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I am reading Bott Tu Differential Forms in Algebraic Topology, and I am having some troubles to do some of its exercises. In chapter 13 (monodromy), exercise 13.7 page 152:

The universal covering $\pi : \mathbb{R} \rightarrow S^1$ given by $\pi (x)=e^{2\pi ix}$ is a fiber bundle with fiber a countable set of points. The action of the loop downstairs on the homology $H_0(fiber)$ is translation by $1 : x \rightarrow x + 1$. In cohomology a loop downstairs sends the function on the fiber with support at $x$ to the function with support at $x + 1$.

Firstly, I dont understand the bold line, why is it true?

Later, the exercise ask you to compute the cohomology of $\mathbb{R}$ by this process:

With $\mathfrak{U}$ the good cover below of $S^1$, $H^*(\mathbb{R})=H^*_D\{C^*(\pi^{-1}\mathfrak{U}, \Omega^*)\}=H_{\delta}H_{d}=H^*(\mathfrak{U},\mathscr{H}^0)$. Compute $H^*(\mathfrak{U},\mathscr{H}^0)$ directly.

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I know the answer must be $\mathbb{R}$ in dimension $0$ and $0$ otherwise. I was capable to do the previous exercise with help from this post.

The problem now is that the fiber is a non finite set, so I was not able to compute neither $C^0(\mathfrak{U},\mathcal{H}^0)$, $C^1(\mathfrak{U},\mathcal{H}^0)$ nor the differential operator between them $\delta$. How can I proceed in this case?

Jotabeta
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1 Answers1

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I'll first outline a bit of the calculation and then deal with the statement in bold.We have the universal covering map $\pi : \mathbb{R} \to S^{1} \simeq \mathbb{R}/\mathbb{Z}$. Let $\{U_{0},U_{1},U_{2} \}$ be the good cover given in the picture above.

We have that $C^{0}(\mathfrak{U},\mathfrak{H}^{0})=\mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}}$ where $\phi_{0},\phi_{1},\phi_{2} : \mathbb{Z} \to \mathbb{R}$ are arbitrary components of each $\mathbb{R}^{\mathbb{Z}}$ component defined by values in each component/interval of $\pi^{-1}(U_{i})$ for each $i=0,1,2$.

Similarly, $C^{1}(\mathfrak{U},\mathfrak{H}^{0})= \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}}$ corresponding to $U_{0} \cap U_{1},U_{0} \cap U_{2},U_{1} \cap U_{2}$

The boundary operator $\delta : C^{0}(\mathfrak{U},\mathfrak{H}^{0}) \to C^{1}(\mathfrak{U},\mathfrak{H}^{0})$ is then given by $\delta(\phi_{0},\phi_{1},\phi_{1}):=(f_{0,1},f_{0,2},f_{1,2})$ where the $f_{i}$ are defined by:

$f_{0,1}(n):=\phi_{1}(n)-\phi_{0}(n),f_{0,2}(n):=\phi_{2}(n)-\phi_{0}(n-1),f_{1,2}(n):=\phi_{2}(n)-\phi_{1}(n)$

Try and see why this is the case from the definition of the boundary operator. Next, I leave it as an exercise to show the surjectivity of the $\delta$.

$Ker(\delta)$ is clearly just $\{(\phi_{0},\phi_{1},\phi_{2}) | \phi_{0}=\phi_{1}=\phi_{2}=r\}$ for some constant $r\in \mathbb{R}$. So, $Ker(\delta) \simeq \mathbb{R}$.

Now, clearly $H^{i}(\mathfrak{U},\mathfrak{H}^{0}) = \mathbb{R}$ when $i=0$ and $0$ otherwise.

Also, note that actually $\mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \oplus \mathbb{R}^{\mathbb{Z}} \simeq \mathbb{R}^{\mathbb{Z}}$

Rithvik Reddy
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