Let $A = \mathbb{C}[x,y,z]/(x y - z^k)$. In fact $A$ is the ring of $\mu_k$ invariants: $A = \mathbb{C}[u,v]^{\mu_k}$ where $g \in \mu_k$ acts by $g(u,v) = (g u, g^{-1} v)$.

This allows one to understand vector bundles on the smooth surface $Spec\ A - (0,0,0)$ as $\mu_k$ equivariant bundles on $Spec\ \mathbb{C}[u,v]$.

However I would like to understand how many of these vector bundles extend to all of $Spec\ A$.


It's well known for $k = 2$ that $Pic(Spec\ A)$ is trivial. But what about higher $k$ and higher rank?

One approach is to consider the natural map $\mathbb{C}[x,y]\to A$ which is finite and flat and then (identifying vector bundles with projective modules) work with modules over $\mathbb{C}[x,y]$ equipped with an endomorphism $\phi$ such that $\phi^k = x y$. But I didn't get very far with this.

  • 1
    $\begingroup$ Why don't you just consider the $\mu_k$-eigendecomposition of $\mathbb{C}[u,v]$ as a finite module over the subring $\mathbb{C}[u^k,v^k,uv] = \mathbb{C}[x,y,z]/\langle xy-z^k \rangle$? You are just asking which of these eigenmodules has rank $1$ at the maximal ideal $\langle x,y,z \rangle$. $\endgroup$ – Jason Starr Aug 13 '14 at 9:59

The only invertible sheaf on $U := \text{Spec}A \setminus \langle x,y,z\rangle$ that extends to all of $\text{Spec} A$ is the trivial invertible sheaf. As you correctly surmise, the invertible sheaves on $W$ are precisely the $\mu_k$-linearized invertible sheaves on $V := \text{Spec}\mathbb{C}[u,v] \setminus \langle u,v \rangle$. Of course every invertible sheaf on $V$ is trivial. So these invertible sheaves are just the characters of $\mu_k$. Since the characteristic is $0$, the character group is $\mathbb{Z}/k\mathbb{Z}$ (I guess even in characteristics dividing $k$, this is true). Thus, denoting by $$q:V\to U,$$ the natural quotient morphism, the invertible sheaves on $U$ are just the $\mu_k$-eigensheaves of the locally free sheaf $q_*\mathcal{O}_U$. The reflexive extensions of these sheaves to all of $\text{Spec} A$ are just the $\mu_k$-eigensheaves of $\mathbb{C}[u,v]$ considered as an $A$-module.

For every $r=0,\dots,k$, denote by $B_r \subset \mathbb{C}[u,v]$ the $A$-submodule of elements that have weight $r$ with respect to the given action. For every $r=1,\dots,k$, both $u^r$ and $v^{k-r}$ give linearly independent, nonzero elements in $B_r/\langle u^k,v^k,uv \rangle B_r.$ Thus, since the "fiber" of $B_r$ at $\langle u^k, v^k, uv \rangle$ is not rank $1$, the module $B_r$ is not projective of rank $1$. Thus, the only eigenmodule that is projective is $B_0$, i.e., $A$.

  • $\begingroup$ Great! The point I was missing is if a locally free on $U$ is to extend to a locally free on $Spec A$ it must be the reflexive extension. $\endgroup$ – solbap Aug 13 '14 at 22:01
  • $\begingroup$ If I understand correctly then the conclusion should be the same for higher rank: any vector bundle on $V$ is trivial so $Spec\ A$ you will see $B_{r_1} \oplus ... \oplus B_{r_n}$ and the fiber at $(u^k,v^k, u v)$ will be bigger than $n$ unless all $r_i = 0$. So $Spec\ A$ has only trivial vector bundles. $\endgroup$ – solbap Aug 13 '14 at 22:04
  • $\begingroup$ I believe this argument can be adapted to show that on $Spec\ A$ every principal $G$ bundle is trivial whenever $G$ is a split reductive group. $\endgroup$ – solbap Aug 14 '14 at 5:24
  • $\begingroup$ The idea is that if $P$ is a nontrivial $G$ bundle on $U$ and $W$ is any etale neighborhood of $(u^k,v^k,u v)$ then $P$ restricted to $W \cap U$ remains nontrivial. $\endgroup$ – solbap Aug 14 '14 at 5:26

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