text:
\section{Introduction}
Let $X$ be a compact $n$-dimensional complex manifold endowed with a Gauduchon metric $g$ whose fundamental form is $\omega$, and let $V$ be a fixed holomorphic vector bundle on $X$. Consider a pair $(E,\phi)$ consisting of a holomorphic vector bundle $E$ on $X$ and a holomorphic section $\phi \in H^0(X,\End E \otimes V)$ such that $\phi \wedge \phi = 0$. The section $\phi$ is  called a {\em Higgs field}. The pair $(E,\phi)$ is then said to be {\em $g$-stable} if for any proper subsheaf $\mc{S} \subset E$ such that $\phi(\mc{S}) \subset \mc{S} \otimes V$ (that is, $\mc{S}$ is {\em $\phi$-invariant}), we have
 \[ \frac{\deg_g(\mc{S})}{\rank \mc{S}} < \frac{\deg_g(E)}{\rank E},\]
 in which case $\phi$ is a {\em $g$-stable Higgs field}. Note that the degree of a torsion-free coherent sheaf $\mc{F}$ on a Gauduchon manifold $(X,g)$ is defined as
 \[ \deg_g(\mc{F}) = \int_X F_h \wedge \omega^{n-1},\]
 where $F_h$ is the curvature of the Chern connection of any Hermitian metric $h$ on $\det \mc{F}$ \cite{LuTe}.
 Moreover, the endomorphism bundle $\End E$ of $E$ naturally splits as $\End E \simeq  \mc{O}_X \oplus \End_0 E$,
 where the sections of $\mc{O}_X$ correspond to multiples of $\id_E$ and the sections of $\End_0 E$ are the trace-free endomorphisms of $E$. We thus have
 \[ H^0(X,\End E \otimes V) = H^0(X,V) \oplus H^0(X,\End_0 E \otimes V).\]
 Given that sections of $H^0(X,V)$ can easily be computed, we focus our analysis on the non-zero trace-free Higgs fields, that is, sections $\phi \in H^0(X,\End_0 E \otimes V)$ with $\phi \neq 0$ and $\phi \wedge \phi = 0$, which we call {\em non-trivial}.

In this paper, we are interested in the case $V = \mc{T}_X$ or $\mc{T}^*_X$, where $\mc{T}_X$ and $\mc{T}^*_X$ are, respectively, the holomorphic tangent and cotangent bundles of $X$. If $V = \mc{T}_X$, the pairs $(E,\phi)$ are called {\em co-Higgs bundles} and correspond to generalized holomorphic vector bundles over the complex manifold $X$ \cite{Gualtieri,Hit-B,Rayan}. Stable rank-2 co-Higgs bundles have been studied on curves \cite{Rayan, Rayan-P1} and on certain projective surfaces \cite{Rayan-P2, Vicente, Biswas-Rayan}. Rank-2 co-Higgs bundles are also related to holomorphic Poisson structures \cite{Polishchuk, Rayan, Matviichuk}. Indeed, given a rank-2 co-Higgs bundle $(E,\phi)$, if the Higgs field $\phi$ is trace-free it induces a non-trivial holomorphic Poisson structure on $\mathbb{P}(E)$.  On the other hand, if $V = \mc{T}^*_X$, the pairs $(E,\phi)$ are called {\em Higgs bundles}. Higgs bundles were first introduced by Hitchin \cite{Hit-Higgs} on algebraic curves and generalised to higher dimensional varieties by Simpson \cite{Simpson}. They also correspond to a special class of solutions of the Kapustin-Witten equations on compact K\"ahler surfaces \cite{Tanaka}. Although Higgs bundles over curves have been extensively studied over the past thirty-five years, less is known in the higher dimensional case. Some results have nonetheless been obtained in higher dimensions by Biswas and Bottacin \cite{Biswas2, Biswas, Bottacin-Higgs}, among others.

 In this article, we consider trace-free co-Higgs and Higgs bundles over non-K\"ahler elliptic surfaces, which correspond to non-trivial principal bundles over a complex curve $B$ with structure group an elliptic curve. We first prove that non-trivial stable co-Higgs bundles only exist if the genus of $B$ is zero, in which case $X$ is a Hopf surface, and non-trivial stable Higgs bundles only exist if the genus of $B$ is greater than or equal to 2. This is done in sections \ref{Existence} and \ref{elliptic}.
 Note that this is consistent with the case of curves: it is known that on a curve $C$ of genus $g(C)$, non-trivial stable Higgs bundles only exist if $g(C) \geq 2$ \cite{Hit-Higgs}, whereas non-trivial stable co-Higgs bundles only exist if $g(C)=0$ \cite{Rayan}. The result for non-K\"ahler elliptic surfaces is, however, a consequence of the following more general result:

 \begin{proposition*}
 Let $(X,g)$ be a Gauduchon manifold such that $\mc{T}_X$ (resp. $\mc{T}_X^*$) is given by an extension of vector bundles $$\begin{tikzcd}0 \arrow[r] & V_1 \arrow{r}{i} & \mc{T}_X \arrow{r}{p} & V_2 \arrow[r] & 0.
\end{tikzcd}$$
If there are no non-trivial $g$-stable pairs $(E, \vp_1:E \to E\otimes V_1)$ or $(E, \vp_2:E \to E\otimes V_2)$, then $X$ does not admit non-trivial $g$-stable co-Higgs bundles (resp. Higgs bundles).
 \end{proposition*}

 In section \ref{Hopf}, we classify all non-trivial rank-2 co-Higgs bundles. The results we obtain can be summarised as follows:

\begin{theorem*}
Let $X$ be a Hopf surface and let $c_2$ be a non-negative integer. For any integer $c_2\geq 0$, there is a non-trivial stable Higgs bundle $(E,\phi)$ on $X$ such that $c_2(E)=c_2$.
\end{theorem*}
As a direct consequence of this theorem, we also have:
\begin{corollary*}
There exist non-trivial holomorphic Poisson structures on $\mathbb{P}(E)$ for filtrable rank-2 bundles $E$ on Hopf surfaces.
\end{corollary*}

Finally, we study rank-2 Higgs bundles on non-K\"ahler elliptic surfaces in section \ref{geq2}. We show, in particular, that when the genus of the base curve $B$ is at least two, there exist non-trivial Higgs pairs $(E,\phi)$ with $c_2(E) = c_2$ for every $c_2 \geq 0$ as long as the underlying bundle $E$ is filtrable. We then use this result to prove:
\begin{theorem*}
Let $\pi:X\to B$ be a principal elliptic surface with base curve $B$ of genus at least two. Let $\mc{M}_{2,\delta,c_2}(X)$ be the moduli space of stable rank-2 torsion-free sheaves on $X$, determinant bundle $\delta$, and second Chern class $c_2$. Then $\mc{M}_{2,\delta,c_2}(X)$ is smooth if and only if every rank-2 vector bundle $E$ such that $\det(E)=\delta$ and $c_2(E)=c_2$ is irreducible.
\end{theorem*}

Note that moduli spaces of stable bundles on non-K\"ahler elliptic surfaces were studied in \cite{BrMor2}, where the authors showed that the moduli spaces were always smooth when the base curve of the elliptic fibration has genus 0 or 1 (see Proposition 4.1). However, no explicit conditions on the Chern classes  were given for smoothness when the genus is at least 2. The above theorem settles this question.

\section{A Necessary Condition for Trace-Free Higgs Fields}\label{Existence}

Let $X$ be a compact complex manifold with Gauduchon metric $g$ and $V$ be a fixed holomorphic vector bundle on $X$. In this section, we derive some facts about pairs $(E,\phi)$ where $E$ is a holomorphic vector bundle on $X$ and $\phi \in H^0(X,\End E \otimes V)$.

We first consider the case where $V$ is a line bundle.

\begin{proposition}\label{tracefree}
Let $(X,g)$ be a compact Gauduchon manifold. If $V$ is a line bundle on $X$ and $(E,\phi: E \rightarrow E \otimes V)$ is a $g$-stable pair with $\phi \neq 0$, then $\deg V \geq 0$.
\end{proposition}

\begin{proof}
We first note that the Higgs field $\phi: E \rightarrow E \otimes V$ on $E$ induces the Higgs field $\phi' := \phi \otimes \id_V: E \otimes V \rightarrow (E \otimes
V) \otimes V$ on $E \otimes V$. Moreover, since $V$ is a line bundle, $P$ is a $\phi$-invariant subsheaf of $E$ if and only if $P \otimes V$ is a $\phi'$-invariant subsheaf of $E \otimes V$. Therefore, $(E,\phi)$ is stable if and only if $(E \otimes V,\phi')$ is stable. Also note that $\ker \phi$ is a $\phi$-invariant subsheaf of $E$ and $\im \phi$ is a $\phi'$-invariant subsheaf of $E \otimes V$. Moreover, $\rank(E) = \rank(E \otimes V)$ so that
\[ \mu(E \otimes V) = \mu(E) + \deg(V).\]
Suppose that $\mu(\im \phi) = \mu(E \otimes V)$. By stability of $(E \otimes V,\phi')$, we must then have
\[\rank(\im \phi) = \rank(E \otimes V) = \rank(E).\]
However, $\im \phi \simeq E / \ker \phi$ and $\ker \phi$ is a torsion-free subsheaf of $E$ if $\phi$ is not injective, in which case $\rank(\im \phi) < \rank(E)$. Hence, $\ker \phi = 0$, implying that $\im \phi \simeq E$. Thus, $\mu(E) = \mu(\im \phi)$ and $\deg V = 0$ in this case.

Let us now assume that $\mu(\im \phi) \neq \mu(E \otimes V)$. Then, $\im \phi$ is a nonzero torsion-free subsheaf of $E \otimes V$ since $\im \phi \neq 0$ by assumption. If $\rank(\im \phi) = \rank(E \otimes V)$, then $\mu(\im \phi) \leq \mu(E \otimes V)$ so that $\mu(\im \phi) < \mu(E \otimes V)$. Otherwise, $\im \phi$ is a $\phi'$-invariant proper subsheaf of $E \otimes V$, which means that $\mu(\im \phi) < \mu(E \otimes V)$ by stability of $(E \otimes V,\phi')$. Thus, $\mu(\im \phi) < \mu(E \otimes V)$ in both cases. Consider the exact sequence
\[ 0 \rightarrow \ker \phi \rightarrow E \rightarrow E/\ker \phi \simeq \im \phi \rightarrow 0.\]
If $\ker \phi = 0$, then $E \simeq \im \phi$ and $\mu(E) = \mu(\im \phi)$. Hence,
\[ \mu(E) = \mu(\im \phi) < \mu(E \otimes V) = \mu(E) + \deg V, \]
implying that $\deg V > 0$. If instead $\ker \phi \neq 0$, then it is a proper $\phi$-invariant subsheaf of $E$ with $\rank(\ker \phi) < \rank(E)$. Indeed, $\ker \phi \neq E$ and $\im \phi \neq 0$ since $\phi \neq 0$. And if $\rank(\ker \phi) = \rank E$, then $\im \phi \simeq E/\ker \phi$ is a non-zero torsion subsheaf of $E$, which is impossible. Thus, $\rank(\ker \phi) < \rank(E)$ and $\mu(\ker \phi) < \mu(E)$ by stability of $(E,\phi)$. Moreover, given the above exact sequence, we obtain $\mu(E)<\mu(\im \phi)$, implying again that $\mu(E) < \mu(E \otimes V)$ and $\deg V > 0$. Putting all cases together gives $\deg V \geq 0$.
\end{proof}

\begin{remark}
This is result similar to what is known about (co)-Higgs bundles on curves: if $X$ is a curve, then stable (co)-Higgs bundles with non-zero Higgs field exist on $X$ if and only if $\mc{T}_X^*$ (resp. $\mc{T}_X$) has non-negative degree \cite{Hit-Higgs, Rayan}.
\end{remark}

In the case where $V$ is the trivial line bundle on $X$, we can say more about $g$-stable pairs:

\begin{proposition}\label{endo}
Let $(X,g)$ be a compact Gauduchon manifold and $(E,\phi)$ be a $g$-stable pair with $\phi \in H^0(X,\End E)$. Then, $\phi = \lambda \id_E$ for some $\lambda \in \mathbb{C}$.
\end{proposition}
\begin{remark}
This proposition gives a slight generalisation of \cite[Theorem 2.1]{Biswas}.
\end{remark}
\begin{proof}
Suppose that $\phi \neq 0$ so that $\im \phi \neq 0$. If $\ker \phi = 0$, then $E \simeq \im \phi$ and $\im \phi$ is a subbundle of $E$ with $\rank(\im \phi) = \rank(E)$, implying that $\phi$ is an automorphism. Let us assume instead that $\ker \phi \neq 0$ so that $\ker \phi$ is a torsion-free subsheaf of $E$ and $\rank(\im \phi) = \rank(E/\ker \phi) < \rank(E)$. Moreover, since $\im \phi \neq 0$, it is a torsion-free subsheaf of $E$, implying that $\rank(\ker \phi) < \rank(E)$. Hence, $\ker \phi$ is a $\phi$-invariant proper subsheaf of $E$ that fits into the exact sequence
\[ 0 \rightarrow \ker \phi \rightarrow E \rightarrow E/\ker \phi \simeq \im \phi \rightarrow 0.\]
By stability of $(E,\phi)$, we have $\mu(\ker \phi) < \mu(E)$, and so $\mu(E)<\mu(\im \phi)$. On the other hand, $\im \phi$ is a non-zero $\phi$-invariant subsheaf of $E$. Moreover, if $\rank(\im \phi) = \rank(E)$, then $\mu(\im \phi) \leq \mu(E)$; and if $\rank(\im \phi) < \rank(E)$, then $\im \phi$ is a proper subsheaf of $E$, implying that $\mu(\im \phi) < \mu(E)$ by stability of $(E,\phi)$. Hence, $\mu(E) < \mu(\im \phi) \leq \mu(E)$, leading to a contradiction. Thus, $\ker \phi = 0$ and $\phi$ is an automorphism of $E$.

