Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Consider the set $\mathcal{C} = C^{\infty}(\mathbb{C}^*, \mathbb{C}^*)$, where $\mathbb{C}^* = \mathbb{C}\backslash\{0\}$.

Both $f(z) = z$ and $g(z) = \bar{z}$ can be seen as elements in $\mathcal{C}$.

Question: Is there a (smooth, not necessarily analytic) homotopy $H : [0, 1] \times \mathbb{C}^* \rightarrow \mathbb{C}^*$ between $f$ and $g$?

I tried what seemed to me like natural choices, such as deforming the imaginary part, but the problem is to avoid producing some function which maps a non-zero complex number to zero.

Motivation: In case anyone is wondering, this problem arises in showing that two complex line bundles over the $2$-sphere are (smoothly) isomorphic. The bundles are $L_g^*$ and $L_{1/g}$, where $g : \mathbb{C}^* \rightarrow \mathbb{C}^*$ is the gluing cocycle (there is only one, since the $2$-sphere is covered by two stereographic projections).

Thanks.

share|improve this question
4  
Maybe I'm being silly, but: wouldn't this yield a homotopy $h$ between the identity and $z \mapsto z^{-1} = \bar z$ on $S^1$ by setting $h(t,z) = \frac{H(t,z)}{\lvert H(t,z)\rvert}$? This couldn't be because $z \mapsto z$ and $z \mapsto z^{-1}$ are distinguished by the degree. Probably such a degree argument can be done directly on $\mathbb{C}^\ast$. –  t.b. Sep 16 '12 at 19:25
6  
No, there is no such homotopy since the fundamental group of $\mathbb{C}^*$ is $\mathbb{Z}$ and such a homotopy would deform the unit circle parametrized clockwise into the unit circle parametrized counter-clockwise. –  Michael Sep 16 '12 at 19:31
    
@t.b.,Michael thanks, I guess you're right, there exists no such thing. My reference must be wrong, I'll check it again. –  student Sep 16 '12 at 19:58

1 Answer 1

up vote 3 down vote accepted

Community verdict (from comments by t.b. and Michael): $z$ and $\bar z$ are not homotopic in $C(\mathbb C^*,\mathbb C^*)$.

More precisely, the set of homotopy classes of continuous maps $h\in C(\mathbb C^*,\mathbb C^*)$ is $\{[z\mapsto z^n]:n\in\mathbb Z\}$, and all classes $[z\mapsto z^n]$ are distinct (distinguished by their action on the fundamental group). The complex conjugation belongs to $[z\mapsto z^{-1}]$.

At least tangentially related reference: Stein Manifolds and Holomorphic Mappings: The Homotopy Principle in Complex Analysis by Franc Forstnerič.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.