I've been studying the torus and the first cohomology group $H^1_{dR}(T)$ for a couple of weeks now. I finally had a breakthrough of understanding and would like to kindly request the community to check my understanding for correctness.

It was my goal to find two inequivalent, closed, non-exact $1$-forms on $T=S^1 \times S^1$.

Here is what I did:

Step 1: Recall the basic example of $$d\theta = {x \,dy - y \,dx\over x^2 + y^2}$$ on $S^1$ and recall that it is closed and not exact.

Step 2: Since $T = S^1 \times S^1$ this suggests that $d\theta, d\varphi$ are closed and not exact where $\varphi$ and $\theta$ denote the angle functions $S^1 \to (0,2\pi)$.

Step 3: To prove that they are not exact we have to calculate $\oint_{S^1}$ for each. This is a line integral and the way to calculate line integrals is by choosing a parametrisation and then substituting it into the integral. We choose the parametrisations

$$ \gamma(t)= (t,0)$$


$$ \gamma'(t) = (0,t)$$

for $t \in (0,2\pi)$ and calculate

$$ \oint d\theta = \int_0^{2\pi} (1,0)\cdot (1,0) \,dt = \int_0^{2\pi}dt = 2 \pi$$ and similarly $$ \oint d \varphi = 2\pi$$

So as claimed both are not exact!

Step 4: $\theta$ and $\varphi$ are not equivalent if they do not differ by a closed $1$-form. Assume there was a closed $1$-form $d\eta$ such that $d \theta = d\varphi + d\eta$. Then integrating on both sides yields

$$ \eta = 0$$

therefore $d\theta$ and $d\varphi$ are not equivalent.

Please, if you read my work, be nitpicky and meticulous. I really really want to understand. If there is anything above that is missing detail then it is probably not due to my being lazy but rather due to my lack of understanding. Any feedback is appreciated.

  • $\begingroup$ I don't seem to follow your proof that $d\theta$ and $d\varphi$ are inequivalent. For instance, could you elaborate on the part of step 4 where you claim "integrating on both sides yields $\eta = 0$"? $\endgroup$ – mollyerin Jan 23 '15 at 3:21
  • $\begingroup$ @mollyerin I meant that if $d\theta = d\varphi + d \eta$ then taking the integral on both sides we get $$ \oint d \theta + \oint d \varphi + \oint \varphi d \eta$$ which equals $$ 2 \pi = 2 \pi + eta$$ hence $\eta = 0$. But maybe I am wrong and $\oint d \eta$ does not equal $\eta$. $\endgroup$ – a student Jan 23 '15 at 3:41
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    $\begingroup$ The fact that you're not being careful about saying which path you're integrating over suggests to me that you're thinking about $\oint d\theta$ as being an integral over $\mathbb{S}^1 \times \lbrace p \rbrace$ and $\oint d\varphi$ as an integral over $\lbrace p \rbrace \times \mathbb{S}^1$. If that's so, then there's no coherent operation which is "integrate both sides" and gives the integrals you want. Also, $\oint d\eta = \eta(b) - \eta(a) = 0$, where $b$ and $a$ are endpoints of the path (which the notation suggests is closed). $\endgroup$ – mollyerin Jan 23 '15 at 4:06
  • $\begingroup$ @mollyerin Thank you! I understand. $\endgroup$ – a student Jan 24 '15 at 1:52
  • $\begingroup$ Why the down vote? $\endgroup$ – a student Jan 24 '15 at 13:44

To be (as requested) nitpicky and (I hope) meticulous: Step $2$ suggests that $d\theta$ is the differential of an angle function $S^1 \to (0, 2\pi)$, but this isn't true for two reasons: First, it is not defined at the point $(1, 0) \in S^1$. More seriously, being closed but not exact, $d\theta$ is not actually the differential of any real-valued function on $S^1$. In this sense, using $d\theta$ to denote the desired $1$-form is potentially confusing.

The second statement is almost true, however (and this arguably justifies the notation $d\theta$ maligned above): Consider the coordinate chart $$\theta_1: (S^1 - \{(1, 0)\}) \to (0, 2\pi)$$ given by the usual angle function. Since $\theta_1$ is a differentiable real-valued function, it determines a perfectly good differential $d\theta_1$ on the domain of the chart. Now, we can cover $S^1$ with the two open arcs $(S^1 - \{(1, 0)\})$ and $(S^1 - \{(-1, 0)\}$, and we can define a second coordinate chart $$\theta_2 : S^1 - \{(-1, 0)\} \to (-\pi, \pi)$$ given by the angle function that takes values in $(-\pi, \pi)$, and again we get a differential $d\theta_2$ on the domain of $\theta_2$.