Since $\mathbb{C}$ is algebraically closed and $\phi \neq 0$, there exists $\lambda \in \mathbb{C}^*$ such that $\ker(\phi - \lambda\id_E) \neq 0$. Moreover, $(E,\phi)$ is stable if and only if $(E,\phi' = \phi - \lambda\id_E)$ is stable (because any subsheaf $P$ of $E$ is $\phi$-invariant if and only if it is $\phi'$-invariant). Therefore, as before, $\phi'$ is an automorphism if $\phi' \neq 0$. However, $\phi'$ cannot be an automorphism since $\ker(\phi - \lambda\id_E) \neq 0$. Hence, $\phi' = 0$ implying that $\phi = \lambda \id_E$ for some $\lambda \in \mathbb{C}^*$.
\end{proof}

Finally, if $V$ is an extension of vector bundles, we have the following:
\begin{proposition}\label{extension}
Let $(X,g)$ be a compact Gauduchon manifold. Suppose that $V$ is an extension of holomorphic vector bundles $$\begin{tikzcd}0 \arrow[r] & V_1 \arrow{r}{i} & V \arrow{r}{p} & V_2 \arrow[r] & 0
\end{tikzcd}$$
on $X$. If $X$ has no non-trivial $g$-stable pairs of the form $(E, \vp_1:E \to E\otimes V_1)$ or $(E, \vp_2:E \to E\otimes V_2)$, then it has no non-trivial $g$-stable pairs of the form $(E,\phi: E \to E\otimes V)$.

\end{proposition}
\begin{proof}
Suppose that every trace-free $g$-stable pair $(E, \vp_i:E \to E\otimes V_i)$ has $\vp_i=0$, where $i=1,2$. Let $\phi:E \to E\otimes V$ be a trace-free stable pair, and set $\vp_2:=(\id_E\otimes p)\circ \phi$. Suppose $P_2$ is a $\vp_2$-invariant proper subsheaf of $E$. Then $\vp_2(P_2)\subset P_2\otimes V_2$, so $\phi(P_2)\subset P_2\otimes V$, meaning that $P_2$ is also $\phi$-invariant. By stability of $(E,\phi)$, we have $\mu(P_2)<\mu(E)$ so that $(E,\vp_2)$ is also stable. Since $\vp_2$ is clearly trace-free, $\vp_2=0$, implying that $\phi=\vp_1\otimes i$ for some trace-free $\vp_1:E \to E\otimes V_1$. Suppose $P_1$ is a $\vp_1$-invariant proper subsheaf of $E$. Then $\vp_1(P_1)\subset P_1\otimes V_1\subset P_1\otimes V$, so $\mu(P_1)<\mu(E)$ by stability of $(E,\phi)$. This implies that $(E,\vp_1)$ is stable, so $\phi=\vp_1\otimes i=0$.
\end{proof}

\section{Principal Elliptic Surfaces}\label{elliptic}
Recall that every principal elliptic surface can be described in the following way \cite{teleman}:

Let $B$ be a compact Riemann surface, and $\Theta\in \Pic^d(B)$. We denote the total space of the holomorphic frame bundle corresponding to $\Theta$ by $\Theta^*$. (As a complex manifold, $\Theta^*$ is the complement of the zero section in the total space of $\Theta$.) If $\tau$ is a complex number with $|\tau|>1$, then $X:=\Theta^*/(\tau)$ is a principal elliptic surface with projection induced from the projection of $\Theta$, where $(\tau)$ acts as a subgroup of the structure group $\C^*$ of $\Theta^*$.

Furthermore, the above description uniquely determines the biholomorphism class of the resulting elliptic surface up to swapping $\Theta$ and $\Theta^\vee$, and $X$ is K\"ahler if and only if $d:=c_1(\Theta)=0$. (Up to homeomorphism the elliptic surface is uniquely determined by $|d|$.)

\begin{proposition}\label{cotangent}
Let $\pi:X\to B$ be a principal elliptic surface. Then the holomorphic cotangent bundle $\T_X^*$ is given by $\pi^*(\mathcal{A}_B)$, where $\mathcal{A}_B$ is an extension $$\begin{tikzcd}0 \arrow[r] & K_B \arrow[r] & \mc{A}_B \arrow[r] & \mc{O}_B \arrow[r] & 0.\end{tikzcd}$$ If $X$ is K\"ahler then the extension splits, and otherwise $\mathcal{A}_B$ is the unique non-split extension.
\end{proposition}
\begin{proof}
Let $\{U_\alpha, \varphi_\alpha\}$ be an atlas of $B$ such that $\{U_\alpha, \psi_\alpha\}$ is a local trivialization for $\Theta^*$ and set $g_{\alpha\beta}:\varphi_\beta(U_\alpha\cap U_\beta)\to \varphi_\alpha(U_\alpha\cap U_\beta)$ to be the corresponding transition functions. We also set $V_0=\{[w]\in \C^*/(\tau):1<|w|<|\tau|\}$ and $V_1=\{[w]\in \C^*/(\tau):|\tau|^{-1/2}<|w|<|\tau|^{1/2}\}$. Let $f_{\alpha\beta}$ be transition maps for $\Theta$. Then $\{\psi_\alpha^{-1}(\vp_\alpha(U_\alpha)\times V_i)\}$ gives an atlas for $X$ with transition functions $\eta_{ij,\alpha\beta}(z,w)=(g_\alpha\beta(z), f_\alpha\beta\tau^k w)$, where $z$ is a local coordinate in $B$, $w$ is a coordinate of $\C^*$, and $$k=\begin{cases}
1 & i=0,j=1,|w|<1\\
-1 & i=1,j=0, |w|>|\tau|^{1/2}\\
0 & \text{otherwise}
\end{cases}.$$
Note that $\diff z$ and $\frac{1}{w}\diff w$ give a local frame for the open set $\psi_\alpha^{-1}(\varphi_\alpha(U_\alpha)\times V_i).$ If we now compute the pullbacks $\eta_{ij,\alpha\beta}^*(\diff \tilde{z})$ and $\eta_{ij,\alpha\beta}^*(\frac{1}{\tilde{w}}\diff \tilde{w})$, we get
\begin{align*}
    \eta_{ij,\alpha\beta}^*(\diff \tilde{z})&=\diff(g_{\alpha\beta}(z))\\
    &=g_{\alpha\beta}' \diff z,\\
    \eta_{ij,\alpha\beta}^*(\frac{1}{\tilde{w}}\diff \tilde{w})&=\frac{1}{f_{\alpha\beta}\tau^k w}\diff(f_{\alpha\beta}\tau^k w)\\
    &=\frac{1}{w}\diff w+ \frac{f_{\alpha\beta}'}{f_{\alpha\beta}}\diff z,
\end{align*}
so $T^*_X$ has transition maps $\begin{bmatrix}g_{\alpha\beta}' & \frac{f_{\alpha\beta}'}{f_{\alpha\beta}}\\ 0 & 1\end{bmatrix}.$ Since the transition maps of $\T^*_X$ depend only on $z$ and since $g_{\alpha\beta}'$ are the transition maps of $\pi^*(\T^*_B)=K_X$, we find that $\T^*_X$ is the pullback of an extension in $\Ext^1(\mc{O}_B, K_B)$. If $c_1(\Theta)=0$, then without loss of generality $f_\alpha\beta$ may be chosen to be locally constant, so the extension splits. If instead $c_1(\Theta)\neq 0$, it is a non-split extension. By Serre duality, $$\Ext^1(\mc{O}_B, K_B)=\Hom(K_B, \mc{O}_B\otimes K_B)^*=\Hom(K_B,K_B)^*\cong \C,$$
so this extension is unique.
\end{proof}
\begin{remark}
In the case where $\pi:X\to \mb{P}^1$ is a Hopf surface, $\mc{A}_B\cong \mc{O}_{\mb{P}^1}(-1)\oplus \mc{O}_{\mb{P}^1}(-1)$. If $\pi:X\to B$ is a Kodaira surface, then $\mc{A}_B$ is the Atiyah bundle on $B$.
\end{remark}

The above proposition also determines the holomorphic tangent bundle $\T_X$, as $\T_X\cong \T_X^*\otimes K_X^\vee=\T_X^*\otimes\pi^*(K_B^\vee)$.

If we use Propositions \ref{tracefree}, \ref{endo}, and \ref{extension} with this description of the (co)-tangent bundle, we get the following:

\begin{proposition}
Let $\pi:X\to B$ be a principal elliptic surface. If $X$ admits a non-zero stable trace-free (co)-Higgs bundle, then so does $B$. In other words, if $X$ has a non-zero stable trace-free Higgs bundle then $B$ has genus at least 2, and if $X$ has a non-zero stable trace-free co-Higgs bundle then $B$ has genus 0.
\end{proposition}

\begin{proof}
Suppose first that $X$ has a non-trivial stable co-Higgs bundle. Then $\mc{T}_X\cong (\pi^*\mc{A}_B)^\vee$ by Proposition \ref{cotangent}, so it is an extension of the anti-canonical bundle by $\mc{O}_X$. As shown in Proposition \ref{endo}, there are no non-zero stable pairs $(E,\phi)$ with $\phi \in \End_0(E)$, so we conclude that $X$ admits a non-zero trace-free stable pair $(E,\phi:E \to E\otimes K_X^{-1})$ by Proposition \ref{extension}. This is only possible if $B$ has genus $0$.
If we now assume that that $X$ has a non-trivial stable Higgs bundle, a similar argument shows that $B$ has genus at least $2$.
\end{proof}

\begin{remark}
If $\pi:X\to B$ is a principal elliptic surface, any Gauduchon metric $g$ on $X$ can be normalized so that $\deg_g(\pi^*(H))=\deg(H)$ for all $H \in \Pic(B)$ \cite{LuTe}.
\end{remark}

\section{Co-Higgs bundles in the Hopf surface case}\label{Hopf}
In this section, $X$ is a Hopf surface so that $B = \mathbb{P}^1$ and $\mc{T}_X\cong \mc{O}_X(F)\oplus \mc{O}_X(F)$ with $F$ a fibre of $\pi$. A vector bundle $E$ on $X$ then admits a trace-free co-Higgs bundle if and only if $H^0(X, \End_0(E)\otimes \mc{O}_X(F))\neq 0$.
\begin{proposition}\label{non-filt-co-Higgs}
Let $E$ be a rank-2 vector bundle with determinant $\delta$ on a Hopf surface $X$ that is not filtrable. Then every trace-free Higgs field on $E$ is 0.
\end{proposition}
\begin{proof}
Since $E$ is non-filtrable, it has generic fibre $E|_{\pi^{-1}(b)}=\lambda_b\oplus \left(\delta|_{\pi^{-1}(b)}\otimes \lambda_b^{-1}\right)$, where $\lambda_b \in \Pic^0(T)$ and $\lambda_b\not\cong \delta|_{\pi^{-1}(b)}\otimes \lambda_b^{-1}$. Then $\End_0E|_{\pi^{-1}(b)}\cong \mc{O}_T\oplus \lambda_b'\oplus (\lambda_b')^{-1}$, where $\lambda_b':=\delta|_{\pi^{-1}(b)}\otimes \lambda_b^{-2}$. Since the generic fibre of $\End_0E(F)$ contains a trivial sub-line bundle, there is a line bundle $L \in \Pic(\mb{P}^1)$ such that $\pi_*(\End_0E(F))\cong L$.  Then there is an injective map from $\pi^*(L)$ to $\End_0E(F)$, so there is a coherent sheaf $\mc{F}$ such that $$\begin{tikzcd}0 \arrow[r] & \pi^*(L)\arrow[r] & \End_0E(F) \arrow[r] & \mc{F} \arrow[r] & 0
\end{tikzcd}$$
is an exact sequence. Note that $\mc{F}|_{\pi^{-1}(b)}\cong \lambda_b'\oplus (\lambda_b')^{-1}$ for general $b$, so the torsion-free part of $\mc{F}$ is non-filtrable.

If we now consider the long exact sequence

$$\begin{tikzcd}0 \arrow[r] & L \arrow[r] & \pi_*(\End_0E(F)) \arrow[r] & \pi_*(\mc{F})\arrow[lld] \\ & L \arrow[r] & R^1\pi_*(\End_0E(F)) \arrow[r] & R^1\pi_*(\mc{F}) \arrow[r] & 0
\end{tikzcd}$$
induced by pushforward, we note that the map from $\pi_*(\mc{F})$ to $R^1\pi_*(\pi^*L)\cong L$ is injective, as $L$ and $\pi_*(\End_0E(F)$ are by definition isomorphic. Since $\mc{F}$ is non-filtrable, $\pi_*(\mc{F})=\pi_*(\mathrm{tors}(\mc{F}))=0$ as $\pi_*(\mc{F})$ is a sub-sheaf of a line bundle.
Since $\pi_*(\mc{F})=0$, we have a second short exact sequence $$\begin{tikzcd}
0 \arrow[r] & L \arrow[r]&R^1\pi_*(\End_0E(F)) \arrow[r] & R^1\pi_*(\mc{F}) \arrow[r] & 0.
\end{tikzcd}$$
Using relative Serre duality, we see that $$R^1\pi_*(\End_0E(F))^\vee=\pi_*(\End_0E(-F)\otimes \omega_{X/\mb{P}^1})=\pi_*(\End_0E(F))\otimes \mc{O}_{\mb{P}^1}(-2)\cong L(-2),$$
since $\omega_{X/\mb{P}^1}=\mc{O}_X$. In addition, since $\mb{P}^1$ is a curve we have that $$R^1\pi_*(\End_0E(F))=R^1\pi_*(\End_0E(F))^{\vee\vee}\oplus \mc{S}$$ for some torsion sheaf $\mc{S}$, so since there is an injective map from $L$ to $R^1\pi_*(\End_0E(F))$, there must also be a non-zero map from $L$ to $R^1\pi_*(\End_0E(F))^{\vee\vee}\cong L^{-1}(2).$ In particular this implies that $L=\mc{O}_{\mb{P}^1}(m)$ with $2m\leq 2.$ Note that since $H^0(X,\End_0E)=0$, we must have $H^0(\mb{P}^1,L(-1))=H^0(\mb{P}^1, \mc{O}_{\mb{P}^1}(m-1))=0$, and therefore $m\leq 0$.

