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Let $T$ be an algebraic torus and $G$ a diagonalizable group; both are over an algebraically closed field $k$ of characteristic $0$ (take $k=\mathbb C$, if you like).

I am trying to understand principal $G$-bundles $P\to T$ over $T$, but I'm getting conflicting conclusions from two perspectives.


From what I've read (e.g. pg 11-13 here: https://arxiv.org/pdf/2009.08675.pdf), the total space $P$ is given by $$P=\operatorname{Spec}_T\left(\bigoplus_{m\in\Gamma} \mathcal L^{\otimes m}\right)$$ where $\Gamma$ is the character group of $G$ (so $\Gamma$ is a finitely generated abelian group) and $\mathcal L$ is a product of line bundles on $T$ indexed by generators for $\Gamma$ (some possibly torsion, if $\Gamma$ has torsion).

But line bundles on $T$ are all trivial, so $\color{red}{\text{I think we just have (?)}}$

$$\begin{align}\operatorname{Spec}_T\left(\bigoplus_{m\in\Gamma} \mathcal L^{\otimes m}\right) = \operatorname{Spec}_T\left(\bigoplus_{m\in\Gamma} \mathcal O_T^{\otimes m}\right) = \operatorname{Spec}(k[T][\Gamma]) &= \operatorname{Spec}(k[T]\otimes_k k[\Gamma])\\&= T\times G.\end{align}$$

This seems to imply that every principal $G$-bundle over $T$ is trivial.


On the other hand, consider the following example. Take $G=\{1,-1\}$ (i.e. $\mu_2$) as a subgroup of $T=\mathbb G_m$ (rank $1$ torus). Then we have an exact sequence $$1 \to G \to T \xrightarrow{t\mapsto t^2} T \to 1$$ where the map $T\to T$ via $t\mapsto t^2$ is a principal $G$-bundle over $T$. This should not be trivial, since $T\not\cong T\times G$ in this case (i.e. the exact sequence does not split).

This seems to contradict the above approach.


My question is: can anyone explain where I've gone wrong? Is one of the approaches correct and the other incorrect?

Edit: I think in the first approach the $\mathcal O_T$-algebra structure will depend on the torsion in $\Gamma$. For example, if $\Gamma=\mathbb Z/2$, then this $\mathcal O_T$-algebra structure will depend on a choice of isomorphism $\mathcal O_T^{\otimes 2}\to\mathcal O_T$. If this is right, then I guess it comes down to understanding these isomorphisms and how they give different $\mathcal O_T$-algebra structures, which is still confusing to me.

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  • $\begingroup$ I think you might benefit from more carefully studying what torsors are. For instance, a $\mu_n$-torsor on some $X$ corresponds (in reasonable situations) to equivalence classes of pairs $(\mathscr{L},\iota)$ where $\mathscr{L}$ is a line bundle on $X$ and $\iota\colon \mathscr{L}^{\otimes n}\to \mathcal{O}_X$ is an isomorphism, where two pairs $(\mathscr{L}_i,\iota_i)$ are equivalent if there exists an isomorphism $f\colon \mathscr{L}_1\to \mathscr{L}_2$ such that $\iota_2\circ f^{\otimes m}=\iota_1$. $\endgroup$ Feb 23, 2023 at 10:18
  • $\begingroup$ In particular, if $\mathrm{Pic}(X)=0$, then $\mathscr{L}$ is abstractly isomorophic to $\mathcal{O}_X$. But, note that the isomorphisms $\iota$ can then just be identified with $\mathcal{O}_X(X)^\times$, and the equivalence relation then means that $\mu_n$-torsors are classified by $\mathcal{O}_X(X)^\times/(\mathcal{O}_X(X)^\times)^n$. In particular, if $X=\mathbb{G}_{m,k}$ where $k$ is algebraically closed, then $\mathrm{Pic}(X)=0$, but $\mathcal{O}(X)^\times/(\mathcal{O}_X(X)^\times)^n=(k^\times T^\mathbb{Z})/(k^\times T^\mathbb{Z})^n$ which is precisely $\mathbb{Z}/n\mathbb{Z}$. $\endgroup$ Feb 23, 2023 at 10:20
  • $\begingroup$ This shouldn't be shocking as (at least in reasonable situations) $\mu_n=\mathbb{Z}/n\mathbb{Z}$ and so $H^1(X,\mu_n)=H^1(X,\mathbb{Z}/n\mathbb{Z})=\mathrm{Hom}(\pi_1(X),\mathbb{Z}/n\mathbb{Z})$. We see then that $\mathbb{Z}/n\mathbb{Z}$ is coming from the automorphisms of the $n$-fold cover $\mathbb{G}_{m,k}\xrightarrow{t\mapsto t^n}\mathbb{G}_{m,k}$ as you already observed. $\endgroup$ Feb 23, 2023 at 10:23
  • $\begingroup$ By the way, the total space is of the $G$-bundle is not what you write -- what is the fiber of that map look like? For instance while it is true that the group of $\mathbb{G}_m$-torsors is isomorphic to $\mathrm{Pic}(X)$ the isomorphism doesn't take $\mathscr{L}$ to $\mathscr{L}\to X$ (i.e. doesn't take the invertible sheaf to the associated geometric line bundle). Indeed, the latter is actually a $\mathbb{G}_a$-bundle. Instead, the total space of the associated $\mathbb{G}_m$-bundle is $\underline{\mathrm{Isom}}(\mathscr{L},\mathcal{O}_X)$ (the 'frame bundle') and is isomorphic to $\endgroup$ Feb 23, 2023 at 10:28
  • $\begingroup$ $\mathscr{L}-\{0\}\to X$ (i.e. the geometric line bundle minus the zero section). $\endgroup$ Feb 23, 2023 at 10:28

1 Answer 1

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Just to get this off the unanswered list, I will elaborate on my comment above.

