# Scheme of Higgs reductions

I'm reading the Bruzzo and Graña Otero's paper Semistable and Numerically Effective Principal (Higgs) Bundles; here: $X$ is a smooth, complex, projective variety; $G$ is a connected, complex, reductive, (affine) algebraic group; $E$ is a principal $G$-bundle on $X$.

A Higgs field $\varphi$ on $E$ is a global section of $Ad(E)\otimes_{\mathcal{O}_X}\Omega_X^1$ such that $[\varphi,\varphi]=0$. A principal Higgs $G$-bundles on $X$ is a pair $\mathfrak{E}=(E,\varphi)$.

Let $K$ be a closed subgroup of $G$ and let $\sigma:X\to E_{\displaystyle/K}$ be a reduction of the structure group of $E$ to $K$; that is, there exists a principal $K$-bundle $F_{\sigma}$ on $X$ and an injective $K$-equivariant bundle morphism $i_{\sigma}:F_{\sigma}\to E$. Let $\Pi_{\sigma}:Ad(E)\otimes_{\mathcal{O}_X}\Omega^1_X\to\left(Ad(E)_{\displaystyle/Ad(F_{\sigma})}\right)\otimes_{\mathcal{O}_X}\Omega^1_X$ be the induced projection; $\sigma$ is called a Higgs reduction of $\mathfrak{E}$ if and only if $\varphi\in\ker\Pi_{\sigma}$.

Let $E_{K}$ denotes the principal $K$-bundle $E\to E_{\displaystyle/K}$. Because: \begin{gather*} T_{E/K,X}\cong E_{\displaystyle/K}\left(Ad_K,\mathfrak{g}_{\displaystyle/\mathfrak{k}}\right),\\ \pi_K^{*}Ad(E)=\pi_K^{*}\left(E(Ad_G,\mathfrak{g})\right)=E_{\displaystyle/K}(Ad_K,\mathfrak{g}), \end{gather*} there exists a natural morphism $\eta:\pi_K^{*}Ad(E)\to T_{E/K,X}$ and $\varphi$ determines a global section $\eta(\varphi)\equiv(\eta\otimes Id)\left(\pi_K^{*}\varphi\right)$ of $T_{E/K,X}\otimes_{\mathcal{O}_{E/K}}\Omega^1_{E/K}$; the zero locus $\mathfrak{R}_K(\mathfrak{E})$ of $\eta(\varphi)$ is called scheme of Higgs reduction of $\mathfrak{E}$ (definition 3.5).

They state (page 6):

By construction, $\sigma:X\to E\left(G_{\displaystyle/K}\right)\cong E_{\displaystyle/K}$ is a Higgs reduction if and only if it takes value in the subscheme $\mathfrak{R}_K(\mathfrak{E})\subset E_{\displaystyle/K}$.

But I had check, via computation, that for any reduction $\sigma$ of $E$ one has $\mathfrak{R}_K(\mathfrak{E})$ is contained in the scheme-theoretic image of $\sigma$. Am I wrong?

• I suspect you made a mistake. For instance, suppose $G=GL(2)$ and $K$ is the subgroup of upper-triangular matrices. Then $E$ is basically a rank two vector bundle on $X$, and $\phi$ is an $\Omega^1_X$-valued endomorphism of $E$ (satisfying $[\phi,\phi]=0$, but the condition is unimportant here). Now $E_{/K}$ is the ${\mathbb P}^1$-bundle of lines in $E$. All the paper is saying is that the `lower left entry' of $\phi$ (with respect to a $K$-reduction of $E$) can be viewed as a section of some vector bundle on $E_{/K}$, and that the reduction is Higgs iff the lower left entry vanishes. Jun 22, 2016 at 14:21
• @t3suji Yes, thank you: I had find a mistake; and your example is clear. Jun 23, 2016 at 12:30

## 1 Answer

Generalising t3suji's comment, let us consider $GL(n)$ Higgs bundles, i.e., rank $n$ Higgs vector bundles $\mathcal E$. The Higgs Grassmannian of rank $m$ quotients is a closed subscheme of the usual Grassmannian bundle; in general it is a proper subset, as there are quotients of $\mathcal E$ that are not Higgs quotients. The images of corresponding sections of the Grassmannian bundle are not contained in the Higgs Grassmannian. This is what the paper states. (In this case $K$ is a suitable parabolic subgroup of $GL(n)$).

• Ok, I understand the ideas and your generalization; but I don't see the obviousness of this from the computation. Can you give me a hint? Jun 23, 2016 at 12:32
• Equation (4) in the paper cited in your question and the discussion around it show that, in the vector bundle case, the Higgs Grassmannian is given by the vanishing of an algebraic morphism, and is therefore a closed subscheme of the usual Grassmannian bundle. The definition of this morphism implies that a quotient is Higgs if and only if the corresponding section takes values in the Higgs Grassmannian. Examples are given in U. Bruzzo, D. Hernandez Ruiperez, Diff. Geom. Appl. 29 (2011) 147-153. Jun 23, 2016 at 13:30