Suppose that $L$ has a non-zero section (i.e. $L=\mc{O}_{\mb{P}^1}$). Then there is an injective map $\vp:E \to E(F)$ which induces an elementary modification $$\begin{tikzcd}0 \arrow[r] & E \arrow{r}{\vp} & E(F) \arrow[r] & \mc{Q} \arrow[r] & 0,
\end{tikzcd}$$

where $\mc{Q}$ is is a torsion sheaf given by $\mc{Q}_b=E_b$ for any point $b \in D$, where $D$ is an effective divisor on $\mb{P}^1$ linearly equivalent to $L$. Since $L=\mc{O}_{\mb{P}^1}$ this implies that $\mc{Q}=0$, so $\vp$ is an isomorphism. However, this is a contradiction since $\deg(E(F))=\deg(E)+1\neq \deg(E)$, so $L$ must not have a non-zero section. Since $H^0(\mb{P}^1, L)\cong H^0(X,\End_0E(F))$, $E$ does not admit a non-trivial Higgs field.
\end{proof}

When $E$ is a filtrable bundle with two (possibly equal) maximal destabilizing bundles $K_1$ and $K_2$, we make use of the fact that $K_1\otimes K_2\otimes \det(E)^{-1}$ is the pullback of a line bundle on $\mb{P}^1$ of degree $m:=\deg(K_1\otimes K_2\otimes \det(E)^{-1})$.

\begin{proposition}\label{hopf-reg}
Let $E$ be a filtrable rank-2 bundle over a Hopf surface $X$ that is regular on the generic fibre of $\pi$. Then the trace-free Higgs fields on $E$ are parameterized by $$H^0(X,\End_0E\otimes \mc{T}_X)=\begin{cases}
\C^{4+2m}, & m \geq -1,\\
0, & \text{otherwise},
\end{cases}$$
where $m$ is as defined in the previous paragraph. Furthermore, the corresponding co-Higgs bundles are stable if and only if $E$ is stable.
\end{proposition}
\begin{proof}
Since $E$ is filtrable with maxmimal destabilizing bundle $K_1$, it fits into the exact sequence $$\begin{tikzcd}0 \arrow[r] & K_1 \arrow{r}{i} & E \arrow{r}{p} & \delta\otimes K_1^{-1}\otimes \mc{I}_Z \arrow[r] & 0,
\end{tikzcd}$$
where $\delta=\det(E)$ and $Z$ is a set of points in $X$ (counting multiplicity). In addition, given that $E$ is regular on the generic fibre of $\pi$, we have $\pi_*(K_1^{-1}\otimes E)=\pi_*(K_2^{-1}\otimes E)=\mc{O}_{\mb{P}^1}$.

After taking the tensor product of our exact sequence with $E^\vee(F)=\delta^{-1}\otimes E(F)$, we have the left-exact sequence in cohomology
$$\begin{tikzcd}0 \arrow[r] & H^0(X,K_1\otimes \delta^{-1}\otimes E(F)) \arrow[r] & H^0(X,\End E(F)) \arrow[r] & H^0(X, K_1^{-1}\otimes E(F) \otimes \mc{I}_Z) \arrow[r] & 0.
\end{tikzcd}$$
Let $s$ be a section of $\mc{O}_X(F)$, and consider the map $p\circ (\id_E\otimes s)$. Since $p$ is surjective, this map is zero if and only if $s=0$, so $$h^0(X, K_1^{-1}\otimes E(F)\otimes \mc{I}_Z)\geq h^0(X, \mc{O}_X(F))=2.$$
However, $K_1^{-1}\otimes E(F)\otimes \mc{I}_Z$ is also a subsheaf of $K_1^{-1}\otimes E(F)$, so $$h^0(X, K_1\otimes E(F)\otimes \mc{I}_Z)\leq h^0(X,K_1^{-1}\otimes E(F))=h^0(\mb{P}^1, \mc{O}_{\mb{P}^1}(1))=2,$$
so $H^0(X, K_1^{-1}\otimes E(F)\otimes \mc{I}_Z)=\C^2$, and the trace-free sections of $\End E(F)$ are parameterized by $$H^0(X,K_1\otimes \delta^{-1}\otimes E(F))=H^0(X, K_2^{-1}\otimes E((1+m)F))=H^0(\mb{P}^1, \mc{O}_{\mb{P}^1}(1+m))=\C^{2+m}.$$

This gives the desired result, as $\End_0E\otimes \mc{T}_X\cong \End_0E(F)\oplus \End_0E(F)$.

Suppose for a contradiction that $E$ has a stable Higgs field $\phi$, but that $\mu(K_1)\geq \mu(E)$. Then $K_1$ is not $\phi$-invariant, so $(p\otimes \id_{\mc{O}_X(F)})\circ\phi\circ i:K_1 \to \delta\otimes K_1^{-1}(F)\otimes \mc{I}_Z$ is non-zero. In particular, this implies that there is a non-zero map from $K_1$ to $K_2((1-m)F)$. If $K_2\neq K_1$ this is impossible, so $K_1=K_2$ and we must have $m\leq 1$. Consider the short exact sequence $$\begin{tikzcd}0 \arrow[r] & \ker(\phi)\otimes K_1^{-1} \arrow[r] & K_1^{-1}\otimes E \arrow[r] & \mathrm{im}(\phi)\otimes K_1^{-1} \arrow[r] & 0.
\end{tikzcd}$$
Since $K_1$ is the unique maximal destabilizing bundle, either $\pi_*(\ker(\phi)\otimes K_1^{-1})$ is a line bundle, or $\ker(\phi)=0$. If $\pi_*(\ker(\phi)\otimes K_1^{-1})$ is a line bundle $L$, then the left exact sequence induced by pushforward is $$\begin{tikzcd}0 \arrow[r] & L \arrow[r] & \mc{O}_{\mb{P}^1} \arrow[r] & \pi_*(\mathrm{im}(\phi)\otimes K_1^{-1}),
\end{tikzcd}$$
where each entry is torsion-free. This implies that $L\cong \mc{O}_{\mb{P}^1}$, so $\deg(\ker(\phi))\geq \deg(K_1)$, a contradiction to the stability of $\phi$.

Therefore we must have that $\phi$ is injective, so there is an induced exact sequence $$\begin{tikzcd}0 \arrow[r] & E \arrow{r}{\phi} & E\otimes \mc{T}_X^* \arrow[r] & \mc{Q} \arrow[r] & 0,
\end{tikzcd}$$
and $\mc{Q}$ is torsion-free away from a fibre $F$. Let $b \in \mb{P}^1$ be such that $\pi^{-1}(b)$ is not in the support of $\mc{Q}$ and $E$ is regular at $b$. Then $\phi|_{\pi^{-1}(b)}$ is an injection from the Atiyah bundle $\mc{A}$ into $\mc{A}\oplus \mc{A}$ such that the restriction to each fibre is trace-free. However, every injective map from $\mc{A}$ to itself has non-zero trace, giving a contradiction. Therefore $E$ has a stable Higgs field only if $E$ is stable.
\end{proof}

If instead $E$ is not regular on the generic fibre, then it has a unique maximal destabilizing bundle $K$, and $\pi_*(K^{-1}\otimes E)=\mc{O}_{\mb{P}^1}\oplus \mc{O}_{\mb{P}^1}(-\ell)$ for some $\ell\geq 0.$
\begin{proposition}\label{hopf-not-reg}
Let $E$ be a stable bundle over a Hopf surface $X$, and assume that $E$ is not regular on the generic fibre. If $n,\ell$ are integers such that $K^2\otimes \det(E)^{-1}=\pi^*(\mc{O}_{\mb{P}^1}(n))$ and $\pi_*(K^{-1}\otimes E)=\mc{O}_{\mb{P}^1}\oplus \mc{O}_{\mb{P}^1}(-\ell)$, then the trace-free Higgs fields on $E$ are parameterized by $$H^0(X, \End_0E\otimes \mc{T}_X)=\begin{cases}
\C^2 & n=-1, \ell>0\\
\C^4 & n=-1, \ell=0\\
0 & \text{otherwise}
\end{cases}.$$
\end{proposition}
\begin{proof}
Since $\pi_*(K^{-1} \otimes E) = \mathcal{O}_{\mathbb{P}^1} \oplus
 \mathcal{O}_{\mathbb{P}^1}(-l)$ for some $l \geq 0$, we have
 \[ H^0(X,K\otimes \det(E)^{-1} \otimes E(F)) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(n+1) \oplus
 \mathcal{O}_{\mathbb{P}^1}(n+1-l))\]
and
\[ H^0(X,K^{-1} \otimes E(F)) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1) \oplus
 \mathcal{O}_{\mathbb{P}^1}(1-l)) = \mathbb{C}^2 \oplus H^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(1-l)).\]

 We also have a natural exact sequence \begin{equation}
     \label{End-exact}\begin{tikzcd}0 \arrow[r] & K\otimes \otimes \det(E)^{-1} E(F) \arrow[r] & \End E(F) \arrow[r] & K^{-1}\otimes E(F)\otimes \mc{I}_Z \arrow[r] & 0
     \end{tikzcd}.
 \end{equation}
 We now consider three cases:
 \begin{itemize}
 \item
 If $l \geq 2$, then $h^0(\mathbb{P}^1, \mathcal{O}_{\mathbb{P}^1}(1-l)) = 0$ so that
 \[ H^0(X,K^{-1} \otimes E(F)) = \mathbb{C}^2 .\]
 This means again that $H^0(X,K^{-1} \otimes E(F) \otimes I_Z) = \iota^*(H^0(X,\mathcal{O}_X(F)))$, and elements of this space correspond to Higgs fields of the form $\id_E \otimes s$ with $s \in H^0(X,\mathcal{O}_X(F))$. Furthermore, the short exact sequence \eqref{End-exact} yields
\[ H^0(X,\End_0 E (F)) \simeq H^0(X,K \otimes \det(E)^{-1}\otimes E(F)).\]
Note that $n < 0$ and $l \geq 0$ so that $n+1 \leq 0$ and $n+1-l \leq -2$. Therefore, we have $h^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(n+1-l)) = 0$ and
\[ H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(n+1))  = \left\{ \begin{array}{ll}
\mathbb{C} & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$}.
\end{array} \right.\]
Thus,
\[ H^0(X,\End_0 E (F)) \simeq H^0(X,K \otimes E(F)) = \left\{ \begin{array}{ll}
\mathbb{C} & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$}.
\end{array} \right.\]

\item
If $l = 1$, then
\begin{align*}
H^0(X,K \otimes \det(E)^{-1}\otimes E(F)) &= H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(n+1) \oplus
 \mathcal{O}_{\mathbb{P}^1}(n)) \\
 &= \left\{ \begin{array}{ll}
\mathbb{C} & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$},
\end{array} \right.
 \end{align*}
and
\[ H^0(X,K^{-1} \otimes E(F)) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1) \oplus
 \mathcal{O}_{\mathbb{P}^1}) = \mathbb{C}^2 \oplus \mathbb{C}.\]
 The terms in the first factor of $H^0(X,K^{-1} \otimes E(F))$ are Higgs fields of the form $\id_E \otimes s$, $s \in H^0(X,\mathcal{O}_X(F))$. The terms in the second factor are injective maps of sheaves $\iota': K(-F) \rightarrow E$, which means that $E$ also fits in an exact sequence of the form
 \[0 \rightarrow K(-F)  \stackrel{\iota'}{\rightarrow} E \stackrel{p'}{\rightarrow} K^{-1}(F) \otimes \det(E)\otimes I_Z \rightarrow 0.\]
 This induces the exact sequence
 \[ 0 \rightarrow K \otimes \det(E)^{-1}\otimes E \longrightarrow \End E(F) \longrightarrow K^{-1} \otimes E(2F) \otimes I_Z \rightarrow 0.\]
 Furthermore, $\iota'$ corresponds to a non-zero global section of $\End_0E(F)$ if and only if $$H^0(X,K \otimes \det(E)^{-1}\otimes E) \neq 0.$$ But $\pi_*(K\otimes \det(E)^{-1} \otimes E) = \mathcal{O}_{\mathbb{P}^1}(n) \oplus \mathcal{O}_{\mathbb{P}^1}(n-1)$ so that
 \[ H^0(X,K\otimes \det(E)^{-1} \otimes E) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(n) \oplus \mathcal{O}_{\mathbb{P}^1}(n-1)) = 0\]
 since $n<-1$.
Hence, no non-zero traceless Higgs fields come from $H^0(X,K^{-1} \otimes E(F) \otimes I_Z)$, proving that we again have
\[ H^0(X,\End_0 E (F)) \simeq H^0(X,K\otimes \det(E)^{-1} \otimes E(F)) = \left\{ \begin{array}{ll}
\mathbb{C} & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$}.
\end{array} \right.
\]

\item
Finally, if $l = 0$, then $\pi_*(K^{-1} \otimes E) = \mathcal{O}_{\mathbb{P}^1} \oplus
 \mathcal{O}_{\mathbb{P}^1}$, which means that the injections $\iota: K \rightarrow E$ are parametrised by
 \[ H^0(X,K^{-1} \otimes E) = H^0(\mathbb{P}^1, \pi_*(K^{-1} \otimes E)) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1} \oplus
 \mathcal{O}_{\mathbb{P}^1}) = \mathbb{C}^2.\]
 Moreover,
\[ H^0(X,K\otimes \det(E)^{-1} \otimes E(F)) = H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1+n) \oplus
 \mathcal{O}_{\mathbb{P}^1}(1+n)) = \left\{ \begin{array}{ll}
\mathbb{C}^2 & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$}.
\end{array} \right.\]
and
\begin{align*}
H^0(X,K^{-1} \otimes E(F)) &= H^0(\mathbb{P}^1,\mathcal{O}_{\mathbb{P}^1}(1) \oplus
 \mathcal{O}_{\mathbb{P}^1}(1)) = \mathbb{C}^2 \oplus \mathbb{C}^2\\
 &= H^0(X,K^{-1} \otimes E) \otimes H^0(X,\mathcal{O}_X(F))\\
 &= \{ p \otimes \iota: \mbox{$\iota:K \rightarrow E$ is an injection}, s \in H^0(X,\mathcal{O}_X(F) \} \\
 &= \{ \iota^*(\id_E \otimes s): \mbox{$\iota:K \rightarrow E$ is an injection}, s \in H^0(X,\mathcal{O}_X(F) \}.
 \end{align*}
 Consequently, none of the non-zero elements in $H^0(X,K^{-1} \otimes E(F))$ correspond to traceless Higgs fields.
Hence,
\[ H^0(X,\End_0 E (F)) \simeq H^0(X,K \otimes \det(E)^{-1}\otimes E(F)) = \left\{ \begin{array}{ll}
\mathbb{C}^2 & \mbox{if $n=-1$} ,\\
0 & \mbox{if $n < -1$}.
\end{array} \right.\]
\end{itemize}

