Explicit calculation of Euler class for $TS^2$

Question

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?

Answer 1

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 $$ \chi(S^2)=\frac{1}{2\pi}\int_{S^2} \operatorname{Pf}\Omega =\frac{1}{2\pi}\int_{S^2} \sin\phi \,d\phi\wedge d\theta = \frac{1}{2\pi} 4\pi = 2. $$

Answer 2

I must be too much into such kind of calculations. Here is my second calculation, using the orthonormal frames throughout (instead of the coordinate vector fields), and so more in the spirit of the original question.

As suggested in the post (but with the $\phi$ and $\theta$ switched as more standard in math), we use the orthonormal frame $$e_1=\frac{\partial}{\partial \phi},\quad e_2=\frac{1}{\sin\phi}\frac{\partial}{\partial \theta}.$$ The dual orthonormal coframe is $$\epsilon^1 = d\phi,\quad \epsilon^2=\sin\phi\,d\theta. $$

As suggested in a comment, we apply the general Kozsul formla \begin{align} g(\nabla_X Y, Z) &= \frac 1 2 \big(X g(Y, Z) + Yg(X, Z)-Zg(X, Y)\\ &+ g([X, Y], Z) - g([X, Z], Y) - g([Y, Z], X\big). \end{align}

Note for the current frame, we do have $$ [e_1, e_2] = -\frac{\cos\phi}{\sin\phi}e_2. $$

I used Maple to compute the following connection forms $$ \omega_{kj}=\sum_{i=1}^2 g(\nabla_{e_i}e_j, e_k)\,\epsilon^i. $$ Note the positon of the indices, since this is how you write the matrix for a linear transformation.

The answer is simple, surprisingly or not. We have $$ \omega = \begin{pmatrix} 0 & -\cos\phi\,d\theta\\ \cos\phi\,d\theta & 0 \end{pmatrix}. $$ (In the orthonormal frame, the endomorphism matrix is supposed to be skew-symmetric.) Then Cartan's structure equation gives $$ \Omega = d\omega + \omega\wedge \omega = \begin{pmatrix} 0 & \sin\phi\,d\phi\wedge d\theta\\ -\sin\phi\,d\phi\wedge d\theta & 0 \end{pmatrix}. $$ Again, we reproduce the answer that $$ \operatorname{Pf}\Omega = \sin\phi\,d\phi\wedge d\theta. $$ This is the volume form of $S^2$, and we get $\chi(S^2)=2$ upon integration.

Answer 3

I must have not fully understood Cartan's method of moving frames. It is actually even quicker.

So following the standard notation, start with the orthonormal frame $$ \omega^1=\epsilon^1=d\phi,\quad \omega^2=\epsilon^2=\sin\phi\,d\theta. $$ Cartan's structure equations are \begin{align} d\omega^i + \omega^i_j\wedge \omega^j &= 0, \\ d\omega^i_j + \omega^i_k\wedge \omega^k_j &= \Omega^i_j. \end{align} Here $(\omega^i_j)$ is the skew-symmetric matrix of connection 1-forms.

In our case, the two connection equations are \begin{align} 0 + \omega^1_2\wedge (\sin\phi\,d\theta)=0,\\ \cos\phi\,d\phi\wedge d\theta + \omega^2_1\wedge d\phi=0. \end{align} This gives that $$ \omega^2_1 = \cos\phi\, d\theta,\quad \omega^1_2=-\omega^2_1=-\cos\phi\, d\theta. $$ This gives back the connection form matrix before. But this is way faster, and doesn't need the Koszul's formula as mentioned in the comment of Ted Shifrin.

Also note that $\Omega^1_2 = d\omega^1_2 = K\omega^1\wedge \omega^2$, and this can be treated as one definition of the Gaussian curvature. Therefore, $$ \operatorname{Pf}(\Omega)=\Omega^1_2 = K\omega^1\wedge \omega^2 = K\,d\mu, $$ where $d\mu$ is the volume form (since $\omega^1$ and $\omega^2$ are orthonormal). This holds true for any surface $M$. Then we are just doing the Gauss-Bonnet theorem $$ \int_M \operatorname{Pf}(\Omega) = \int_M K\,d\mu = 2\pi\chi(M). $$