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Let $\mathbb{G}_m$ be the multiplicative group and $T$ a maximal torus of a semisimple group. Let $X^*(T)=\{ \phi: T \to \mathbb{G}_m \}$ be the set of characters and $X_*(T)=\{ \phi^{\vee}: \mathbb{G}_m \to T \}$ the set of co-characters. Let \begin{align} L(T) = \{ \phi: \mathbb{G}_m \to T \}, \end{align} and \begin{align} L_0(T) = \{ \phi \in L(T): \forall \mu \in X^*(T), \mu \circ \phi \in L_0(c) \}, \end{align} where $L_0(c)$ is the set of all invertible Taylor series in $c$.

In the paper Geometric and unipotent crystals, Section 2.4, page 7, Lemma 2.6, it is said that the multiplicative map $X_*(T) \times L_0(T) \to L(T)$ is a bijection.

How to show that an element $\phi$ in $L(T)$ can be decompose as a product $\phi = \phi' \phi''$, where $\phi' \in X_*(T)$ and $\phi'' \in L_0(T)$? Thank you very much.

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Although the notation and explanation of the paper is terrible, it's not supposed to be a very deep fact.

Indeed, let's assume for simplicity that $T = \mathbb G_m^n$ is a split torus. Then arbitrary morphisms of schemes $\mathbb G_m \to T$ correspond to $n$-tuples $(f_1,\ldots,f_n) \in (k[x^{\pm 1}]^\times)^n$. Under this correspondence, the characters correspond to the monomials $$\{(x^{a_1},\ldots,x^{a_n})\ |\ a_1,\ldots,a_n \in \mathbb Z \}.$$ What the authors mean by formal loops $\phi \colon \mathbb G_m \to T$ is morphisms $\operatorname{Spec} k((x)) \to T$, where $\operatorname{Spec} k((x))$ can be thought of as $\operatorname{Spec}\left(k[x^{\pm 1}] \otimes_{k[x]} k[[x]]\right)$ (i.e. $\mathbb G_m \times_{\mathbb A^1} \hat{\mathbb A}^1$; the 'formal completion of $\mathbb G_m$ at the origin'). This is an algebraic geometry version of looking at very small loops.

Algebraically, the elements of $L(T)$ are given by $n$-tuples $(f_1,\ldots,f_n) \in (k((x))^\times)^n$. The condition on $L_0(T)$ is that the $f_i$ are invertible elements of the power series ring $k[[x]]$.

Finally, the claim they make just boils down to the statement that any $(f_1,\ldots,f_n) \in (k((x))^\times)^n$ can be written uniquely as a product $$(f_1,\ldots,f_n) = (g_1,\ldots,g_n) \times (h_1,\ldots,h_n),$$ where the $g_i$ are monomials (coming from $X_*(T)$) and the $h_i$ are invertible elements of $k[[x]]$ (coming from $L_0(T)$). Explicitly, if $$f_i = \sum_{j\in\mathbb Z} a_j x^j,$$ then $g_i = x^{j_0}$ for the minimal $j_0$ such that $a_{j_0} \neq 0$, and $$h_i = \frac{f_i}{g_i} = \sum_{j = 0}^\infty a_{j+j_0} x^j$$ is invertible in $k[[x]]$ since its constant term $a_{j_0}$ is nonzero.

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  • $\begingroup$ thank you very much. I have question about $Spec k((x))$. I think that $k((x))$ is a field. Therefore $Spec k((x))$ is just $\{0\}$? $\endgroup$ Commented May 7, 2018 at 14:14
  • $\begingroup$ @JianrongLi: well, set-theoretically it has just one point, but writing $\{0\}$ is a bit suggestive (one might think that the point is a $k$-point, which is very false). $\endgroup$ Commented May 7, 2018 at 16:11
  • $\begingroup$ thank you very much. How to think Spec $k((x))$ as $k[x^{\pm 1}] \otimes_{k[x]} k[[x]]$? $\endgroup$ Commented May 9, 2018 at 8:36
  • $\begingroup$ The ring $k((x))$ is the fibre product: to tensor something over $k[x]$ with $k[x^{\pm 1}]$ just means inverting $x$. $\endgroup$ Commented May 9, 2018 at 14:13
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Your definitions are incorrect --- $L(T)$ should be the set of formal loops and $L_0(T)$ the set of formal disks. With the correct definitions, the statement is obvious.

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