Timeline for Expressing properties of graded algebras in terms of the $\mathbb{G}_m$action
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2017 at 22:19 | comment | added | Dylan Wilson | Julian- this is why I said "from the point of view of functions" but you're absolutely right that saying transitive is probably too misleading. | |
| Jan 13, 2017 at 22:16 | comment | added | Denis Nardin | For example it could mean that $V/\mathbb{G}_m$ is proper (take $V$ to be the punctured cone on a projective variety) | |
| Jan 13, 2017 at 21:56 | comment | added | Julian Rosen | Assuming the grading is non-negative, the action extends to the monoid $\mathbb{A}^1$. The action of $0\in\mathbb{A}^1$ on $A$ is projection onto $A_0$, so $A_0=k$ iff the action of $0\in\mathbb{A}^1$ on $V$ maps everything to a point. | |
| Jan 13, 2017 at 21:38 | comment | added | Saal Hardali | @JulianRosen Oh I see. What about "no orbit is a divisor"? | |
| Jan 13, 2017 at 21:31 | comment | added | Julian Rosen | Consider $V=\mathbb{A}^n$ with the standard action. The action is very far from transitive, and most of the orbits are infinite, but the invariant are still just the constants. | |
| Jan 13, 2017 at 21:27 | comment | added | Saal Hardali | @JulianRosen Could it be interpreted as being transitive after we throw away all the finite orbits? | |
| Jan 13, 2017 at 21:24 | comment | added | Julian Rosen | One has to be cautious about picturing the action as transitive. The condition isn't that $V/\mathbb{G}_m$ is a point, just that this quotient doesn't have any non-constant global functions. For instance, $V$ can have arbitrarily high dimension, so the action might be very far from transitive. | |
| Jan 13, 2017 at 21:18 | vote | accept | Saal Hardali | ||
| Jan 13, 2017 at 21:18 | comment | added | Saal Hardali | This is so simple! So basically the conditions together mean that $V$ has a "transitive" action of $\mathbb{A}^1$ s.t. the natural map to the vector space dual to linear functions is an embedding. Thank you! | |
| Jan 13, 2017 at 21:08 | comment | added | Dylan Wilson | I should mention that another justification for viewing $A_0$ as a model for $V/\mathbb{G}_m$ is that $A_0$ is precisely the primitives for the coaction of the Hopf algebra representing $\mathbb{G}_m$; and more generally the primitives of a Hopf algebra action are like a model for geometrically taking the quotient in a well-behaved way. | |
| Jan 13, 2017 at 21:07 | comment | added | Dylan Wilson | looks like some people in the comments beat me to numbers 1 and 2 | |
| Jan 13, 2017 at 21:06 | history | answered | Dylan Wilson | CC BY-SA 3.0 |