Since in all cases $H^0(X, \End_0E\otimes \mc{T}_X)=H^0(X, \End_0E(F))\oplus H^0(X, \End_0E(F))$, this gives the result.
\end{proof}

\begin{proposition}\label{hopf-filt-not-stable}
If $E$ is not stable, then a co-Higgs bundle $(E, \phi)$ is stable only if $E$ is an extension of line bundles.
\end{proposition}
\begin{proof}
We know by Proposition \ref{hopf-reg} that if $(E,\phi)$ is a stable co-Higgs bundle with underlying bundle not stable, then $E$ is not regular on the generic fibre, so it fits into the exact sequence $$\begin{tikzcd}0 \arrow[r] & K \arrow{r}{i} &E \arrow{r}{p} & K^{-1}\otimes \delta\otimes \mc{I}_Z \arrow[r] & 0
\end{tikzcd}$$
with $K$ the unique maximal destabilizing bundle of $K$. In addition, there exist non-negative integers $n,\ell$ such that $\pi_*(K^{-1}\otimes E)=\mc{O}_{\mb{P}^1}\oplus \mc{O}_{\mb{P}^1}(-\ell)$ and $K^2\otimes \delta^{-1}=\pi^*(\mc{O}_{\mb{P}^1}(n)).$

Suppose that $\phi$ is a stable Higgs field on $E$. Then since $\deg(K)\geq \mu(E)$, $K$ is not $\phi$-invariant and the map $$(p\times \id_{\mc{O}_X(F)})\circ \phi \circ i:K \to K^{-1}\otimes \delta(F) \otimes \mc{I}_Z$$ is non-zero. These maps are parameterized by $$H^0(X, K^{-2}\otimes \delta(F)\otimes \mc{I}_Z)=H^0(X, \pi^*(\mc{O}_{\mb{P}^1}(1-n))\otimes \mc{I}_Z)$$
and in order for this map to be non-zero we must have that either $Z=\emptyset$ or $n=0$. When $Z=\emptyset$ the bundle $E$ is an extension of line bundles, so we restrict to the case where $n=0$ and $Z \neq \emptyset.$

First consider the case when $\phi$ has non-zero kernel. Then since $\ker(\phi)$ and $\mathrm{im}(\phi)$ are $\phi$-invariant subbundles of $E$ and $E(F)$ respectively, we have $$\deg(\ker(\phi))<\mu(E)<\deg(\mathrm{im}(\phi))<\mu(E(F))=\mu(E)+1.$$
Furthermore, by the natural exact sequence relating $E$ to $\ker(\phi)$ and $\mathrm{im}(\phi)$ we have $$\deg(\mathrm{im}(\phi))=\mu(E)-\deg(\ker(\phi)),$$ so $\mu(E)-1<\deg(\ker(\phi))<\mu(E).$ Since $K$ is the unique maximal destabilizing bundle of $E$, $\deg(K)-\deg(\ker(\phi)$ is a non-negative integer. However, this gives a contradiction as $\mu(E)=\deg(K)$, so $\phi$ is injective. Then we have an exact sequence
\begin{equation}\label{elementary-modification3}
0 \rightarrow E \stackrel{\phi}{\rightarrow} E \otimes \mathcal{O}_X(F) \rightarrow Q \rightarrow 0,
\end{equation}
where $Q$ is a torsion sheaf supported on a single fibre $\pi^{-1}(b)$, $b \in B$. Note that
\[ Q|_{\pi^{-1}(b)} = E|_{\pi^{-1}(b)} \otimes \mathcal{O}_X(F) =  E|_{\pi^{-1}(b)}.\]
Moreover, all extensions of the form \eqref{elementary-modification3} are parametrised by
\[ H^0(F, Hom(Q|_{\pi^{-1}(b)}, E|_{\pi^{-1}(b)})) = H^0(X,\End E|_{\pi^{-1}(b)}).\]
But given such a $Q$, all the elementary modifications of $E \otimes \mathcal{O}_X(F)$ by $Q$ are parametrised by $H^0(X,\End E|_{\pi^{-1}(b)})$ so that $\phi$ just the identity map away from $\pi^{-1}(b)$, contradicting the fact that $\phi$ is traceless.

Therefore we must have that $Z=\emptyset$, so that $E$ is an extension of line bundles.
\end{proof}

Putting all this together, we obtain the following:

\begin{theorem}
Let $X$ be a Hopf surface and $c_2$ be a non-negative integer. Moreover, let $(E,\phi)$ be a trace-free rank-2 co-Higgs bundle on $X$ with $c_2(E) = c_2$. Then:
\begin{enumerate}
    \item If $E$ is non-filtrable, in which case $c_2 > 0$, then $\phi = 0$.
    \item If $E$ is filtrable, then one can have $\phi \neq 0$ for all $c_2 \geq 0$.
    \item If $c_2 > 0$, then $(E,\phi)$ is stable if and only if $E$ is stable.
    \item If $c_2 = 0$, then $E$ admits a non-trivial stable Higgs field if and only if $E = L \otimes (\mc{O}_X \oplus \mc{O}_X(-F))$ for some $L \in \Pic(X)$.
\end{enumerate}
Consequently, there exists a non-trivial stable rank-2 co-Higgs bundle $(E,\phi)$ on $X$ with $c_2(E) = c_2$ for all $c_2 \geq 0$.
\end{theorem}

\begin{proof}
Part 1 follows immediately from Proposition \ref{non-filt-co-Higgs}, so we can restrict to the case where $E$ is filtrable. We thus assume that $E$ is filtrable. Then, there exist non-trivial Higgs fields $\phi$ for all $E$ with $c_2 \geq 0$ by Propositions \ref{hopf-reg} and \ref{hopf-not-reg}. In the case that $c_2>0$, Propositions \ref{hopf-reg} and \ref{hopf-filt-not-stable} imply that $(E,\phi)$ is stable if and only if $E$ is stable. We are therefore left with considering the case where $c_2 = 0$ and $E$ is an extension of line bundles. Suppose that $E$ is an extension of $L'$ by $L$ for some $L,L' \in \Pic(X)$. Then, $E$ corresponds to an element of $H^1(X,(L')^{-1} \otimes L)$, which is non-zero if and only if $(L')^{-1} \otimes L \simeq \mc{O}_X(mF)$ for some integer $m$. Thus, if $(L')^{-1} \otimes L \not\simeq \mc{O}_X(mF)$ for all $m \in \mathbb{Z}$, the extension splits and $E = L \oplus L'$ with $H^0(X,(L')^{-1} \otimes L) = H^0(X,L^{-1} \otimes L') = 0$, implying that $L$ and $L'$ are both $\phi$-invariant and $(E,\phi)$ is not stable. Whereas, if $(L')^{-1} \otimes L \simeq \mc{O}_X(mF)$ for some integer $m$, then $L' \simeq L \otimes \mc{O}_X(-mF)$ and
\[ H^1(X,(L')^{-1} \otimes L) = H^0(\mathbb{P}^1,R^1\pi_*(\mc{O}_X(mF)) \oplus H^1(\mathbb{P}^1,\pi_*(\mc{O}_X(mF))). \]
Note that the non-zero elements of $H^0(\mathbb{P}^1,R^1\pi_*(\mc{O}_X(mF))$ correspond to non-trivial extensions of $L'$ by $L$ that are regular on the generic fibres of $\pi$. Hence, $E$ corresponds to an element of
\[ H^1(\mathbb{P}^1,\pi_*(\mc{O}_X(mF))) = H^1(\mathbb{P}^1,\mc{O}_{\mathbb{P}^1}(m)). \]
In other words, $E$ is the pullback to $X$ of an extension of $\mc{O}_{\mathbb{P}^1}(-m)$ by $\mc{O}_{\mathbb{P}^1}$ up to tensoring by $L$. Since all bundles on $\mathbb{P}^1$ are direct sums of line bundles, this means that $E = L \otimes (\mc{O}_X \oplus \mc{O}_X(m'F))$ for some integer $m'$. Moreover, a direct computation shows that $(E,\phi)$ is stable only if $m'=-1$.

To construct a stable co-Higgs bundle with second Chern class $c_2>0$, let $L_1$ and $L_2$ be line bundles in $\Pic(X)$ such that $-1<\deg(L_1)-\deg(L_2)<1$. Set $E'=L_1\oplus L_2$ and let $E$ be the elementary modification of $E'$ along a fibre $F$ of $\pi$ by a line bundle $H$ on $F$ of degree $c_2$. Then, $\det(E)=\det(E')\otimes \mc{O}_X(-F)$, $c_2(E)=c_2$, and since $H$ has positive degree, the destabilizing line bundles of $E$ are $L_1(-F)$ and $L_2(-F)$. We now have $\mu(E)=\frac{\deg_g(L_1)+\deg_g(L_2)-1}{2}$, and since $-1<\deg_g(L_1)-\deg_g(L_2)<1$,

$$\deg_g(L_1(-F))=\deg_g(L_1)-1<\frac{\deg_g(L_1)+\deg_g(L_2)}{2}-\frac{1}{2} = \mu(E)$$
and

$$\deg_g(L_2(-F))=\deg_g(L_2)-1<\frac{\deg(L_1)+\deg(L_2)}{2}-\frac{1}{2} = \mu(E),$$ so $E$ is stable. Furthermore, $m=\deg\left(L_1(-F)\otimes L_2(-F) \otimes (L_1\otimes L_2\otimes \mc{O}_X(-F))^{-1}\right)=-1$. Recall that the trace-free Higgs fields of $E$ are parameterized by $H^0(\mb{P}^1, \mc{O}_{\mb{P}^1}(m+1))$ by Proposition \ref{hopf-reg}, so there is a one-dimensional family of trace-free Higgs fields on $E$. Together with the non-trivial stable Higgs fields on $\mc{O}_X\oplus \mc{O}_X(-F)$, this shows that there is a non-trivial stable co-Higgs bundle with second Chern class $c_2$ for all $c_2\geq 0$.
\end{proof}

\begin{remark}
Note that $\mc{O}_X \oplus \mc{O}_X(-F) = \pi^*(\mc{O}_{\mathbb{P}^1} \oplus \mc{O}_{\mathbb{P}^1}(-1))$. Moreover, referring to Proposition \ref{cotangent}, $\mc{T}_X$ fits into the exact sequence
$$\begin{tikzcd}0 \arrow[r] &  \mc{O}_X \arrow[r] & \mc{T}_X \arrow[r] & \pi^* \mc{T}_{\mathbb{P}^1} \arrow[r] & 0.\end{tikzcd}$$
A direct computation then gives us that non-trivial $g$-stable Higgs fields on $E = L \otimes (\mc{O}_X \oplus \mc{O}_X(-F))$ are in one-to-one correspondence with stable non-trivial Higgs fields $\psi: \mc{O}_{\mathbb{P}^1} \oplus \mc{O}_{\mathbb{P}^1}(-1) \rightarrow (\mc{O}_{\mathbb{P}^1} \oplus \mc{O}_{\mathbb{P}^1}(-1)) \otimes  \mc{T}_{\mathbb{P}^1}$ on $\mathbb{P}^1$ for any line bundle $L \in \Pic(X)$. In other words, stable co-Higgs bundles on $X$ with $c_2=0$ are pull-backs of stable co-Higgs bundles on $\mathbb{P}^1$, whose moduli spaces were studied in \cite{Rayan,Rayan-P1}.
\end{remark}

\section{Higgs bundles in the Kodaira dimension 1 case}\label{geq2}

Let $\pi:X\to B$ be a non-K\"ahler principal elliptic surface with base $B$ of genus $g\geq 2$. In this case, $\deg(K_X)=2g-2>0$. Note that for any trace-free stable Higgs bundle $(E, \phi)$ on $X$, $\phi$ is of the form $\vp\otimes i$ for some trace-free $\vp:E \to E\otimes K_X$ with $(E,\vp)$ stable by Proposition \ref{extension}, where $i:K_X \to \mc{T}_X^*$ is the injection given in Proposition \ref{cotangent}. Therefore it suffices to study trace-free stable pairs of the form $(E, \vp:E \to E\otimes K_X)$.

\begin{proposition}\label{regular}
Let $E$ be a rank-2 filtrable vector bundle with (possibly equal) maximal destabilizing bundles $K_1$ and $K_2$ and that is regular on the generic fibre of $\pi$. A trace-free Higgs field $\phi$ on $E$ is then stable if and only if $E$ is, and the trace-free Higgs fields on $E$ are parameterized by $H^0(B, K_B\otimes H)$, where $$H:=\pi_*(\det(E)^{-1}\otimes K_1\otimes K_2).$$
\end{proposition}
\begin{remark}
By \cite[Proposition 3.4]{BrMor2}, the bundle $\det(E)^{-1}\otimes K_1\otimes K_2=\pi^*(H)$ whenever $E$ is filtrable.
\end{remark}
\begin{proof}
Since $E$ is filtrable with maxmimal destabilizing bundle $K_1$, $E$ fits into an exact sequence $$\begin{tikzcd}0 \arrow[r] & K_1 \arrow{r}{i} & E \arrow{r}{p} & \delta\otimes K_1^{-1}\otimes \mc{I}_Z \arrow[r] & 0,
\end{tikzcd}$$
where $\delta=\det(E)$ and $Z$ is a zero-dimensional subset of $X$. In addition, since $E$ is regular on the generic fibre we have $\pi_*(K_1^{-1}\otimes E)=\pi_*(K_2^{-1}\otimes E)=\mc{O}_B$.

In order to show that stability of $(E,\phi)$ implies stability of $E$, we first assume that $K_1\neq K_2$, $\mu(K_1)\geq \mu(E)$, and $\phi$ is a non-zero $\mu$-stable Higgs field on $E$. Note that $E$ fits into an exact sequence $$\begin{tikzcd}0 \arrow[r] & K_1 \arrow{r}{i} & E \arrow{r}{p} & K_2\otimes \pi^*(H^{-1})\otimes \mc{I}_Z \arrow[r] & 0
\end{tikzcd}.$$ in this case we must have $p\otimes \id_{K_X}\circ \phi \circ i\neq 0$, as otherwise $K_1$ would be a $\phi$-invariant subsheaf with slope at least equal to the slope of $E$. However, since $K_1$ and $K_2$ are distinct maximal destabilizing bundles, $$\Hom(K_1, \delta\otimes K_1^{-1}\otimes K_X\otimes \mc{I}_Z)=\Hom(K_1, K_2\otimes \pi^*(H^{-1})\otimes K_X \otimes \mc{I}_Z)=0,$$ giving a contradiction.

