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Let $G$ a semisimple group over $k=\bar{k}$.

Let the $k$-indgroup, $LG^{-}\subset G(k[t^{-1}])$ be the kernel of the reduction. We know by Faltings that the multiplication map:

$LG^{-}\times G(k[[t]])\rightarrow G(k((t)))$ is an open immersion.

Is it still true, if we replace $LG^{-}$ by $LG^{-n}\subset G(k[t^{-1}])$, which is the kernel of the réduction mod $t^{-n}$?

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  • $\begingroup$ Why would we expect this to be true? Thinking, as naively as possible, of an element of $LGL_n$ as a sequence of matrices $(g_i)_{i \in \mathbbZ}$ where $g_i = I$ for almost all negative $i$, and there is some inverse sequence $(h_i)$, then Faltings's result tells us "most" such sequences should be of the form $((g_i)_{i < 0}, (g_i)_{i \ge 0})$, where there are respective inverses for the two parts: $(h'_i)_{i < 0}$ and $(h''_i)_{i \ge 0}$. But if this intuition is reasonable, there is no reason most sequences should be allowed to have a gap, i.e. $g_i=0$ for $i$ from $-n$ to $-1$ $\endgroup$
    – jackson
    Commented Aug 25, 2020 at 21:14

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