Recently, I'm reading the paper "Analytic structures on the space of flat vector bundles over a compact Riemann surface" by Gunning. In the introduction of this paper, Gunning says that the set $H^1(M,PL(n, \mathbb{C}))$ of projectively flat bundles over a compact Riemann surface $M$ has $n$ components, which can be put in one-to-one correspondence with one another. Then he claims that this property follows directly from the short exact sequence(the details are omitted) $$0 \rightarrow \mathbb{Z}_n \rightarrow SL(n, \mathbb{C}) \rightarrow PL(n, \mathbb{C}) \rightarrow 0,$$ where $\mathbb{Z}_n = \mathbb{Z}/(n\mathbb{Z})$.

I think we could obtain the following exact sequence of cohomology groups from the above one $$0 \rightarrow H^0(M,\mathbb{Z}_n) \rightarrow H^0(M, SL(n, \mathbb{C})) \rightarrow H^0(M,PL(n,\mathbb{C})) \rightarrow H^1(M,\mathbb{Z}_n) \rightarrow H^1(M, SL(n, \mathbb{C})) \rightarrow H^1(M,PL(n,\mathbb{C})) \rightarrow H^2(M,\mathbb{Z}_n)$$ since $\mathbb{Z}_n$ is in the center of $SL(n, \mathbb{C})$. From the connectness of $M$, we have (right?) $$H^0(M,\mathbb{Z}_n) = \mathbb{Z}_n, H^0(M, SL(n, \mathbb{C}))=SL(n, \mathbb{C}), H^0(M,PL(n,\mathbb{C}))=PL(n,\mathbb{C}), H^2(M,\mathbb{Z}_n)=\mathbb{Z}_n.$$

Therefore, we get an exact sequence $$0 \rightarrow H^1(M,\mathbb{Z}_n) \rightarrow H^1(M, SL(n, \mathbb{C})) \rightarrow H^1(M,PL(n,\mathbb{C})) \xrightarrow{\delta} \mathbb{Z}_n$$

I guess these $n$ components arise from the preimages of $\delta$. But how to prove

  1. The map $\delta$ is surjective.
  2. For any $a_1, a_2 \in \mathbb{Z}_n$, there exists a natural one-to-one correspondence between $\delta^{-1}(a_1)$ and $\delta^{-1}(a_2)$.

Furthermore, we know that for any projective bundle $\phi$ over a compact Riemann surface $M$, there always exists a holomorphic vector bundle $E$ such that $\phi = \mathbb{P}(E)$. Hence, if $\phi \in H^1(M,PL(n,\mathbb{C}))$, we can find a holomorphic vector bundle $E$ with $\phi = \mathbb{P}(E)$ over $M$. But we don't know whether $E$ is flat or not (or $\deg(E)$ is zero or not). I think

  1. $\delta(\mathbb{P}(E)) = [\deg(E)] \in \mathbb{Z}_n$.

Who can help me prove them or give a counterexample? Thank you very much!

