# $S^2$ Tangent Bundle Stably Trivial?

I know that the tangent bundle $$TS^2$$ is stably trivial: if $$\nu$$ denotes the normal bundle of the embedding of $$S^2$$ in $$\mathbb{R}^3$$ as the unit sphere, then $$\nu$$ is a trivial line bundle. But then the sum of these is trivial, since it just gives the restriction of the tangent bundle on $$\mathbb{R}^3$$. If I wanted, I could just tack on an extra trivial line bundle to have $$TS^2 \oplus \varepsilon^2 \cong \varepsilon^4$$.

Now $$S^2 \cong \mathbb{CP}^1$$ is also a complex manifold, so $$TS^2$$ is a complex line bundle. If this complex bundle were stably trivial, then the product theorem for Chern classes would yield that the total Chern class of $$TS^2$$ is trivial. However, $$c(TS^2) = 1 + c_1(TS^2) = 1+e(TS^2)$$ where the Euler class $$e(TS^2)$$ is twice a generator of $$H^2(S^2) \cong \mathbb{Z}$$, a contradiction. Doesn't this now contradict the last line of the previous paragraph, since rank-$$2k$$ real trivial bundles are the same as rank $$k$$-complex trivial bundles?

The difference between a rank $$2k$$ real bundle and a rank $$k$$ complex bundle is that in a complex bundle one is using $$\mathbb C$$ as the scalar field, whereas in a real bundle one is only using $$\mathbb R$$ as the scalar field.
Now let's suppose we are given a complex $$n$$-dimensonal bundle $$E \mapsto X$$.
It is true that if this bundle is trivial over $$\mathbb C$$, i.e. if there is a $$\mathbb C$$-linear isomorphism of bundles $$E \mapsto X \times \mathbb C^n$$ over $$X$$, then one immediately obtains an $$\mathbb R$$ linear isomorphism of bundles $$E \mapsto X \times \mathbb R^{2n}$$ over $$X$$: just restrict the scalar field from $$\mathbb C$$ to $$\mathbb R$$.
But the converse does not hold: if there is an $$\mathbb R$$ linear isomorphism of bundles $$E \mapsto X \times \mathbb R^{2n}$$ over $$X$$, you don't know how to extend this to a $$\mathbb C$$-linear isomorphism; roughly speaking, you don't know how to define multiplication by $$i$$ in $$X \times \mathbb R^{2n}$$ in such a way that makes the given map of bundles into a $$\mathbb C$$-linear isomorphism.
• Ah I see. So even though on the top $TS^2$, $\varepsilon^2$, and $\varepsilon^4$ all have some pretty obvious complex structures, the identification $TS^2 \oplus \varepsilon^2 \cong \varepsilon^4$ doesn't respect those complex structures. I suppose the (real) isomorphism really does have to be mixing up $TS^2$ and the first factor of $\varepsilon^2$, while leaving the second factor of $\varepsilon^2$ totally untouched. Apr 9, 2020 at 21:13