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I'm first putting the question into it's context, so probably you can see if i'm asking the wrong question to get what i want.

The Task is to show that the Möbius (Moebius) strip is a Vector bundle with base space $S^1$ and typical fiber $\mathbb{R}$. The strip is defined as follows: Let $R$ be the open square $R = (-\pi, 2 \pi) \times (-1,1) \subset \mathbb{R}^2$. On $R$ we define the equivalence relation $\sim$ to be: $(x_1,y_1) \sim (x_2,y_2)$ if $\{x_1=x_2, y_1=y_2\}$ or $\{|x_1-x_2|=2\pi,y_1=-y_2\}$. The Möbius strip is now defined as $E = R / \sim$.

As projection $\pi: E \to S^1$ i chose simply the projection on the first component. The fiber over every point of $S^1$ is therefore the interval $(-1,1)$ which needs to be isomorphic to $\mathbb{R}$ as vectorspace.

My question is now: What is the vector space structure on the fiber $(-1,1)$? Or: How can i turn (-1,1) into a vector space?

My first try was to define vector addition as $x \oplus y = \tanh\left( \tanh^{-1}(x) + \tanh^{-1}(y)\right)$ but then i couldn't find a sensible scalar multiplication with $\mathbb{R}$.

Any help and comments are appreciated.

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Choose your favorite homeomorphism $f\colon\mathbb R \to (-1, 1)$, for example $\tanh$, and define $$v + w = f(f^{-1}(v) + f^{-1}(w))$$ $$a\cdot v = f(a\cdot f^{-1}(v))$$ All you're doing here is transplanting the vector space structure of $\mathbb R$ onto $(-1, 1)$ so there isn't really anything to check as far as it being a well defined vector space.

You will, on the other hand, have to check that your transition maps are linear. This should hold as long as you don't vary the homeomorphism $f$ from fiber to fiber.

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  • $\begingroup$ Thanks! I tried that with the addition and it worked but somehow made some mistakes to verify the vextor space axioms for the scalar multiplication. $\endgroup$
    – user9784
    Mar 14, 2013 at 17:47

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