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?


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 $S^{3}$.

  • $\begingroup$ Sure, just choose $\ell$ to be a constant isotropic one-dimensional vector subspace of $\mathbb{C}^2$ with its standard inner product: $\langle (a_1,a_2),(b_1,b_2) \rangle = a_1b_1 + a_2b_2$. Or were you thinking of something other than the standard inner product on $\mathbb{C}^2$? $\endgroup$ – Jason Starr Oct 26 '15 at 18:37
  • $\begingroup$ @JasonStarr I consider $a_{1}\bar{b_{1}}+a_{2} \bar{b_{2}}$. Could you please explain what is the constant isotropic one dimensional space? $\endgroup$ – Ali Taghavi Oct 26 '15 at 18:41
  • $\begingroup$ If the line is constant, how a vector is orthogonal to itself? $\endgroup$ – Ali Taghavi Oct 26 '15 at 18:44
  • $\begingroup$ The standard inner product is the one used to define, for instance, the complex Lie group $\textbf{SO}_2(\mathbb{C})$. If you want a Hermitian inner product, then it is best to specify this. $\endgroup$ – Jason Starr Oct 26 '15 at 18:47
  • $\begingroup$ @JasonStarr Thank you I revise it. $\endgroup$ – Ali Taghavi Oct 26 '15 at 18:49

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.

  • $\begingroup$ @Sebastien Thank you for your interesting answer. using this idea can one give a shorter proof of the Borsuk Ulam theorem for $S^{2n+1}$? For this: is it true to say that every line bundle over odd spheres is trivial? $\endgroup$ – Ali Taghavi Oct 26 '15 at 19:49
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    $\begingroup$ @AliTaghavi: I haven't really thought about higher dimensional Borsuk Ulam. But I can say at least that complex line bundles are classified by $c_1$, so over odd spheres, they are all trivial. $\endgroup$ – Sebastian Goette Oct 26 '15 at 20:27
  • $\begingroup$ @Sebastien Assume that $M$ is a manifold (not necessarily compact) with non trivial second homology? Does this imply that there is a nontrivial complex line bundle over $M$? The motivation is the following:Since we did not need to know the homology of $S^{3}$, we can have the following generalization: $\endgroup$ – Ali Taghavi Oct 27 '15 at 6:23
  • $\begingroup$ Assume that $M$ is a topological space with a continuous involution such that M contains S^2 and the involution restrict to antipodal map. If every line bundle over M is trivial then there is no a Z_{2} equivariant map from M to S^2. $\endgroup$ – Ali Taghavi Oct 27 '15 at 6:25
  • $\begingroup$ I edited the answer to adress your first question. For the second, you want to take a nontrivial line bundle over $S^2$ and pull it back to $M$? It could become trivial on the way. E.g., take a nontrivial map $S^3\to S^2$ (the Hopf fibration, say). Extend it to $S^3\to S^2\subset\mathbb C P^2$ and it becomes trivial (because $H^2(S^3)=0$). $\endgroup$ – Sebastian Goette Oct 27 '15 at 7:48

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