# Why $T\mathbb{S}^{2}\otimes T\mathbb{S}^{2}$ is trivial?

I encountered this problem in an homework problem set of algebraic topology. Naturally I thought about Bott Periodicity which implies $K(\mathbb{S}^{2})\cong \mathbb{Z}$. But I am beware that we are working in $KO$ theory instead of $K$ theory, so this is not applicable. After checking online I found $KO(\mathbb{S}^{2})\cong \mathbb{Z}_{2}$, however I do not know a simple proof of it. My thoughts are:

1) Would it be appropriate to work this via Chern class of $\mathbb{S}^{2}$? Since we have ch($\mathbb{S}^{2}\otimes \mathbb{S}^{2}$)=ch($\mathbb{S}^{2}$)+ch($\mathbb{S}^{2}$), I only need to show $ch(\mathbb{S}^{2})$ has characteristic 2. So we have $ch(\mathbb{S}^{2})=ch(\mathbb{C}\mathbb{P}^{1})$. Geometrically it is "clear" that the tautologous bundle should have order 2 because of the half-twisting, but I encountered the same problem in here: namely the complex algebra can be different from the real algebra. Since Chern class is defined over hermitian vector bundles I do not know how to carry out this any further.

2) On the other hand it should be simple to solve this problem without using characteristic classes, as the author did not introduce the concept in that section. But I do not know how to work out the trivilization nicely. To prove it indeed trivialize I would need four independent sections; I do not know how to find them.

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What about the fact that vector bundles over $S^2$ are classified by clutching construction, and the map classifying $TS^2$ is twice the generator of $\pi_1(SO(2))$. If you think about this as a path of matrices $A(t)$, the tensor square bundle is classified by the path $A(t) \otimes A(t)$ in $SO(4)$, which is then trivial in $\pi_1 SO(4)$ (being divisible by 2, for example).
Could you elaborate how the map classifying $TS^{2}$ be twice the generator of $\pi_{1}(SO(2))$? Also the step from $SO(2)\otimes SO(2)$ to $SO(4)$ seems unclear to me (do you just put them on diagonal blocks?) Anyway this is very helpful. – Kerry Oct 29 '11 at 0:29
The computation follows from the fact that $SO(3)$ is $RP3$ (en.wikipedia.org/wiki/SO%283%29#Topology) and the fundamental group is stable after that (en.wikipedia.org/wiki/Orthogonal_group#Homotopy_groups, look at the fiber bundle and use the homotopy long exact sequence). The generator in $\pi_1(SO(n))$ is hence induced (under inclusion) from the generator in $RP3=SO3$ which is a circle of rotations around the same axis. – Max Oct 29 '11 at 15:08