# The scheme-theoretic flow-in locus

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 $X$.

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 $X^T$.

Let $R^+ \subset R$ be the subring generated by elements of non-negative $\mathbb{Z}$-degree, and let $V = \operatorname{Proj} \left(R\;\otimes_{R^+}S\right)\subset X$.

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.

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.

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).

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:

Suppose that $V$ is irreducible and generically reduced. Does it follow that $V$ is reduced?

• Might you hope that your $V$ is always Cohen-Macaulay? (I expect it's no coincidence that you and Ben are asking similar questions...) – Dave Anderson Feb 16 '11 at 15:52
• Also, the requirement that all nonzero-degree elements vanish in $S$ seems a little confusing. E.g., one version of the "standard" action on ${\Bbb P}^1$ has $R=k[x,y]$, with degrees $(1,-1)$. What could $S$ be? I think you need to take the irrelevant ideal of $S$ into account. – Dave Anderson Feb 16 '11 at 15:56
• @Dave: As you say, let $R = k[x,y]$, where $x$ and $y$ both have $\mathbb{N}$-degree 1, so that $X = \operatorname{Proj} R = \mathbb{P}^1$. To define a $T$-action on $X$, we need to choose the $\mathbb{Z}$-degrees of $x$ and $y$. If you want to get the standard action, you should let $x$ have degree 1 and $y$ have degree $0$. Then you can let $S = k[t]$, where $t$ has $\mathbb{N}$-degree 1 and $\mathbb{Z}$-degree 0, and the map from $R$ to $S$ sending $x$ to $0$ and $y$ to $t$ induces the inclusion of one of the two fixed points. – Nicholas Proudfoot Feb 16 '11 at 17:16
• @Dave continued: If you want to get the inclusion of the other fixed point, you should let $x$ have degree 0 and $y$ have degree $-1$, and then define your map from $R$ to $S$ by sending $x$ to $t$ and $y$ to $0$. These two choices of $\mathbb{Z}$-grading on $R$ induce the same $T$-action, but with a different equivariant structure on the tautological line bundle. With my conventions, the restriction of the tautological line bundle from $X$ to $Y$ will have the trivial $T$-action on every fiber. – Nicholas Proudfoot Feb 16 '11 at 17:20
• @Dave continued again: If you choose the $\mathbb{Z}$-degrees of $x$ and $y$ to be 1 and $-1$ as you say, then you get the square of the standard action. Furthermore, you get it with a $T$-structure on the tautological line bundle such that $T$ acts nontrivially on the fibers at both fixed points. Instead, you should assign degrees $(2,0)$ or $(0,-2)$, depending on which of the two fixed points you want to look at. Finally, you're right, it's no coincidence that Ben and I are asking similar questions! – Nicholas Proudfoot Feb 16 '11 at 17:23