By trial and error I found that $dx,dy$ are generators of $H^1_{dR}$ of $T=S^1\times S^1$. Verifying that they generate the first cohomology group is not difficult.

My problem is: I found them by trying some $1$-forms and verifying their properties but I would really like to have geometric insight and be able to derive why these generate the first cohomology group.

My work/idea and the problems I ran into:

Parameterise $S^1 \times S^1$ with the angles $(x, y)\in [0,2\pi]^2$. Normally we choose local coordinates because global coordinates do not exist but in this case they do exist.

Problem: It is not clear to me that if a $1$-manifold $M$ is parameterized by a global coordinate $x$ and another $1$-manifold $M'$ is parameterised by a coordinate $y$ then the product $M \times M'$ is parameterized (again globally) by $(x,y)$.

But for the sake of this question let's assume that indeed the product is parameterized by the product of the coordinates.

My real problem is:

How do I obtain the differential $1$-forms $dx,dy$ from the angles $x,y$?

My idea was to take the total derivative of the angle functions. But if we denote the angle by $\theta$ instead of $x$ then the total derivative turns out to be $$ d\theta = -{y \over x^2 + y^2}dx + {x \over x^2 + y^2}dy$$

which leaves me confused.


1 Answer 1


The coordinates $x$ and $y$ are not global. If they were, the torus would be diffeomorphic to $\mathbb{R}^2$. If you are thinking of $S^1$ as $[0, 2\pi]$ where the endpoints are identified, then $x$ and $y$ are coordinates on $(0, 2\pi)\times(0, 2\pi)$ which is the complement of the red and purple circles in the image below.

$\hspace{7mm}$enter image description here

The question still remains, how do you obtain the one-forms $dx$ and $dy$? They are obtained in the same way: $dx$ comes from the first factor and $dy$ comes from the second factor, so we just need to see how to obtain $dx$ on $S^1$. Consider $S^1$ as $\mathbb{R}/\mathbb{Z}$, the quotient of $\mathbb{R}$ by the equivalence relation $p \sim p' \Leftrightarrow p - p' \in \mathbb{Z}$; we will denote the element of $\mathbb{R}/\mathbb{Z}$ corresponding to $p \in \mathbb{R}$ by $[p]$.

Now, $dx$ is a well-defined one-form on $\mathbb{R}$, but it also descends to a one-form on $\mathbb{R}/\mathbb{Z}$. We define the one-form $dx$ on $\mathbb{R}/\mathbb{Z}$ by $dx_{[p]} := dx_p$. This is well-defined because if $p \sim p'$, then $dx_p = dx_{p'}$ (in fact $dx_p = dx_q$ for any $p$ and $q$). Despite the fact that the one-form we obtain on the torus is denoted $dx$, it is not an exact form - I discuss this further in my answer to your new question.

If you would prefer to think of $S^1$ as $[0, 2\pi]$ where $0$ and $2\pi$ are identified (i.e. $S^1$ parameterised by angle), then the above discussion holds for $\mathbb{R}/2\pi\mathbb{Z}$.

The final formula you use is when $\theta$ is the second polar coordinate on $\mathbb{R}^2\setminus\{0\}$ and $x$ and $y$ are the usual Euclidean coordinates. Here $x$ is a coordinate of the first copy of $S^1$ and $y$ is a coordinate on the second copy.

  • $\begingroup$ Thank you! Do there not exist global coordinates for the torus at all or is just that the ones I chose aren't global? $\endgroup$
    – a student
    Jan 5, 2015 at 23:20
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    $\begingroup$ There are no global coordinates. An $n$-dimensional manifold has a global coordinate system if and only if it is diffeomorphic to $\mathbb{R}^n$ (the coordinates providing the diffeomorphism). $\endgroup$ Jan 5, 2015 at 23:22
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    $\begingroup$ Given a chart $(U, \varphi)$ on $M$ and a chart $(V, \psi)$ on $N$, $(U\times V, (\varphi, \psi))$ is a chart on $M\times N$. However, there are other charts on $M\times N$ that are compatible with these. $\endgroup$ Jan 5, 2015 at 23:32
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    $\begingroup$ No it doesn't. The definition you link to is talking about a $k$-form at a point $p$. A one-form has to be linear over smooth functions, that is $dx(fv) = fdx(v) = f$. I think you need to spend some time understanding differential forms; you should have done this before trying to deal with cohomology. $\endgroup$ Jan 6, 2015 at 18:20
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    $\begingroup$ The one-form $dx$ is the exterior derivative of the first coordinate function where it is defined, but the form $dx$ is defined on the entire torus despite the fact that coordinate function is not. $\endgroup$ Jan 8, 2015 at 5:58

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