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Suppose we have $n$ planes $H_1, \ldots, H_n$ in $\mathbb{R}^m$ of codimension $q$, or equivalently of dimension $d=m-q$. I want to choose a vector which does not belong to the planes in a continuous way. There are two versions of this problem, depending on how we parameterize the planes, and the answer can be actually different.

  1. Unframed version. Let $Gr_q(m) $ be the grassmannian, that is the space of codimension q planes in $\mathbb{R}^m$. Does there exist a function

$$c : Gr_q(m) ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

  1. Framed version. Let $V_{d,m}$ be the Stiefel manifold, that is the space of orthonormal systems in $\mathbb{R}^m$ of cardinality $d$. Does there exist a function

$$c : V_{d,m} ^n \to \mathbb{R}^m$$

Such that $c(H_1, \ldots, H_n) \not \in H_i$ for all $i$?

Note. I slightly changed the notation to agree with Chris one; now $d$ denote the dimension of the planes and $q$ the codimension.

Andrea Marino
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  • Let me also remark that I probably have a proof of this, but it is overcomplicated... It is a question that raised when I was studying configuration spaces, and I am interested in knowing if there is a direct proof like "consider this invariant" or something like that. – Andrea Marino Mar 27 '21 at 14:30
  • What if we relax the codimensione condition, so that the space is not disconnected anymore? – Andrea Marino Mar 27 '21 at 14:40
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    Sorry, I deleted my comment because I thought I had fundamentally misunderstood your question, but I only trivially misunderstood the question! If you reduce the dimension and enforce a codimension of strictly greater than $1$, then I imagine that such a function probably exists, but it likely won't be pretty. – Theo Bendit Mar 27 '21 at 14:43
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    Let us think to the case $d=2, m=3$, that is lines in the space. You can equivalently formulate the problem as: given $n$ points in the projective space $\mathbb{P}^2(\mathbb{R})$, you want to find an $(n+1)$-th point which is distinct from the others. I don't see how this is trivial, even if in this case we relaxed the codimension constraint... To be honest, I think it cannot exist at all! – Andrea Marino Mar 27 '21 at 14:49
  • Actually, yeah, it's not clear to me that such a function exists at all. I'm sure that we could find a path as we continuously change the elements of the Grassmannian, but it's not clear that this path is generated by a function. By the way, is it OK if $c$ is not symmetric? That is, if we simply permute the elements of the Grassmannian, do you care if $c$ generates a different value? – Theo Bendit Mar 27 '21 at 15:00
  • No, I don't care. It can be non symmetric! – Andrea Marino Mar 27 '21 at 15:00
  • Even with $n=1$, you're asking for a nowhere-zero section of the tautological quotient bundle on the Grassmannian. This is equivalent to asking for a nowhere-zero section of the tautological subbundle. The Euler class of this bundle is nonzero, I believe. So you won’t even solve your problem with only one plane, let alone more. – Ted Shifrin Mar 27 '21 at 17:26
  • Uhm, I am not sure. In the case $d=1, n=1$ you can just take the vector that defines the hyperplane: it will always be perpendicular to the whole hyperplane, in particular it will always be outside it. However, this kind of reasoning (topological obstructions) is exactly what I am looking for! – Andrea Marino Mar 27 '21 at 17:31
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    In the case $d=1$ it's trivial if you use oriented hyperplanes, but false with non-oriented. This is a nice question, by the way. – Ted Shifrin Mar 27 '21 at 17:49
  • I think Ted's point is that if you have, say, a line $l$ in the plane, and $c(l)$ is on one half of it, when you rotate that line around $\pi$ radians until it is back to itself, $c(l)$ must be on the opposite side. This makes $c$ not well defined. (But this is fixed by using orientated planes.) I think in the $n \geq 2 $ its always false when $d = 1$, oriented or not, because you can take the cell where $c(l)$ lives and move the planes until that cell becomes empty (you may have to allow the hyperplanes to coincide). – Elle Najt Mar 27 '21 at 17:52
  • In the case of a tuple of points $p = (p_1, \ldots, p_n)$ in the plane, I think you can construct $c(p)$ by hand, by taking some fixed unit vector $v$ and letting $c(p) = \lambda( p) v$, where $\lambda(p)$ is some continuous function of the norms of $p = (p_1, \ldots, p_n)$ with $\lambda(p) \geq 2 \max( ||p||) + 1$. Maybe similar reasoning would work for the other codimension $\geq 2 $ cases. – Elle Najt Mar 27 '21 at 17:59
  • @LorenzoNajt: thank you for explaining me the Ted comment, now I take it. I was identifying an hyperplane with "the" vector of norm 1 which is perpendicular to it, but you actually have 2 of them and you must choose an orientation. So this boils down to the fact that the $\mathbb{Z}/2Z$ bundle $\text{oriented hyperplanes}\to \text{unoriented hyperplanes}$ is non trivial, as you have shown with the monodromy action. Concerning $d=1, n\ge 2$ yes, you are right: it was an observation by Theo he then removed. There's no need to make hyperplanes coincide (it's a connectedness stuff) – Andrea Marino Mar 27 '21 at 18:05
  • @LorenzoNajt: actually, I am thinking about planes passing through zero (subspaces) , isn't that the definition of grassmannian? However, I think your example works because the configuration space in dimension 1 is trivial. Specifically [continues] – Andrea Marino Mar 27 '21 at 18:08
  • If we think to $\mathbb{R}^2\setminus {0}$ as $S^1 \times \mathbb{R}$, you are ignoring the $S^1$ component and putting a point at the rightmost part of the configuration on the line. In general I have the impression that unboundedness of planes of positive dimension makes projection onto a line surjective. On the other hand, we could recycle this projecting idea to reduce to smaller cases. Thank you for your good observation. – Andrea Marino Mar 27 '21 at 18:13
  • @TedShifrin Initially I had the same thought, but I am not so sure now. It is not clear to me how to extract a section of the the $n=1$ case from the $n>1$ case. It seems to me that such a task requires that we can perturb the diagonal of $\mathbb{R}P(m)^n$ away from the fat diagonal. I think this might be equivalent to calculating some differentials of some spectral sequences commonly used to study configuration spaces. – Connor Malin Mar 27 '21 at 21:07
  • What you can guarantee is a section for the $n=1$ case away from $m-1$ points. – Connor Malin Mar 27 '21 at 21:13

