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7
votes
Accepted
Is there a $2$ dimensional foliation tangent to this particular $3$ dimensional distribution?
There is no co-dimension 2 foliation of the desired kind in a neighborhood of any point on the line $L\subset\mathbb{R}^4$ defined by $x-w=y=z=0$. Thus, if one wants to find such a foliation, it will …
4
votes
Accepted
A strongly non-integrable distribution
Now that this other question as been answered, we can answer this original question: The answer is 'yes' there exists a non-integrable tangent $3$-plane field on $\mathbb{R}^4$ that has no subordinat …
8
votes
Accepted
On certain 2 dimensional foliation of $Gl(2,\mathbb{R})$ deleted by scalar matrices
Since $A^2 = \mathrm{tr}(A)\,A - \det(A)\,I_2$, and $\det(A)\not=0$ for $A\in M$, the leaves of your foliation are the same as the leaves of the foliation determined by the vector fields $X(A) = A$ an …
2
votes
Accepted
Nonlinear PDE for a 2D foliation
It's easy to derive a third-order (nonlinear) differential equation for $u(x,y)$ that satisfies your conditions (1) and (2): Namely, set $\theta(x,y) = \arctan\bigl(u_y(x,y)/u_x(x,y)\bigr)$ and then …
2
votes
Accepted
A cohomology associated with a codimension one foliation
First, it's not finite dimensional, even in the case of a torus. Just let $x,y$ be the $2\pi$-periodic functions on the torus and take $\alpha = \mathrm{d} x$, and you'll see that $H^0$ is all the fun …
4
votes
Accepted
A possible sub-Riemannian structure associated to a non-symmetric matrix
First, the condition that the kernel of $\alpha$ be integrable is not just that the matrix $A$ be symmetric although, of course, that is sufficient. For example, when the dimension $n$ is equal to $2 …
8
votes
Accepted
Rank of a distribution
(1) The boxed sentence would be better written as "A distribution $S$ of rank $r$ on a manifold $M$ is an assignment of an $r$-dimensional subspace $S_p$ of $T_pM$ to each point $p\in M$." Confusing …
8
votes
Accepted
Can we foliate the space $\mathbb{R}^3$ with Frenet curves whose tangent and normal vectores...
Then there is much more that can happen with the foliations. …
11
votes
Accepted
Which submanifolds are leaves of a foliation?
If the normal bundle of $\Sigma$ in $M$ is orientable, then there always exists such a foliation. The idea is that one can construct a smooth function $f$ on $M$ such that $\Sigma$ is the set of zero …
7
votes
Examples and non-examples of Riemannian foliations
For your second kind of example, consider the foliation of the unit $3$-sphere that is the integral curves of the vector field
$$
X = p\left(x^1\frac{\partial\ }{\partial x^0 } -x^0\frac{\partial\ }{\ …
10
votes
Accepted
Anti_symplectic 2-forms
For your first question: The condition for integrability of $\mathrm{ker}\alpha$ when $\alpha$ is decomposable is that $\mathrm{d}\alpha = \lambda\wedge \alpha$ for some $1$-form $\lambda$.
For your …
5
votes
Orthonormal frame on $\mathbb{S}^3$ orthogonal to foliations
The Qualitative Theory of Foliations, CBMS Regional Conference Series in Mathematics, Volume 27 (1977), AMS/CBMS.) …
8
votes
Accepted
The concept of convex foliation
If $n>1$, and a smooth diffeomorphism $f:U\to V$ (where $U$ and $V$ are, say, convex, open subsets of $\mathbb{R}^n$) carries convex sets to convex sets, then it is easy to show that it must carry eac …
5
votes
Foliations by holomorphic curves on complex surfaces
Update: I have had a little time this weekend to think more deeply about the construction I outlined below and I have concluded that, unfortunately, the construction that I thought would work to produ …
8
votes
Accepted
Complementary integrable vector fields
If you are looking for information about codimension $1$ foliations of manifolds, you should, perhaps, look up references that discuss secondary characteristic classes, such as the Godbillon-Vey class, … Blaine Lawson's article Foliations (Bulletin of the AMS 80 (1974), 369–418). …