Good question.

Every abstract algebraic group over a perfect field (such as ) is an extension of an abelian variety by an affine (=linear) algebraic group (Theorem of Barsotti–Chevalley). On the other hand, metaplectic covers of groups such that don’t surject on non-trivial compact Lie groups.

Now abelian varities are projective so their sets of real or complex points are compact. It follows that if a metaplectic group was the set of real points of an algebraic group it would be the set of real points of an affine algebraic group.

]]>As written for Example 2, I’m not sure the adjoint representation is the identity as you claim:

while Ad X indeed has the matrix representation of X, it looks like

Ad Y actually has the matrix representation of -Y (since [Y, X] = -Y and [Y, Y] =[Y, Z] = 0, the matrix representation for Ad Y should have a -1 where the matrix of Y has 1)

and similarly Ad Z has the matrix representation of -alpha Z.

That, though, would be fixed if we replaced Y with -Y and Z with -alpha Z in the original definitions, though: is this right?

*[Corrected, thanks – T.]*

The circle group S^1, which **has** R as its Lie algebra…

*[Corrected, thanks – T.]*

*[Corrected, thanks – T.]*

Thanks Allen! I have one stupid question; I thought it was only the affine algebraic groups that necessarily had finite-dimensional representations; does the above argument rule out the metaplectic group being an abstract algebraic group rather than an affine one?

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