Your mistake was in the understanding of the total space of a $\mu_{n,X}$-torsor as $\underline{\mathrm{Spec}}\left(\bigoplus_{i=0}^{n-1}\mathscr{L}^{\otimes i}\right)\to X$ where $\mathscr{L}$ is an $n$-torsion element of $\mathrm{Pic }(X)$. Namely, how are you thinking of a $\bigoplus_{i=0}^{n-1}\mathscr{L}^{\otimes i}$ as a quasi-coherent $\mathscr{O}_X$-algebra? Namely, the multiplication here should be obtained via the natural maps

$$\mathscr{L}^{\otimes a}\otimes \mathscr{L}^{\otimes b}\to \mathscr{L}^{\otimes(a+b)},$$

but, of course, as we only consider $\mathscr{L}^{\otimes i}$ for $i=0,\ldots,n-1$ what you really need is a map like

$$\mathscr{L}^{\otimes a}\otimes \mathscr{L}^{\otimes b}\to \mathscr{L}^{\otimes(a+b\mod n)}.$$

This is doable as $\mathscr{L}^{\otimes n}\cong \mathscr{O}_X$, but not canonically so. In fact, you need to fix an isomorphism $\iota\colon \mathscr{L}^{\otimes n}\xrightarrow{\approx}\mathscr{O}_X$ and then from this structure one can actually give $\bigoplus_{i=0}^{n-1}\mathscr{L}^{\otimes i}$ the structure of a quasi-coherent $\mathscr{O}_X$-algebra.

Let us call this algebra structure $\mathscr{A}(\mathscr{L},\iota)$. Then, one can check that the isomorphism class of $\mathscr{A}(\mathscr{L},\iota)$ only depends on the following equivalence relation on the pair $(\mathscr{L},\iota)$: $(\mathscr{L}_1,\iota_1)\sim (\mathscr{L}_2,\iota_2)$ is there exists an isomorphism $f\colon \mathscr{L}_1\xrightarrow{\approx}\mathscr{L}_2$ such that $\iota_2\circ f^{\otimes n}=\iota_1$. In this way, using $H^1(X,\mu_{n,X})$ to denote the isomorphism classes of $\mu_{n,X}$-torsors, one gets a map

$$\{(\mathscr{L},\iota)\}/\sim\, \longrightarrow\, H^1(X,\mu_{n,X}).$$

This is actually an isomorphism of groups where one defines a group structure on the source in the obvious way: $(\mathscr{L}_1,\iota_1)\otimes (\mathscr{L}_2,\iota_2)=(\mathscr{L}_3,\iota_3)$ where $\mathscr{L}_3=\mathscr{L}_1\otimes\mathscr{L}_2$, and $\iota_3$ is the composition of

$$(\mathscr{L}_1\otimes\mathscr{L}_2)^{\otimes n}\xrightarrow{\text{natural}}\mathscr{L}_1^{\otimes n}\otimes \mathscr{L}_2^{\otimes n}\xrightarrow{\iota_1\otimes\iota_2}\mathscr{O}_X\otimes\mathscr{O}_X\xrightarrow{\text{natural}}\mathscr{O}_X.$$

There are several ways to prove this assertion, let me list two here.

  1. You can first prove that the association of $\mathscr{L}$ of the sheaf $\underline{\mathrm{Isom}}(\mathscr{O}_X,\mathscr{L})$ (which associates to an $X$-scheme $T$ the isomorphisms $\mathscr{O}_T\to \mathscr{L}_T$) is actually an isomorphism $\mathrm{Pic}(X)\to H^1(X,\mathbf{G}_{m,X})$ -- this is classical, for instance see here. One can then deduce the result for $\mu_{n,X}$ by tracing through the long exact sequence in cohomology obtained from the Kummer sequence $1\to \mu_{n,X}\to\mathbf{G}_{m,X}\to\mathbf{G}_{m,X}\to 1$.
  2. A more conceptual reason (which also handles the case of $\mathbf{G}_{m,X}$-torsors) is to see that $\mathscr{A}(\mathscr{L},\iota)$ represents the scheme $\underline{\mathrm{Isom}}((\mathscr{O}_X,\iota_0),(\mathscr{L},\iota))$ where $\iota_0\colon \mathscr{O}_X^{\otimes n}\xrightarrow{\approx}\mathscr{O}_X$ is the obvious map. One then wins from the general yoga that if $\mathcal{S}$ is any stack over a site $\mathcal{X}$, and any $x$ is any object of $\mathcal{C}_T$ (for $T$ an object of $\mathcal{X}$) then the association $y\mapsto \underline{\mathrm{Isom}}(x,y)$ is an isomorphism $\mathrm{Tw}(x)\to H^1(T,\underline{\mathrm{Aut}}(x))$. Here $\mathrm{Tw}(x)$ is the (isomorphism classe) of objects of $\mathcal{C}_T$ which are locally isomorphic to $x$ (i.e. the 'twists'). Why is this relevant here? Well, line bundlesform a stack, and therefore so do pairs $(\mathscr{L},j)$ where $j\colon \mathscr{L}^{\otimes m}\to \mathscr{O}_X$. Note then that $\{(\mathscr{L},\iota)\}/\sim$ is precisely $\mathrm{Tw}((\mathscr{O}_X,\iota_0))$. But, what is $\underline{\mathrm{Aut}}(\mathscr{O}_X,\iota)$? It's precisely $\mu_{n,X}$!
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