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Notation

Let $\mathfrak g$ be a the Lie algebra of an algebraic group $G\subseteq GL(V)$ over a(n algebraically closed) field $k$ (I'm actually thinking $G=GL_n$, so $\mathfrak g=\mathfrak{gl}_n$). Then any element $X$ of $\mathfrak g$ can be uniquely written as the sum of a semi-simple (diagonalizable) element $X_s$ and a nilpotent element $X_n$ of $\mathfrak g$, where $X_s$ and $X_n$ are polynomials in $X$. The nilpotent cone $\mathcal N$ is the subset of nilpotent elements of $\mathfrak g$ (elements $X$ such that $X=X_n$).

People often talk about the nilpotent cone as having the structure of a subvariety of $\mathfrak g$, regarded as an affine space, but usually don't say what the scheme structure really is. To really understand a scheme, I'd like to know what its functor of points is. That is, I don't just want to know what a nilpotent matrix is, I want to know what a family of nilpotent matrices is (i.e. what a map from an arbitrary scheme $T$ to $\mathcal N$ is). Since any scheme is covered by affine schemes, it's enough to understand what an $A$-valued point (a map $\mathrm{Spec}(A)\to \mathcal N$) is for any $k$-algebra $A$. So my question is

What functor should $\mathcal N$ represent?

A guess

Well, an $A$-point of $\mathfrak g$ is "an element of $\mathfrak g$ with entries in $A$" (again, I'm really thinking $\mathfrak g=\mathfrak{gl}_n$, so just think "a matrix with entries in $A$"), so I would expect that such an $A$-point happens to be in $\mathcal N$ exactly when the given matrix is nilpotent. That is, $\mathcal N(\mathrm{Spec}(A))=\{X\in \mathfrak{g}(\mathrm{Spec}(A))| X^N=0$ for some $N\}$.

However, this is wrong. That functor isn't even an algebraic space, even for the nilpotent cone of $\mathfrak{gl}_1$. If it were, the identity map on it would correspond to a nilpotent regular function $f$ (a nilpotent $1\times 1$ matrix), and this would be the universal nilpotent regular function; every other nilpotent regular function anywhere else would be a pullback of this one. But whatever the degree of nilpotence of this function (say $f^{17}=0$), there are some nilpotent regular functions which cannot be a pullback of it (something with nilpotence degree bigger than 17). If this version of the nilpotent cone were representable, you can show that the $\mathfrak{gl}_1$ version would be too.

Another guess

I think the answer might be that an $A$ point of $\mathcal N$ is a matrix ($A$ point of $\mathfrak g$) so that all the coefficients of the characteristic polynomial vanish. This is a scheme and it has the right field-valued points, but why should this be the nilpotent cone? What is the meaning of having all coefficients of the characteristic polynomial vanish for a matrix with entries in $A$?

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    $\begingroup$ By way, this is very late, but one thing to consider about the characteristic poly definition: this is really saying that you should look at matrices whose minimal polynomial is $x^n$; that's quite different from being nilpotent, as considering the 1 x 1 matrix with a nilpotent entry shows. You were getting confused because you were getting nilpotence from the ring, and nilpotence from the matrix mixed up with each other (which is understandable, since most people think about matrices over fields, and so only have to think about the latter). $\endgroup$
    – Ben Webster ♦
    Commented Dec 8, 2016 at 3:34

2 Answers 2

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The ideal which defines the nilpotent cone is generated by the homogeneous elements of positive degree in $\mathbb{C}[\mathfrak g]^G$. In the case of $GL_n$, this ideal is generated by the functions $\mathrm{tr}\;X^k$ for $k$ up to the dimension, and also by the coefficients of the characteristic polynomial. See, for example, [Springer, T. A. Invariant theory. Lecture Notes in Mathematics, Vol. 585. Springer-Verlag, Berlin-New York, 1977. iv+112 pp. MR0447428]

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    $\begingroup$ Just to tease this out a little bit more, in case it helps anyone: the question This is a scheme and it has the right field-valued points, but why should this be the nilpotent cone? is really "do these equations generate a radical ideal"? Mariano's answer says "Yes, they do", or even more specifically "there aren't any other schemes lying between the reduced nilpotent cone and the scheme defined by the vanishing of the characteristic polynomial". Incidentally, it's easy to prove; this N is a complete intersection hence Cohen-Macaulay hence S1, so generic reducedness implies reducedness. $\endgroup$
    – Allen Knutson
    Commented Jan 8, 2010 at 18:47
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    $\begingroup$ Thanks for the clarification Allen, but why should N be reduced? The naive definition is so horribly non-reduced that it's not representable. It feels like a bit of a hack to take the reduced thing with the same field-valued points, but I get the impression that there's a good reason for it, and that reason is what I'm after. For example, why not define N as the subscheme cut out by X^n=0 (same n as in gl_n)? $\endgroup$
    – Anton Geraschenko
    Commented Jan 8, 2010 at 19:08
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    $\begingroup$ because "the variety of nilpotent matrices" is the wrong way to think of the nilcone. That's not how it usually appears in practice. It really should be thought of as "matrices with all their eigenvalues 0." $\endgroup$
    – Ben Webster ♦
    Commented Jan 8, 2010 at 19:40
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    $\begingroup$ * why not define N as the subscheme cut out by X^n=0 (same n as in gl_n)* Your use of "gl_n" instead of "M_n" is the giveaway here: the equation "X^n = 0" doesn't make sense in a general reductive Lie algebra, but the definition Ben gives does. So there's a sense in which you're thinking about the "wrong" equations. I can't resist mentioning a famous open problem: what are the equations that define the space (X,Y): XY=YX? Maybe those are the right ones; nobody knows. $\endgroup$
    – Allen Knutson
    Commented Jan 8, 2010 at 19:50
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    $\begingroup$ The famous open problem is whether the ideal generated by those $n^2$ quadratic equations $(AB-BA)_{ij} = 0$ is radical. $\endgroup$
    – Allen Knutson
    Commented Jun 10, 2019 at 23:07
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Another way of saying Mariano's answer is that the nilcone is the scheme-theoretic fiber of the projection map $\mathfrak{g}\to \mathfrak{g}// G$ where $\mathfrak{g}// G$ is the categorical quotient of the adjoint action. This is usually how it comes up anyways.

Yet another way to say this is that the nilcone is the classical limit of the quotient of the universal enveloping algebra modulo a maximal central ideal. This might sound complicated, but I certainly can't think of a more important appearance of the nilcone than this fact.

Well, there's also its appearance as the affinization of $T^*G/B$, which also gives the reduced scheme structure.

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  • $\begingroup$ I see. My confusion comes from thinking about varieties of certain nilpotent representations. A tangential remark: The (scheme-theoretic) "fiber of the quotient map" description agrees with the (set theoretic) Chevalley-Jordan decomposition description only in the case of a reductive algebraic group, but both make sense for non-reductive algebraic groups. I assume the nilpotent cone of the Lie algebra of a non-reductive group should be the fiber product with the nilpotent cone of gl_n over any faithful representation, but that nobody cares. $\endgroup$
    – Anton Geraschenko
    Commented Jan 8, 2010 at 20:38
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    $\begingroup$ Scheme-theoretic fibre over $0$, right? $\endgroup$
    – LSpice
    Commented May 19, 2019 at 15:52

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