Consider the trivial bundle $\epsilon\_{2}=S^{2}\times \mathbb{C}^{2}$ with the standard Hermitian inner product $<(a,b), (c,d)>=a\bar{c}+b\bar{d}$.


Assume that $\ell$ is a sub line bundle of $\epsilon\_{2}$ such that $f^{\*} (\ell)$ is orthogonal to $\ell$, where $f$ is the antipodal map.



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**Motivation:**


If the answer is negative(and is not actually based on computation with second $\mathbb{Z}\_{2}-$cohomology of $S^{2}$ or $\mathbb{R}P^{2}$ we would give an alternative proof for the Borsuk-Ulam theorem in dimension $3$ as follows:(Since the proof of the Borsuk Ulam theorem is based on the above cohomology).


We need to prove that: There is no a continuous odd function $g:S^{3}\to S^{2}$. But $S^{2}$ can be identified with all projections ($A=A^{\*}=A^{2}$) in $M\_{2}(\mathbb{C})$ via $(x,y,z)\mapsto 1/2\pmatrix{1-z&x+yi \\ x-yi&1+z}$(As we learned from page 21 of the book of Alain Connes, Non commutative geometry). With this identification, the antipodal map $x\mapsto -x$ of $S^{2}$ can be read as $A\mapsto 1-A$ for projection $A\in M\_{2}(\mathbb{C})$.


**Note that the range of projection $A$ is orthogonal to the range of $1-A$**.


On the other hand every continuous map $g:S^{3}\to \text{projections of} M\_{2}(\mathbb{C})$ defines a line bundle over $S^{3}$. It can be easily shown that every line bundle over $S^{3}$ is trivial(using clutching functions and the fact that $\pi\_{2}(GL\_{1}(\mathbb{C})$ is trivial.


So if $g$ satisfies $g(-x)=1-g(x)$ we actually obtain a line bundle $\ell$ over $S^{3}$(as a subbundle of 2-trivial bundle $\epsilon\_{2}$) such that $\text{antipodal}^{\*}(\ell)$ is orthogonal to $\ell$. Now we restrict to $S^{2}$ to avoid needing the homology or cohomology of $S^{3}$.



A question on complex line bundle over $S^{2}$
Regard the line bundle $\ell\subset S^2\times\mathbb C^2$ as a map $f$ from $S^2$ to the space of complex lines in $\mathbb C^2$, which is $\mathbb C P^1=S^2$ as well. You ask that antipodal pairs are mapped to antipodal pairs, so you get an induced map $\bar f\colon\mathbb R P^2\to\mathbb R P^2$,
which acts as id on $\pi\_1(\mathbb R P^2)$. Hence, it also acts as id on $H^1(\mathbb R P^2;\mathbb Z/2)$, and because $H^1(\mathbb R P^2;\mathbb Z/2)$ generated the cohomology ring, also on $H^2(\mathbb R P^2;\mathbb Z/2)$.
But this implies that $\bar f$ has odd degree. Because $p^\*\bar f^\*=f^\*p^\*$,
where $p\colon S^2\to\mathbb R P^2$ is the projection, the degree of $f$ is odd, too.


Complex line bundles on a space $X$ can be classified by homotopy classes of maps $X\to\mathbb C P^\infty=BU(1)$. On the other hand, $\mathbb C P^\infty$ is also a $K(\mathbb Z,2)$, so second integral cohomology classes also correspond to homotopy classes of such maps. For a given line bundle $\ell$ on $X$,
the first Chern class $c\_1(\ell)\in H^2(X)$ is the corresponding class.
Because $\mathbb C P^\infty$ has a CW structure where the 3-skeleton is $\mathbb C P^1$, each map $S^2\to\mathbb C P^\infty$ is homotopic to exactly one homotopy class of maps $S^2\to\mathbb C P^1=S^2$, and its degree equals $c\_1(\ell)[S^2]$. In the case at hand, $\ell$ is classified by $f$, which is of nonzero degree, so $\ell$ is nontrivial.


