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The following observation between the spaces of global sections of line bundles on the nilpotent cone and the Hall-Littlewood polynomials is made in a recent physics preprint 1403.0585. Is this a known mathematical fact?

Pick a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$. Let $\mathcal{N}$ be the nilpotent cone of $\mathfrak{g}$. Given an element $w$ in the positive Weyl chamber of the weight lattice of $\mathfrak{g}$, there is a natural ad $\mathfrak{g}$-equivariant line bundle $L_w$ on $\mathcal{N}$. Let $V_w=H^0(\mathcal{N},L_w)$. $V_w$ has a natural action of $G\times \mathbb{C}^\times$ (where $G$ is a Lie group for $\mathfrak{g}$).

Then the character of $V_w$ as a $G\times \mathbb{C}^\times$ representation is (essentially) given by the Hall-Littlewood polynomial $P_w(x,q)$ of type $\mathfrak{g}$ where $(x,q)\in G\times \mathbb{C}^\times$.

I must confess that I don't know how to construct $L_w$ for general $\mathfrak{g}$. For type $A$, it goes as follows. Let $\mathfrak{g}=\mathfrak{sl}(N)$. The nilpotent cone $\mathcal{N}$ is a holomorphic symplectic quotient of a linear space (in particular, a quiver variety in the sense of Nakajima). In particular, it has the form

$\mathcal{N}=X / GL(1)\times GL(2) \times \cdots GL(N-1)$

Then the line bundle $L_w$ is

$L_w = (X\times \mathbb{C} )/GL(1)\times GL(2) \times \cdots GL(N-1)$

where $GL(1)\times GL(2) \times \cdots GL(N-1)$ acts on $\mathbb{C}$ by a one-dimensional representation $\chi_w: GL(1)\times GL(2) \times \cdots GL(N-1)\to \mathbb{C}^\times$ determined by $w$ (a weight vector in the positive Weyl chamber of the weight lattice of $\mathfrak{sl}(N)$.

I would also appreciate if you could tell me how $L_w$ can be constructed for other $\mathfrak{g}$, as part of the question.

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    $\begingroup$ I don't think you really mean the nilcone, as your construction of the line bundle doesn't make sense in this case (what's the fiber over 0? by your description, it should be $\mathbb{C}$ modulo a non-trivial action of the product of GL's, which isn't a line); these line bundles will only make sense on $T^*(SL_n/B)$. Of course, you might prefer to think about their pushforward to the nilcone, but that's emphatically not a line bundle. $\endgroup$
    – Ben Webster
    Mar 12, 2014 at 20:35

2 Answers 2

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I took notes on a lecture that Mark Haiman gave explaining how to do a similar thing on the cotangent bundle on the flag variety: (see 2.1 of http://math.berkeley.edu/~monks/seminars/Notes.pdf )

This is close to what you are asking if we consider the Springer resolution of the cotangent bundle to the nilpotent cone, but I do not know how to get your construction on the nose.

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  • $\begingroup$ Thanks, that should be the correct mathematical statement of what I wanted to say :-p $\endgroup$ Mar 13, 2014 at 4:34
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I think Steven's answer addresses whether this was already known (yes), but let me try to clear up what's going with these line bundles.

In Nakajima's picture, you're describing the nilcone in an inherently Springery way: you consider the quiver $\mathbb{C}^1\to \mathbb{C}^2 \to \mathbb{C}^3\to \cdots \to \mathbb{C}^N$, double it and consider the action of $GL(1)\times \cdots \times GL(N-1)$ acting by pre- and post-composition. If $X$ is the map going right, and $Y$ the map going left (let's just take the sum of all the arrows; we can distinguish them by what space they act on), the moment map condition says that $[X,Y]=0$. In particular, the composition $XY|_{\mathbb{C}^N}$ is nilpotent, since $(XY)^{N+1}=X^{N+1}Y^{N+1}=0$. Note that this map is unchanged by the acton of $GL(1) \times \cdots \times GL(N-1)$. So, this is what you mean by getting the nilcone in the Nakajima way.

Your line bundles on the other hand, correspond to the determinants of the tautological bundles associated to the smaller spaces $\mathbb{C}^j$. So, the fundamental problem with talking about line bundles on the nilcone is that for those to make sense, you would have to start with the nilpotent and find the smaller spaces (whatever that means). Which is impossible.

This is a sign that you are doing the wrong thing. The problem is you never say what kind of quotient you're taking of the space X; if you want the nilcone, it's a very non-free quotient, which doesn't bode well for the line bundles. Instead, you have to throw out "bad points," specifically, you must require that $X$ is injective at each step. Thus, I can think of the quiver as actually a flag, and the action of $GL(1)\times \cdots \times GL(N-1)$ as relating all the different ways of parametrizing the spaces in the flag. Thus, modding out by them leaves us with a flag, and we still have $XY$, which is now a nilpotent transformation preserving the flag (and as it happens, all the other $Y$'s are determined by this one; they're just its restriction to the steps of the flag). Thus, we've found the Springer resolution of the nilcone.

In arbitrary types, the line bundles on $T^*G/B$ have a nice description; they're in bijection with integral weights of the torus. In the case of $SL(N)$, the tautological bundles in the quiver become tautological bundles in the flag variety, so you can describe one in terms of the other. It certainly looks to me like Steven's story becomes yours in this case.

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  • $\begingroup$ Thank you very much, Ben, and sorry for not distinguishing the nilcone and T^*G/B ... $\endgroup$ Mar 13, 2014 at 4:35

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