We want a consistent way of perturbing a submanifold away from itself. For $0$-dimensional submanifolds, this is the same data as a nowhere-vanishing vector field: we may flow the points off themselves. For $r>0$, a nowhere-vanishing vector field may flow a submanifold along itself, which would be a disaster. The setup below takes this into account, exploiting the extra data we have at each point. For example, notwithstanding the Hairy Ball Theorem, the 2-sphere can still flow oriented 1-dimensional submanifolds (see Examples section below).

A Precise Question

Let $M$ be an oriented, closed, connected Riemannian manifold of dimension $n$. Define $Gr_+(TM,r)$, the ``universal oriented Grassmannian bundle'': the fiber over a point $m$ is the space of oriented $r$-subspaces of the tangent space $T_mM$. Points on $Gr_+(TM,r)$ can be written $(m,\alpha)$ where $\alpha \in \Omega^r_mM$ is a non-zero wedge product of tangent vectors at $m$ considered up to positive rescaling.

The space $Gr_+(TM,r)$ comes with a natural vector bundle of rank $n-r$: the fiber over the point $(m,\alpha)$ is the oriented subspace of $T_mM$ determined by $\star \alpha$, the ``oriented orthogonal complement'' of the oriented subspace determined by $\alpha$. (Here the Hodge star $\star$ is taken with respect to the chosen orientation on $M$)

Let $Gr_+^{\perp}(TM,r)$ denote this vector bundle, the ``oriented orthogonal complement of the universal oriented Grassmannian bundle.'' This is a rank $n-r$ vector bundle on $Gr_+(TM,r)$.

When does $Gr_+^{\perp}(TM,r)$ admit a nowhere-vanishing global section?


When $r=0$, $Gr_+^{\perp}(TM,0)$ has a nowhere-vanishing global section exactly when the Euler characteristic of $M$ is zero: Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field?

The case $Gr_+^{\perp}(S^2, 1)$ is fun. Given a point on $S^2$ (considered as a unit vector in $\mathbb{R}^3$) and a tangent vector at that point (considered as another unit vector othogonal to the first) we are trying to determine a unit vector at the point which is orthogonal to the tangent vector. In other words, given two unit vectors in $\mathbb{R}^3$ we need to continuously determine a third unit vector that is perpendicular to the first two. We may use the usual cross product in $\mathbb{R}^3$! A similar construction applies to $Gr_+^{\perp}(S^6, 1)$; see http://en.wikipedia.org/wiki/Seven-dimensional_cross_product .


Minor observations:

  • If $n-r=1$, this is the same as an orientation of $M$.

  • If $r=1$, then it is enough to have an almost complex structure on $M$. (I.e. a section $J$ of $\mathrm{End}(T_{\ast} M)$ which squares to $-1$.) A real matrix which squares to $-1$ has no real eigenvectors, so we can just multiply your tangent vector by $J$. Conversely, if there is a section $K$ of $\mathrm{End}(T_{\ast} M)$ which has no real eigenspaces, I claim that there is another section $J$ with square $-1$. For $x \in X$, let $\lambda_1(x)$, ..., $\lambda_n(x)$ be the complex eigenvalues of $K(x)$. By Lagrange interpolation, there is a polynomial $f_x$ with real coefficients so that $f_x(\lambda_j(x)) = \pm i$, where the sign is the same as the sign of $\mathrm{Im}(\lambda_j)$. It's pretty straightforward that the coefficients $f_x$ vary continuously with $x$; take $J_x = f_x(K(x))$. Unfortunately, it is not true that $\mathrm{Diff}(S^{n-1})$ retracts onto $O(n)$ so I don't know whether linear automorphisms of the sphere are a good model for thinking about all automorphisms.

There can be local obstructions. Here are two examples:

  • If $n$ is odd and $r=1$, this can never be done. Just looking at what is required at a single point, you are asking for an automorphism of $S^{n-1}$ with no fixed points. But $\chi(S^{n-1}) =2$, so there are no such automorphisms.

  • If $n \geq 4$ is even and $n-r=2$, this can never be done. At a single point, of $X$, you asking for a section of a certain circle bundle over $Y:=G_{+}(2, \mathbb{R}^n)$. Restricting to $\mathbb{CP}^{n/2-1}$ inside $Y$, this becomes the nontrivial line bundle $\mathcal{O}(-1)$ so there is no section of this circle bundle over $Y$. If I knew more about characteristic classes, I suspect that I could show such a obstruction for $n$ odd as well.

I think you should first work out when there are characteristic class obstructions for the local problem.


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