  • 1
    $\begingroup$ Inside $\textbf{PGL}(n,\mathbb{C})$, consider the torus $T'$ that consists of equivalence classes of diagonal $n\times n$ matrices where the first $n-1$ diagonal entries are equal, say $s$, and the last diagonal entry can be different, say $t$. Via the map that sends such a matrix to $st^{-1}$, this torus is equivalent to $\mathbb{C}^\times$. The inverse image $T$ of $T'$ in $\textbf{SL}(n,\mathbb{C})$ is a copy of $\mathbb{C}^\times$ via the entry $s$, where now $t^{-1}$ equals $s^{n-1}$. Thus the map between tori is $s\mapsto s^n$ . . . $\endgroup$ Jun 13, 2017 at 11:37
  • 1
    $\begingroup$ . . . Consider the long exact sequence of cohomology coming from the short exact sequence of Abelian groups, $$1 \to \mu_n \to T \to T' \to 1,$$ $$H^1(M,\mathbb{C}^\times)\xrightarrow{(-)^n} H^1(M,\mathbb{C}^\times) \xrightarrow{\delta} H^2(M,\mu_n).$$ Via first Chern classes, this is $$H^2(M,\mathbb{Z}(1)) \xrightarrow{n\cdot (-)} H^2(M,\mathbb{Z}(1)) \to H^2(M,\mathbb{Z}/n\mathbb{Z}(1)).$$ This is clearly surjective, since $H^3$ is zero. Starting from a $T'$-torsor whose first Chern class is not $n$-divisible, the associated $\textbf{PGL}(n,\mathbb{C})$-torsor generates $H^2(M,\mu_n)$. $\endgroup$ Jun 13, 2017 at 11:44
  • $\begingroup$ @JasonStarr Thanks! Do you mean the weight zero Hodge structure by $H^2(M,\mathbb{Z}(1))$? I still have something confused in your argument. By your construction, we know that the exact sequences $$0 \rightarrow \mu_n \rightarrow T \rightarrow T' \rightarrow 0$$ and $$0 \rightarrow\mu_n \rightarrow \mathbb{C}^{\times} \xrightarrow{(-)^n} \mathbb{C}^{\times} \rightarrow 0$$ are equivalent. $\endgroup$
    – swalker
    Jun 14, 2017 at 8:40
  • $\begingroup$ Note that in the sequence $ H^1(M, \mathbb{C}^{\times}) \xrightarrow{(-)^n}H^1(M, \mathbb{C}^{\times}) \xrightarrow{\delta} H^2(M, \mu_n)$, the map $\delta = 0$, i.e. for any $L' \in H^1(M, \mathbb{C}^{\times}), \delta(L')=0$. As a consequence, the image of any $T'$-torsor is just one point in $\mu_n$. So how to construct the associated $\mathbf{PGL}(n,\mathbb{C})$-torsor and prove this torsor generates $H^2(M, \mu_n)$(It should use the last sequence to prove it, but how to use it?) And I want to know what gives the one-to-one correspondence between the components(Question 2). Thank you. $\endgroup$
    – swalker
    Jun 14, 2017 at 8:41
  • $\begingroup$ The map $\delta$ is not the zero map. The self-map on $H^1(M,\mathbb{C}^\times)$ of raising to the $n^{\text{th}}$ power corresponds to multiplication by $n$ of the first Chern class. The multiplication by $n$ map is not surjective on $H^1(M,\mathbb{Z}(1))$. The raising to the $n^{\text{th}}$ power map is not surjective on $H^1(M,\mathbb{C}^\times)$. $\endgroup$ Jun 14, 2017 at 10:19

1 Answer 1


Edit. The OP has clarified that all of the sheaves are sheaves of locally constant functions, rather than sheaves of continuous functions or $C^\infty$ functions. In that case, it is better to consider principal bundles that are induced from an Abelian subgroup of the normalizer of the maximal torus that is not contained in the maximal torus. For instance, let $\Gamma'$ be the subgroup of $\textbf{PGL}(n,\mathbb{C})$ generated by the equivalence classes of the following two matrices, $$ A=\left[\begin{array}{cccccc} 0 & 1 & 0 & \dots & 0 & 0\\ 0 & 0 & 1 & \dots & 0 & 0\\ 0 & 0 & 0 & \dots & 0 &0\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 &\dots & 0 & 1 \\ 1 & 0 & 0 & \dots & 0 & 0 \end{array} \right], \ \ B=\left[\begin{array}{cccccc} 1 & 0 & 0 & \dots & 0 & 0\\ 0 & z & 0 & \dots & 0 & 0 \\ 0 & 0 & z^2 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 &\dots & z^{n-2} & 0 \\ 1 & 0 & 0 & \dots & 0 & z^{n-1} \end{array} \right], $$ where $z$ is a primitive $n^{\text{th}}$ root of unity. The group $\Gamma'$ is isomorphic to $(\mathbb{Z}/n\mathbb{Z})\times \mu_n$. The inverse image $\Gamma$ of $\Gamma'$ in $\textbf{SL}(n,\mathbb{C})$ is a non-Abelian group. It is a central extension of $\Gamma'$ by $\mu_n\cdot \text{Id}_{n\times n}$. The commutation relation is $ABA^{-1}B^{-1} = z\cdot \text{Id}_{n\times n}$. The central extension of finite groups, $$ 1 \to \mu_n\cdot \text{Id}_{n\times n} \to \Gamma \to (\mathbb{Z}/n\mathbb{Z})\times \mu_n \to 1,$$ gives a long exact sequence of non-Abelian cohomology with a connecting map, $$\delta:H^1(M;(\mathbb{Z}/n\mathbb{Z})\times \mu_n) \to H^2(M;\mu_n).$$ This connecting map is what is usually called the Weil pairing. There are many ways to construct the Weil pairing; the construction above is just one (probably not the best, since it is a little unclear how this behaves when you replace the integer $n$ by a multiple of $n$).

