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Let $(M,g)$ be a Riemannian manifold and let $U\subset M$ be an open subset. Suppose that the tangent bundle over $U$ splits into two orthogonal distributions $TU=\mathcal{E}\oplus \mathcal{F}$.

Is it possible that the two $C^{\infty}(U)$-bilinear maps

\begin{align*}I:\mathcal{E}\times\mathcal{E}&\to \mathcal{F}\\ &(X,Y)\mapsto pr_{\mathcal{F}}(\nabla_X Y) \end{align*}


\begin{align*}I:\mathcal{F}\times\mathcal{F}&\to \mathcal{E}\\ &(X,Y)\mapsto pr_{\mathcal{E}}(\nabla_X Y) \end{align*}

are both antisymmetric in $X$ and $Y$ without vanishing?

If $\mathcal{E}$ and $\mathcal{F}$ both were integrable, both maps would be symmetric. So is this in some sense the most non-integrable way, distributions can be?

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up vote 5 down vote accepted

Yes, this can happen. A little experimentation with the structure equations shows that there is a $3$-parameter family of homogeneous examples in dimension $4$: Let $c_1,\ldots,c_4$ be nonzero constants satisfying $c_1c_2=c_3c_4$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis of left-invariant $1$-forms $\omega_1,\ldots,\omega_4$ that satisfy the structure equations \begin{aligned} d\omega_1 &= 2c_1\ \omega_2\wedge\omega_3 + 2c_3\ \omega_4\wedge\omega_3\ ,\\\\ d\omega_2 &= 2c_1\ \omega_3\wedge\omega_1 \ ,\\\\ d\omega_3 &= 2c_2\ \omega_4\wedge\omega_1 + 2c_4\ \omega_2\wedge\omega_1\ ,\\\\ d\omega_4 &= 2c_2\ \omega_1\wedge\omega_3 \ . \end{aligned}

Now endow $G$ with the Riemannian metric $g$ for which the $\omega_i$ define an orthonormal coframing, let $e_1,\ldots,e_4$ be the dual ($g$-orthonormal) vector fields, and let $\mathcal{E}$ be the $2$-plane field spanned by $e_1$ and $e_2$ while $\mathcal{F}$ is the $2$-plane field defined by $e_3$ and $e_4$.

One easily checks that this is an example of the desired type: If $\nabla$ is the Levi-Civita connection of this metric, then $$ \nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad \nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0 \mod \mathcal{E} $$ and $$ \nabla_{e_3}e_3\equiv0\ ,\quad\nabla_{e_3}e_4\equiv c_3e_1, \quad \nabla_{e_4}e_3\equiv-c_3e_1\ ,\quad\nabla_{e_4}e_4\equiv 0 \mod \mathcal{F}, $$ as desired.

There are non-homgeneous examples in dimension $4$ as well. A little more work with the structure equations shows that there exists a $4$-parameter family of examples of cohomogeneity $2$. (I don't know how many of these are complete.) If there is interest, I can give the structure equations of these examples as well.

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Thank you! Are there any references for that? – Klaus Kröncke Apr 13 '12 at 8:40
@Klaus: I don't know any references that specifically address the problem you raised. There are many references to the structure equations and the method of the moving frame, which is what I used to find the homogeneous examples and the cohomogeneity $2$ examples. I suspect that, in dimension $4$, there are no other examples, but I would need to finish checking a couple of cases to be sure of that. – Robert Bryant Apr 13 '12 at 12:03
Ok, thank you again. Maybe you can also answer this question: Is it possible, that the tangent bundle over even-dimensional spheres splits in two subbundles of same dimension? Do you now any characterization of compact manifolds admitting such a splitting? – Klaus Kröncke Apr 16 '12 at 8:12
@Klaus: The tangent bundle of the $2n$-sphere does not admit any splitting at all, much less into same-dimensional bundles. The reason is that $e(TS^{2n})\not=0$, while $e(E)=0$ for any vector bundle over $S^{2n}$ whose dimension is not $0$ or $2n$. – Robert Bryant Apr 16 '12 at 17:25
@RobertBryant So is it true to say that every compact even dim manifold with vanishing intermediate cohomology satisfies in the property that the tangent bundle is indecomposible? – Ali Taghavi Oct 1 '14 at 19:18

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