10
$\begingroup$

Consider the complex projective variety given by $X^n = 0$, where $X\in \mathrm{M}_n(\mathbb{C})$ and, say, $n\geq 3$. Some basic properties of it are already mentioned in this question:

https://math.stackexchange.com/questions/405291/variety-of-nilpotent-matrices

I would like to know if its geometry has been studied in more detail in the sense of complex geometry (algebraic, differentiable, analytic).

References are appreciated since the question above mentions only Jantzen's "Lie Theory".

Improve this question
$\endgroup$
1
  • 17
    $\begingroup$ This is one of the most studied varieties in all of algebraic geometry; a phrase to to google is "nilpotent orbit." See e.g. Ch.3 of Chriss-Ginzburg for an introduction. $\endgroup$
    – dhy
    Commented Apr 6, 2019 at 20:58

1 Answer 1

Reset to default
9
$\begingroup$

Consider $PGL_n$ acting by conjugation on the space of $n\times n$ matrices $M_n$, and let the GIT quotient map be $\pi:M_n\to M_n//PGL_n$. I think you are asking about the geometry of the fibre of zero.

Regardless, I believe this is a special case of the nullcone for a reductive group action on a vector space $V$.

Two papers that come to mind about the geometry of the nullcone are:

  1. A Stratification of the Null Cone Via the Moment Map by Linda Ness (appendix by David Mumford), American Journal of Mathematics, Vol. 106, No. 6 (Dec., 1984), pp. 1281-1329
  2. Irreducible components of the nullcone by Richardson, R. W., Invariant theory, 409–434, Contemp. Math., 88, Amer. Math. Soc., Providence, RI, 1989.

The work of these two authors is worth reading if you are interested in these kinds of problems (and the references therein).

Improve this answer
$\endgroup$
7
  • 1
    $\begingroup$ Is it really the geometry of the fibre of zero? I think that the fiber of zero is just zero. Instead, I think that the OP wants the fibre over the union of all Jordan normal forms with diagonal all zero. The PGL$_n$ equivalence classes of these are then just the permutations of the various Jordan blocks. So the OP's variety will have a number of different components, of varying dimensions. $\endgroup$
    – Joe Silverman
    Commented Apr 7, 2019 at 13:57
  • 1
    $\begingroup$ @JoeSilverman $\pi$ is the map $\pi(A)=(c_1(A),...,c_n(A))$ where $c_i$ are the coefficients of the characteristic polynomial. So if $\pi(A)=(0,...,0)$ then $A^n=0$. $\endgroup$
    – Sean Lawton
    Commented Apr 7, 2019 at 14:58
  • 1
    $\begingroup$ Conversely, in the conjugation class of any nilpotent matirx, there is a sequence that limits to the zero matrix. Thus, if $A^n=0$, then $\pi(A)=(0,...,0)$. $\endgroup$
    – Sean Lawton
    Commented Apr 7, 2019 at 15:08
  • 1
    $\begingroup$ Thanks. So you're mapping the quotient $M_n//PGL_n$ to $\mathbb A^n$ by taking the coefficients of the minimal polynomial. That's a well-defined map of sets, and then, as you say, the OP's variety is $\pi^{-1}(0,0,\ldots,0)$. But now I'm a bit worried about the GIT quotient bit. My recollection is that the $PGL_n$-stable locus in $M_n$ is the set of diagoanlizable matrices (or maybe it's better to take the action by $SL_n$, since then we can use $\mathcal{O}(1)$). But in either case, does the GIT quotient of $M_n$ exists as a variety? $\endgroup$
    – Joe Silverman
    Commented Apr 7, 2019 at 20:33
  • 4
    $\begingroup$ It is not a geometric quotient, but it is categorical (and it is a quotient in the sense of Mumford's book). And yes, it is definitely not in 1-1 correspondence with orbit space. But it is in 1-1 correspondence (and even homeomorphic in the strong topology) to the polystable quotient (orbit space of points with closed orbits). Moreover, it is homotopic to the orbit space (again in the strong topology). $\endgroup$
    – Sean Lawton
    Commented Apr 7, 2019 at 22:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .