Question 1: What is a complete classification of all positive integers $m,n$ with the following property:

There is a continuous map $f:S^n \to \mathbb{C}P^m$ such that $f$ maps antipodal points to orthogonal lines. Namely for every $x\in S^n$ we have $\;f(x) \perp f(-x)$. Here the later perpendicularity is meant as $f(x)$ is orthogonal to $f(-x)$ with respect to the standard inner product of $\mathbb{C}^{m+1}$

In particular, is it true to say that such map does not exist if $n>2m$?


One can prove the three dimensional Borsuk Ulam theorem without any explicit or implicit use of homology-cohomology as follows:

(An Equivalent formulation of )Borsuk_Ulam in dimension $3$: There is no an odd continuous function $f:S^3\to S^2$.

Proof: We identify $S^2$ with $\mathbb{C}P^1$. Then, as I learned from Sebastian Goette via his MO comment, the antipodal points of $S^2$ corresponds to orthogonal lines in $\mathbb{C}P^1$. So we have to prove that there is no a continuous map $f:S^3\to \mathbb{C}P^1$ with the property that $f$ maps antipodal points to orthogonal lines. For the contrary assume that such $f$ exist. Every map $f:S^3 \to \mathbb{C}P^1$ determines a complex line bundle $\ell$ over $S^3$ where $\ell$ is athe pull back of the tautological line bundle over $\mathbb{C}P^1$ so is a subbundle of the trivial bundles $S^3 \times \mathbb{C}^2$. Obviousely every line bundle over $S^3$ is a trivial bundle because the corresponding clutching function $K:S^2 \to GL(1,\mathbb{C})$ is null homotp because any such $K$ has a logarithm by the lifting lemma in the covering space theory. We take a non vanishing section $S:S^3 \to \mathbb{C}^2$ for the line bundle $\ell$. Put $\omega= dx\wedge dy$, the natural determinant $2\_$ form on $\mathbb{C}^2$. Then $\omega(S(x), S(-x)):S^3 \to \mathbb{C} \setminus \{0\}$ is an odd continuous function, a contradiction by the 2 dimensional BU where the later has an elementary proof. Because every continuous function $S^2\to S^1$ has a logarithm so obviously it can not be an odd map $\;\blacksquare$

Remark 1: Instead of identification $S^2$ with $\mathbb{C}P^1$, one can identify $S^2$ with the space of projections of $M_2(\mathbb{C})$. In this case antipodal maps correspond to orthogonal projections. The precise identification is the following: $(x,y,z)\mapsto 1/2\begin{pmatrix} 1-z&x+yi\\x-yi&1+z \end{pmatrix}$

Remark 2 The above proof, which is independent of homology or cohomology, and involves the orthogonality of lines in the projective space or projections of the matrix algebra, not only motivates the question $1$ above but also motivates the following question:

Question 2: Can one generalize the above proof to find new proof of the higher dimensional BU, without involving Homology-Cohomology?

  • 3
    $\begingroup$ There is a combinatorial proof of Borsuk-Ulam which does not involve homology-cohomology via Tucker's Lemma $\endgroup$ – Tony Huynh Aug 15 '17 at 15:11
  • $\begingroup$ @TonyHuynh Thank you so much for this link. $\endgroup$ – Ali Taghavi Aug 16 '17 at 7:26
  • $\begingroup$ @Malkoun Thank you. I try to understand your first comment. $\endgroup$ – Ali Taghavi Aug 16 '17 at 17:46

Such a map $S^n\to \mathbb CP^m$ exists if and only if either $n<2m$ or $n=2m=2$.

To see this, first note that such a map is the same as a $\mathbb Z/2$-equivariant map from $S^n$ to a certain subspace of $\mathbb CP^m\times \mathbb CP^m$, namely the space of pairs $(L,M)$ such that $L\perp M$. Now note that the latter subspace is an equivariant deformation retract of the space of pairs $(L,M)$ such that $L\neq M$. So the question is, for $X=\mathbb CP^m$, is there a map $f:S^n\to X$ such that $f(-x)\neq f(x)$ for all $x$?