Now suppose instead that $K_1=K_2$, and consider the exact sequence $$\begin{tikzcd}0 \arrow[r] & K_1^{-1}\otimes \ker(\phi) \arrow[r] & K_1^{-1}\otimes E \arrow[r] & K_1^{-1}\otimes \mathrm{im}(\phi) \arrow[r] & 0.
\end{tikzcd}$$
Since $\phi\neq 0$, $\ker(\phi)$ is a subsheaf of $K_1$, and therefore either $\pi_*(K_1^{-1}\otimes \ker(\phi))$ is a line bundle or $\ker(\phi)=0$. In the first case, taking pushforward by $\pi$ gives a left-exact sequence $$\begin{tikzcd}0 \arrow[r] & \pi_*(K_1^{-1}\otimes \ker(\phi)) \arrow[r] & \mc{O}_B \arrow[r] &\pi_*(K_1^{-1}\otimes \mathrm{im}(\phi))
\end{tikzcd}$$
where each entry is torsion-free. Thus the map from $\pi_*(K_1^{-1}\otimes \ker(\vp))$ to $\mc{O}_B$ has torsion-free cokernel, so in particular $\deg(\ker(\phi))\ge \deg(K_1)$, a contradiction. Therefore we must have that $\phi$ is injective, so there is an induced exact sequence $$\begin{tikzcd}0 \arrow[r] & E \arrow{r}{\phi} & E\otimes \mc{T}_X^* \arrow[r] & \mc{Q} \arrow[r] & 0,
\end{tikzcd}$$
and $\mc{Q}$ is torsion-free away from a divisor $D$ such that $2D$ is linearly equivalent to $2K_X$. let $b \in B$ be such that $\pi^{-1}(b)$ is not in the support of $\mc{Q}$ and $E$ is regular at $b$. Then $\phi|_{\pi^{-1}(b)}$ is an injection from the Atiyah bundle $\mc{A}$ into $\mc{A}\oplus \mc{A}$ such that the restriction to each fibre is trace-free. However, every injective map from $\mc{A}$ to itself has non-zero trace, giving a contradiction. Therefore $E$ has a stable Higgs field only if $E$ is stable.

We now determine the dimension of the space of trace-free Higgs fields on $E$.  If we take the tensor product of our exact sequence with $E^\vee=\delta^{-1}\otimes E\otimes K_X$ and look at cohomology, we get the left-exact sequence $$\begin{tikzcd}0 \arrow[r] & H^0(X,\delta^{-1}\otimes K_1\otimes E\otimes K_X) \arrow[r] & H^0(X,\End(E)\otimes K_X) \arrow[r] & H^0(X,K_1^{-1}\otimes E\otimes K_X\otimes \mc{I}_Z).
\end{tikzcd}$$
Note that for any element of $H^0(X,\End(E)\otimes K_X)$ of the form $\id\otimes s$ with $s \in H^0(X,K_X)$, its image in $H^0(X, K_1^{-1}\otimes E\otimes K_X\otimes \mc{I}_Z)$ is zero if and only if $s=0$, so $h^0(X,K_1^{-1}\otimes E\otimes K_X \otimes \mc{I}_Z)\geq g$. Since $K_1^{-1}\otimes E\otimes K_X \otimes \mc{I}_Z$ is a subsheaf of $K_1^{-1}\otimes E\otimes K_X$, we also have $$h^0(X,K_1^{-1}\otimes E\otimes K_X \otimes \mc{I}_Z)\leq h^0(X, K_1^{-1}\otimes E\otimes K_X)=h^0(B, \pi_*(K_1^{-1}\otimes E)\otimes K_B)=h^0(B, K_B)=g,$$
so $H^0(X, K_1^{-1}\otimes E\otimes K_X\otimes \mc{I}_Z)=\C^g$, and \begin{align*}H^0(X,\End_0(E)\otimes K_X)&\cong H^0(X, \delta^{-1}\otimes K_1\otimes E\otimes K_X)\\
 &= H^0(X, K_2^{-1}\otimes \pi^*(H)\otimes E\otimes K_X)\\
 &= H^0(B, \pi_*(K_2^{-1}\otimes E)\otimes H\otimes K_B)\\
 &= H^0(B, H\otimes K_B).\end{align*}
\end{proof}

We now consider the case of bundles which are not regular on the generic fibre. General computations are much more difficult in this situation than in Proposition \ref{regular}, but we can give a lower bound for the dimension of the space of Higgs fields for a bundle of this type.

\begin{proposition}
Let $E$ be a rank-2 vector bundle which is not regular on the generic fibre, and let $K$ be its maximal destabilizing bundle. Then $h^0(X,\End_0(E)\otimes K_X)\geq h^0(B,H\otimes K_B)$, where $H:=\pi_*(\det(E)^{-1}\otimes K^2)$.
\end{proposition}
\begin{proof}
Suppose that $E$ is not regular on the generic fibre so that $\pi_*(K^{-1}\otimes E)$ has rank 2.  In particular, this allows us to conclude that $\pi_*(K^{-1}\otimes E)$ is given by an extension of the form $$\begin{tikzcd}0 \arrow[r] & \mc{O}_B \arrow[r] & \pi_*(K^{-1}\otimes E) \arrow[r] & L\arrow[r] & 0 \end{tikzcd}$$ for some $L \in \Pic(B)$. Since $K$ is the maximal destabilizing bundle for $E$, there is an exact sequence \begin{equation}\label{split}\begin{tikzcd}0 \arrow[r] & \mc{O}_X \arrow[r] & K^{-1}\otimes E \arrow[r] & K^{-2}\otimes \delta \otimes \mc{I}_Z \arrow[r] & 0
\end{tikzcd},\end{equation}
where $\delta:=\det(E)$ and $Z$ is a finite set. Notice that $K^{-2}\otimes \delta=\pi^*(H^{-1})$, so taking the long exact sequence induced from pushforward by $\pi$ gives $$\begin{tikzcd} 0 \arrow[r] & \mc{O}_B \arrow[r] & \pi_*(K^{-1}\otimes E) \arrow[r] & \pi_*(\mc{I}_Z)\otimes H^{-1} \arrow[lld]\\
& \mc{O}_B \arrow[r] & R^1\pi_*(K^{-1}\otimes E) \arrow[r] & R^1\pi_*(\mc{I}_Z)\otimes H^{-1} \arrow[r] & 0.
\end{tikzcd}$$
Since $R^1\pi_*(K^{-1}\otimes E)$ has rank 2 by the Grothendieck-Riemann-Roch theorem, the connecting homomorphism from $\pi_*(\mc{I}_Z)\otimes H^{-1}$ to $\mc{O}_B$ must be zero, and $L\cong \pi_*(\mc{I}_Z)\otimes H^{-1}$. By taking the tensor product of the exact sequence \eqref{split} with $K\otimes E^\vee\otimes K_X=K\otimes \delta^{-1}\otimes E \otimes K_X$, we obtain $$\begin{tikzcd}0 \arrow{r}{i} & K\otimes \delta^{-1}\otimes E \otimes K_X \arrow[r] & \End(E) \otimes K_X \arrow{r}{p} & K^{-1}\otimes E \otimes K_X \otimes \mc{I_Z} \arrow[r] & 0,
\end{tikzcd}$$
and as in the previous cases the multiples of the identity in $\End(E)\otimes K_X$ are mapped injectively by $p$, so the space of trace-free Higgs fields contains a subspace isomorphic to $$H^0(X, K\otimes \delta^{-1}\otimes E \otimes K_X)\cong H^0(B, \pi_*(K^{-1}\otimes E)\otimes H\otimes K_X).$$ Since $K^{-1}\otimes E$ has the trivial bundle as a  subsheaf, we get $$h^0(X,\End_0(E)\otimes K_X)\geq h^0(B, \pi_*(K^{-1}\otimes E)\otimes H\otimes K_X)\geq h^0(B, H\otimes K_X).$$
\end{proof}

While in general the case of a bundle which is not regular on the generic fibre is intractable, a classification is possible for the case of an extension of line bundles.

\begin{proposition}
Let $E$ be a rank-2 vector bundle that is an extension of line bundles, and suppose that $E$ is not regular on the generic fibre. Then, there is a line bundle $\delta \in \Pic(X)$ and a rank-2 vector bundle $F$ on $B$ such that $E\cong \delta\otimes \pi^*(F)$. Furthermore, the Higgs fields on $E$ are precisely the pullbacks of Higgs fields on $F$ twisted by $\delta$.
\end{proposition}
\begin{proof}
Since $E$ is not regular on the generic fibre, it must satisfy the exact sequence $$\begin{tikzcd}
0 \arrow[r] & \mc{O}_X \arrow[r] & K^{-1}\otimes E \arrow[r] & \pi^*L \arrow[r] & 0,
\end{tikzcd}$$
where $K$ is the maximal destabilizing bundle of $E$ and $L \in \Pic(B)$ is a line bundle. Extensions of the above form are parameterized by $H^1(X,\pi^*L^{-1})$, and by the Leray spectral sequence $H^1(X,\pi^*L^{-1})\cong H^0(B, L^{-1})\oplus H^1(B, L^{-1}).$ Furthermore, any extension whose representative has non-zero first factor in this decomposition will be regular on the generic fibre, so those which are not regular on the generic fibre are parameterized by $H^1(B, L^{-1})$. This group also parameterizes extensions of $L$ by $\mc{O}_B$ on $B$ and the pullback of such a bundle is an extension which is not regular on the generic fibre, so every extension of $\pi^*L$ by $\mc{O}_X$ which is not regular on the generic fibre is a pullback. Let $F$ be the rank-2 bundle on $B$ such that $K\otimes \pi^*F\cong E$. Then $H^0(X,\End E\otimes K_X)=H^0(B,\pi_*(\End E)\otimes K_B)=H^0(B, \End F\otimes K_B)$, so there is an isomorphism between the spaces of Higgs fields on $F$ and Higgs fields on $E$.
\end{proof}

By using the descriptions of the trace-free Higgs fields in the above cases, we can show that the Chern classes of a filtrable rank-2 bundle with a non-trivial Higgs field are only restricted in that the Chern classes must admit filtrable bundles.

\begin{proposition}\label{filt}
Let $E$ be a rank-2 filtrable bundle with $c(E)=(1,c_1,c_2)$. Then, there is a stable rank-2 vector bundle $F$ with $\det(F)=\det(E)$ and $c_2(F)=c_2(E)$ such that $F$ has a non-zero trace-free Higgs field.
\end{proposition}
\begin{proof}
Since $E$ is filtrable, there are line bundles $L_1$ and $L_2$ and a finite set $Z$ such that $E$ is fits into the exact sequence $$\begin{tikzcd}0 \arrow[r] & L_1 \arrow[r] & E \arrow[r] & L_2\otimes \mc{I}_Z \arrow[r] & 0
\end{tikzcd}.$$
We split the proof into three cases: when $E$ has a jump, when $E$ has no jumps but has a spectral curve with non-trivial self-intersection, and when the spectral curve has self-intersection $0$.
We first consider the case where $Z\neq \emptyset$, meaning that $E$ has a jump. Let $L_a$ be a line bundle with constant factor of automorphy such that $$-1<\deg(L_1)-\deg(L_2)+2\deg(L_a)<0.$$ Since the degree function maps line bundles with constant factor of automorphy surjectively onto $\R$, such an $L_a$ always exists. Since $\deg(L_1)-\deg(L_2)+2\deg(L_a)$ is not an integer, there is a point $b \in B$ such that $(L_1\otimes L_a)|_{\pi^{-1}(b)} \not\cong (L_2\otimes L_a^{-1})|_{\pi^{-1}(b)}$. Define $F'=(L_1\otimes L_a\otimes \pi^*(\mc{O}_B(b)))\oplus (L_2\otimes L_a^{-1})$. Let $H$ be any line bundle on $\pi^{-1}(b)$ such that $\deg(H)=|Z|$, and define $F$ to be the elementary modification $$\begin{tikzcd} 0 \arrow[r] & F \arrow[r] & F' \arrow[r] &\iota_{b,*}H \arrow[r] & 0
\end{tikzcd},$$
where $\iota_b$ is the inclusion of $\pi^{-1}(b)$ into $X$. Then $\det(F)=\det(E)$, $c_2(F)=c_2(E)$, and $F$ fits into the exact sequences $$\begin{tikzcd} 0 \arrow[r] & L_1\otimes L_a \arrow[r] &F \arrow[r] & L_2\otimes L_a^{-1} \otimes \mc{I}_Y \arrow[r] & 0 \\ 0 \arrow[r] & L_2\otimes L_a^{-1}\otimes \pi^*(\mc{O}_B(-b)) \arrow[r] & F \arrow[r] & L_1\otimes L_a\otimes \pi^*(\mc{O}_B(b)) \otimes \mc{I}_Y \arrow[r] & 0
\end{tikzcd}$$
for some finite set $Y \subseteq X$. In particular, since $|Z|>0$, $E$ has destabilizing bundles $L_1\otimes L_a$ and $L_2\otimes L_a^{-1}\otimes \pi^*(\mc{O}_B(-b))$. By assumption, $\deg(L_1\otimes L_a)<\deg(L_2\otimes L_a)-\frac{1}{2}$ and $$\deg(L_2\otimes L_a^{-1}\otimes \pi^*(\mc{O}_B(-b)))<\deg(L_1\otimes L_a\otimes \pi^*(\mc{O}_B(b)),$$ so $F$ is a stable bundle. Finally, by Proposition \ref{regular} the trace-free Higgs fields on $F$ are parameterized by $H^0(B, K_B(-b))$ and by Riemann-Roch $$h^0(B,K_B(-b))=h^0(B,\mc{O}_B(b))+g-2\geq g-1,$$ so $F$ has a non-zero trace-free Higgs field.

