Balanced metrics on twisted Higgs bundles

Abstract

A twisted Higgs bundle on a Kähler manifold X is a pair $(E,\phi )$ consisting of a holomorphic vector bundle E and a holomorphic bundle morphism $\phi :M \otimes E \rightarrow E$ for some holomorphic vector bundle M. Such objects were first considered by Hitchin when X is a curve and M is the tangent bundle of X, and also by Simpson for higher dimensional base. The Hitchin–Kobayashi correspondence for such pairs states that $(E,\phi )$ is polystable if and only if E admits a hermitian metric solving the Hitchin equation. This correspondence is a powerful tool to decide whether there exists a solution of the equation, but it provides little information as to the actual solution. In this paper we study a quantization of this problem that is expressed in terms of finite dimensional data and balanced metrics that give approximate solutions to the Hitchin equation. Motivation for this study comes from work of Donagi–Wijnholt (JHEP 05:068, 2013) concerning balanced metrics for the Vafa–Witten equations.

Appendix: Weakly geometric metrics

We include here the justification for the terminology used for weakly geometric metrics.

Proposition 8.1

Fix a hermitian metric H on E and let $\Vert \cdot \Vert _k$ be the geometric metric on $H^0(E(k))$ induced by $H_k:=H\otimes h_L^k$. Then if $\phi :M\otimes E\rightarrow E$ is non-zero then $\Vert \cdot \Vert _k$ is weakly geometric with respect to $\phi$. Moreover the constant $c'$ can be chosen uniformly as H varies in a bounded set of metrics on E.

Lemma 8.2

Let $M'$ and E be hermitian vector bundles and

$$\begin{aligned} m:H^0(E(k)) \otimes H^0(M')\rightarrow H^0(M'\otimes E(k)) \end{aligned}$$

be the natural multiplication map. Then there exists a constant C independent of k such that

$$\begin{aligned} \Vert m\Vert ^2\le C \end{aligned}$$

where all the vector spaces are endowed with the induced $L^2$-metric. In fact one can take

$$\begin{aligned} C= h^0(M') \sup \left\{ |t(x)|^2_{\infty } : t\in H^0(M'), \Vert t\Vert =1\right\} . \end{aligned}$$

Proof

Let $s_{\alpha }$ be an orthonormal basis for $H^0(E(k))$ and $t_{\beta }$ an orthonormal basis for $H^0(M')$. Any $v\in H^0(E(k))\otimes H^0(M')$ can be written as $v = \sum _{\beta } v_{\beta }$ where $v_{\beta }= \sum _{\alpha } a_{\alpha \beta } s_{\alpha }\otimes t_{\beta }$ for some coefficients $a_{\alpha \beta }$. So $\Vert v\Vert ^2 =\sum _{\beta } \Vert v_{\beta }\Vert ^2$ and $\Vert v_{\beta }\Vert ^2 = \Vert \sum _{\alpha } a_{\alpha \beta } s_{\alpha }\Vert ^2$.

Now let $C' = \sup _{\beta }\{ \Vert t_{\beta }\Vert ^2_{\infty }\}$. Then

$$\begin{aligned} |m(v_{\beta }(z))|^2 = |t_{\beta }(z)|^2 \left| \sum _{\alpha } a_{\alpha \beta } s_{\beta }(z)\right| ^2 \end{aligned}$$

and so

$$\begin{aligned} \Vert m(v_{\beta })\Vert \le C' \int _X \left| \sum _{\alpha } a_{\alpha \beta } s_{\beta }(z)\right| ^2 \frac{\omega ^n}{n!} = C' \Vert v_{\beta }\Vert ^2. \end{aligned}$$

Hence using Cauchy–Schwarz,

$$\begin{aligned} \Vert m(v)\Vert ^2 \le h^0(M') \sum _{\beta } \Vert m(v_{\beta })\Vert ^2 \le C \sum _{\beta } \Vert v_{\beta }\Vert ^2 = C \Vert v\Vert \end{aligned}$$

as claimed.$\square$

Proof of Proposition 8.1

We shall show that

$$\begin{aligned} \frac{c'}{{\text {rk}}_E}k^n \le {\left| \!\left| \!\left| \phi _{*} \right| \!\right| \!\right| }^2, \qquad \Vert \phi _{*}\Vert \le c'. \end{aligned}$$

(8.1)

We first deal with the operator norm. The map $\phi _*$ is the composition of the multiplication map $H^0(M)\otimes H^0(E(k)) \rightarrow H^0(M\otimes E(k)$ and the pushforward $H^0(M\otimes E(k))\rightarrow H^0(E(k))$. The norm of this multiplication map is bounded independent of k by Lemma 8.2 applied with $M'=M$. The norm of the pushforward is clearly bounded, as $\phi$ is continuous and the vector spaces are endowed with their $L^2$-metrics. Thus we have $\Vert \phi _*\Vert = O(k^0)$ as claimed.

