Principal bundle over Riemann surfaces

Question

I have come across a paper which does not make explicit the notation they're following and I am a little bit confused. The paper in question is the following one https://www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0778-04.pdf. In this paper the author considers a principal bundle $P\rightarrow R$ where $R$ is a Riemann surface and in section $2$, it defines a $k$-th order trivialization of $P$ at a point $Q$ as an isomorphism $$O_{R}(P)\otimes O_{C,Q}/\mathfrak{m}_{Q}^{k+1}\simeq G(\mathbb{C}[\xi]/\xi^{k+1})$$ My question is: what does $O_{R}(P)$ mean? And what is the notation $G(\mathbb{C}[\xi]/\xi^{k+1})$? Thank you very much for your time and support.

Answer 1

Here are guesses (didn't look into the paper). $G$ is the Lie group (for the definition of a principal $G$-bundle one should fix a Lie group $G$). But probably here $G$ is a nice matrix group/algebraic group, say something like $SL_n (\mathbb{R})$. Then you can define $G(A)$, the $A$-valued points of $G$, for an $\mathbb{R}$-algebra $A$, so in the example, those are $n\times n$-matrices over $A$ with determinant $1$. So concretely, those are like "jets" for $A = \mathbb{C} [\zeta] / \zeta^{k+1}$ (giving a Taylor expansion up to degree $k$, at a point, roughly). Then $O_R (P)$, I guess, is the frame bundle (I think it is called like that) - depending on the definitions, it is simply the principal $G$-bundle (or a "$G$-torsor" etc., everything depends on the definitions). So basically, the expression on the left is like sections of your frame bundle, at the "$k$-th formal neighbourhood" of your point $Q$, and the isomorphism with the thing on the right "trivializes" this section.

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For example of $G(\mathbb{C}[\zeta] / \zeta^{k+1})$, consider $G = GL_n$. Then to imagine an element of $G(\mathbb{C}[\zeta] / \zeta^{k+1})$ explicitly, you imagine an $n \times n$-matrix of complex polynomials in $\zeta$, but where you disregard powers of $\zeta$ which are $k+1$ and above (i.e. you consider two entries the same if their difference is divisible by $\zeta^{k+1}$, so you work in the quotient algebra), and such that the determinant (a complex polynomial in $\zeta$) is not divisible by $\zeta$. Plugging in $\zeta := 0 $, you obtain simply a complex invertible matrix - that is your "zeroth Taylor coefficient". But this whole matrix gives you a whole Taylor expansion to order $k$.

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I think it is good to first understand $k = 0$. You imagine $O_R (P)$ as consisting of sections of a geometric object over $R$, each fiber of which is a $G$-torsor (a free and transitive $G$-space). Then the tensor product on the left will be simply the fiber over $Q$, i.e. a single $G$-torsor. Then isomorphism with $G(\mathbb{C})$ on the right will mean that you trivialized your $G$-torsor, i.e. picked a "base point", picked a point. So for $k = 1$, a trivialization at the point $Q$ is simply choosing a point in the fiber over $Q$ of the prinipal $G$-bundle at question.

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Maybe you can proceed as follows ($G$ is some nice complex algebraic group). First, understand what is a principal $G$-bundle on $R$. It is a complex manifold $E$ with an holomorphic map to $R$, and a holomorphic $G$-action on $E$, such that locally it is isomorphic to the projection $G \times R \to R$... Then $O_R (P)$ you can interpret as the holomorphic sections of $E \to R$. Then the left hand side you can interpret as sections locally at $Q$, modulo an equivalence relation that two sections are equal if their value, as well as derivatives up to $k$-th, at $Q$, are equal. The right hand side should be better called $G(O_{R,Q}/ \mathfrak{m}^{k+1}_Q)$, and it is the same as the left hand side, but for the trivial principal $G$-bundle, i.e. for the projection $G \times R \to R$. The isomorphism should respect the $G$-action.