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(may be silly question but), Does there exist two fiber bundles (or in particular vector bundles) whose total spaces are the same but base spaces are different?

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4 Answers 4

up vote 3 down vote accepted

It seems to me that the question would be more interesting if we require the fiber, as well as the total space, to be the same (which might be hidden in the question). So let me present such an example.

The fiber is the disjoint union of two points, let us say $G=\{\pm 1\}$, and the total space is the disjoint union of two circles, $P=S^1\times\{\pm 1\}$. Now consider two following actions of the group $G=\{\pm 1\}$ on $P$ (I only descibe the action of $-1$, and $1$ acts trivially) :

$\mathcal{A}_1(-1) : S^1\times\{\pm 1\} \rightarrow S^1\times\{\pm 1\}$; $(\theta,a)\mapsto(\theta,-a)$ and

$\mathcal{A}_2(-1) : S^1\times\{\pm 1\} \rightarrow S^1\times\{\pm 1\}$; $(\theta,a)\mapsto(\theta+\pi, a)$

where $a$ can ben $1$ or $-1$, and $S^1$ is seen as the real line $\mathbf{R}$ quotiented by $2\pi\mathbf{Z}$.

Now this gives two principal $G$-bundles, one with base space $S^1$ (the one coming from $\mathcal{A}_1$) and one with base space $S^1\times\{\pm 1\}$ (the one coming from $\mathcal{A}_2$).

Unfortunately, I cannot think of an example with a connected fiber for the moment.

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I just posted an answer about the connected fiber case, if you're interested. –  Jason DeVito Apr 10 '12 at 14:23

Sure, consider the space $\mathbb{S}^1 \times \mathbb{R}$ with the projections $\pi_1: \mathbb{S}^1 \times \mathbb{R} \to \mathbb{S}^1$ (a trivial fiber bundle with fiber $\mathbb{R}$) and $\pi_2: \mathbb{S}^1 \times \mathbb{R} \to \mathbb{R}$ (a trivial fiber bundle with fiber $\mathbb{S}^1$). For an example of (trivial) vector bundles, take the total space to be, say, $\mathbb{R}^3$, written as a product space $\mathbb{R}^2 \times \mathbb{R}$. Again, the natural projections give two vector bundles over different base spaces.

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oh of course. thnx. –  wqr Apr 7 '12 at 16:30

Let $G$ be a topological group, $H$ and $K$ closed subgroups admitting a local cross section (see Steenrod, section 7). Then $H\rightarrow G\rightarrow G/H$ and $K\rightarrow G\rightarrow G/K$ are fibre bundles.

In Martin Wanvik's example, our topological group is $\mathbb{R}^3$, which shares the very nice property that $\mathbb{R}^3/\mathbb{R}^1\cong \mathbb{R}^2$ and vice versa.

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To complement Samuel T's example, and to give an example of you's construction, here's a family of examples which came up a lot in my thesis.

First, a general fact: If $H$ is a compact Lie group acting a manifold $M$ freely (i.e., the only element of $H$ which fixes any point of $M$ is the identity element), then the orbit space $M/H$ has the structure of a smooth manifold in such a way as the projection $M\rightarrow M/H$ is a principal fiber bundle with fiber $H$.

If $M = S^3\times S^3$, and $H = S^1\times S^1$, then there are linear actions of $H$ on $M$ with quotient $S^2\times S^2$, or $\mathbb{C}P^2\sharp - \mathbb{C}P^2$ or $\mathbb{C}P^2\sharp \mathbb{C}P^2$ (and no other quotients are possible), where $-X$ denotes $X$ with reversed orientation. So, we can get three different bases while the total spaces and fibers are the same (and fibers connected, as Samuel T wanted).

If $M = S^3\times S^3\times S^3$ and $H = S^1\times S^1\times S^1$, then Totaro has shown that there are infinitely many different bases of the form $M/H$, even up to rational homotopy equivalence. So this is an example of infinitely many different bases for the same total space, same fiber, and with the fiber connected.

If you insist the the fibers be simply connected, then block embedding of $SU(2)$ into $SU(4)$ and the diagonal block embedding of $SU(2)$ into $SU(4)$ gives rise to 2 different (up to homotopy) bases, giving fiber bundles with total space $SU(4)$ and fibers $SU(2)$ (2 connected), and different bases.

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