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I wonder whether or not ${\bf R}$ is a topological group.

In fact ${\bf R}$ is a group with addition structure.

So is it topological group ?

In fact any ${\bf R}^n$ is a topological group. Note that any element in ${\bf R}^n$ is mapped into diagonal matrics in $GL_n({\bf R})$

From this we conclude that any fiber bundle with ${\bf R}^n$-fiber is a principal fiber bundle.

This implies that definition of any fiber bundle is equal to that of the principal fiber bundle.

Am I right ?

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1 Answer 1

up vote 5 down vote accepted

You are correct that $\mathbb{R}^n$ is naturally a topological group for all choices of $n$ (though I'm not sure why $\mathbb{R}^n\subseteq GL_n(\mathbb{R}$) is important), but it does not follow that all bundles with fiber $\mathbb{R}^n$ are principal bundles.

For $P\rightarrow M$ to be a principal $\mathbb{R}^n$ bundle, you must have that $\mathbb{R}^n$ acts on the fibers by translation in such a way that the quotient space $P/\mathbb{R}$ is naturally $M$.

Now, consider the Möbius band over a circle (also known as the tautological bundle over $\mathbb{R}P^1$). Draw a picture of it! Now, $\mathbb{R}$ must act by translation on each fiber. So pick a fiber, and pick a direction and tell yourself "the element $1\in\mathbb{R}$ moves points in that direction. Now, for nearby points, you must pick the same direction. But if you keep applying this argument as you go around the Möbius, when you come back to where you started, you learn that $1$ must move the fiber in the opposite direction. This means there is no consistent choice of $\mathbb{R}$ action on all the fibers, so it's not a principal bundle.

The situation is actually worse:

Theorem: A smooth fiber bundle $P\rightarrow B$ (with $B$ "nice enough") with fiber $\mathbb{R}^n$ is a principal bundle iff it's trivial.

Proof: If it's trivial, $P = \mathbb{R}^n\times B$ and we just let $\mathbb{R}^n$ act on itself in the usual way.

Conversely, note the bundle has a global section (the $0$ section!) and every principal bundle with a global section is trivial. (The section lets you pick a point as the identity in each fiber and then the action of the group on itself determines the rest of the points).

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Thak you for your good explanation. –  Hee Kwon Lee Nov 13 '12 at 3:00

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