Now suppose that $Z=\emptyset$ and that $c_1(L_1\otimes L_2^{-1})^2\neq 0.$ Again choose $L_a$ to be a line bundle with constant factor of automorphy such that $$-1<\deg(L_1)-\deg(L_2)+2\deg(L_a)<0.$$
Extensions $$\begin{tikzcd} 0 \arrow[r] & L_1\otimes L_a \arrow[r] & F \arrow[r] & L_2\otimes L_a^{-1}\arrow[r] & 0
\end{tikzcd}$$
are parameterized by $\Ext^1(L_2\otimes L_a^{-1},L_1\otimes L_a)\cong H^0(B, R^1\pi_*(L_2^{-1}\otimes L_1\otimes L_a^2),$ where $R^1\pi_*(L_2^{-1}\otimes L_a\otimes L_a^2)$ is a torsion sheaf supported on points $b \in B$ such that $$\Sigma_{L_1\otimes L_a}]\cap \Sigma_{L_2\otimes L_a^{-1}}\cap \pi^{-1}(b)\neq \emptyset.$$
(By the assumption that $c_1(L_2^{-1}\otimes L_1)^2\neq 0$ there is at least one such point in the support.) In addition, for any choice of section $s \in H^0(B, R^1\pi_*(L_2^{-1}\otimes L_1\otimes L_a^2)$ the corresponding extension has maximal destabilizing bundles $L_1\otimes L_a$ and $L_2\otimes L_a^{-1}\otimes \pi^*\mc{O}_B(-D_s)$, where $D_s$ is the divisor on which $s$ is supported. Therefore if we choose $s$ so that it is supported on a single point $b$, the corresponding extension $F$ will have maximal destabilizing bundles $L_1\otimes L_a$ and $L_2\otimes L_a^{-1}\otimes \pi^*\mc{O}_B(-b)$. The bundle $F$ is stable and clearly has the same determinant and second Chern class as $E$. Furthermore, the trace-free Higgs fields on $F$ are parameterized by $H^0(B,K_B(-b))\neq 0.$

Finally we consider the case where $Z=\emptyset$ and $c_1(L_1\otimes L_2^{-1})^2=0.$ Then there are line bundles $\delta \in \Pic(X)$ and $\lambda \in \Pic(B)$ such that $\det(E)=\delta^2\otimes \pi^*\lambda$. Let $F$ be a rank-2 stable bundle on $B$ with determinant $\lambda$. Since $F$ is a stable bundle on $B$, it must have a trace-free Higgs field which lifts to a trace-free Higgs field on $\delta\otimes\pi^*F.$
\end{proof}

Finally, we consider the case where $E$ is not filtrable.
\begin{proposition}
If $E$ is a rank-2 non-filtrable bundle on $X$, then $E$ has a non-zero trace-free Higgs field if and only if the spectral curve of $E$ is a bisection whose normalization is an unramified cover of $B$.
\end{proposition}
\begin{proof}
Suppose $E$ is a rank-2 non-filtrable bundle on $X$. Then on a general fibre $T$, $\End_0(E)\otimes K_X|_T$ has the form $\mc{O}_T\oplus \lambda_1\otimes \lambda_2^{-1} \oplus \lambda_1^{-1}\otimes \lambda_2$ with $\lambda_1,\lambda_2\in \Pic^0(T), \lambda_1\neq\lambda_2$. Since general fibres contain a factor of $\mc{O}_T$ and $E$ is non-filtrable, there is a line bundle $L \in \Pic(B)$ so that $\pi_*(\End_0(E)\otimes K_X)=L$. Note that $\pi_*(\End_0(E))=L\otimes K_B^{-1}$, and $H^0(X, \End_0(E))=0$, so $\Hom(K_B, L)=0$. Thus we have an exact sequence $$\begin{tikzcd}0 \arrow[r] & \pi^*L \arrow[r] & \End_0(E)\otimes K_X \arrow[r] & \mc{F}\arrow[r] & 0
\end{tikzcd},$$
for some rank-2 sheaf $\mc{F}$ whose torsion-free component is non-filtrable. Taking the long exact sequence induced by pushforward gives $$\begin{tikzcd} 0\arrow[r] & L \arrow[r] & L\arrow[r] & \pi_*(\mc{F})\arrow[llld] \\ R^1\pi_*(\pi^*L)\arrow[r] & R^1\pi_*(\End_0(E)\otimes K_X)\arrow[r] & R^1\pi_*(\mc{F})\arrow[r] & 0
\end{tikzcd}.$$
Notice that $R^1\pi_*(\pi^*L)=R^1\pi_*(\mc{O}_X)\otimes L=L$. Since the torsion-free part of $\mc{F}$ is non-filtrable, $\pi_*(\mc{F})$ is a torsion sheaf which injects into $L$, and we must in fact have $\pi_*(\mc{F})=0$. Making this substitution in the second row of the exact sequence and looking at the dual sequence results in $$\begin{tikzcd}
0 \arrow[r] & R^1\pi_*(\End_0(E)\otimes K_X)^\vee\arrow[r] & L^{-1}
\end{tikzcd}.$$
By relative Serre duality $R^1\pi_*(\End_0(E)\otimes K_X)^\vee\cong \pi_*(\End_0(E)\otimes K_X^{-1}\otimes \omega_{X/B})\cong \pi_*(\End_0(E)\otimes K_X)\otimes K_X^{-2}=L\otimes K_X^{-2}$. Since there is an injective map from $L\otimes K_X^{-2}$ to $L^*$, we must have $\Hom(L^2, K_X^2)\neq 0$.
Taking cohomology of the original exact sequence gives that $H^0(B, L)\cong H^0(X, \End_0(E)\otimes K_X)$.

Suppose that $H^0(B, L)\neq 0$, so that there is a non-zero trace-free Higgs field $\phi$ on $E$. Since $E$ is non-filtrable, $\phi$ is injective, so there is an elementary modification $$\begin{tikzcd}0 \arrow[r] & E \arrow{r}{\phi} & E\otimes K_X \arrow[r] & \mc{Q} \arrow[r] &0
\end{tikzcd},$$
where $Q$ is a torsion sheaf supported on an effective divisor $D$ for $\pi^*L$ and for every $b \in B$ with $\pi^{-1}(b) \in D$, $\mc{Q}|_{\pi^{-1}(b)}=E|_{\pi^{-1}(b)}$. Note that the determinant relations for elementary modifications give that $\det(E)\otimes \pi^*L^2=\det(E\otimes K_X)=\det(E)\otimes K_X^2$, so $L=K_B\otimes \lambda_0$, where $\lambda_0 \in \Pic^0(B)$ satisfies $\lambda_0^2=\mc{O}_B$. Since $\Hom(K_B, L)=0$, we must have $\lambda_0\neq \mc{O}_B$.

Returning to the exact sequence $$\begin{tikzcd} 0 \arrow[r] & L \arrow[r] & R^1\pi_*(\End_0(E)\otimes K_X) \arrow[r] & R^1\pi_*(\mc{F}) \arrow[r] & 0
\end{tikzcd},$$
recall that $R^1\pi_*(\End_0(E)\otimes K_X)^{\vee\vee}=L^{-1}\otimes K_B^2=L$ since $L^2=K_B^2$, so we have $R^1\pi_*(\End_0(E)\otimes K_X)\cong L\oplus R^1\pi_*(\mc{F})$. Note that the torsion component of $R^1\pi_*(\End_0(E)\otimes K_X)$ is the skyscraper sheaf corresponding to the set of points $b\in B$ where $E|_{\pi^{-1}(b)}$ is not regular, weighted in the following manner: the skyscraper sheaf has rank 2 at points where $E|_{\pi^{-1}(b)}=\lambda_0\oplus \lambda_0$, and rank $2m$ when the allowable elementary modification of $E$ at $b$ decreases the second Chern class by $m$. Since $\mc{F}$ is the quotient of $\End_0(E)\otimes K_X$ by a pullback line bundle, its Chern character is given by $2+(c_1^2(E)-4c_2(E))w_X=2-8\Delta(E)w_X$, where $w_X$ is the generator of $H^4(X, \Z)$. If we now consider Grothendieck-Riemann-Roch for $\pi_*(\mc{F})$, we have $\pi_*(\chern(\mc{F})\cdot\todd(X))=-8\Delta(E)w_B$, where $B$ is the generator of $H^2(B, \Z)$, and $$\chern(\pi_!(\mc{F}))\cdot \todd{B}=(\chern(0)-\chern(R^1\pi_*\mc{F})\cdot (1+(1-g)w_B)=-\chern(R^1\pi_*\mc{F}),$$ so the torsion component of $R^1\pi_*(\End_0(E)\otimes K_X)$ has first Chern class $8\Delta(E)$. However, when $E$ is non-filtrable it satisfies the relation $\Delta(E)=\frac{\overline{C}^2}{8}+\frac{\ell}{4}$, where $\overline{C}$ is the bisection piece of the spectral curve of $E$ and $\ell$ is the number of points in $\overline{C}$ lying over a jump, weighted with multiplicity \cite[Theorem 4.1]{BrMor}. Note that every fibre where $E$ is not regular corresponds either to a jump or to a singularity of $\overline{C}$. Adding a jump to the spectral curve of $E$ will increase its discriminant by at least $\frac{1}{4}$, (generically $1/2$), and using the self-intersection formula $$\rho_a(\overline{C})=\frac{\overline{C}^2}{2}+2g-1$$ we see that adding a singularity to $\overline{C}$ must increase the discriminant by at least $\frac{1}{4}$. Let $a$ be the number of ramification points of the normalization $C$ of $\overline{C}$, $s$ the number of points where $\overline{C}$ is singular, and $j_m$ the number of points in $\overline{C}$ lying above a jump with a degree-$m$ allowable elementary modification. Our equation for the torsion component of $R^1\pi_*(\End_0(E)\otimes K_X)$ gives $$2s+\sum\limits_{m\geq 1} mj_m=8\Delta(E)$$
and the discriminant formula gives $$\frac{a}{4}+\frac{s}{4}+\sum\limits_{m\geq 1}\frac{mj_m}{4}\leq \Delta(E).$$
Comparing these two formulas, we see that $a=j_m=0$ for all $m\geq 1$, so the spectral curve of $E$ is a bisection $\overline{C}$ whose normalization is unramified.

Now suppose instead that $E$ is non-filtrable and that the spectral curve of $E$ is a bisection $\overline{C}$ whose normalization is unramified. We once again have an exact sequence $$\begin{tikzcd}0 \arrow[r] & \pi^*L\arrow[r] & \End_0(E)\otimes K_X \arrow[r] & \mc{F}\arrow[r] & 0
\end{tikzcd},$$
where $L=\pi_*(\End_0(E)\otimes K_X)$ and $\mc{F}$ is a rank-2 torsion-free sheaf with $\pi_*(\mc{F})=0$. As before, applying Grothendieck-Riemann-Roch to $\pi_*(\mc{F})$ gives that $R^1\pi_*(\mc{F})$ is a torsion sheaf supported on $8\Delta(E)$ points counting multiplicity, and since the spectral curve of $E$ has no ramification points or jumps, the torsion component of $R^1\pi_*(\End_0(E)\otimes K_X)$ is also supported at $8\Delta(E)$ points counting multiplicity. Because of this, taking the pushforward of the exact sequence gives $L\cong \pi_*(\End_0(E)\otimes K_X)$ and $R^1\pi_*(L)=L\cong R^1\pi_*(\End_0(E)\otimes K_X)$. Taking duals and applying relative Serre duality gives $L^{-1}\cong R^1\pi_*(\End_0(E)\otimes K_X)^\vee\cong \pi_*(\End_0(E)\otimes K_X)\otimes K_B^{-2}\cong L\otimes K_B^{-2}$, meaning that $L\cong K_B\otimes \lambda_0$ for some $\lambda_0 \in \Pic(B)$ with $\lambda_0^2\cong \mc{O}_B$ and $\lambda_0\neq \mc{O}_B$. (Such line bundles exist whenever $B$ has genus greater than $0$.) Since $K_X\otimes L^{-1}\cong \lambda_0$ is a non-trivial element of $\Pic^0(B)$, Riemann-Roch gives $h^0(B,L)=h^0(B, L)-h^0(B,K_X\otimes L^{-1})=g-1.$ Finally, since $\pi_*(\End_0(E)\otimes K_X)\cong L$, $H^0(X,\End_0\otimes K_X)\cong H^0(B, \pi_*(\End_0\otimes K_X))=\C^{g-1}\neq 0$, so $E$ has non-zero trace-free Higgs fields.
\end{proof}

\begin{remark}\label{non-filt}
If the Chern classes of $E$ are in the non-filtrable range and $\Delta(E)>0$, then $E$ is regular away from jumps \cite[Proposition 3.16]{Boulter}, so by the above proposition $E$ does not admit non-zero trace-free Higgs fields.
\end{remark}

\begin{theorem}\label{smooth}
Let $\pi:X\to B$ be a principal elliptic surface with base of genus at least two. Let $\mc{M}_{2,\delta,c_2}(X)$ be the moduli space of stable torsion-free sheaves on $X$ with rank 2, determinant bundle $\delta$, and second Chern class $c_2$. If $\mc{M}_{2,\delta,c_2}(X)$ has positive dimension, then it is smooth if and only if there is no filtrable rank-2 vector bundle $E$ such that $\det(E)=\delta$ and $c_2(E)=c_2$.
\end{theorem}
\begin{proof}
By a standard deformation theory argument, the Zariski tangent space of $\mc{M}_{2,\delta,c_2}(X)$ at a point corresponding a vector bundle $E$ is isomorphic to $H^1(X, \End_0(E)).$ Since $E$ is by assumption stable we have $$h^1(X, \End_0(E))=h^2(X, \End_0(E))-\chi(X, \End_0(E))=h^0(X, \End_0(E)\otimes K_X)-\chi(X, \End_0(E)),$$
so $\mc{M}_{2,\delta,c_2}(X)$ has a singular point at $E$ if and only if $E$ has a non-zero trace-free Higgs field. The result then follows immediately from Proposition \ref{filt} and Remark \ref{non-filt}.
\end{proof}

\begin{remark}
In order to construct smooth moduli spaces of vector bundles using Theorem \ref{smooth}, we can use the result \cite[Theorem 4.5]{BrMor} which says that $\mc{M}_{2,\delta,c_2}(X)$ is a non-empty moduli space that contains no filtrable bundles whenever $-e_\delta\leq 4\Delta(2,\delta,c_2)<4m(2,\delta)$, where $e_\delta$ is an invariant of the determinant bundle bounded in the range $g\leq e_\delta\leq 0.$ Therefore, it is sufficient to find an elliptic surface $X$ with base curve $B$ of genus $g$ and a line bundle $\delta\in \Pic(X)$ such that $4m(2,\delta)\geq g+1$. This is equivalent to finding a genus $g$ curve $B$ and an elliptic curve $T$ such that there is a non-constant map from $B$ to $T$ and every non-constant map from $B$ to $T$ has degree at least $g+1$. One example satisfying this is when $B$ is the hyperelliptic curve of genus 2 given on an affine piece by $$V(45x^6-297x^5+845x^4-1306x^3+1073x^2-360x-45y^2)\subset B$$ and $T$ is the elliptic curve given on an affine peice by $$V(1125x^4-4777x^3+6212x^2-2560x-1125y^2)\subset T.$$
(These curves can be found using \cite[Lemma 5.1]{Shaska} with $a=4,c=5$, which shows that $B$ has a maximal cover of degree 3 over $T$. Any other map factors through an isogeny between elliptic curves in the decomposition of the Jacobian of $B$, so we can check that this is minimal by verifying that the two elliptic curves in the decomposition have no isogenies. A simple way to check this is to note that the two elliptic curves have a different number of $\mathbb{F}_7$-points, meaning they are not isogenous.)
\end{remark}

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text:
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text:
«Провинциалка» () — итальянский драматический фильм 1953 года режиссера Марио Сольдати, экранизация одноименного произведения Альберто Моравиа.

