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This might be a silly question, but bear with me.

Let $X$ be a topological space, $V$ a vector space and consider the vector bundle $E=X\times V$ with $\pi:E\to X$, $(p,v)\mapsto p$. This is the trivial $V$-bundle over $X$.

I had been under the impression that any vector bundle $E'=X\times V$ was trivial, but I realised that I had not been accounting for the fact that for a trivial bundle, $\pi$ must be projection onto the first factor. My question is, can we show that $E\cong E'$ in general, even when $\pi':E'\to X$ is not just projection onto the first factor?

To show this, I would need to find a homeomorphism $\phi:E\to E'$ such $\pi'\circ\phi=\pi$, and $\pi'=\pi\circ\phi^{-1}$. The obvious idea would be to use the local trivialisations of $E'$, but I don't quite see how to do so.

Any help or references would be much appreciated.

CoffeeCrow
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1 Answers1

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As I mentioned in my comment above, the answer to the question is no, as demonstrated by the following example by Igor Belegradek in this MathOverflow answer which I have replicated below:

Consider the pullback $\xi$ of $TS^2$ via the projection of $S^2\times\mathbb R$ onto the first factor. The bundle $\xi$ is a nontrivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$ because its pullback under the inclusion $S^2\to S^2\times\mathbb R$ is $TS^2$, which is nontrivial. On the other hand, its total space is $\mathbb R\times TS^2$ which is diffeomorphic to $S^2\times\mathbb R^3$, which is the total space of the trivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$.

Let me add some details.

Let $\pi : TS^2 \to S^2$ be the vector bundle projection, and $\operatorname{pr}_1 : S^2\times\mathbb{R} \to S^2$ the natural projection map. Define $\xi := \operatorname{pr}_1^*TS^2$ and consider the inclusion $i : S^2 \to S^2\times\mathbb{R}$, $p \mapsto (p, 0)$. Note that

$$i^*\xi = i^*\operatorname{pr}_1^*TS^2 = (\operatorname{pr}_1\circ\ i)^*TS^2 = \operatorname{id}_{S^2}^*TS^2 = TS^2.$$

As the pullback of a trivial bundle is trivial, and $i^*\xi = TS^2$ is non-trivial, we see that $\xi$ is a non-trivial vector bundle over $S^2\times\mathbb{R}$.

On the other hand, by definition, the total space of $\xi$ is given by

$$\xi = \operatorname{pr}_1^*TS^2 = \{((p, t), v) \in (S^2\times\mathbb{R})\times TS^2 \mid \pi(v) = p\}.$$

Now note that, as bundles over $S^2$, we also have

$$\varepsilon^1\oplus TS^2 = (S^2\times\mathbb{R})\oplus TS^2 = \{((p, t), v) \in (S^2\times\mathbb{R})\times TS^2 \mid \pi(v) = p\}$$

where the last equality is the definition of the direct sum of vector bundles.

As $TS^2 \to S^2$ is stably trivial, $\varepsilon^1\oplus TS^2 \cong \varepsilon^3 = S^2\times\mathbb{R}^3$. So the total space of the non-trivial rank two vector bundle $\xi \to S^2\times\mathbb{R}$ is diffeomorphic to $S^2\times\mathbb{R}^3$ and hence to the trivial rank two vector bundle over $S^2\times\mathbb{R}$, i.e. $(S^2\times\mathbb{R})\times\mathbb{R}^2$.

The point is that the total space of the non-trivial vector bundle $\xi \to S^2\times\mathbb{R}$ can also be viewed as the total space of a trivial bundle over $S^2$.

Michael Albanese
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  • That's a very nice example. It seems to me that all one needs to make it work is that $TS^2$ is both non-trivial and stably trivial. In particular, it should work for $TS^{2n}$. – Jason DeVito - on hiatus Apr 03 '23 at 02:23
  • @JasonDeVito: You are correct. In general, if $E \to B$ is a continuous/smooth real rank $m$ vector bundle such that $E\oplus\varepsilon^n \cong \varepsilon^{m+n}$, then the total space of the corresponding bundle $\xi$ is homeomorphic/diffeomorphic to the trivial rank $m$ real vector bundle over $B\times\mathbb{R}^n$. – Michael Albanese Apr 05 '23 at 02:41
  • I see. Do you happen to know of an example of a non-stably trivial bundle $E\rightarrow B$ where $E$ is abstractly diffeomorphic to $B\times \mathbb{R}^n$? – Jason DeVito - on hiatus Apr 05 '23 at 02:44
  • @JasonDeVito: I don't. If you assume that the diffeomorphism $E \to B\times\mathbb{R}^n$ is compatible with the projections to $B$, then all the Stiefel-Whitney classes of $E$ vanish. – Michael Albanese Apr 05 '23 at 19:22
  • To make sure I understand your last comment, the point is that you're not assuming that the diffeo acts linearly on each fiber, right? I'm guessing a proof: Thom has a defintion of SW via the Thom isomorphism and Steenrod powers on the Thom space of a vector bundle. it seems like such a diffeo induces a homeo of the Thom spaces. If this line of reasoning works, I think it also works merely having a homeomorphism $E\rightarrow B\times \mathbb{R}$ compatible with the projections. – Jason DeVito - on hiatus Apr 06 '23 at 14:33
  • @JasonDeVito: Right, I'm not assuming the diffeomorphism is linear on each fiber. That's not the proof I had in mind. Instead $w(TE) = \pi^w(TB)\pi^w(E)$ while $w(T(B\times\mathbb{R}^n)) = \operatorname{pr}_1^w(TB)$. The diffeomorphism $\varphi : E \to B\times\mathbb{R}^n$ being compatible with the projections means $\operatorname{pr}_1\circ\varphi = \pi$, so $\varphi^w(T(B\times\mathbb{R}^n)) = \pi^*w(TB)$. – Michael Albanese Apr 06 '23 at 18:24
  • Makes sense! Thanks – Jason DeVito - on hiatus Apr 06 '23 at 18:42