I'm just starting to learn about vector bundles, I want to compute the transition functions of the bundle $TS^n$.

I started with the stereographic atlas $U_1 = S^n - \{N\}$ and $U_2 = S^n - \{S\}$ where $\phi_i : U_i \rightarrow \mathbb{R}^n$ is the diffeomorphism given by the stereographic projection.

I have computed explicitly these applications and their inverses. My question is, do these applications induce local trivializations for $TS^n$? In that case, the transition map at $x$ is given by $$d_x(\phi_2 \circ \phi_1^{-1})$$ am I right?


The tangent bundle of a manifold $M$ is a vector bundle, denoted $(TM,\pi,M)$ where $\pi(p,V_p)=p$. Given coordinate chart $(U,\varphi)=(U,x^1,\dots x^n)$ on $M$ we get the induced map $$T\varphi:\pi^{-1}(U)\to U\times \mathbb{R}^n \ \ \ \ \ \ \ \ \ \ (p,V^i\frac{\partial}{\partial x^i})\mapsto (p,V^1,\dots,V^n)$$

You can verify that $\pi_1\circ T\varphi=\pi$, where $\pi_1:U\times \mathbb{R}^n\to U$ is the projection, and that for each $p\in M$, the map $T\varphi|_p:T_pM\to\{p\}\times\mathbb{R^n}$ is an isomorphism. That is, each coordinate chart $(U,\varphi)$ is a local trivialization of $TM$.

As for the transition functions if I understand your notation then what you have is correct:

Suppose that $(U,\phi_1)=(U,x^1,\dots,x^n)$ and $(V,\phi_2)=(V,y^1,\dots,y^n)$ are two coordinate charts. Then for arbitrary $(p,W^1,\dots,W^n)\in (U\cap V)\times \mathbb{R}^n$ we have that $$\phi_2\circ\phi_1^{-1}\left(p,W^1,\dots,W^n\right)=\phi_2\left(q,W^i\frac{\partial}{\partial x^i}\right)=\phi_2\left(q,W^i\frac{\partial y^j}{\partial x^i}\frac{\partial}{\partial y^j}\right)=\left(p,\frac{\partial y^1}{\partial x^i}W^i,\dots,\frac{\partial y^n}{\partial x^i}W^i\right)$$so that the transition functions are given by the Jacobian.

  • $\begingroup$ You should take into account that while the image of $q$ by $\phi_1$ is $p$, there's no reason to expect $\phi_2(q)$ to be equal to $p$ as well. Still helpful though! Is this Jon Herman from Waterloo? $\endgroup$ Nov 26 '15 at 5:42
  • $\begingroup$ You're right, but it is true that $\phi_2(\phi_1^{-1}(p))=p$. It follows from the commutative diagram with the trivializations and projection. And maybe!? $\endgroup$
    – JonHerman
    Dec 3 '15 at 20:33
  • $\begingroup$ It think we actually have $\phi_2 \circ \phi_1^{-1}(p) = p/|p|^2.$ Basically $p$ is reflected over the copy of $S^{n-1}$ in $\Bbb R^n$. It's John from Western. $\endgroup$ Dec 3 '15 at 22:01

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