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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?

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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.

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  • $\begingroup$ 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. $\endgroup$ Apr 9, 2020 at 21:13
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    $\begingroup$ In general if you try to use the Hermitian metric to split up a vector bundle into subbundles (which is what you are trying to do with your exact sequence) you are going to have a bad time if you expect it to give a holomorphic decomposition. $\endgroup$
    – Sempliner
    Apr 9, 2020 at 21:19

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