# Principal bundle (or torsor) for a diagonalizable group over a torus

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.

• 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 $\iota_2\circ f^{\otimes m}=\iota_1$. Commented Feb 23, 2023 at 10:18
• 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 $\mathbb{Z}/n\mathbb{Z}$. Commented Feb 23, 2023 at 10:20
• 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. Commented Feb 23, 2023 at 10:23
• 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 Commented Feb 23, 2023 at 10:28
• $\mathscr{L}-\{0\}\to X$ (i.e. the geometric line bundle minus the zero section). Commented Feb 23, 2023 at 10:28

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 $$\mu_{n,X}$$!