## Tag Archives: sphere

### Stereographic projection from center

Suppose we are doing stereographic projection at the center. Namely consider the following map

$\displaystyle \phi:\mathbb{S}^2\rightarrow \mathbb{R}^2$

$\displaystyle (x,y,z)\mapsto\frac{(u_1,u_2,-1)}{\lambda}$

where ${\lambda=\sqrt{u_1^2+u_2^2+1}}$. Then one can see ${u_1=\frac{x}{z}}$, ${u_2=\frac{y}{z}}$. Under this stereographic projection, a strip will be equivalent to some lens domain on the sphere.

Let us pull the standard metric of ${\mathbb{S}^2}$ to the ${\mathbb{R}^2}$. For the following statement, we will always omit ${\phi^*}$ and ${\phi_*}$. Calculation shows,

$\displaystyle dx=\left(\frac{1}{\lambda}-\frac{u_1^2}{\lambda^3}\right)du_1-\frac{u_1u_2}{\lambda^3}du_2=z(-1+x^2)du_1+xyzdu_2$

$\displaystyle dy=-\frac{u_1u_2}{\lambda^3}du_1+\left(\frac{1}{\lambda}-\frac{u_2^2}{\lambda^3}\right)du_2=xyzdu_1+z(-1+y^2)du_2$

$\displaystyle dz=\frac{1}{\lambda^3}(u_1du_1+u_2du_2)=z^2(xdu_1+ydu_2)$

therefore

$\displaystyle dx^2+dy^2+dz^2=z^2(1-x^2)du_1^2-2xyz^2du_1du_2+z^2(1-y^2)du^2_2$

$\displaystyle =\frac{1}{\lambda^2}(\delta_{ij}-\frac{u_iu_j}{\lambda^2})du_idu_j$

here we used the fact that ${(x,y,z)\in \mathbb{S}^2}$. Suppose ${\bar\nabla}$ is the connection on ${\mathbb{S}^2}$ equipped with the standard metric. We want to calculate ${\bar \nabla_{\partial_{u_i}}\partial_{u_j}}$. To that end, it is better to use the ${x,y,z}$ coordinates in ${\mathbb{R}^3}$

$\displaystyle \partial_{u_1}=z(x^2-1)\partial_x+xyz\partial_y+xz^2\partial_z$

$\displaystyle \partial_{u_2}=xyz\partial_x+z(y^2-1)\partial_y+yz^2\partial_z$

Since we know

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=\left(\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}\right)^T=\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}-\langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle\partial_r$

where ${T}$ means the tangential part to ${\mathbb{S}^2}$ and ${\partial_r=x\partial_x+y\partial_y+z\partial_z}$ is the unit normal to ${\mathbb{S}^2}$. Using the connection in ${\mathbb{R}^3}$, we get

$\displaystyle \nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1}=z^2\left[3x(x^2-1)\partial_x+y(3x^2-1)\partial_y+z(3x^2-1)\partial_z\right]$

$\displaystyle \langle\nabla^{\mathbb{R}^3}_{\partial u_1}\partial_{u_1},\partial_r\rangle=z^2(x^2-1)$

One can verify from the above equalities that

$\displaystyle \bar\nabla_{\partial u_1}\partial_{u_1}=2xz\partial_{u_1}$

Similarly

$\displaystyle \bar \nabla_{\partial_{u_1}}\partial_{u_2}=yz\partial_{u_1}+xz\partial_{u_2}$

$\displaystyle \bar\nabla_{\partial u_2}\partial_{u_2}=2yz\partial_{u_2}$