Let $G$ be a simply connected Lie group and $\Gamma$ a cocompact discrete and torsion-free subgroup. Is there a (real or complex) smooth vector bundle $E$ over the manifold $G/\Gamma$, which does not admit a flat connection?
$\begingroup$
$\endgroup$
8
-
3$\begingroup$ You may be missing some hypotheses. Otherwise, there are very simple examples provided by smooth complex line bundles with nontrivial first Chern class over $\mathbb{T}^2 = \mathbb{R}^2/\mathbb{Z}^2$. $\endgroup$– Robert BryantCommented Sep 26, 2023 at 14:05
-
3$\begingroup$ Chern classes are computed in terms of curvature of any connection. A flat connection has zero Chern classes. See Chern, Vector bundles with a connection, Global Differential Geometry. $\endgroup$– Ben McKayCommented Sep 26, 2023 at 14:40
-
1$\begingroup$ I think this old answer is general enough to addresses also this question. Is it not? $\endgroup$– Igor KhavkineCommented Sep 26, 2023 at 23:58
-
1$\begingroup$ @IgorKhavkine: The OP has not asked that $E$ be the tangent bundle of $G/\Gamma$. Indeed, the tangent bundle of $G/\Gamma$ is trivial, so it does have a flat connection. $\endgroup$– Robert BryantCommented Sep 27, 2023 at 8:59
-
2$\begingroup$ @IgorKhavkine: However, that answer has something wrong with it. The statement 'If a vector bundle admits a flat connection, then the rational Pontryagin classes of the tangent bundle vanish..." is just false. I suspect that I.B. meant to start with "If a tangent bundle admits a flat connection...", which would have been a true statement. Also, the final statement that 'most vector bundles don't admit a flat connection' is true, but that says nothing about the vector bundles over particular manifolds, such as $G/\Gamma$. $\endgroup$– Robert BryantCommented Sep 27, 2023 at 11:09
|
Show 3 more comments
1 Answer
$\begingroup$
$\endgroup$
Just so there'll be an answer: Whether every vector bundle over $G/\Gamma$ admits a flat connection depends on the group $G$ and the subgroup $\Gamma$.
For example, if $G=\mathrm{SU}(2)\simeq S^3$ and $\Gamma = \{e\}$, then, since every (real or complex) vector bundle over $G/\Gamma \simeq S^3$ is trivial (because $\pi_2(H)=0$ for any Lie group $H$ (see Steenrod's Topology of Fibre Bundles, Corollary 18.6), it follows, a fortiori that every vector bundle $E$ over $G/\Gamma \simeq S^3$ has a flat connection.
Meanwhile, if $G=\mathbb{R}^2$ and $\Gamma = \mathbb{Z}^2\subset G$, we know that the isomorphism classes of complex line bundles over $G/\Gamma=\mathbb{T}^2$ are parametrized by the elements of $H^2(\mathbb{T}^2,\mathbb{Z})\simeq \mathbb{Z}$, and the curvature of a connection for such a line bundle can only be identically zero for the line bundle that represents $0\in H^2(\mathbb{T}^2,\mathbb{Z})$. Thus, none of the other complex line bundles can have a connection with vanishing curvature.
answered Sep 28, 2023 at 0:13