I'm currently trying to calculate the Euler class $e(TS^2)$ for the sphere $S^2$, but I have some difficulties doing this. I define the Euler class as $$e(TS^2)= \left[\operatorname{Pf}\left(1/2\pi\Omega\right)\right]$$ where $\Omega$ is the curvature matrix of the Levi-Civita connection on $TS^2$.
The Riemannian metric on $S^2$ is given by $$g=d\theta+\sin^2(\theta)d\varphi.$$
What I think I need to do is to find an orthonormal frame $(e_1,e_2)$ and use that to find the dual coframe $(\varepsilon^1,\varepsilon^2)$. Use these somehow to find the connection $1$-forms $\omega^i_j$ and use the second structural equation to find the curvature $2$-forms $\Omega^i_j$.
After I have found these I obtain $\Omega$ and I would need to calculate the Pfaffian $\operatorname{Pf}\left(1/2\pi\Omega\right)$.
I know that $\frac{\partial}{\partial \theta}$ and $\frac{1}{\sin\theta}\frac{\partial} {\partial \varphi}$ form the orthonormal frame. The coframe should then be given by $\varepsilon^1 = d\theta$ and $\varepsilon^2 = \sin\theta d\varphi$. How can I now find $\omega^i_j$'s?
Best Answer
Since I like such kind of calculations (I do them using Maple), I do want to provide a detailed answer.
But before I start, your convention of $\theta$ and $\phi$ is the physicists' one, and I want to switch to the mathematicians' one.
Therefore $\phi$ is the angle with the $z$ axis and ranges in $[0, \pi]$, and $\theta$ is the angle in the $xy-$plane of the projection and ranges in $[0, 2\pi]$. Therefore the coordinates are $$ x^1 = \phi,\quad x^2 = \theta. $$
The metric tensor $$ g=d\phi^2 + \sin^2 \phi\,d\theta^2 $$ in matrix form is $$ (g_{ij}) = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\phi\end{pmatrix}. $$
The formula for the Christoffel symbol, in my used-to convention, is $$ \Gamma_{ij}^k = \frac 1 2 g^{k\ell}(g_{i\ell,j}+g_{\ell j, i}-g_{ij,\ell}), $$ where the Einstein's convention of repeated indices being summed is used. The nontrivial ones in this case are $$ \Gamma_{12}^2=\Gamma_{21}^2 = \frac{\cos\phi}{\sin\phi},\quad \Gamma_{22}^1 = -\sin\phi\cos\phi. $$ Then we compute the Riemann (3, 1)-curvature tensor by $$ R_{ijk}^\ell = \Gamma_{jk,i}^\ell-\Gamma_{ik,j}^\ell+\Gamma_{jk}^m\Gamma_{mi}^\ell - \Gamma_{ik}^m\Gamma_{mj}^\ell. $$ The only nontrivial ones are $$ R_{121}^2=-R_{211}^2=-1,\quad R_{122}^1=-R_{212}^1=\sin^2\phi. $$ This means that the curvature operator $$ R_{12}=R_{\frac{\partial}{\partial \phi}\frac{\partial}{\partial \theta}}=\nabla_{\frac{\partial}{\partial \phi}}\nabla_{\frac{\partial}{\partial \theta}}-\nabla_{\frac{\partial}{\partial \theta}}\nabla_{\frac{\partial}{\partial \phi}} $$ (the Lie bracket is zero) in the basis $(\frac{\partial}{\partial \phi},\frac{\partial}{\partial \theta})$ is represented by $$ \begin{pmatrix} 0 & \sin^2\phi\\ -1 & 0 \end{pmatrix}. $$
To compute the Pfaffian, we use the orthonormal basis instead, as stated in the post, as $(e_1, e_2)=(\frac{\partial}{\partial \phi}, \frac{1}{\sin\phi}\frac{\partial}{\partial \theta})$, and the $R_{12}$ is now $$ \begin{pmatrix} 0 & \sin\phi\\ -\sin\phi & 0 \end{pmatrix}. $$ This is skew-symmetric, as should be. Its Pfaffian is $\sin\phi$.
Therefore we have computed that (since $\Omega=R_{ij}dx^idx^j$ is the End-valued curvature form) $$ \operatorname{Pf}\Omega = \sin\phi\, dx^1\wedge dx^2=\sin\phi\, d\phi\wedge d\theta. $$
This is nothing but the volume form of $S^2$. Therefore, the Euler characteristic is the integral of the Euler class as $$ \chi(S^2)=\frac{1}{2\pi}\int_{S^2} \operatorname{Pf}\Omega =\frac{1}{2\pi}\int_{S^2} \sin\phi \,d\phi\wedge d\theta = \frac{1}{2\pi} 4\pi = 2. $$