Because $\Gamma$ is a nilpotent group of nilpotency class $2$, every group homomorphism, $$\rho:\pi_1(M;\ast) \to \Gamma,$$ factors through the $2$-nilpotent quotient $\Pi(M)$ of $\pi_1(M;\ast)$. This fits into a central extension, $$0 \to \left(\bigwedge^2_{\mathbb{Z}} H_1(M;\mathbb{Z})\right)/H_2(M;\mathbb{Z})\to \Pi \to H_1(M;\mathbb{Z})\to 0,$$ where a generator $\gamma$ of $H_2(M;\mathbb{Z})$ maps to the usual relation in $\bigwedge^2H_1(M;\mathbb{Z})$, $$\gamma \mapsto \alpha_1\wedge \beta_1 + \dots + \alpha_g\wedge \beta_g$$ for the standard presentation, $$\pi_1(M,\ast) = \langle \alpha_1,\dots,\alpha_g,\beta_1,\dots,\beta_g: [\alpha_1,\beta_1]\cdots [\alpha_g,\beta_g] \rangle.$$

So now, for every integer $m$, consider the homomorphism $$\sigma_m: \Pi \to \Gamma',$$ $$\alpha_1 \mapsto [A], \ \beta_1 \mapsto [B^m], \ \alpha_{i>1} \mapsto [\text{Id}_{n\times n}], \ \beta_{j>1}\mapsto [\text{Id}_{n\times n}].$$ This maps $\gamma$ to $z^m\cdot \text{Id}_{n\times n}$. Thus, $\sigma_m$ lifts to a representation $\rho_m:\Pi \to \Gamma$ if and only if $m$ is divisible by $n$. When $m$ is not divisible by $m$, the commutator of $A$ and $B^m$ maps to $z^m\text{Id}_{n\times n}$. Thus the corresponding $\Gamma'$-principal bundle over $M$ corresponding to $\sigma_m$ has image under $\delta$ equal to $m$ times the generator of $H^2(M;\mu_n)$.

Since the long exact sequence in non-Abelian cohomology associated to a central extension is functorial, it follows that for the associated $\textbf{PGL}(n,\mathbb{C})$-principal bundle over $M$, the image under $\delta$ also equals $m$ times the generator. Thus, $\delta$ is surjective.

Original post. I will add some details. Let $M$ be a compact, closed, oriented, real manifold of real dimension $2$. Denote by $\textbf{SL}(n,\mathbb{C})$, resp. $\textbf{PGL}(n,\mathbb{C})$, the real Lie group of $n\times n$ complex matrices with determinant $1$, resp. equivalence classes of $n\times n$ complex matrices up to scaling.

Denote by $T'$ the Lie subgroup of $\textbf{PGL}(n,\mathbb{C})$ consisting of equivalence classes of diagonal $n\times n$ matrices such that the first $(n-1)$ diagonal entries are all equal, say $s$, and the last diagonal entry, $t$, need not equal $s$. Denote by $T$ the inverse image of this Lie subgroup in $\textbf{SL}(n,\mathbb{C})$. This is the set of the same type of matrices, but now under the restriction that $t^{-1}$ equals $s^{n-1}$. There is a short exact sequence of Lie groups, $$1 \to \mu_n \to T \to T' \to 1,$$ where $\mu_n$ denotes the finite subgroup of scalar multiples of the identity matrix by $n^{\text{th}}$ roots of $1$. The morphism $T\to T'$ is a submersion, and even a finite unbranched cover.

