Let $G$ be a semisimple algebraic group, $C$ be a smooth projective curve, and $\omega$ be the canonical line bundle.

The stack $\mathrm{Higgs}_{\omega}$ is defined as the stack associating to each $S$ the groupoid consisting of $(E, \phi)$, where $E$ is a $G$-torsor over $X \times S$ and $\phi \in \Gamma(C \times S, \operatorname{ad}(E) \otimes_C D)$. Here, how, given this torsor $E$ on $C \times S$, does $\operatorname{ad}(E)$ refer to the associated bundle $E \times_G \mathfrak{g}$?

Main question: Given a stack $X$, one can abstractly define its co-tangent stack. How does one show that the abstract definition of $T^* \mathrm{Bun}_G$ can be identified with $\mathrm{Higgs}_{\omega}$?


The tangent complex to $\operatorname{Bun}_G(C)$ can be identified with $T_{\operatorname{Bun}_G(C)}=\mathbf{R}\pi_*{\operatorname{ad} P[1]}$, where $\pi:\operatorname{Bun}_G(C)\times C\rightarrow \operatorname{Bun}_G(C)$ is the natural projection and $P$ is the universal bundle.

Then the cotangent stack is $T^*{\operatorname{Bun}_G(C)} = \operatorname{Spec} \operatorname{Sym} (T_{\operatorname{Bun}_G(C)})$. Maps from $U$ into the total space of the bundle $T^*{\operatorname{Bun}_G(C)}\rightarrow \operatorname{Bun}_G(C)$ are the same as maps $U\rightarrow \operatorname{Bun}_G(C)$ together with a section of the dual sheaf of $T_{\operatorname{Bun}_G(C)}$. Relative Serre duality identifies $\mathcal{Hom}(T_{\operatorname{Bun}_G(C)}, \mathcal{O})$ with $\mathbf{R}\pi_*\mathcal{Hom}(\operatorname{ad} P, \omega_C)\cong\mathbf{R}\pi_*(\operatorname{ad} P\otimes\omega_C)$ using the Killing form.

So, maps $U\rightarrow H^0(T^*{\operatorname{Bun}_G(C)})$ to the underlying ordinary stack are identified with $G$-bundles over $U\times C$ and a section $\phi\in H^0(U\times C, \operatorname{ad} P\otimes \omega_C)$.

  • $\begingroup$ Thanks Pavel! Quick question: how does one derive the description of the tangent complex you mention at the start? What is the universal bundle P on $Bun_G(C)×C$ - is it the pullback of some object on $Bun_G(C)$? $\endgroup$
    – Vinoth
    Dec 22 '12 at 15:54
  • $\begingroup$ I relearned today to my surprise that base change for coherent sheaves doesn't hold in general. So how do you know here that the fiber of the pushforward at some point is the cohomology of the fiber? $\endgroup$ Jan 13 '18 at 14:59
  • $\begingroup$ Base change works for schematic quasi-compact quasi-separated morphisms of prestacks (see e.g. Prop 3.10 in arxiv.org/abs/0805.0157). For instance, $C\times \mathrm{Bun}_G(C)\rightarrow \mathrm{Bun}_G(C)$ satisfies the assumptions. $\endgroup$ Jan 13 '18 at 18:39

$\newcommand\leftnamed[1]{\ \stackrel{#1}\longleftarrow\ }\newcommand\rightnamed[1]{\ \stackrel{#1}\longrightarrow\ }$I meant to leave this as a comment to add to Pavel's excellent answer, answering Vinoth's first question in the comments, but it was too long.

Why is the tangent complex of $\operatorname{Bun}_G(X)$ equal to $\pi_*{\operatorname{ad}P[1]}$? Well, for a mapping stack we have $$Y \leftnamed a X\times\operatorname{Map}(X,Y) \rightnamed b \operatorname{Map}(X,Y)$$ and its mapping stack is $\mathbf{T}_{\operatorname{Map}(X,Y)}=b_*a^*\mathbf{T}_Y$.† Apply this to our case $\operatorname{Bun}_G(X)=\operatorname{Map}(X,BG)$: $$BG\ \stackrel{a}{\longleftarrow}\ X\times\text{Bun}_G(X)\ \stackrel{\pi}{\longrightarrow}\ \text{Bun}_G(X).$$ The tangent complex of classifying space is $\mathbf{T}_{BG}=\mathfrak{g}[1]$ concentrated in degree one. This follows from the distinguished triangle of $\mathbf{T}$ for the composition $$* \rightnamed{} BG \rightnamed{} *$$ Note that this is just $\operatorname{ad}\gamma[1]$, where $\gamma=*\to BG$ is the tautological bundle. Next, the tautological bundle $P$ is defined to be the pullback $\require{AMScd}$ \begin{CD} P @>>> *\\ @V V V@VV V\\ X\times\operatorname{Bun}_G(X) @> a>> BG \end{CD} so $a^*\mathbf{T}_{BG}=\operatorname{ad}P[1]$ definitionally. Pushing forward by $\pi_*$ we are done.

†See Toen's "Derived Algebraic Geometry and Deformation Quantization".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.