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Let $F=\mathbb{C}((t))$. Let $G=GL_n$. Then $G(F)$ acts on $\mathfrak{g}(F)$ by gauge transformation: $$ g.x:=gxg^{-1} + \dot{g}g^{-1},\quad \quad g\in G(F), \quad x\in \mathfrak{g}(F). $$ Here, $\dot{g}$ is the matrix obtained by differentiating the entries of $g$ (with respect to $t$). The gauge equivalence classes of elements of $\mathfrak{g}(F)$ can be understood geometrically as formal connections.

Here is an alternative way of understanding gauge equivalence. Let $\partial:=\frac{d}{dt}$ and let $$ \tilde{\mathfrak{g}}:=\mathfrak{g}(F)\oplus \mathbb{C}.\partial. $$ Then $\tilde{\mathfrak{g}}$ is a Lie algebra over $\mathbb{C}$ with bracket defined by $[x, \partial] = \dot{x}$ for all $x\in \mathfrak{g}(F)$. In the theory of affine Kac-Moody algebras, this is known as an extended loop algebra. Note that $\mathfrak{g}(F)$ identifies with the commutator subalgebra $[\tilde{\mathfrak{g}}, \tilde{\mathfrak{g}}]$. Similarly, we have the group $\tilde{G}$ and $G$ identifies with the commutator subgroup $[\tilde{G}, \tilde{G}]$.

Now consider the adjoint action of $\tilde{G}$ on $\tilde{\mathfrak{g}}$. Let us restrict this to an action of $G(F)$ on $\tilde{\mathfrak{g}}$. Then, it is easy to see that $$ g(x+\partial)g^{-1}=gxg^{-1}+\dot{g}g^{-1}, \quad \quad g\in G(F), \quad x\in \mathfrak{g}(F). $$ Thus we recover the gauge action by simply looking at the adjoint action of $G(F)$ on the set $\partial+\mathfrak{g}(F)$.

We can summarise the above discussion as follows: the adjoint quotient $(\mathfrak{g}(F)+\partial)/G(F)$ is the same as formal connections. Similarly, the set $(\mathfrak{g}(F)+\lambda\partial)/G(F)$ identifies with formal $\lambda$-connections. (In the sense of Deligne and Simpson - here $\lambda\in \mathbb{C}$). The upshot is that we have given a geometric meaning to the adjoint quotient $\tilde{g}/G(F)$.

Now note that $\tilde{G}$ acts on $\tilde{G}$ by the adjoint action. We can restrict this to an action of $G(F)$ on $\tilde{G}$.

Question: What is the geometric meaning of the quotient $\tilde{G}/G(F)$?

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    $\begingroup$ I don't think the Lie algebra you wrote down integrates to a group: the infinitesimal action of $\partial$ on $F$ corresponds to a translation which does not preserve the punctured disk. Usually one defines $\partial = td/dt$. In this case the Lie algebra integrates and $\tilde{G} / G(F)$ is the space of $q$-difference connections on the punctured disk, i.e. $G$-bundles $P$ together with an isomorphism $q^* P\cong P$. Here $q\in\mathbf{C}^\times$ is the parameter integrating $\partial$ and $q^* P$ is the translate of $P$ under the disk automorphism $t\mapsto qt$. $\endgroup$
    – Pavel Safronov
    Feb 11, 2016 at 10:22
  • $\begingroup$ Pavel, the OP indicates to me that if you were to write this as an answer, it would be gladly accepted. $\endgroup$
    – Todd Trimble ♦
    Mar 27, 2017 at 10:34

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