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Suppose $p\colon E\to X$ is a vector bundle of finite rank and let $f\colon Y\to X$ be a continuous map. My professor claimed that subbundles of the pullback $f^*(E)$ are not themselves necessarily pullbacks of subbundles of $E$ along $f$. Why is this case? Can someone provide an example of when the words "subbundle" and "pullback" fail to commute in this manner?

The total space of the pullback is defined to be $$\{(y,v)\in Y\times E\colon f(y)=p(v)\}.$$ The total space of a subbundle $S$ lives inside this set. Is there no canonical way to identify bases of fibers in the pullback $f^*(E)$ with bases of fibers in $E$ itself so that the subbundle $S$ can be identified with a subbundle of $E$?

For context: I was studying Chern classes in the setting of algebraic geometry. I saw that for a vector bundle $E$ on a scheme $X$, a certain pullback of the vector bundle, $f^*(E)$, admits a "Chern decomposition" into a product involving the Chern roots. I believed that this will also imply there exists a Chern decomposition of $E$ itself but apparently this is not the case. The obstruction to this is that subbundles of $f^*(E)$ may fail to be pullbacks of subbundles of $E$. Comments on this topic are also welcome.

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    $\begingroup$ As a hint, the tangent bundle of $S^{2}$ has very few subbundles, so try using it as your $E$ (and $S^{2}$ as your $X$). $\endgroup$
    – Rafi
    Jun 6 at 21:51
  • $\begingroup$ @Rafi is it true the tangent bundle of $S^2$ fails to admit line subbundles? Then if $Y$ is a rectangular region in $\mathbb{R}^2$ and $f\colon Y\to S^2$ is the standard parameterization, then any line subbundle of $f^*(TM)$ fails to be the pullback of a line bundle on $S^2$ because there are no line bundles on $S^2$? $\endgroup$
    – Andrew Paul
    Jun 6 at 22:15
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    $\begingroup$ Yes, this follows from that $S^{2}$ has no nontrivial line bundles and from the hairy ball theorem. I think your counterexample works, but I had in mind just making $Y$ a point and $f$ some inclusion thereof in $S^{2}$ to the same effect. $\endgroup$
    – Rafi
    Jun 6 at 22:47

2 Answers 2

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Consider a subbundle $q\colon F\rightarrow X$ of $p\colon E\rightarrow X$, which can be decomposed as $q\colon F\hookrightarrow E\xrightarrow{p}X$. When looking at the pullback bundles $f^*E$ and $f^*F$ over $Y$, you get an induced map $f^*F\rightarrow f^*E$ due to the universal property of the pullback.

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Since the whole and lower square are both pullbacks, the upper square is a pullback. Therefore $f^*F\hookrightarrow f^*E$ is injective as monomorphisms are stable under pullback. This shows, that pullbacks of subbundles are again subbundles. However, not every subbundle of the pullback bundle arises that way.

Consider the map $S^1\rightarrow S^1,z\mapsto z^n$. The pullback of the trivial bundle $\underline{\mathbb{R}^2}$ with zero twists is again the trivial bundle $\underline{\mathbb{R}^2}$ and the pullback of the Möbius bundle with one twist will be the line bundle with $n\operatorname{mod}2$ twists, so the trivial bundle for $n$ even and again the Möbius bundle for $n$ odd. So for $n$ even, the Möbius bundle is a subbundle of the pullback bundle $\underline{\mathbb{R}^2}$, but cannot arise as a pullback bundle of a subbundle of $\underline{\mathbb{R}^2}$.

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    $\begingroup$ @AndrewPaul Yes, with $\underline{\mathbb{R}^2}$ I mean the vector bundle $S^1\times\mathbb{R}^2$. Yes, exactly, there is no such $k$. $\endgroup$
    – Samuel Adrian Antz
    Jun 6 at 22:45
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    $\begingroup$ @SamuelAdrianAntz How does this account for that the line bundle on $S^{1}$ with $n$ twists is the line bundle on $S^{1}$ with $n'$ twists if(f) $n=n'\pmod{2}$? $\endgroup$
    – Rafi
    Jun 6 at 22:51
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    $\begingroup$ @Rafi I guess if you insist that $n$ is even then $nk\ncong 1\pmod{2}$ for any $k$. $\endgroup$
    – Andrew Paul
    Jun 6 at 22:56
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    $\begingroup$ @Rafi Oddly enough, I've not heard about that yet, even though I've just found it directly in the introduction of Allen Hatcher's "Vector Bundles and K-theory" (pi.math.cornell.edu/~hatcher/VBKT/VB.pdf). Then we have to take an even $n$ for the counterexample to work. Thank you! $\endgroup$
    – Samuel Adrian Antz
    Jun 6 at 23:01
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    $\begingroup$ @Rafi That's true, but I don't need it in my counterexample. I took the trivial bundle $\underline{\mathbb{R}^2}$ of which the Möbius bundle $M$ is a subbundle with $M\oplus M\cong\underline{\mathbb{R}^2}$. $\endgroup$
    – Samuel Adrian Antz
    Jun 6 at 23:33
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Here is a somewhat stupid example: Take $X = \{*\}$ the one-point set, then any vector bundle $E$ on $X$ is trivial, and so is the pull-back $f^*E = \mathbb R^n \times Y$. But in general, $\mathbb R^n \times Y$ might have a non-trivial subbundle¹ $F \subset \mathbb R^m \times Y$, so $F$ cannot be a pull-back from $X$.

¹ as in the other answer take $Y = S^1$ with the Möbius bundle $F \subset \mathbb R^2 \times S^1$.

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  • $\begingroup$ Of course! Seems like I should consider bundles on singletons when searching for these sorts of counterexamples... $\endgroup$
    – Andrew Paul
    Jun 7 at 8:55

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