# Classification of (complex algebraic) vector bundles on punctured affine space

The Quillen-Suslin 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, 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?

• The pullback of the tangent bundle from $\mathbb{P}^n$ is nontrivial for every $n\geq 2$. – Jason Starr Mar 6 '17 at 17:28
• I realize now that I should have phrased question 1. differently. Vector bundles on $\mathbb{P}^n$ giving already a rich moduli problem, I should have probably rather concentrated on asking if there are non-equivariant bundles on $X_n$... – Qfwfq Mar 6 '17 at 17:51
• Edited the question. – Qfwfq Mar 6 '17 at 18:17
• Using weighted actions, you have weighted projective spaces as quotients and shouldn't they give you other vector bundles? – Mohan Mar 6 '17 at 18:17
• Related question: mathoverflow.net/questions/35788/… – Jorge Vitório Pereira Mar 6 '17 at 18:21

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 $\leq n-1$?

• Interesting. What about line bundles? Can $M$ ever have rank $1$? – Qfwfq Mar 10 '17 at 16:03
• The rank of $M$ equals $n$. Since the construction above only works for $n\geq 2$ (so that we know that $Z_{n-1}(\underline{f})$ is reflexive), it does not give examples of line bundles. – Jason Starr Mar 10 '17 at 16:05

Jean-Pierre Serre, Prolongement de faisceaux analytiques coh ́erents, Ann. Inst. Fourier (Grenoble) 16 (1966), no. fasc. 1, 363–374. MR MR0212214 (35 #3088) p. 372 proves that there are infinitely many distinct holomorphic line bundles on $\mathbb{C}^2-0$.

• Wait but, if I understand correctly, Serre does the case of analytic vector bundles, which on a nonprojective variety might be rather different from the algebraic ones. – Qfwfq Mar 6 '17 at 18:14
• Indeed, the analytic and algebraic cases are very different here; any algebraic line bundle on $\mathbb{A}^2 - 0$ is trivial. Both algebraically and analytically, any line bundle on $\mathbb{A}^2 - 0$ trivializes on $\mathbb{C} \times \mathbb{C}^{\ast}$ and $\mathbb{C}^{\ast} \times \mathbb{C}$, so it is determined by the gluing function, a non-vanishing function on $\mathbb{C}^{\ast} \times \mathbb{C}^{\ast}$. – David E Speyer Mar 6 '17 at 18:22
• Algebraically, every such function is of the form $\alpha x^i y^j$ for some scalar $\alpha$, and thus can be written as the product of $\alpha x^i$, which extends to $\mathbb{C}^{\ast} \times \mathbb{C}$, and $y^j$, which extends to $\mathbb{C}^{\ast} \times \mathbb{C}$. Thus, this line bundle is trivial. – David E Speyer Mar 6 '17 at 18:22
• Every algebraic vector bundle on $\mathbb C^2- 0$ extends to an algebraic vector bundle on $\mathbb C^2$ (see Arapura's answer to mathoverflow.net/questions/35788/…) and therefore must be trivial. – Jorge Vitório Pereira Mar 6 '17 at 18:22
• Analytically, $\exp(x^{-1} y^{-1})$ is an example of a non-vanishing holomorphic function on $\mathbb{C}^{\ast} \times \mathbb{C}^{\ast}$ which cannot be written as a product of a function which extends to $\mathbb{C} \times \mathbb{C}^{\ast}$ and one which extends to $\mathbb{C}^{\ast} \times \mathbb{C}$, so it gives a line bundle which is holomorphically nontrivial. – David E Speyer Mar 6 '17 at 18:23