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Consider some Lie algebra $g$. Consider corresponding loop Lie algebra $g((t))$.

Question What can be said about the center of $\tilde U (g((t))$ in terms of the center of $U(g)$ ?

(Where by $\tilde U$ is "informally" universal enveloping of $g((t))$, however technically (guided by example of reductive g) we should allow ourselves some central extension of $g((t))$ and some completion (see details below) ).

Expected answer (which is known to be true for reductive g) assume center of $U(g)$ is generated by some elements $C_i$, then the center of $\tilde U (g((t))$ is generated by elements $C_{i,k}$ k=$-\infty, \infty$. Where symbols of this $C_{i,k}$ are easy to write...

Is this naive answer true/known ?

Some background/evidences

1) If $g$ is reductive Lie algebra then by results of Goodman-Wallach, Hayshi, Feigin-Frenkel one should make central extension of g((t)) by element "c", then put this "c" equal to some constant ("critical level") and corresponding universal enveloping algebra will have the center of expected size.

So basically I want to say that for reductive $g$ the answer to my question is positive.

2) If we consider not $U(g((t)))$, but consider $S(g((t))$ and its invariants. It seems quite easy to see that everything will also be Okay - and we do not need any central extension. (Modula I am not mistaking in some considerations).

3) If we might hope that Duflo map can be extended to loop Lie algebras then (2) will imply positive answer. However it is not clear about whether such extension is possible...

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