Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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*}

and

\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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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
1  
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
1  
@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 9 hours ago

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.