Andrew Dudzik
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Registered User
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15h |
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Why do rigid spaces have “not enough points”? Two more useful examples: 1) A descending sequence of discs with empty intersection gives another such skyscraper sheaf (which also comes from a Berkovich point). 2) There are (non-"overconvergent") examples that don't come from Berkovich points, but do come from points in Huber's adic space. For example, let $F(V)=\mathbb{Z}$ if $V$ contains the open unit disc with finitely many open discs of radius $<1$ removed, and $F(V)=0$ otherwise. One way to think of this second example is: the skyscraper sheaf at the origin of the canonical reduction, $\mathbb{A}^1_{\tilde{k}}$. |
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Dec 5 |
awarded | ● Popular Question |
