Normal Cone as a Quotient Orbit Space

Question

Let $V=V(J)$ be the variety defined by the ideal $J\subset\Bbb{C}[x_0,\dots,x_d]$ generated by the 2x2 minors of $$\begin{pmatrix} x_0 & x_1 &\cdots & x_{d-1} \\ x_1 & x_2 &\cdots & x_d \end{pmatrix}$$ (in other word by the set of equations $x_jx_{k+1}=x_{j+1}x_k$, for $j,k=0,\dots,d$).

Now consider the $\Bbb{Z}_d$-action (cyclic group) on $\Bbb{C}^2$ given by $$ a\cdot (z_1,z_2)=(az_1,az_2), \quad a^d=1$$ where $a\in\Bbb{C}$ is a primitive $d$-rooth of unity. I am trying to prove the following

Claim: There is a 1:1 correspondence between $V$ and the orbit space $\Bbb{C}^2/\Bbb{Z}_d$.

In fact the map $\phi\colon\Bbb{C}^2\rightarrow V$ given by $$(z_1,z_2)\longmapsto (z_1^d,z_1^{d-1}z_2,\dots, z_1z_2^{d-1},z_2^d)$$ is clearly constant on the action orbits. Also it isn't hard to check that the fibers of $\phi$ are precisely the orbits. I am not managing to show surjectivity though...

Fix $(x_0,\dots,x_d)\in V$ and let $z_0\in x_0^{1/d}$ and $z_2\in x_d^{1/d}$. Then we have $$\phi(z_1,z_2)= (x_0,\dots, z_1^{d-j}z_2^j,\dots,x_d)$$ So we just need to prove the identity $x_j=z_1^{d-j}z_2^j$. Clearly we need to use the fact that $x_jx_{k+1}=x_{j+1}x_k$, but I am stuck here...

Can somebody help me?

Answer 1

To answer exactly the question you asked, I'll roughly sketch some ideas to show $ \phi(z_1, z_2) = (x_0,\dots, x_d) $.

Let $ w = x_0^{-1/d} x_d^{1/d} $, then one can write $ \phi(z_1, z_2) = (x_0, \dots, x_0 w^j, \dots, x_0 w^{d}) $. Now, it is not too bad to show inductively that $ w = \frac{x_1}{x_0} $. Basically, since $ x_0 x_d = x_1 x_{d-1} $, $x_d^{1/d} = x_0^{-1/d} x_1^{1/d} x_{d-1}^{1/d} $, and therefore $ x_0^{-1/d} x_d^{1/d} = x_0^{-2/d} x_1^{1/d} x_{d-1}^{1/d} $. Then use the same trick on $ x_{d-1} $.

Note also that $ \frac{x_1}{x_0} = \frac{ x_1 x_1}{x_0 x_1} = \frac{x_0 x_2}{x_0 x_2} = \frac{x_1}{x_0}$, generally, $ w = \frac{x_{k+1}}{x_k} $. Then you are done, since you can write $ w^j = \frac{x_1}{x_0} \frac{x_2}{x_1} \dots \frac{x_{j}}{x_{j-1}} $.

Answer 2

Instead of looking at $\phi$ directly, I think it's better to consider the ring homomorphism associate to $\phi$. Since $V=\operatorname{Spec}\Bbb{C}[x_0,\cdots,x_d]/I$ where $I=(x_j x_{k+1}-x_{j+1}x_k)$ and $\Bbb{C}^2/\Bbb{Z}_d=\operatorname{Spec}\Bbb{C}[z_1,z_2]^{\Bbb{Z}_d}$, $\phi$ induces a ring homomorphism $\sigma:\Bbb{C}[x_0,\cdots,x_d]/I\rightarrow \Bbb{C}[z_1,z_2]^{\Bbb{Z}_d}$ defined by $\sigma(\bar{x_i})=z_1^{d-i}z_2^i$.

Claim: $\sigma$ is a ring isomorphism, hence $\phi$ is an isomorphism.

Proof:

It is clear that $\Bbb{C}[z_1,z_2]^{\Bbb{Z}_d}$ is a subring of $\Bbb{C}[z_1,z_2]$ generated by all degree $d$ homogeneous polynomial of $(z_1,z_2)$, hence $\Bbb{C}[z_1,z_2]^{\Bbb{Z}_d}$ is generated by $\{\sigma(\bar{x_i})\}_{i=0}^d$. As a result, $\sigma$ is surjective.

To prove the injectivity of $\sigma$, we look at $\tilde\sigma: \Bbb{C}[x_0,\cdots,x_d]\rightarrow \Bbb{C}[z_1,z_2]$ instead, where $\tilde\sigma(x_i)=z_1^{d-i}z_2^i$. It suffices to show $\ker(\tilde\sigma)=I$. By the definition of $\tilde\sigma$ we have $I\subset \ker(\tilde\sigma)$.

Since both $\ker(\tilde\sigma)$ and $I$ are generated by homogeneous polynomials, it suffices to show that for a homogeneous polynomial $f\in \ker(\tilde\sigma)$, we have $f\in I$. Let $I_n$ be the degree $n$ part of $I$, $n:=\deg f$. Under the lexicographic order, there exists $g\in f+I_n$ such that $g$ has the lowest leading term among polynomials in $f+I_n$. We will show that $g=0$, hence $f\in I$.

Otherwise, suppose the leading term of $g$ is $x_k^{n_k}x_{k+1}^{n_{k+1}}\cdots x_d^{n_d}$, with $n_k\neq 0$. If there exists $l\geq 2$ such that $n_{k+l}\neq 0$, then $x_k x_{k+l}-x_{k+1} x_{k+l-1}\in I$, hence $x_k^{n_k-1}x_{k+1}^{n_{k+1}+1}\cdots x_{k+l-1}^{n_{k+l-1}+1}x_{k+l}^{n_{k+l}-1}\cdots x_d^{n_d}\in x_k^{n_k}x_{k+1}^{n_{k+1}}\cdots x_d^{n_d}+I_n$, so the leading term of $g$ is not the lowest among polynomials in $f+I_n$, we get a contradiction! Thus $n_{k+2}=\cdots =n_d=0$, which means the leading term of $g$ is $x_k^{n_k}x_{k+1}^{n_{k+1}}$. As a result, the leading term of $g(z_1^d,z_1^{d-1}z_2,\cdots,z_2^d)$ is nonzero, which contradicts to $g\in \ker(\tilde\sigma)$. QED