Connectivity of the space of transverse vector fields

Question

Suppose we have a smooth, closed manifold $M$ of dimension $n$ and connectivity $k$. What can we say about the connectivity of the space of all tangent vector fields on $M$ that are transverse to the zero section? I'm assuming it depends on $n$ and $k$, but what I'm hoping is that it grows with $n$ and $k$.

I know, of course, that this space is dense in the space of all vector fields (its complement has measure zero), but that doesn't say anything about its connectivity.

Actually, what I'm really interested in is a subspace: Suppose that $M$ has positive Euler characteristic and consider only those transverse vector fields where the associated sign at each zero is $+1$. (So the number of zeros is the Euler characteristic.) What can we say about the connectivity of this space?

Knowing the answer to this question might help show that a construction I'm working on does what I hope it does, but I think it's an interesting question in its own right.

Edit: To clear up a possible confusion, when I'm asking about the "connectivity," I mean to ask, for what $m$ can we say that the space of vector fields in question is $m$-connected? And does $m$ grow with $n$ and $k$?

And I think I need to restrict to the subspace mentioned above: Assume that $M$ has positive Euler characteristic and restrict to those transverse vector fields where all the zeros are "positive." If we allow arbitrary transverse vector fields, I believe we'll get an infinite number of components, one for each combination of positive and negative zeros that add up to the Euler characteristic. I want to concentrate on the component where all the zeros are positive.

Answer 1

Let $F(M)$ be this space of vector fields. Let's work this out for an even-dimensional sphere $M=S^n=S^{2p}$, so $n=2p$ and $k=2p-1$. I claim that it is not rationally $3$-connected, if $n\ge 3$.

This space $F(S^{n})$ fibers over the space $Conf_2(S^{n})\sim \mathbb RP^{n}$ of unordered pairs of points in $S^n$, by recording where the field vanishes. Call the fiber (over, say, some pair of antipodal points) $F_0(S^{n})$. Then $F_0(S^n)$ fibers over $GL_n^+(\mathbb R)\times Gl_n^+(\mathbb R)\sim SO_n\times SO_n$ by recording the derivative of the field at those two points. The fiber of $F_0(S^n)\to SO_n\times SO_n$ is homotopy equivalent to the space of all nowhere-vanishing tangent fields on $S^{n-1}\times I$ having some specified restriction to the boundary, and that's the same as the space of all paths in $Map(S^{n-1},S^{n-1})$ between a certain pair of points. So $F(S^n)$ fibers over $A\sim \mathbb RP^n$ and the fiber is fibered over $B\sim SO_n\times SO_n$ and the fiber of that is $C\sim\Omega Map(S^{n-1},S^{n-1})$.

Now $A$ is rationally $(n-1)$-connected, and $C$ is rationally $(n-2)$-connected, but $B$ has rationally nontrivial $\pi_3$ if $n\ge 3$, so the same is true of $F(S^{2p})$.

In fact, I think that looking more closely we will find that it is not even $1$-connected.