Turning to the first equation in (5.2) recall that the leading order asymptotic of the Bergman kernel is given by

$$\begin{aligned} B_k = \sum _{\alpha } s_{\alpha } \otimes s_{\alpha }^{*,H_k} = k^n{\text {Id}}+ O(k^{n-1}) \end{aligned}$$

where $\{s_{\alpha }\}$ is an orthonormal basis for $H^0(E(k))$. Here $B_k$ is considered as an smooth section of ${\text {End}}(E)$ and the error term can be taken in the supremum norm determined by H, and is uniform as H varies over a bounded set. We recall that in this expression the term $s_{\alpha }\otimes s_{\alpha }^{*,H_k}$ denotes taking the fibrewise dual, so should be considered as an element in ${\text {End}}(E\otimes L^k) \simeq {\text {End}}(E)$, and under this identification

$$\begin{aligned} s_{\alpha }\otimes s_{\alpha }^{*,H_k}(\zeta ) = s_{\alpha }\tau ^{-1} (s_{\alpha },\zeta \otimes \tau )_{H_k}\quad \text {for } \zeta \in E_z, 0\ne \tau \in L_z^k. \end{aligned}$$

In particular taking the trace this implies $\sum _{\alpha }|s_{\alpha }(z)|_{H_k}^2 = {\text {rk}}_E k^n + O(k^{n-1})$. So applying the endomorphism $B_k$ to some non-zero $\zeta \in E_x$ gives

$$\begin{aligned} k^n\zeta + O(k^{n-1}) = \sum _{\alpha } s_{\alpha }(z) \tau ^{-1} (s_{\alpha }(z),\zeta \otimes \tau )_{H_k}. \end{aligned}$$

Now tensoring with some $\eta \in M_z$, applying $\phi$ and taking the norm-squared gives

$$\begin{aligned} k^{2n}|\phi (\zeta \otimes \eta )|_H^2&= \left| \sum _{\alpha } \phi (s_{\alpha }(z) \eta \tau ^{-1}) (s_{\alpha }(z),\zeta \otimes \tau )_{H_k} + O(k^{n-1})\right| _H^2\\&= \left| \sum _{\alpha } \phi (s_{\alpha }(z)\eta ) \tau ^{-1} (s_{\alpha }(z),\zeta \otimes \tau )_{H_k} + O(k^{n-1})\right| _H^2\\&\le \sum _{\alpha } |\phi (s_{\alpha }(z)\eta )|_{H_k}^2 \sum _{\alpha } |s_{\alpha }(z)|_{H_k}^2 |\zeta |_{H}^2 + O(k^{2n-1})|\eta |_{H_M}^2|\zeta |_H^2 \end{aligned}$$

where the last inequality uses Cauchy–Schwarz for the sum, and then again for the inner product $(s_{\alpha }(z),\zeta _z\otimes \tau )_{H_k}.$ Thus

$$\begin{aligned} k^{2n} \frac{|\phi (\zeta \otimes \eta )|_H^2}{|\zeta |_H^2} \le {\text {rk}}_E k^n \sum _{\alpha } |\phi _{*}(s_{\alpha }(z)\eta )|_{H_k}^2 + O(k^{2n-1})|\eta |_{H_M}^2. \end{aligned}$$

(8.2)

Now fix an orthonormal basis $t_{\beta }$ for $H^0(M)$. For each $\beta$ let $\phi _{\beta }:E\rightarrow E$ be

$$\begin{aligned} \phi _{\beta }(\zeta ) = \phi (\zeta \otimes t_{\beta }(z))\text { for } \zeta \in E_z \end{aligned}$$

Observe that since M is globally generated, and $\phi \ne 0$, there is at least one $\beta$ for which $\phi _{\beta }$ is non-zero. We let $c'_{\beta }$ be the $L^2$-metric of $\phi _{\beta }$, i.e.

$$\begin{aligned} c'_{\beta }:= \Vert \phi _{\beta }\Vert _H^2 := \int _X \Vert \phi _\beta |_z\Vert _H^2 \frac{\omega ^n}{n!} \end{aligned}$$

where $\Vert \phi _\beta |_z\Vert _H$ is the operator norm of $\phi _{\beta }|_z:E_z\rightarrow E_z$, and set

$$\begin{aligned} c':= \sum _{\beta } c'_{\beta }>0. \end{aligned}$$

Now if $t_{\beta }(z)\ne 0$ then substituting $\eta = t_{\beta }(z)$ into (8.2) gives

$$\begin{aligned} \frac{k^{n}}{{\text {rk}}_E} \frac{|\phi _{\beta }(\zeta )|_{H}^2}{|\zeta |_{H}^2}\le \sum _{\alpha } |\phi _{*}(s_{\alpha }\otimes t_{\beta }(z))|_{H_k}^2 + O(k^{n-1})|t_{\beta }(z)|_{H_M}^2 \end{aligned}$$

and taking the supremum over all non-zero $\zeta \in E_z$ gives

$$\begin{aligned} \frac{k^{n}}{ {\text {rk}}_E} \Vert \phi _{\beta }|_z\Vert _{H}^2 \le \sum _{\alpha } |\phi _{*}(s_{\alpha }\otimes t_{\beta }(z))|_{H_k}^2 + O(k^{n-1})|t_{\beta }(z)|_{H_M}^2. \end{aligned}$$

Moreover this inequality clearly holds if $t_{\beta }(z)=0$. Thus taking the sum over all $\beta$ and then integrating over X gives

$$\begin{aligned} \frac{c' k^{n}}{{\text {rk}}_E} \le \sum _{\alpha ,\beta } \Vert \phi _*(s_{\alpha }\otimes t_{\beta })\Vert _{H_k}^2 + O(k^{n-1}) = {\left| \!\left| \!\left| \phi _* \right| \!\right| \!\right| } + O(k^{n-1}) \end{aligned}$$

which gives (8.1).$\square$