0

Differential Topology Hirsch Chapter 1 Section 2 Problem 12: For each $n \geq 0$ there is a diffeomorphism $$(TS^n) \times \mathbb{R} \approx S^n \times \mathbb{R}^{n+1}$$ [Hint: there are natural isomorphisms $T_xS^n \oplus \mathbb{R} \approx \mathbb{R}^{n+1}$.

Wanted to double check if this was right:

Regarding the hint, I think what it means is that $T_xS^n$ lives in $\mathbb{R}^n$, and moreover, it spans $\mathbb{R}^n$ as well. This is because if we were to define a chart on $\mathbb{S}^n$ we will find that $\frac{\partial }{\partial x_i}$ for a basis for $\mathbb{R}^n$ Therefore, $TS^n$ which is $(S^n, T_xS^n)$ can now be written as $S^n \times \mathbb{R}^n$. Thus, if we "plug in" our expression for $TS^n$ into $T_xS^n \oplus \mathbb{R} \approx \mathbb{R}^{n+1}$ the assertion will follow.

Math_Day
  • 1,468

1 Answers1

2

What is correct about your argument:

Yes, for each $x\in \mathbb S^n$, $T_x\mathbb S^n$ is isomorphic to $\mathbb R^n$. This is true because they are both real vector space of the same dimension, and an explicit isomorphism can be given using a local coordinates as you described.

What is wrong:

In general, the tangent bundle of $\mathbb S^n$ is not trivial: there is no vector bundle isomorphism between $T\mathbb S^n$ and $\mathbb S^n \times \mathbb R^n$. For example when $n=2$, the hairy ball theorem states that any vector fields on $\mathbb S^2$ must have a zero. (Thus $T\mathbb S^2$ is not $\mathbb S^2 \times \mathbb R^2$ since the latter has global fields $x\mapsto (1,0)$). See more here: indeed $T\mathbb S^2$ is not homeomorphic to $\mathbb S^2 \times \mathbb R^2$.

You are actually missing the crucial word in their hint: You need to find a natural isomorphism between $T_x\mathbb S^n \times \mathbb R$ and $\mathbb R^{n+1}$. By natural, you need something independent of coordinates. To do this, try to think $T_x\mathbb S^n$ as the plane in $\mathbb R^{n+1}$ which is perpendicular to $x \in \mathbb S^n \subset \mathbb R^{n+1}$.

Arctic Char
  • 16,972