If $n<2m$ then the answer is yes because you can embed $$S^n\subset\mathbb R^{n+1}\subseteq\mathbb R^{2m}=\mathbb C^m\subset \mathbb CP^m.$$

To get a negative result when $n=2m$, note that for any map $f:S^n\to N$ to a smooth $n$-manifold there is a mod $2$ obstruction to having $f(x)\neq f(-x)$ for all $x$. It can be defined homologically, or it can be defined more geometrically by first putting $f$ in general position and then counting how many unordered pairs $(x,-x)$ satisfy $f(x)=f(-x)$. This element of $\mathbb Z/2$ depends only on the homotopy class of $f$. If $f$ is homotopic to a constant then you can work out (replacing $N$ by $\mathbb R^n$ if you like) that the obstruction is nontrivial. (This is a way of thinking about Borsuk-Ulam.) When $N=\mathbb CP^m$ then every map $S^{2m}\to X$ is homotopic to a constant unless $m=1$. But there are more homotopy classes of maps $S^2\to \mathbb CP^1$, and in fact for the maps of odd degree that obstruction is trivial.

The negative result for $n>2m$ follows from the negative result for $n=2m$ (when $m>1$). The negative result for $S^3\to\mathbb CP^1$ is a case of Borsuk-Ulam, as you observed.

Edit: The geometric idea for defining the mod $2$ invariant is this. An equivariant map $F:S^n\to N\times N$ (such as $x\mapsto (f(x),f(-x)$) is always homotopic through equivariant maps to a smooth map such that $F$ is transverse to the diagonal $\Delta_N\subset N\times N$. Then $F^{-1}(\Delta_N)$ is a finite set (zero-dimensional compact manifold) in $S^n$ invariant under $x\mapsto -x$, giving a finite set in $\mathbb RP^n$. The cardinality of this, reduced mod $2$, is the invariant. It is independent of the choice of $F$ within the homotopy class because if $F_0$ and $F_1$, both smooth and transverse to $\Delta_N$, are equivariantly homotopic then the homotopy $H:S^n\times I\to N\times N$ may be chosen to be transverse to $\Delta_N$, and then $H^{-1}(\Delta_N)$ is a cobordism in $S^n\times I$ giving a cobordism in $\mathbb RP^n\times I$, and a compact one-dimensional manifold must have an even number of boundary points.

  • $\begingroup$ Dear Prof. Goodwillie, Thank you very much for your very interesting answer. May I ask you to hint me about the $\mathbb{Z}/2\mathbb{Z}$ obstructions corresponding to maps with $f(x) \neq f(-x)\;\;\forall x \in s^{2m}$? $\endgroup$ – Ali Taghavi Aug 23 '17 at 1:08
  • 1
    $\begingroup$ I've edited the answer to give a sketch of the more geometric way of defining it. $\endgroup$ – Tom Goodwillie Aug 23 '17 at 3:16
  • $\begingroup$ Thank you again for your very interesting idea. But just a question: Is it obvious this invariant is non zero for null homotopic maps? $\endgroup$ – Ali Taghavi Aug 23 '17 at 11:11
  • $\begingroup$ Yes. Take the image of $f$ to lie in a chart: $S^n\subset \mathbb R^n\times \mathbb R\to \mathbb R^n\cong U\subset N$, where $\mathbb R^n\times \mathbb R\to \mathbb R^n$ is projection. $\endgroup$ – Tom Goodwillie Aug 23 '17 at 11:36
  • $\begingroup$ I corrected a typo in the answer, where $\mathbb RP^n$ should have been $\mathbb R^n$. $\endgroup$ – Tom Goodwillie Aug 23 '17 at 11:43

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