Let $G$ be a complex, connected, simply connected, semisimple group. I'm trying to compare the following two spaces: The free loop space $LG$ of $G$, and the $\mathbb C((z))$-valued points of $G$, $G(\mathbb C((z)))$, $$LG = \{ f:S^1 \to G : f \substack{\text{is the restriction of a} \\ \text{ holomorphic map }\mathbb C^\times \to G}\}.$$ $$ G( \mathbb C((z))) = \hom_{\text{Sch}_{\mathbb C}}(\operatorname{Spec} \mathbb C((z)), G).$$

In the way they are currently defined, they seem like they might be troublesome to work with, so let's fix an embedding $G \hookrightarrow GL_n(\mathbb C)$. The elements of $f\in LG$ then look like functions with Fourier series $$ f(z) = \sum_{k=-\infty}^\infty A_k z^k, \qquad A_k \in M_{n\times n}\mathbb C, \text{ and } f(z) \in G \text{ for all }z \in \mathbb C^\times;\tag{1}$$ that is, there is an analytic convergence condition on the element $f(z)$.

Is there a sense in which the (Lie group?) embedding $G \hookrightarrow GL_n(\mathbb C)$ corresponds to a map of affine group schemes $G \hookrightarrow GL_n$ which agrees with $G(\mathbb C) \hookrightarrow GL_n(\mathbb C)$?

If this is the case, then we should get a corresponding map $G(\mathbb C((z))) \hookrightarrow GL_n(\mathbb C((z)))$, and we can then write elements $f\in GL_n(\mathbb C((z)))$ in a similar sense to those in (1); namely, $$f(z) = \sum_{k=-\infty}^\infty A_k z^k, \qquad A_k \in M_{n\times n}\mathbb C. \tag{2}$$ Such sums in this context seem to be taken in a strictly formal context. Nonetheless, it seems as though there should be some ``spiritual'' notion of convergence to guarantee that the above sum lives in not only $GL_n(\mathbb C((z)))$ but actually in $G(\mathbb C((z)))$ rather than simply $M_{n\times n}\mathbb C((z))$.

Is there a notion of convergence for elements of the form (2)? How does one write this down? Does it agree with that of (1)?

Edit: @S. Carnahan has been good enough to point out an algebraic condition which implies precisely when elements of $GL_n(\mathbb C((z)))$ are in $G(\mathbb C((z)))$.

However, I still have a strange suspicion there should be some sort of analytic condition. To corroborate this, let us consider Ginzburg's paper Perverse sheaves on a loop group and Langlands' duality. On page 6, he discusses a finite dimensional analogue $G(\mathbb C[z,z^{-1}])$ where elements, given the embedding above, have the form $$f(z) = \sum_{k=-m}^m A_k z^k, \qquad A_k \in M_{n\times n}\mathbb C, \text{ and } f(z) \in G(\mathbb C) \text{ for all }z \in \mathbb C^\times.$$ How are these (finite) sums still endowed with a convergence property? In particular, we note that $z$ is taken to be in $\mathbb C^\times$ rather than just a formal sum satisfying some polynomial conditions which come from the embedding. Is it really the finiteness of the summation which somehow makes this possible? Why would that break in an infinite sum?

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    $\begingroup$ You may choose your embedding to be a closed immersion defined by polynomial relations with coefficients in $\mathbb{C}$. Then formal series for an element of $GL_n(\mathbb{C}((t)))$ corresponds to an element of $G(\mathbb{C}((t)))$ if and only if it satisfies the same polynomial relations. $\endgroup$ – S. Carnahan Jan 21 '14 at 23:12
  • $\begingroup$ For $\mathbb{C}[z,z^{-1}]$, you get the same polynomial relations. Indeed, the properties of your Lie group imply algebrizability, and the functor of points gives the same relations for any commutative ring over $\mathbb{C}$. $\endgroup$ – S. Carnahan Jan 23 '14 at 7:14
  • $\begingroup$ @S.Carnahan Forgive me if I am misunderstanding you, but it is not clear to me that this answers my question. I believe I understand what you mean by the polynomial relations, but where does this manifest in the $G(\mathbb C[z,z^{-1}])$ example? Is it subtly the $f(z) \in G(\mathbb C)$ condition? $\endgroup$ – Tyler Holden Jan 27 '14 at 17:37
  • $\begingroup$ The finite sums in $G(\mathbb{C}[z,z^{-1}])$ are unconstrained by any convergence condition - convergence is a property of infinite sums. I'm afraid this doesn't answer your main question. $\endgroup$ – S. Carnahan Feb 1 '14 at 22:55
  • $\begingroup$ There is an obvious thing that breaks in passing from $\mathbb{C}[z^\pm]$ to $\mathbb{C}((z))$: the former has ring homomorphisms to $\mathbb{C}$ and the latter doesn't. On the other hand $G(\mathbb{C}((z))) = G(\mathbb{C}[z^{-1}])T(\mathbb{C}[z^\pm])G(\mathbb{C}[[z]])$. You can talk about convergence for $G(\mathbb{C}[[z]])$ in the sense that you can ask if $g \in GL_n(\mathbb{C}[[z]])$ lands in $G(\mathbb{C}[z]/z^n) \subset GL_n(\mathbb{C}[z]/z^n)$ for all $n$. $\endgroup$ – solbap Apr 14 '14 at 7:12

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