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 $n>2m$?

Motivation:

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?