Orbit Type Strata of C^3: 2-torus Action (a,b)

  • Thread starter Thread starter HMY
  • Start date Start date
  • Tags Tags
    Orbit Type
HMY
Messages
13
Reaction score
0
a 2-torus action on C^3 can be defined by
(a,b).(x,y,z)= (abx, a^-1by, bz)

What are the orbit type strata of C^3 here?

2-torus can be thought of (S^1)^2.
0 is the only fixed point I can tell, so it's one strata.
I just don't understand this seemingly simple action.
 
Physics news on Phys.org
Orbits are not fixed points. (0,0,0) is an orbit, agreed. But there is at least one other - every point lies in an orbit.

So fix a point (u,v,w) and look at where the torus maps it. What is the resulting space? It is a quotient space of the 2-torus, but by what? I.e.e when is the map (a,b)-->(abx,by/a,bz) not injective? (this is a constraint on x,y,z) Where it is injective impleis the orbits are 2-toruses, and where it isn't they are something else.
 
am I properly making sense of this?

Call this map f: (a,b)-->(abx,by/a,bz)

f is not injective when you look at (a, b) with b=0 &
a not= 0.

eg.
take another point (c,d) with d=0 & c not= a & c not= 0
So (a,b) not= (c,d). But f(a,b) = (0,0,0) & f(c,d) = (0,0,0)


matt grime said:
So fix a point (u,v,w) and look at where the torus maps it. What is the resulting space? It is a quotient space of the 2-torus, but by what? I.e.e when is the map (a,b)-->(abx,by/a,bz) not injective? (this is a constraint on x,y,z) Where it is injective impleis the orbits are 2-toruses, and where it isn't they are something else.
 
That f is a map from where to where? What is the alleged image? The set of all points (abx,by/a,bz) with x,y,z in C^3?

I don't see what that map has to do with the problem.

Here's a point in C^3: (1,0,0). What is the orbit of that point under group action?
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top