This is for HW 7, oh comedy hologram

Question 1: Line integrals of vector fields

Answer:

Part A: If \(X\) is the gradient of a function, applying stokes theorem to the form \(X_x \cdot v\) when integrating along the path \(\gamma\) will cause the integral to be 0.

If \(X\) is conservative, this means that it is exact (recall Ian's Talk), hence there is a function \(f\) whose derivative is exactly \(X_{\gamma(t)} \cdot \gamma'(t) dt\)

Part B: If \(X\) is conservative on this star shaped subset \(U\), then from our calc 3 knowledge, we know that the curl of the gradient of a function is equal to 0.

Now, if the curl of \(X\) is equal to 0, then this means that \(X\) was the gradient of some function \(f\), hence, it is conservative.

Question 2: First cohomology of the torus

Answer:

If \(w\) is exact then the associated vector field to \(w\), call it \(X\), is conservative. From the previous exercise, we know that this means integrating it over any pw smooth closed curve \(\gamma\) will return 0. Conversely, if \(\int_{\gamma_j} w = 0\) then we can interpret our 1-form as a vector field. It is conservative along any \(\gamma_j\), hence this is the gradient of a function, thus making \(w\) exact.

I don't see how the hint is really helpful for me. The more obvious thing to do for myself was just work with the definition of exactness of \(w\).

Question 3: Cup Products

Answer: Given \(w \in \Omega^p(M)\) and \(\eta \in \Omega^q(M)\) closed forms. We will show that \(w \wedge \eta \in \Omega^{p+q}(M)\) is closed. Observe that

\(d(w \wedge \eta) = dw \wedge \eta + (-1)^p w \wedge d \eta = 0 \)

So closed forms get wedged and are also closed forms. Wedging a closed form with an exact form is also a closed form. So then if we picked representatives from the cohomology classes of \([w]\) and \([\eta]\), since these classes were exact, the elements of the wedge of their class are too. Hence wedging these elements also returns another closed element. Thus, the cohomology class of \(w \wedge \eta\) depends solely on what the cohomology class of \(w\) and \(\eta\) were. Because of this, we can define our cup product to be the function which takes pairs of classes in the different degree cohomologies of M and returns the class of the wedge of the representatives of the original classes. Meaning:

\(\cup: H^p _{dR} (M) \times H^q _{dR} (M) \rightarrow H^{p+q}_{dR}(M)\)

given by \([w] \cup [\eta] = [w \wedge \eta]\)

Checking billinearity: \(\begin{split} [\alpha w + \beta m] \cup [\eta] &= [(\alpha w + \beta m) \wedge \eta] \\ &= [(\alpha w \wedge \eta) + (\beta m \wedge \eta)] \\ &= [\alpha w \wedge \eta] + [\beta m \wedge\eta] \end{split}\)

This makes sense since \(\alpha w \wedge \eta\) and \(\beta m \wedge \eta\) are both \(p+q\) forms. So when we add them together, it amounts to simply adding together the obvious representatives from their cohomology classes. Hence the map is billinear. The other side of billinearity is similar.

Question 4: More punctures

Answer: Puncturing \(\mathbb{R}\) twice yields 3 copies of \(\mathbb{R}\), thus \(H_{dR}^0(\mathbb{R}) \setminus \{e_0,e_1\} \cong \mathbb{R}^3\)

If we puncture \(\mathbb{R}^2\) twice, this can be retracted to a wedge of two circles. Which has cohomology group isomorphic to \(\mathbb{R}\) So then \(H_{dR}^1 \mathbb{R}^2 \setminus \{e_1,e_2 \} \cong H_{dR}^1 (S^1 \wedge S^1) \cong H_{dR}^1 (S^1) \times H_{dR}^1 (S^1) \cong \mathbb{R}^2\)

This is also true for the 0th cohomology group. But any other cohomology group is zero for this.

Lastly, since we can describe the \(n\)-sphere as a punctured \(n+1\)-space. We can see that twice punctured \(n+1\) space will have cohomology isomorphic to the wedge of two \(n-1\) spheres. Which has cohomology \(H_{dR}^k(S^n) \cong \mathbb{R}\) if \(k=0\) or \(k=n\).

So then \(H_{dR}^k(\mathbb{R}^n \setminus \{e_1,e_2\}) \cong H_{dR}^k(S^{n-1}) \times H_{dR}^k(S^{n-1}) \cong \mathbb{R}^2\)

Whenever \(k=0\) or \(k=n\).

The differential forms that give a basis from their cohomology classes are exactly those that would correspond to the notion of "going around". So in the case of the plane, we would need two differential forms which correspond to the equation of going around the points removed. Not sure where to start calculating this, feels like its on the tip of my tongue.

Question 5: Poincare Duality Part 1

Answer: See the attached image I'll send with this homework. To be honest, I'm pretty lost on how to show the this map descends into a linear map between cohomologies. I feel like with the picture I have attached, it should be something akin to diagram chasing. But then I just can't write the map down. The current PD map is solid since it knows how to map the \(p\) forms. What I want to do here in the cohomology case is to pick representatives from the class and integrate them. ie. \(\int_M w \wedge \eta\) Where \(w \in [w] \in H^p_{dR}(M)\) and \(\eta \in [\eta] \in H^{n-p}_{dR}(M)\)

What if the map for \(w \in \Omega^p(M)\) went \(\ast(w)\) (the Hodge star), followed by quotienting by our exact forms to land us in the cohomology, then finally dualize that. To then define the map between the \(pth\) cohomology and the dual \(n-p\) cohomology, just keep track of the exact forms in \(\Omega^p(M)\) as you move them through the various maps.