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The set $\pi^{-1}(\mathcal{O}^\mu)$ is always a smooth submanifold of $\mathcal{O}_\lambda$ (though it may well be empty). When it is not empty, it is a single orbit of $\mathrm{U}(n)\subset \mathrm{ …

For all odd $n=2k{+}1>3$ one can write the volume form $\alpha$ on $S^n$ as a wedge product $\alpha=\beta\wedge\gamma$ with $\mathrm{deg}(\beta)$ and $\mathrm{deg}(\gamma)$ both greater than $1$. Jus …

The answers to your questions are 'no' and 'yes'. In the first place, there is no finite dimensional manifold whose Lie algebra of vector fields contains all of the Lie algebras ${\frak{sl}}(n,\mathb …

The answer is 'no'. I don't have access to the right references right now, but I think an argument goes as follows. (Also, as Will points out below, you can use Stiefel-Whitney classes to get the sa …

Yes, one can always do this. Start with a $p$-centered local coordinate system $(x^\sigma,y^\rho)$ on an open $p$-neighborhood $U\subset M$ such that $S\cap U$ is given by $y^\rho=0$ $(1\le\rho\le r) …

No. The simplest case is $M = S^1$ and $N = \mathbb{R}$. Then any nonconstant map $f:M\to N$ has rank at most 1, but there is no smooth map from $M$ to $N$ that has constant rank $1$. The questio …

No. If such a pair existed, they would be tangent to a codimension $1$ foliation of $S^3$. Such a foliation must have a Reeb component, in particular, a compact leaf, $T\subset S^3$, which would be …

A simple example is to let $M=S^2$ and let $L$ be the nonabelian Lie algebra of dimension $2$. If such an $\alpha$ existed, its range would be a rank-1 subbundle $L\subset TS^2$, but this cannot exis …

Let $M$ be an $n$-manifold endowed with a nonvanishing $n$-form $\Omega$, let $\mathrm{Diff}(M,\Omega)$ denote the group of $\Omega$-preserving diffeomorphisms of $M$, and, for $x\in M$, let $\mathrm{ …

The answer to your first question depends on the $2$-dimensional subbundle of $TM$. (I don't have an answer to your second question, which is harder.) Suppose that $\Delta$ is a contact structure on …

This notion of 'the sphere at infinity' is commonly encountered in hyperbolic geometries. Gromov, in particular, has used it in studying the behavior of discrete transformation groups on hyperbolic m …

Most of Cartan's isoparametric hypersurfaces in $S^4$ have your desired properties: They have constant principal curvatures (in fact, they are homogeneous), nearly all of them have all three principa …

The only thing one really needs to define the operation $\mathrm{Div}:{\frak{X}}(S^3)\to C^\infty(S^3)$, i.e., mapping vector fields on $S^3$ to functions on $S^3$, is a volume form on $S^3$. (One ca …

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 …

The answer is 'no, in general': For example, if $(M^4,J)$ is any real-analytic almost complex $4$-manifold, one can easily construct (locally, in a neighborhood of any point of $M$) a pair $(A,B)$ of …