For every Lie group $G$, denote by $\underline{G}$ the corresponding sheaf on $M$ of $C^\infty$ functions from open subsets of $M$ to $G$ (it would be fine to work with continuous functions). Then the short exact sequence of Lie groups above gives a short exact sequence of sheaves of Abelian groups. That short exact sequence of sheaves of Abelian groups gives a long exact sequence of cohomology. The relevant portion of that sequence is as follows, $$H^1(M,\underline{T}) \to H^1(M,\underline{T'}) \xrightarrow{\delta} H^2(M,\mu_n).$$ Both $T$ and $T'$ are isomorphic to the Lie group $\mathbb{C}^\times$ via the isomorphisms, $$T\xrightarrow{\cong} \mathbb{C}^\times, \ \ \text{Diag}(s,\dots,s,s^{-(n-1)}) \mapsto s, $$ $$T'\xrightarrow{\cong} \mathbb{C}^\times, \ \ [\text{Diag}(s,\dots,s,t)] \mapsto st^{-1}.$$ Via these isomorphisms, both $H^1(M,\underline{T})$ and $H^1(M,\underline{T'})$ are isomorphic to $H^1(M,\underline{C}^\times)$. The diagram becomes, $$H^1(M,\underline{\mathbb{C}}^\times) \xrightarrow{(-)^n} H^1(M,\underline{\mathbb{C}}^\times) \xrightarrow{\delta} H^2(M,\mu_n).$$ Using the exponential sequence of Lie groups, $$0 \to \mathbb{Z}(1) \to \mathbb{C} \xrightarrow{\text{Exp}} \mathbb{C}^\times \to 0,$$ there is a long exact sequence of cohomology. The connecting maps of that long exact sequence give a commutative diagram, $$\begin{array}{ccccc} H^1(M,\underline{\mathbb{C}}^\times) &\xrightarrow{(-)^n} & H^1(M,\underline{\mathbb{C}}^\times) & \xrightarrow{\delta} & H^2(M,\mu_n) \\ \downarrow & & \downarrow & & \downarrow \\ H^2(M,\mathbb{Z}) &\xrightarrow{n(-)} & H^2(M,\mathbb{Z}(1)) & \xrightarrow{} & H^2(M,\mathbb{Z}/n\mathbb{Z}(1)) \end{array}.$$ The last vertical map is an isomorphism; the others are the first Chern class maps. The first Chern class maps are surjective. Finally, because $H^3(M,\mathbb{Z}(1))$ is zero, it follows that for every element of $H^2(M,\mu_n)$, there exists a principal bundle for $T'$ whose associated class in $H^1(M,\underline{T'})$ maps to the specified element of $H^2(M,\mu_n)$. Since $T'$ is a Lie subgroup of $\textbf{PGL}(n,\mathbb{C})$, there is an associated principal bundle for $\textbf{PGL}(n,\mathbb{C})$. By the functoriality of non-Abelian cohomology, this principal bundle for $\textbf{PGL}(n,\mathbb{C})$ has the same image in $H^2(M,\mu_n)$.

Finally, the long exact sequence of non-Abelian cohomology is part of the basic setup of non-Abelian cohomology. In general it is not true that there is any canonical isomorphism of the fibers of the connecting map with the pointed set preceding in the sequence. What is true is that there is a "twisting" construction. There is an excellent discussion of this in Serre's "Galois cohomology". You can also find some notes on my webpage: http://www.math.stonybrook.edu/~jstarr/papers/Escola_07_08d_nocomments.pdf

  • $\begingroup$ I apologize for my ambiguous notation, I probably should give the precise definitions for ${\Bbb C}^{\times}, SL(n,{\Bbb C})\cdots$ at first. Here ${\Bbb C}^{\times}$ denotes the constant sheaf, i.e. a sheaf of constant maps from open subsets to ${\Bbb C}^{\times}$. And $H^1(M, {\Bbb C}^{\times})$ is the set of all flat complex line bundles. The definitions of $SL(n,{\Bbb C})$ and $H^1(M, SL(n,{\Bbb C}))$ are similar. My first question is for any $\beta \in H^1(M, {\Bbb Z}_n)$, can we find a holomorphic vector bundle $E$, which is projectively flat, such that $\delta({\Bbb P}(E)) = \beta$. $\endgroup$
    – swalker
    Jun 15, 2017 at 13:19

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