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Suppose $\gamma: \mathbb{R}\rightarrow \mathbb{S}^{n}$ is a smooth curve. Let $\gamma(t)=(x^{1}(t)...x^{n+1}(t))$.

Let $\mathbb{D}^{n}$ be embedded into $\mathbb{R}^{n+1}$ by viewing $\mathbb{R}^{n+1}$ as $\mathbb{R}^{n}\times \mathbb{R}$ and introduce coordinate embedding $$f: y\rightarrow \langle y, (1-|y^{2}|)^{1/2}\rangle$$ of the ball of radius 1 in $\mathbb{R}^{n}$ into $\mathbb{S}^{n}$. The round metric on $T\mathbb{S}^{n}$ identify $$\langle (x,v), (x,w)\rangle=\langle v, w\rangle$$ with $x\in \mathbb{S}^{n},v,w\in T\mathbb{S}_{x}$. Now in order to find a geodesic on $\mathbb{S}^{n}$ I need to find the Christoffel symbols $\Gamma^{i}_{jk}$. And to find Christoffel symbols I need to find the pull-back metric (from $\mathbb{D}^{n}\rightarrow \mathbb{S}^{n}$). Taubes now assert that we have $$g_{ij}=\delta_{ij}+y_{i}y_{j}(1-|y|^{2})^{1/2}$$

I am wondering why this is true. The pull-back metric for map between manifolds $\psi:M\rightarrow N$ and vector bundle $E\rightarrow N$ with a given fibre-wise metric is defined by $\langle (p,v), (p,w)\rangle=v\cdot w,p\in M, v,w\in E$. So in our case we are working with $\mathbb{D}^{n}\rightarrow \mathbb{S}^{n}$, with the bundle $T\mathbb{S}^{n}\rightarrow \mathbb{S}^{n}$ (endowed with round metric) pulled back.

By definition of metric we would be expecting $g_{ij}(x)=\langle \partial y_{i},\partial y_{j}\rangle, \partial y_{i},\partial y_{j}\in T\mathbb{S}^{n}_{x}$. Now since $x=(y,(1-y^{2})^{1/2})$, by definition above we can pull it back to $f^{*}T\mathbb{S}^{n}_{x}$. But the evaluation seems to be the same and I could not understand how he get his formula. Since the sphere case is the simplest possible, I feel I need to ask for help.

I also thought about differentiating directly. Then we would have $g_{ij}=\delta_{ij}+y_{i}y_{j}(1-|y|^{2})^{-1}$. This still does not match Taube's formula and I do not know what is wrong.

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