Сюжет

Франко Вануцци и его жена Джемма собираются переезжать в Рим, где он получил должность профессора в университете. И неожиданно во время ужина, когда графиня Эльвира Кочеану начала описывать преимущества большого города, Джемма накинулась на нее с ножом.

В ролях

 Мэрилин Бьюфёрд — Анна Сертори

Примечания

Ссылки
 «Провинциалка» Festival de Cannes

Фильмы Италии 1953 года
Чёрно-белые фильмы Италии
Экранизации романов Италии


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Internet portal jumps into downloads of digital songs and says more musical maneuvers are yet to come.
Yahoo said Tuesday that it is acquiring digital music specialist Musicmatch for $160 million in cash.
The Sunnyvale, Calif.-based company said the buyout would help it expand its presence in the simmering online music space, where Apple Computer's iTunes service has proven the segment's potential by selling over 125 million songs, typically at 99 cents apiece. CNET News.com first reported acquisition talks between Yahoo and Musicmatch in February.
"Yahoo is committed to being a major player in digital music," Terry Semel, chief executive of Yahoo said in a statement Tuesday. "This combination bolsters our strategy to capture the largest audience of consumers as they make the shift to digital music."
But the company faces formidable competition. Besides early leader Apple, the field also includes a recent entry by software giant Microsoft, a feisty presence in RealNetworks, a reinvigorated Napster and low-price specialist Wal-Mart Stores.
Yahoo said the Musicmatch deal is one of several new initiatives in which it will invest to expand its music portfolio this year.
Founded in 1997, Musicmatch markets digital music software, dubbed Jukebox, and maintains a download service that boasts over 225,000 subscribers.
Included in those numbers are customers of Musicmatch On Demand, a subscription service that offers streaming music, and the company's online music download store. Musicmatch, which is based in San Diego, says that approximately 700,000 songs are available through its online catalog.
Jumping into music downloads signifies a switch from Yahoo's past music strategy, which has focused on streaming audio and video through Launch, a subsidiary that Yahoo bought for $12 million in 2001. Yahoo also distributes the Napster music service, which lets customers buy music downloads and sign up for subscriptions.
Industry analysts said that while the Musicmatch acquisition helps to round out Yahoo's array of digital music services, it will be hard for the company to compete with more established players such as Apple, since the computer maker markets both downloads and one of the most popular devices for playing tunes, the iPod.
"The challenge for Yahoo is how to differentiate, since all the music services basically offer the same content, in the same format, for the same price," said Michael Gartenberg, analyst with New York-based JupiterResearch. "Since Musicmatch doesn't support iPod any more, you wonder how it can maintain itself as a proprietary service."
Gartenberg pointed out that despite all the hype in computing circles, digital music downloads still represent only a small fraction of overall music industry sales. Jupiter has projected that music downloads will become an $800 million business by 2009, while the music industry racked up over $11 billon in CD sales in 2003 alone.
In July, Musicmatch launched a new version of Jukebox that added a search engine for browsing tracks and that also lets customers rank music searches by popularity. The feature was designed to let people buy and centrally manage music that's downloaded, recorded from CD or streamed on demand. Musicmatch allows subscribers to share complete playlists or individual songs, which recipients can listen to up to three times without officially subscribing to its download or streaming services.
Musicmatch currently charges $9.95 per month for its on-demand music service, offers individual downloads for 99 cents per song, and offers an online radio service free of charge.


text:
Esuli (titolo originale in inglese: Exiles) è l'unica opera teatrale dello scrittore irlandese James Joyce, scritta nel 1915 mentre viveva a Trieste. Rifiutata da W. B. Yeats per l'Abbey Theatre di Dublino e da diversi teatri americani nonostante le pressioni di Ezra Pound, la pièce ebbe la sua prima assoluta a Monaco di Baviera nel 1919. Stroncata da critica e pubblico, la rappresentazione fu un clamoroso insuccesso ed Esuli viene generalmente considerato il lavoro meno riuscito di Joyce; il giudizio negativo del dramma è stato parzialmente rivalutato grazie a messe in scena recenti di maggiore successo, a partire dal revival diretto da Harold Pinter per il Mermaid Theatre di Londra nel 1970. Lo stile e la struttura dell'opera sono chiaramente ispirati al teatro di Ibsen, di cui Joyce era un grande estimatore, mentre la trama è liberamente ispirata a I morti, il racconto conclusivo di Gente di Dublino, e ad alcuni elementi autobiografici: durante la permanenza a Trieste Joyce decise infatti di soprassedere e anzi incoraggiare il flirt tra il giornalista veneziano Roberto Prezioso e la compagna Nora Barnacle.

Trama

Primo atto
Richard Rowan è tornato a Dublino dopo nove anni passati in Italia, il Paese in cui era scappato in una fuga d'amore con Bertha: la relazione tra i due sarebbe stata biasimata in Irlanda, sia per la giovane età della ragazza che per la differenza sociale tra i due. Pur non essendosi mai sposati, Bertha e Richard hanno avuto un figlio, il piccolo Archie, di otto anni, e dopo la morte della madre Richard ha portato la famiglia di nuovo a Dublino. Mentre Berta e Archie sono fuori, Beatrice Justice, l'insegnante di piano del bambino, arriva a casa Rowan, da cui manca da diverse settimane. La donna infatti si è allontanata da Dublino dopo che Richard la ha dichiarato i propri sentimenti e i due parlano brevemente del passato mentre aspettano il ritorno dello studente di musica con la madre. Beatrice e Richard si conoscono da bambini e hanno continuato a scriversi lunghissime lettere durante il volontario esilio dell'uomo; tra i due c'era stato anche un breve flirt quando erano ragazzi, interrotto bruscamente e sostituito da Beatrice con una relazione con il cugino, il giornalista Robert Hand. I due vengono interrotti dall'arrivo di Bertha e Archie, che precedono di poco Robert: il giornalista è venuto per parlare con il vecchio amico Richard di un'interessante proposta di lavoro.

Mentre Beatrice va ad insegnare musica ad Archie al piano di sopra e Richard si allontana brevemente, Robert rimane solo con Bertha, a cui aveva passato un biglietto alcuni giorni prima in cui le dichiarava il suo amore. Durante il loro breve momento insieme, Bertha lascia che Robert le tocchi la mano, le baci gli occhi, le accarezzi i capelli e alla fine i due si baciano appassionatamente. Robert scrive un indirizzo alla donna, chiedendole di raggiungerlo quella sera alle otto. I due vengono interrotti dall'arrivo di Richard, che congeda la compagna per parlare con l'amico. Robert ha buone notizie: il vice-cancelliere ha invitato Richard a cena per quella sera alle otto, per offrirgli una cattedra di letteratura inglese all'università. Negli anni dell'assenza di Richard, Robert si è prodigato per salvaguardare la reputazione dell'amico, danneggiata dallo scandalo, e chiede all'uomo di non rilasciare commenti sui pettegolezzi su di lui. Quando Robert lascia casa Rowan, Bertha rivela al compagno tutto quello che è successo tra lei ed il giornalista durante la sua assenza: la notizia non sconvolge Richard, che anzi ha chiesto alla donna di assecondare Robert e di tradirlo pure se è quello che vuole. Bertha desidera che il padre di suo figlio le impedisca di andare all'appuntamento, ma Richard rifiuta sprezzante di imporre la sua autorità su di lei, lasciandola libera di fare quello che vuole al riguardo. Bertha lo accusa di volerle far commettere adulterio per portare la loro relazione a termine, dato che sa che Richard è innamorato di Beatrice.

Secondo atto
Nel suo cottage a Ranelagh, il covo da scapoli che un tempo divideva con Richard, Robert aspetta l'arrivo di Bertha. Ma, quando apre la porta, è Richard a entrare: il vecchio amico gli rivela che Bertha l'ha informato di ogni scambio di parola e ogni contatto tra lei e il giornalista. Robert si sente tradito e ridicolizzato, ma supplica perdono e dice che l'umiliazione lo curerà da future infedeltà. Ma Richard non è arrabbiato e dice a Robert che Bertha potrà ancora andare a letto con lui se lo vorrà, dato che ha lasciato alla compagna la libertà di fare quello che vuole a riguardo. I due parlano della monogamia e Richard confessa di aver tradito Bertha ripetutamente, cosa di cui si vergogna. Ma del resto Richard non è convinto che due persone siano destinate insieme ad amarsi appassionatamente per tutta la vita e non vuole che un giorno Bertha rimpianga di non avere avuto amori e avventure al di fuori di lui. Robert e Richard decidono quindi di andare avanti con il doppio corteggiamento e vedere se Bertha andrà davvero a letto con il giornalista o se resterà fedele al padre di suo figlio. Quando Bertha bussa alla porta ed entra è stupita ma non del tutto sorpresa di trovarvi Richard, dato che sospettava che l'uomo fosse troppo geloso per lasciarla andare così avanti con la sceneggiata. Robert li lascia e i due parlano della loro relazione: Richard nega di essere geloso e nonostante Bertha gli chiede di proibirle di fare qualcosa con il giornalista Richard rifiuta e le ribadisce che deve fare come meglio crede. Mentre Richard si prepara ad andarsene Bertha lo accusa di stare andando a incontrare Beatrice, un'insinuazione che manda l'uomo su tutte le furie.

Robert rientra dal giardino zuppo di pioggia e chiede a Bertha perché ha rivelato tutto su di loro al marito. Bertha sostiene che non ci sia stata malizia nella decisione e che non l'ha fatto come un esperimento o per umiliare Robert, ma solo perché Richard glielo ha chiesto. I due parlano del passato, di quando erano più giovani prima che Richard e Bertha fuggissero a Roma. Robert era già innamorato di Bertha e aveva incoraggiato Richard a fuggire da solo da Dublino per mettere alla prova i suoi sentimenti per la ragazza, così che lui avrebbe potuto approfittare della sua assenza per sedurre la ragazza. Evidentemente Richard non accolse il suggerimento e fuggì con Bertha, ma Robert non l'aveva mai dimenticata. Anche Bertha ammette di non essere stata del tutto indifferente al giornalista e rivela che Richard e lei parlavano spesso di lui durante il loro esilio. Mentre i due restano in silenzio ad ascoltare la pioggia, Bertha si alza per andarsene ma Robert la ferma e la supplica di dirgli che lo ama. Bertha resta immobile e non risponde.

Terzo atto
La mattina successiva a casa Rowan Brigid, la cameriera, si stupisce di trovare la padrona di casa in piedi e alla finestra così presto. Bertha è chiaramente pensierosa e dopo un malinconico momento con Brigid aiuta il piccolo Archie a prepararsi per uscire: il lattaio ha infatti promesso al bambino di insegnargli a guidare il suo calesse delle consegne e sta venendo a prenderlo. Dopo che Archie è uscito, Beatrice viene in visita per mostrare a Bertha l'articolo su Richard che Robert ha fatto pubblicare sul giornale quella mattina. Beatrice spiega a Bertha che il cugino ha scritto l'articolo in ufficio fino a notte fonda e che non è tornato a casa prima delle due di notte, quando poi si è messo a preparare le valigie. L'insegnante di piano sospettava infatti che l'articolo potesse essere contro Richard invece che a suo favore, dato che ha notato una certa freddezza tra i due. Bertha invece insinua che Beatrice sia molto intima con casa Rowan, un'affermazione che secca l'insegnante, che potrebbe ancora avere sentimenti per il cugino. Richard arriva dallo studio per leggere l'articolo e Bertha manda a chiamare Robert con un biglietto da Brigid. Bertha confessa di essersi sempre sentita a disagio davanti al mondo per la sua relazione con Richard, credendo di essere per lui un ostacolo a causa della sua poca educazione e classe sociale inferiore. Decide di accantonare ogni animosità con Beatrice e le due decidono di diventare amiche prima che l'insegnante lasci la casa per evitare di incontrare il cugino. Richard accusa Bertha di aver allontanato Beatrice da lui come ha fatto con tutti i suoi amici, ma la donna nega indignata e rinfaccia al compagno di non averle neanche chiesto cosa è successo la notte prima con Robert. Quando lei si offre di dirglielo Richard rifiuta, non volendo sapere nulla e ribattendo che Bertha è libera.

Robert entra in abito da viaggio, pronto a partire per il Surrey. Parla brevemente con Bertha prima ma non è chiaro se i due abbiano dormito insieme la notte prima e la donna presto tronca la conversazione per chiamare Richard. Rimasto solo con l'amico, Robert gli dice che non è successo niente con Bertha la notte prima, che la donna se n'è andata poco dopo di lui ed è tornata a casa, anche se Richard, impegnato con la scrittura, non l'ha sentita rientrare. Robert giura di non avere toccato la donna, che ritiene di appartenere a Richard esattamente come nove anni prima. Il colloquio tra i due uomini viene interrotto dal ritorno di Archie e prima di partire Robert insinua vagamente che il bambino potrebbe essere suo. Rimasti soli, Berta e Richard cercano di trovare un senso nella vicenda, ma Richard non riesce a convincersi che non sia successo niente tra la compagna e l'amico, si confessa geloso e ammette che il suo unico desiderio è sempre stato quello di essere unito a Bertha al di là di vincoli legali e convenzioni sociali.

