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Let $(E,\phi)$ be a $G$-Higgs bundle $\phi\in H^{0}(X,ad(E)\otimes D)$ where $D$ is a divisor on X.

I suppose that $(E,\phi)\in \mathcal{M}^{ani}$ the anisotropic locus.

In particuler, this bundle is stable as a Higgs bundle because, it doesn't have any reduction to a parabolic.

Does it imply that the underlying bundle $E$ is itself stable?

More generally, when a stable Higgs bundle has a stable underlying bundle.

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A detailed answer to your question in the case of Riemann surfaces is given in Proposition 3.3 of Hitchin's "Self-duality equations on a Riemann surface". –  Sebastian Dec 6 '12 at 7:42

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First, in the standard definition $D = K_X$, so I will give an example in this case. Let $X$ be a curve of genus 2 and $E = O \oplus O(P)$ for a point $P \in X$. Clearly $E$ is unstable with $O(P)$ being the only destabilizing subbundle. Define $\phi$ to be the composition $$ O \oplus O(P) \to O(P) \to O(K_X) \to O(K_X) \oplus O(K_X + P), $$ where the first map is the projection, the second is the embedding given by the point $P' \in |K_X - P|$, and the third is the embedding into the first summand. It is clear that $O(P)$ does not extend to a Higgs subbundle, so $(E,\phi)$ is stable.

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