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.

Is there an example of this situation such that $\ell$ is a trivial line bundle?

**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 $S^{3}$.

Hermitian inner product, then it is best to specify this. $\endgroup$ – Jason Starr Oct 26 '15 at 18:47