1 Answers1

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This is a partial answer, though its not particularly elementary. (EDIT: I am using d to denote the dimension of the planes, rather than the codimension as stated in the question, so things need to be re-indexed). If $2d+1\geq m$, $d\neq m$ then for any $n$, we cannot find such a map. Assume such a map exists, and consider the real steifel manifolds $V_{m,d}$. These are the space of orthonormal $d$ tuples in $\mathbb{R}^n$, and we have a natural surjective map $V_{m,d}\xrightarrow{\pi} V_{m,d-1}$ given by forgetting the last vector in our list. By restricting a hypothetical $c$ map to the first coordinate (with other coordinates some fixed spaces), it suffices to show that no map can exist when $n=1$.

We claim that a map of the form described will yield a section of the map $$V_{m,d+1}\xrightarrow{\pi}V_{m,d}$$

To see this, first normalise the map $c$ to land in unit one vectors in $\mathbb{R}^n$. Then for a given orthonormal $d$ frame, map it to the orthonormal $d+1$ frame with new vector $c(Span(v_1,..v_{d}))$. This is clearly a section, so now consider mod $2$ cohomology, and in particular the cup product structure.

By results of Borel we know this graded ring, which I read at https://www.researchgate.net/publication/254207624_The_cohomology_rings_of_real_Stiefel_manifolds_with_integer_coefficients

The cohomology ring of $V_{m,d}$ is generated by classes of generators $z_i$ of degree $i$ for $m-d\leq i\leq m-1$, subject to the relation $z_i^2=z_{2i}$ when $2i\leq m-1$, and $z_i^2=0$ else. The map $\pi$ induces the natural inclusion of these rings, so consider $z_{m-d-1}^2=z_{2m-2d-2}$ in $H^*(V_{m,d+1},\mathbb{Z}/2\mathbb{Z})$. On one hand, our section is the identity on this element, since $z_{2m-2d-2}$ is in the image of the natural inclusion, but on the other, the minimal positive degree nonzero cohomology group of $V_{m,d}$ is in degree $m-d$, so $z_{m-d-1}$ must be in the kernel. So since the induced map on cohomology rings is a map of graded rings, no such map $c$ can exist in this case.

On the other side, the case of $d=1$, $m> 2$ is also not possible. Similarly, reduce to the $n=1$ case, and thus we obtain a map $\mathbb{RP}^{m-1}\rightarrow S^{d-1}$ such that when we compose with the natural quotient $S^{d-1}\rightarrow \mathbb{RP}^{m-1}$, the induced endomorphism of $\mathbb{RP}^{m-1}$ has no fixed points. Then consider the trace of this endomorphism on mod $2$ cohomology. Since it has no fixed points, this trace is $0$, and thus this endomorphism is nonzero on some nonzero cohomology group (since its an isomorphism on $H^0$), so induces an isomorphism on $H^{m-1}(\mathbb{RP}^{m-1},\mathbb{Z}/2\mathbb{Z})$, since all the other maps factor through $0$. But a generator for this top cohomology is the $m-1$ fold cup product of the class in $H^1$, so this must be the zero map (since $S^{m-1}$ is simply connected). Thus, no map can exist in this case either.

Chris H
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  • Hi Chris H, thank you for your interesting points. Regarding the second case, you are formally right, though this can be solved by considering the oriented hyperplanes. In this case the $n=1$ case is possible, while the $n\ge 2$ is impossible because more than one hyperplane disconnects the space. Regarding the first obstruction, in principle $c(Span(v_1, \ldots, v_d)) $ is not orthogonal to $v_1, \ldots, v_d$, so you have to graham-schmidt somehow. Does this process preserve continuity? – Andrea Marino Mar 28 '21 at 07:52
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    Ahh yep, that’s my bad, one does indeed need to use GS to orthogonalise, but this is fine, this a continuous process, the algorithm is just additions/scalar multiplications. One can also avoid it too, if one works with the noncompact steifel manifold, with no orthonormality relations imposed. The continuity of GS yields that these spaces are homotopy equivalent. – Chris H Mar 28 '21 at 08:12
  • There is also a little misunderstanding on the $d$: in your solution is the dimension of the planes, while I stated d as the codimension. I think you inequality becomes then $2d-1 \le m$. And also $d > 0$, otherwise the problem is trivial impossible. I like your formulation with stiefel manifolds because parameterize planes with an ortonormale basis. This prevents non-geometric obstruction given by monodromy, like the one you found in the second paragraph. I will add this "framed version" to the question. – Andrea Marino Mar 28 '21 at 08:15
  • I edited to address the indexing issue, and one interesting facet of this question is whether this problem is always reducible to the n=1 case, that is, if the obstruction ever actually depends on n. – Chris H Mar 28 '21 at 08:28
  • The framed version in codimension one depends on whether $n\ge 2$ or $n=1$. In other cases, I actually agree that since the things are "thin" and we have excluded monodromy, the problem that can arise is that we cannot add a new "canonical" point to some space (as you demonstrated, and I still can't understand what is really going on). Let's see if someone other can settle the $n=1$ case by looking at the fibration you mentioned! – Andrea Marino Mar 28 '21 at 11:22
  • Sorry, I changed the notation to agree with yours, so that now everyone can freely use dimension or codimension. I thought that d is more natural for dimension – Andrea Marino Mar 28 '21 at 17:23
  • Even if it's not a definitive answer, I awarded you the bounty not to make it expire. It's definitely a step toward the general solution! – Andrea Marino Apr 10 '21 at 12:20