Timeline for Has the geometry of the variety of nilpotent matrices over $\mathbb{C}$ been studied?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 9, 2019 at 15:59 | vote | accept | M.G. | ||
| Apr 7, 2019 at 22:10 | comment | added | Sean Lawton | 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). | |
| Apr 7, 2019 at 22:07 | comment | added | Joe Silverman | Okay, fair enough. But your GIT quotient is not, for example, a geometric quotient (nor even a categorical quotient) in the sense of Mumford's book. So for example, the $\mathbb C$ points of your quotient $M_n//PGL_n\cong\text{Spec}(k[c_1,\ldots,c_n])\cong\mathbb A^n$ are most definitely not in one-to-one correspondence with the points of the set-theoretic quotient $M_n(\mathbb C)/PGL_n(\mathbb C)$ of the geometric points on $M_n$. | |
| Apr 7, 2019 at 21:57 | comment | added | Sean Lawton | @JoeSilverman To answer your question, with respect to a $G$-linearized line bundle, the GIT quotient by a reductive group $G$ on a quasi-projective $G$-variety is always a quasi-projective variety. For the affine quotient, which what I am referring to in this case, you just take $Spec$ of the ring of invariants. It is not hard to see that the ring of invariants is exactly $k[c_1,...,c_n]$. So the GIT quotient is $\mathbb{A}^n$ since $c_1,...,c_n$ are algebraically independent. | |
| Apr 7, 2019 at 20:33 | comment | added | Joe Silverman | 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? | |
| Apr 7, 2019 at 15:08 | comment | added | Sean Lawton | 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 $\pi(A)=(0,...,0)$. | |
| Apr 7, 2019 at 14:58 | comment | added | Sean Lawton | @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 $A^n=0$. | |
| Apr 7, 2019 at 13:57 | comment | added | Joe Silverman | 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. | |
| Apr 7, 2019 at 13:21 | history | edited | Sean Lawton | CC BY-SA 4.0 |
Small edit.
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| Apr 6, 2019 at 20:56 | history | answered | Sean Lawton | CC BY-SA 4.0 |