Origini, stampa e debutto
Joyce scrisse Esuli nel 1915, mentre si trovava a Trieste con la moglie e non aveva ancora pubblicato Ritratto dell'artista da giovane (1916). Amante del teatro e di Ibsen, Joyce inviò una copia del manoscritto ad Ezra Pound nell'autunno del 1915, che scrisse un lungo articolo celebrativo dell'opera per la rivista specializzata Drama, in cui apparve nel febbraio 1916. In realtà, Pound aveva delle perplessità sull'opera, ritenendola troppo complicata per essere seguita facilmente dal pubblico a teatro. Nonostante le parole lusinghiere del poeta, il dramma faticò a trovare un impresario e un teatro che la mettessero in scena. I tentativi di Pound e Joyce furono respinti da teatri di Manchester, Torino, Berna, Zurigo e Dublino, dato che Yeats non lo ritenne adatto per la programmazione dell'Abbey Theatre, oltre a considerare l'opera sincere ed interessante ma non all'altezza del Ritratto dell'arista da giovane. Allo stesso modo, la London's Stage Society bocciò la richiesta di mettere in scena Esuli, che aveva comunque suscitato degli apprezzamenti da Thomas Sturge Moore e George Bernard Shaw; dopo essere stata rifiutata dalla Stage Society per la stagione del 1916, il dramma fu nuovamente bocciato l'anno successivo per la programmazione 1917-1918. Data la difficoltà nel mettere in scena l'opera, Joyce optò per la pubblicazione ed Esuli venne dato alle stampe e pubblicato contemporaneamente a Londra e New York il 25 maggio 1918 da Grant Richards e B.W. Huebsch. Stefan Zweig ricevette una copia del libro a Vienna e ne divenne immediatamente un sostenitore, tanto da scrivere immediatamente una lunga lettera congratulatoria a Joyce e battersi personalmente perché l'opera - da lui ritenuta una grande rivelazione artistica - raggiungesse le scene. Fu infatti grazie a Zweig che l'opera fu tradotta in tedesco e portata in scena dalla compagnia di Max Reinhardt a Monaco, con Hermine Körner nel ruolo di Bertha. Joyce non ottenne il visto e non poté viaggiare in Germania per la prima della sua opera, che andò in scena il 7 agosto 1919 al Münchner Kammerspiele di Monaco di Baviera, e attese a Zurigo un telegramma che lo informasse delle reazioni del pubblico. Il giudizio finale fu lapidario e l'opera venne considerata un autentico fiasco. Il testo stampato ottenne però recensioni più favorevoli e Joyce fu lodato per il modo in cui portava in scena (o, almeno, sulla pagina) l'adulterio e le ansie della vita domestica.

Commento

I personaggi
Joyce scrisse alcuni appunti su Esuli in un quaderno vuoto con la copertina blu, ora conservato all'Università di Buffalo. In esso, il drammaturgo ha scritto alcune pagine di appunti sui personaggi e le loro motivazioni. In particolare, Joyce si è soffermato sul personaggio di Bertha, che ha forti richiami femminili legati alla Luna e al suo ciclo: la donna ha 28 anni — come 28 giorni è il ciclo lunare ma anche il ripetersi delle mestruazioni — e Robert nella loro prima scena insieme la paragona alla luna per il colore del suo vestito. Per Bertha, tradire Richard è incomprensibile non tanto perché follemente innamorata dell'uomo, quanto più perché non ha l'energia mentale di ricominciare da zero con un altro uomo e ridiventare vergine: la sua paralisi nasce dall'incapacità di gettarsi ancora nell'oceano dell'anima di un altro ("into the ocean of another's soul"). Joyce dà anche indicazione a un'immaginaria attrice che interpreta il ruolo di Bertha, indicando che lo stato d'animo della donna quando si sente abbandonata da Robert deve essere portato in scena come uno stato catatonico o di ipnosi:

La gelosia
Joyce considera lOtello di Shakespeare uno studio inclompleto della gelosia, dato che il punto di vista del tragediografo elisabettiano, così come quello di Baruch Spinoza, si concentra esclusivamente sugli aspetti sensoriali di questo stato d'animo. Per Richard la gelosia diventa un ostacolo al raggiungimento della libertà individuale che, non riuscendo a raggiungere, proietta su Bertha. Tuttavia, i suoi stessi desideri e le sue passioni gli impediscono il raggiungimento dell'agognata libertà e costringendo Bertha a raggiungere la libertà che lui immagina si scopre tradito dal proprio bisogno di amore ed amicizia.

Il titolo ed il concetto di esilio
Pur trovandosi a Trieste del tutto volontariamente, Joyce era solito considerarsi in esilio lontano dall'Irlanda. Il tema dell'esilio nella pièce non è però solo geografico — Bertha e Richard erano esuli a Roma — ma è soprattutto uno stato mentale. Sempre nei suoi appunti, Joyce scrive che una nazione riscuote una penale da coloro che hanno osato lasciarla insaldata al loro ritorno, evidenziando quindi il prezzo da pagare nel tornare in patria: per Richard e Bertha sono i pregiudizi e i pettegolezzi che li aspettano a Dublino per il loro comportamento contrario alle norme sociali che prevedono il matrimonio. Ma i due protagonisti maschili sono esuli anche perché la scelta di Bertha inevitabilmente costringerà l'altro (in questo caso Robert) all'esilio, non potendo rimanere nella vita della donna che lo ha rifiutato né dell'uomo che ella ha scelto al posto suo. La condizione iniziale di esuli ricoperta da Bertha e Richard passa quindi a Robert alla fine del dramma.

Elementi autobiografici
Steven Helming ha evidenziato che in un certo senso Esuli si contrappone a Ritratto dell'artista da giovane perché se in Stephen Dedalus Joyce riscopre ed esplora momenti della sua infanzia, adolescenza e giovinezza, nella pièce mostra un ritratto dell'artista al momento presente. In Esuli, Joyce confronta se stesso come artista mettendosi idealmente in competizione con Ibsen, ma si relaziona anche con gli eroi ibseniani per libertà sessuale ed indipendenza emotiva. Durante il periodo triestino, Joyce desiderava sperimentare un matrimonio aperto nel triangolo amoroso tra lui, Nora e l'amico Roberto Prezioso, di cui incoraggiò le advances alla moglie, per poi impedirle di consumare il tradimento. La preoccupazione di Joyce per l'adulterio pervade lUlisse nella figura di Leopold Bloom, ma lo stesso scrittore disse per lettera a Frank Budgen che avrebbe voluto che Nora lo tradisse per avere qualcosa di cui scrivere.

Così come Bertha e Richard (l'alter ego di Joyce nella pièce), lo scrittore e Nora Barnacle vivevano insieme in Italia senza essere sposati e, come i protagonisti di Esuli, si considerava in esilio. Oltre all'esperienza vissuta con Prezioso, il personaggio di Robert Hand è ispirato anche ad Oliver St. John Gogarty e Vincent Cosgrave, con cui condivide della caratteristiche, mentre si ipotizza che la parte di Beatrice Justice possa ispirarsi alla cugina di Joyce, Elizabeth Justice, morta nel 1912.

Storia delle rappresentazioni

I primi allestimenti
La prima in lingua inglese dell'opera avvenne sei anni dopo il debutto tedesco, alla Neighborhood Playhouse di New York, dove Esuli rimase in cartellone per quarantuno repliche dal 19 febbraio 1925. Le cinque recensioni del dramma furono contrastanti: due favorevoli, due contrarie e una più ambigua. La recensione del New York Times evidenziò che ci sarebbe aspettata una maggiore originalità stilistica e formale da Joyce, il cui Ulisse era ormai già stato pubblicato da tre anni. La prima londinese andò in scena il 14 febbraio 1926 al Regent Theatre per la regia W.G. Fay ed il pubblicò mostrò un maggiore apprezzamento del testo, soprattutto nei suoi primi due atti; l'ambiguità del terzo fu accolta più freddamente. Ancora una volta però le recensioni si rivelarono ostili. Nel maggio 1950 la pièce tornò sulle scene londinesi, al Q Theatre, per la regia di Esmé Percy: le recensioni furono un po' più favorevoli, anche se criticarono l'eccessivo attaccamento stilistico e formale ad Ibsen e la struttura stessa dell'opera, che rivela le vere intenzioni dei personaggi in un modo che rende difficile la loro comprensione. Quasi tre decenni dopo il debutto sulle scene, la critica teatrale aveva cominciato ad ammorbidirsi nei confronti di Esuli e l'influente critico Brooks Atkinson scrisse di aver apprezzato la pièce nonostante i suoi difetti quando il dramma tornò sulle scene newyorchesi in una produzione dell'Off Broadway al Renata Theatre: pur trovando Esuli poco drammatico lo considerò comunque "coraggioso e stimolante". Un revival londinese del 1967 passò quasi del tutto inosservato, se non per la recensione in The Times che sottolineò come il talento di Joyce fosse imbrigliato male nella camicia di forza della struttura ibseniana.

La rivalutazione critica
Un punto di svolta nella ricezione critica di Esuli può essere rintracciato nella produzione londinese del 1970, diretta da Harold Pinter per il Mermaid Theatre; il cast includeva John Wood (Richard), Vivien Merchant (Bertha), Timothy West (Robert) e Lynn Farleigh (Beatrice). Le recensioni evidenziarono che l'opera fosse più di un esercizio ibseniano, ma che anzi il testo fosse pervaso da i dilemmi morali tipicamente joyciani. L'influente critico di The Times Benedict Nightingale lodò l'opera, considerandola un capolavoro realista superiore allL'anitra selvatica. Il 7 ottobre 1971 la Royal Shakespeare Company ripropose la versione di Pinter di Esuli all'Aldwych Theatre del West End londinese, questa volta con T.P. McKenna nel ruolo di Robert; pur essendo virtualmente invariata, la produzione ottenne recensioni leggermente meno favorevoli, ma l'allestimento fu comunque un successo. Il Peacock Theatre di Dublino ospitò la prima irlandese dell'opera il 21 febbraio 1973 e le recensioni locali fecero notare che per quanto i critici londinesi avessero attribuito gran parte del successo recente dell'opera alla regia di Pinter, è nel testo di Joyce che risiede il valore artistico di Esuli. Recensioni favorevoli accolsero anche il ritorno del dramma sulle scene newyorchesi nel 1977, in un revival diretto da Rob Thirkield per il Circle Repertory Theatre. Un'altra produzione di successo è stata diretta da James Macdonald per il Royal National Theatre di Londra nel 2006, con un cast che comprendeva Adrian Dunbar, Dervla Kirwan e Peter McDonald; le recensioni sottolinearono che, pur non essendo un capolavoro, Esuli è stato ingiustamente ignorato dalla critica.

Marco Sciaccaluga tradusse e diresse la prima italiana di Esuli, andata in scena al Teatro Ariosto di Reggio Emilia dal 13 novembre 1986, con un cast che annoverava Aroldo Tieri, Giuliana Lojodice e Mino Bellei. La produzione fu ricevuta positivamente dalla critica e Maria Grazie Gregori dell'Unità definiì Esuli "un testo bellissimo, ambiguo e inquietante dove l'oggettività della situazione, dell'intreccio, si squarcia all'improvviso su interrogrativi destinati a rimanere senza risposta". Scaccaluga curò anche una nuova produzione del dramma per la Sala Duse del Teatro Stabile di Genova nel novembre del 2010, con un cast che annoverava Antonio Zavatteri, Lisa Galantini, Aldo Ottobrino, Barbara Moselli, Federica Granata e Giacomo Costella; il revival è stato accolto positivamente dalle principali testate nazionali.

Note

Bibliografia

L'opera
 In italiano

 In inglese

Apparato critico

Collegamenti esterni

Opere di James Joyce
Opere teatrali ambientate a Dublino
Opere teatrali del 1918


Le rêve Canadien: French immigrants find Canada the land of opportunity - Canada Immigration and Visa Information. Canadian Immigration Services and Free Online Evaluation.
For many Canadians, France is a dream country with the name of the country bringing to mind the fantasy of rain-soaked Paris cobblestones, Monet-like fields of lavender and kisses in the moonlight. It has become almost a right-of-passage for young people fresh out of university to stop over as they backpack through Europe.
However, it may come as a surprise that for the French, especially for French youth who've grown disenchanted with their country's brutally competitive job market, Canada has become a dream destination too.
Since the economic meltdown of 2008, French immigration to Canada has exploded. Between 2008 and 2012, there were 34,619 new permanent residents from France to Canada — a 38 per cent increase over the five-year period immediately before the crash. Meanwhile, temporary foreign workers who work in Canada nearly doubled with 78,267 workers entering the country between 2008-12. French student visas to study in Canada have been issued at a steadier rate, with only an 18-per-cent increase since 2008. However, since 2003, Canada has had a 50 percent increase in students from France.
Coming to Canada has become so popular that the French magazine, L'Express Reussir publishes an annual edition on immigrating to Canada: Moving to Canada. Laurence Pivot, once an immigrant herself, has been the editor of this special edition since 2007. The Canada issue of L'Express is part travel brochure, part immigration crash course. Although it paints a pretty rosy picture of Canada as a multicultural haven where individual differences are celebrated, it also offers a fairly exhaustive explanation of the complex immigration process.
Of the more than 30,000 French citizens who applied for Canadian permanent residence and immigrated to Canada in 2012, about 80 per cent settled in Quebec, where French is the predominant spoken language. Although it may be easier for French people to live in Quebec, because of the language laws, many still experience a culture shock. In Quebec, Laurence Pivot says, people speak directly and candidly, while in France, conversations about everything from politics to coffee can quickly develop into long, circuitous arguments.
For many, however, the difference between the two cultures is what makes the province so attractive. Anne-Laure Piaraly and her best friend Lisa Renault moved to Montreal in 2007, hoping to practise their English and enjoy a "friendlier" scene. When asked if Quebecers are more similar to people from Vancouver or France, Renault says: "I think more like people from Canada … I like that. I don't want to speak badly about my home … but the people are very aggressive, while here they are very friendly," she says.
Because of the language, many French citizens don't think about moving to other places, like Toronto or Vancouver. In Ottawa, for example from 2006 – 11, only 305 French immigrants came to the city. In order to encourage francophones from across the world to settle outside Quebec, the federal government created 'The Francophone Significant Benefit Program' in 2011. The program aimed to give skilled francophones a fast-track to a temporary work permit if they wanted to settle in cities with francophone minority communities.
Leafing through the l'Express's pages, peppered with adds for RBC and immigration services, one sees Canada through new eyes. In France's Canada, economic opportunity, wide open spaces and yes, Rob Ford, all offer French citizens the chance to stretch their legs and be someone new.
Youth unemployment in France is at an all-time high, climbing to 26 per cent in 2013, up about 45 per cent since just before the market crashed in 2008. In Paris people must live in the city's more affordable suburbs, and commute long distances every day just to stay afloat.