Example 24.4 Tu's An Introduction to Manifolds

Question

I'm trying to understand the example 24.4 (De Rham Cohomology of a circle), which also contains a lemma.

I'll write down the all example and I'll write my questions

Let $S^1$ be the unit circle in the $xy$-plane. By proposition 24.1. because $S^1$ is connected $H^0(S^1) = \mathbb{R}$ and because $S^1$ is one dimensional $H^k(S^1) = 0$ for all $k \geq 2$. It remains to compute $H^1(S^1)$.

I assume by $H^k(S^1) = 0$ is actually meant $H^k(S^1) = \left\{ 0 \right\}$, right?

Recall from subsetion 18.7 the map $h : \mathbb{R}^2 \to S^1, h(t) = (\cos(t),\sin(t))$. Let $i : [0,2\pi] \to \mathbb{R}$ be the inclusion map. Restricting the domain of $h$ to $[0,2\pi]$ gives a parameterization $F := h \circ i : [0,2\pi] \to S^1$ of the circle. In example 17.16 we found a nowhere-vanishing 1-form $\omega = -ydx + xdy$ on $S^1$ and showed that $F^* \omega = i^*h^*\omega = i^*dt = dt$. Thus, $$ \int_{S^1} \omega = \int_{F([0,2\pi])} \omega = \int_{[0,2\pi]} F^* \omega = \int_0^{2\pi} dt = 2\pi $$ Since the circle has dimension 1, all the 1-forms on $S^1$ are closed, so $\Omega^1(S^1) = Z^1(S^1)$. The integration of 1-forms on $S^1$ defines a linear map $$ \varphi :Z^1(S^1) = \Omega^1(S^1) \to \mathbb{R}, \;\;\varphi(\alpha) = \int_{S^1} \alpha $$ Because $\varphi(\omega) = 2\pi \neq 0$, the linear map $\varphi : \Omega^1(S^1) \to \mathbb{R}$ is onto.

Question 2: Why does $\varphi(\omega) \neq 0$ implies such map is onto?

By Stoke's theorem every, the exact 1-forms on $S^1$ are in $\ker \varphi$

Question 3 : How exactly is the Stoke's theorem used to prove the exact 1-forms are in $\ker \varphi$?

Conversely we will show that all 10forms in $\ker \varphi$ are exact. Suppose $\alpha = f\omega$ is a smooth 1-form on $S^1$ such that $\varphi(\alpha) = 0$. Let $\bar{f} = h^*f = f\circ h \in \Omega^0(\mathbb{R})$. Then $\bar{f}$ is periodic of period $2\pi$ and $$ 0 = \int_{S^1} \alpha = \int_{F([0,2\pi])} \alpha = \int_{[0,2\pi]} F^*\alpha = \int_{[0,2\pi]} (i^*h^*f)(t)F^* \omega = \int_0^{2\pi} \bar{f}(t)dt $$

The following lemma is stated, and proved, but I'll just state it

Lemma 24.5 Suppose $\bar{f}$ is a $C^{\infty}$ periodic function of period $2^\pi$ on $\mathbb{R}$ and $\int_0^{2\pi} \bar{f}(t)dt = 0$. Then $\bar{f}dt = d\bar{g}$ for a $C^{\infty}$ periodic function $\bar{g}$ of period $2\pi$ on $\mathbb{R}$.

After the proof...

Let $\bar{g}$ be the periodic function of period $2\pi$ on $\mathbb{R}$ from lemma 24.5. By proposition 18.12, $\bar{g} = h^*g$ for some $C^{\infty}$ function $g$ on $S^1$. It follows that $$ d\bar{g} = dh^*g = h^*(dg) $$ On the other hand, $$ \bar{f}dt = (h^*f)(h^*\omega) = h^*(f\omega) = h^*\alpha $$ since $h^* : \omega(S^1) \to \omega^1(\mathbb{R})$ is injective, $\alpha = dg$. This proves $\ker \varphi$ consists of exact forms. Therefore, integration induces an isomorphism $$ H^1(S^1) = \frac{Z^1(S^1)}{B^1(S^1)} = \mathbb{R} $$

here it is...

Question 4 : Where does $H^1(S^1) = \mathbb{R}$ comes from? My only guess is since $B^1(S^1) = \ker \varphi$ there must something subtle I'm probably missing...

Thank you

Answer 1

I assume by $H^k(S^1) = 0$ is actually meant $H^k(S^1) = \{0\}$, right?

Yes. It is a common abuse of notation. The first $0$ denotes the zero vector space and the second zero denotes the zero vector inside the zero vector space $\{0\}$.

Why does $\varphi(\omega)\neq 0$ implies such map is onto?

This is linear algebra. If a linear functional $f\colon V \to \Bbb K$ is non-zero, it is surjective. Take $\lambda \in \Bbb K$. Since there is $v \in V$ with $f(v) \neq 0$, we have that $f(\lambda v/f(v)) = \lambda$, so the range of $f$ is the whole $\Bbb K$.

How exactly is Stokes' theorem used to prove that exact $1$-forms are in $\ker \varphi$?

If $M^n$ is compact and without boundary, and $\beta \in \Omega^{n-1}(M)$, then $$\int_M {\rm d}\beta = \int_{\partial M} \beta = \int_\varnothing \beta = 0.$$

Where does $H^1(S^1) = \Bbb R$ comes from?

Integration is a surjective map $\int_{S^1}\colon Z^1(S^1) \to \Bbb R$ with kernel $B^1(S^1)$. This is what he proved. So the first isomorphism theorem says that $H^1(S^1) = Z^1(S^1)/B^1(S^1) \cong \Bbb R$.