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Let $E$ be a $G$-bundle over a compact Calabi-Yau 3-fold, $G$ is semi-simple,compact Lie group. If $E$ is a stable bundle, is the first Chern class of $E$ always zero?

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    $\begingroup$ What do you call a $G$-bundle? For most of us it means a $G$-principal bundle, but this has no natural Chern classes. $\endgroup$
    – abx
    Commented Jul 3, 2016 at 17:17
  • $\begingroup$ Yes,your are right, $E$ is a principal $G$-bundle.But I can't understand why this has no natural Chern classes? $\endgroup$
    – user94640
    Commented Jul 4, 2016 at 6:32
  • $\begingroup$ Well, how do you define them? $\endgroup$
    – abx
    Commented Jul 4, 2016 at 7:23

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If you just look at $U(1)$ bundles, so that there is a $c_1$, then clearly the answer is no, because Calabi-Yau 3-folds are algebraic, so have projective embeddings, for which $c_1$ of the tautological line bundle $O(1)$ pulls back to be positive, so $c_1>0$.

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  • $\begingroup$ But if $G$ is semisimple, then yes because semisimple Lie groups have no 1-dimensional nontrivial reps. $\endgroup$
    – Ben McKay
    Commented Jul 3, 2016 at 19:26
  • $\begingroup$ Do I contradict myself? Very well, then I contradict myself, I am large, I contain multitudes. Walt Whitman $\endgroup$
    – Ben McKay
    Commented Jul 3, 2016 at 20:11
  • $\begingroup$ Through pull back the bundle $E$ over CY-3 fold to $\pi^{\ast}E$ over the $G_{2}$ manifold $CY^{3}\times S^{1}$. I had proved this assumed, but I can't belive my result. $\endgroup$
    – user94640
    Commented Jul 4, 2016 at 6:36
  • $\begingroup$ It is not that fancy; you don't need to think about $G_2$ manifolds or Calabi-Yau manifolds. If $G$ is a Lie group and $E \to X$ is a principal $G$-bundle over any Hausdorff topological space $X$, not necessarily a manifold, then the associated line bundles are $L=E \times^{\rho} \mathbb{C}$ where $\rho \colon G \to \mathbb{C}^{\times}$ is a representation of $G$. But if $G$ is semisimple, then $G$ has only on 1-dimensional representation: the trivial one. So $\rho(g)=1$ and therefore $L$ is trivial and admits a flat connection, and so $c_1(L)=0$. $\endgroup$
    – Ben McKay
    Commented Jul 4, 2016 at 7:55

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