Now, as $\theta_1$ and $\theta_2$ are both angle functions, they differ by a constant on each component $X$ of the overlap $W := S^1 - \{(1, 0), (-1, 0)\}$ of their domains, and (on each $X$) we have $\theta_2\vert_X = \theta_1\vert_X + c$ for some constant $c$ and thus $d\theta_1\vert_X = d\theta_2\vert_X$. Hence, we can patch $d\theta_1$ and $d\theta_2$ into a single smooth $1$-form sometimes (as in the question) denoted $d\theta$ but which I'll denote $\alpha$. Now, both $d\theta_1$ and $d\theta_2$ are differentials of functions, so they are exact and hence closed. But unlike exactness closedness is a local property, and so $\alpha$ is closed. The argument using Stokes' Theorem in the question shows that $\alpha$, however is not exact.

On the other hand, we can dispense with coordinate charts altogether by exploiting the formula $$\alpha = \frac{-y \,dx + x \, dy}{x^2 + y^2}$$ for $\alpha$ (strictly speaking, $\alpha$ is the pullback of the $1$-form on $\mathbb{R}^2 - \{0\}$ defined by the r.h.s. via the usual embedding $S^1 \hookrightarrow \mathbb{R}^2 - \{0\}$). Then, we can describe the desired $1$-forms on $T = S^1 \times S^1$ as follows: As for any Cartesian products, there are natural projection maps $\pi_i: T \to S^1$, $i = 1, 2$ onto the first and second factors of $S^1$. Then, we can define $1$-forms $\alpha_i$ on $T$ by pulling back our $1$-form $\alpha$ via these maps, that is, setting $$\alpha_i := \pi_i^* \alpha.$$ By naturality of the exterior derivative, we have $d\alpha_i = d\pi_i^* \alpha = \pi_i^* d\alpha = 0$, so these are closed, but integrating each over an appropriate closed path (which, with our new characterization of these forms, is very quick) shows that they are not exact.

Finally, no form of the argument in (4) cannot be correct, as $\eta = 0$ implies that the two $1$-forms are actually the same. (Also, there is a mistype: Two closed forms define the same cohomology class if they differ by an exact form, not a closed form.) Anyway, we can approach this part very concretely: The two $1$-forms will be inequivalent if we can find some $1$-cycle (closed path) $\gamma$ such that $\oint_{\gamma} \alpha_1 \neq \oint_{\gamma} \alpha_2$, but direct checking shows that exactly the paths considered in the integral (which, roughly speaking, wrap around one factor circle in the torus but not the other) in the question have this property.

  • $\begingroup$ Thank you! Just to see if I understand your feedback to (4): I made a mistake, I really meant not exact. So what we have to show is that $d\theta - d \varphi$ is not exact. And this we can do by applying Stokes' theorem (according to which the integral is zero because the torus has an empty boundary). Is that correct? $\endgroup$ – a student Jan 23 '15 at 10:38
  • $\begingroup$ No answer = no? $\endgroup$ – a student Jan 23 '15 at 11:02
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    $\begingroup$ You're welcome, I hope it was useful. What form do you want to apply Stokes' Theorem to? If $d\theta - d\varphi$ were exact, then by Stokes' Theorem for all curves $\gamma$ we would have $\oint_{\gamma} (d\theta - d\varphi) = 0$. Contrapositively, if we can find a curve $\gamma$ such that $\oint_{\gamma} (d\theta - d\phi) \neq 0$, then $d\theta - d\varphi$ cannot be exact, i.e., $[d\theta] \neq [d\varphi]$. $\endgroup$ – Travis Willse Jan 23 '15 at 11:12
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    $\begingroup$ Great, I'm glad you found it helpful. Feel free to ask follow-up questions if any come up. Cheers! $\endgroup$ – Travis Willse Jan 23 '15 at 11:27
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    $\begingroup$ You're right about $d\theta_1$ and the $d\theta_2$, of course, I've fixed the above post. By "local property" in this setting I mean that if you patch together two (or more) closed forms the resulting form is also closed (continuity is local in this same sense). On the other hand, this example demonstrates that exactness is not local in this sense, i.e. when you patch together two exact forms the resulting form might not be exact. $\endgroup$ – Travis Willse Jan 24 '15 at 2:17

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