion, regardless of its non-algebraic nature. $\endgroup$
    – David Roberts ♦
    Commented Aug 4, 2015 at 1:41
  • $\begingroup$ @David Roberts: I think the holonomy map is almost never a submersion. The exponential map is not an open map, even for compact Lie groups e.g. $SU(2)$, (see math.stackexchange.com/questions/301504/…), and not surjective in general for noncompact Lie groups (e.g. $SL_2(\mathbb C)$). $\endgroup$
    – Sam Gunningham
    Commented Aug 8, 2015 at 13:10
  • $\begingroup$ @SamGunningham in the smooth case it is! Namely, $\mathcal{A}^{sm} \to G$ is a surjective submersion for a Milnor regular Lie group (infinite-dimensional even!). I took Scott's use of the exponential map to be merely illustrating what holonomy looks like on a particular subspace of $\mathcal{A}$. $\endgroup$
    – David Roberts ♦
    Commented Aug 10, 2015 at 0:02

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