Notation
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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 $X=X\_n$).




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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?
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> 
> 


### 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 $A$?



What does the nilpotent cone represent?
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](http://www.ams.org/mathscinet-getitem?mr=MR0447428)]


