The [Quillen-Suslin](https://en.wikipedia.org/wiki/Quillen%E2%80%93Suslin_theorem) theorem asserts that there are no nontrivial vector bundles over the affine space $\mathbb{A}^{n+1}$, $n\geq 0$.
Let's work over the complex numbers. What can be said about vector bundles on the *punctured* affine space $X\_n=\mathbb{A}^{n+1}\smallsetminus\{0\}$?
According to [this paper](https://arxiv.org/pdf/1303.0575.pdf), there seem to be room for nontrivial vector bundles.


Let $\mathbb{C}^{\*}$ act on $X\_n$ by the action $\lambda.(x\_0,\dots,x\_n):=(\lambda x\_0,\lambda x\_1,\dots, \lambda x\_n)$ whose quotient is $\mathbb{P}^n$.
Notice that equivariant v.b. on $X\_n$ are in bijection -via pullback- with v.b. on $\mathbb{P}^n$, and the latter form already a rich moduli problem on its own. In this question we concentrate on the specificity of $X\_n$



> 
> **1.** Is there some sort of classification of v.b. on $X\_n$, taking as a starting base -say- the "classification" of stable v.b. on $\mathbb{P}^n$ given by the corresponding moduli spaces?
> 
> 
> What about particular ranks, for example the case of **line bundles**?
> 
> 
> **2.** Are there vector bundles on $X\_n$ that are not pullbacks of v.b. on $\mathbb{P}^n$, that is, v.b. on $X\_n$ that do not admit an equivariant structure?
> 
> 
> 



Classification of (complex algebraic) vector bundles on punctured affine space
I spoke to my colleague Song Sun, and he reminded me of a discussion that he and I had about Question 2 some time ago. For $n\geq 2$, there are many examples of locally free sheaves on $X\_{n} = \mathbb{A}^{n+1}\setminus\{0\}$ that admit no equivariant structure. Denote by $S$ the polynomial ring $\mathbb{C}[x\_0,\dots,x\_n]$. Denote by $S\_+$ the maximal ideal $\langle x\_0,\dots,x\_n \rangle$. Let $\underline{f}=(f\_0,\dots,f\_n)$ be a regular sequence of elements in $S\_+$. 


Associated to this regular sequence there is a Koszul complex $(K\_\bullet(\underline{f}),d\_\bullet)$ of $S$-modules where $K\_0(\underline{f})$ equals $S$, where $K\_1(\underline{f})$ equals $S^{\oplus(n+1)}$, where $K\_r(\underline{f})$ is the $r^{\text{th}}$ exterior power of $K\_1(\underline{f})$, and where $d\_r:K\_r(\underline{f}) \to K\_{r-1}(\underline{f})$ is the unique sequence of $R$-module homomorphisms such that $d\_1(g\_1,\dots,g\_n) = f\_1g\_1 + \dots + f\_ng\_n$ and such that $(K\_{\bullet}(\underline{f}),d\_\bullet)$ is a differential graded $R$-algebra. To be precise, $d\_{r-1}\circ d\_r$ equals the zero homomorphism, and $d\_{r+s}(\alpha \wedge \beta) = d\_r(\alpha)\wedge \beta + (-1)^r\alpha\wedge d\_s(\beta)$ for every $r,s\geq 0$, for every $\alpha\in K\_r(\underline{f})$, and for every $\beta\in K\_s(\underline{f})$. 


For every integer $r\geq 1$, denote by $Z\_r(\underline{f})$ the kernel of $d\_r$. Since $\underline{f}$ is a regular sequence, this is the same as the image of $d\_{r+1}$. Thus, also define $Z\_0(\underline{f})$ to be the image of $d\_1$, i.e., the ideal $I$ generated by $\underline{f}$. In particular, $Z\_{n+1}(\underline{f})$ is the zero module, and $Z\_n(\underline{f})$ is $K\_{n+1}(\underline{f})$. 


**Fact 1.** For all $r$ with $1\leq r \leq n$, $Z\_r(\underline{f})$ is a reflexive $S$-module.


**Proof.** By construction $K\_r(\underline{f})$ is a free $S$-module, hence reflexive, and $K\_{r-1}(\underline{f})$ is torsion-free (even free). The kernel of every $S$-module homomorphism from a reflexive module to a torsion-free module is reflexive. **QED**


**Fact 2.** For $M=Z\_{n-1}(\underline{f})$, the first Fitting ideal $\text{Fitt}\_1(M)$ equals the ideal $I$ generated by $(f\_0,\dots,f\_n)$.


**Proof.** The Fitting ideal can be computed from any finite presentation of $M$. For $Z\_{n-1}(\underline{f})$, one such presentation is $d\_{n+1}:K\_{n+1}(\underline{f}) \to K\_n(\underline{f})$. By self-duality of the Koszul complex, the Fitting ideal of $d\_{n+1}$ is $I$. **QED**


Since the Fitting ideal is intrinsic, if $M$ is equivariant, then $I$ is a homogeneous ideal. However, there are many $S\_+$-primary ideals $I$ that are generated by a regular sequence, yet are not homogeneous ideals. For instance, one example is $$\underline{f}=(x\_0,x\_1,\dots,x\_{n-2},x\_{n-1}-x\_n^2,x\_n^3).$$ For $n\geq 2$, for such $\underline{f}$, the module $M=Z\_{n-1}(\underline{f})$ is reflexive, the restriction of $\widetilde{M}$ to $\mathbb{A}^{n+1}\setminus\{0\}$ is locally free (visibly it is locally free on each $D(f\_i)$ for $i=0,\dots,n$). Yet $\widetilde{M}$ is not equivariant since $I$ is not a homogeneous ideal.


**Edit.** Song Sun asked the following variant of the question. For a reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ and whose Fitting ideals are all homogeneous, is the module equivariant? Also, note that every module as constructed above (also allowing other syzygy modules of the complexes) has rank $\geq n$. So here is a second variant: is every reflexive $S$-module that is locally free on $\mathbb{A}^{n+1}\setminus\{0\}$ equivariant provided that the rank is $\leq n-1$?


