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Feb 26, 2010 at 14:20 comment added Anatoly Preygel Vague as I was, I realize I was wrong on how to topologize it (and this might help?). It should be topologized as with the subspace topology (viewed as pairs $(b,b^{-1})$) inside $\mathbb{C}((t))^2$, where this latter is (I think) given the topology I was describing. I can't tell if this helps, but it might.
Feb 26, 2010 at 14:04 comment added Anatoly Preygel The algebraic loop space is not the base change but the restriction of scalars from $\mathbb{C}$ to $\mathbb{C}((t))$. So, its $\mathbb{C}$ points are by definition $\mathbb{G}_m(\mathbb{C}((t))=\mathbb{C}((t))^{\times}$: the non-zero Laurent series. As to how we topologize it, my guess is as a direct limit of product topologies. Both $L \mathbb{G}_m$ and this have a subspace homeomorphic to $\mathbb{Z} \times \mathbb{G}_m$: loops of the form $z \mapsto t z^n$ for some $(n,t)$. The inclusion is an equiv. for the topological one, what about the alg. geom. one?
Feb 26, 2010 at 13:47 history edited Chris Schommer-Pries CC BY-SA 2.5
Added a few more insights.
Feb 26, 2010 at 13:19 history answered Chris Schommer-Pries CC BY-SA 2.5