The scheme-theoretic flow-in locus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T20:58:22Z http://mathoverflow.net/feeds/question/55569 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55569/the-scheme-theoretic-flow-in-locus The scheme-theoretic flow-in locus Nicholas Proudfoot 2011-02-16T00:45:11Z 2011-02-17T00:16:09Z <p>Let $R$ be a ring with an $\mathbb{N}\times\mathbb{Z}$-grading. The $\mathbb{N}$-grading allows you to construct the scheme $X = \operatorname{Proj} R$, and the $\mathbb{Z}$-grading defines an action of $T = \mathbb{G}_m$ on $X$.</p> <p>Let $S$ be an $\mathbb{N}\times\mathbb{Z}$-graded quotient of $R$ such that every element of $R$ whose $\mathbb{Z}$-degree is nonzero vanishes in $S$, and let $Y = \operatorname{Proj} S$. Then we have an inclusion from $Y$ into the fixed locus $X^T$. $T$ acts trivially on the tautological bundle of $Y$, which is equal to the restriction of the tautological bundle of $X$. Assume that $X$ and $Y$ are smooth and that $Y$ is a connected component of $X^T$.</p> <p>Let $R^+ \subset R$ be the subring generated by elements of non-negative $\mathbb{Z}$-degree, and let $V = \operatorname{Proj} \left(R\;\otimes_{R^+}S\right)\subset X$.</p> <p>Morally, $V$ should be thought of as the locus of points $x$ with the property that $\lim_{t\to 0}t\cdot x \in Y$, though in fact the support of $V$ may be strictly bigger than this set. For example, let $X= T^*\mathbb{P^1}$, where the action of $T$ on $X$ is induced from the standard action on $\mathbb{P^1}$. Then $X$ has two fixed points which I'll call $N$ and $S$ (for the North and South poles). If we take $Y$ to be $N$, then $V$ will be the cotangent fiber over $N$. If we take $Y$ to be $S$, then $V$ will be the union of the cotangent fiber over $N$ with the zero section (even though the set-theoretic flow-in locus consists only of the zero-section minus $N$). In particular, $V$ need not be irreducible.</p> <p>More generally, if $X = T^{*}(G/B)$ and $T$ acts on $X$ via a generic cocharacter of $G$, $V$ will be a union of conormal varieties to Schubert strata that appear with various multiplicities. In particular, $V$ need not be reduced.</p> <p>The example that most interests me is when $X$ is the family over $\mathbb{A}^1$ with special fiber $T^{*}(G/B)$ and general fiber $G/H$, where $H$ is a maximal torus of $G$. This family is $U(1)$-equivariantly diffeomorphic (but not isomorphic) to $T^{*}(G/B)\times \mathbb{A}^1$, where $U(1)\subset T$ acts trivially on $\mathbb{A}^1$. Thus the components of $X^T$ are images of sections of the map from $X$ to $\mathbb{A}^1$. In this example $V$ will always be irreducible and reduced, even though its intersection with $T^{*}(G/B)$ is neither irreducible nor reduced (see the previous paragraph). </p> <p>What I would like to know is that if I take $X$ to be a variety like this (a twistor families for a "nice" symplectic variety on which $T$ acts hamiltonianly with isolated fixed points), then $V$ will be reduced. In an attempt to get this, I'll try posing the following general question:</p> <blockquote> <p>Suppose that $V$ is irreducible and generically reduced. Does it follow that $V$ is reduced?</p> </blockquote>