If $\gamma$ is a closed Loop in $S^2$ and $p\in S^2$, where $\gamma$ is the boundary curve of some region $X$ in $S^2$ (and $\gamma$ satisfied some regularity conditions), someone told me that the holonomy map $H_\gamma:TS^2_p\rightarrow TS^2_p$ is just rotation by the area of $X$. I tried to obtain a reference/proof for this. Could someone help me out?
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If $\gamma$ is piece wise geodesics, the Gauss-Bonnet theorem states $\int_X dA+\theta=2\pi$ – Xipan Xiao Jul 31 '15 at 17:56
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I figured it out :)) – Urs Aug 03 '15 at 16:51
1 Answers
As the Gaussian curvature is $K=1$ on $S^2$, the general Gauss-Bonnet theorem in our setting (assuming that $\gamma$ is positively oriented w.r.t. $X$) becomes
$$ \int_\gamma \, k_g \, \mathrm{d}s + \mathrm{area}\,(X) + \sum \theta_{\text{ext}} = 2\pi \qquad \qquad (*) $$
where $k_g$ is the geodesic curvature of $\gamma$ and $\sum \theta_\text{ext}$ is the sum of its external angles (which is possibily nonvanishing when $\gamma$ is piecewise smooth).
Now, fixing any parallel vector field $E$ along $\gamma$, we interpret $k_g$ as the rate of change of the oriented angle $-\alpha$ between the tangent vector field $\gamma'$ and $E$, that is
$$ k_g = -\frac{\mathrm{d}\alpha}{\mathrm{d}t} $$
in such a way that
$$ \Delta \alpha = - \left( \int_\gamma \, k_g \, \mathrm{d}s + \sum \, \theta_\text{ext}\right) $$
Substituting into $(*)$ we finally get
$$ \Delta \alpha = \mathrm{area}\,(X) - 2\pi $$
Remarks:
- If $\gamma$ is everywhere smooth, of course $\sum \, \theta_\text{ext} = 0$
- One may observe that the term $-2\pi$ gives no contribution in the last formula as long as we deal with angles. In general, the extra information we get by considering $\Delta \alpha$ not mod $2\pi$ is of course the "rotation number" of parallel transport along $\gamma$.
References:
M. do Carmo, Differential Geometry of Curves and Surfaces. (See in particular Proposition 4-4.3 and the discussion following the local version of Gauss-Bonnet theorem in section 4-5.)
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