Torus: Period Integrals & De Rham Cohomology Explained

Hey guys! Ever found yourself lost in the abstract world of Riemann surfaces, desperately trying to make sense of cohomology and period integrals? Well, you're not alone! Today, we're going to break down a fascinating exercise involving tori, lattices, and all that good stuff. So, buckle up and let's dive in!

Understanding the Torus $T = \mathbb{C}/L$

First things first, let's set the stage. Imagine you've got a lattice $L$ sitting pretty in the complex plane $\mathbb{C}$. Think of it as a grid of points, generated by two complex numbers that are linearly independent over the reals. Now, construct a torus $T$ by taking the quotient of $\mathbb{C}$ by this lattice, denoted as $T = \mathbb{C}/L$. Intuitively, this means we're identifying points in the complex plane that differ by a lattice point. Picture taking a parallelogram defined by your lattice and gluing opposite sides together – voilà, you've got a torus!

This construction might seem abstract, but it's incredibly powerful. It allows us to study the geometry and topology of tori using the tools of complex analysis. In this article, we are going to explore the exercise of the torus. The exercise is composed of the following parts:

1. Prove that $dx$ and $dy$ generate $H^1(T, \mathbb{R})$.
2. Express a basis of $H^1(T, \mathbb{Z})$ using $dx$ and $dy$.
3. Relate period integrals to De Rham cohomology.

Let's tackle these one by one, shall we?

i) Proving $dx$ and $dy$ Generate $H^1(T, \mathbb{R})$

Okay, so we want to show that the real-valued 1-forms $dx$ and $dy$ generate the first De Rham cohomology group $H^1(T, \mathbb{R})$. What does this even mean? Well, De Rham cohomology is a way of studying the topology of a manifold (in our case, the torus) using differential forms. A 1-form is basically something you can integrate along a curve, and $H^1(T, \mathbb{R})$ captures information about 1-forms that are closed (their exterior derivative is zero) but not exact (they're not the exterior derivative of a 0-form).

The key idea here is to use the fact that the torus is obtained by quotienting the complex plane by a lattice. This means we can lift any differential form on the torus to a differential form on the complex plane that is invariant under the action of the lattice. Specifically, let $\omega$ be a closed 1-form on $T$. Since $\mathbb{C}$ is the universal cover of $T$, we can lift $\omega$ to a 1-form $\tilde{\omega}$ on $\mathbb{C}$ such that $\tilde{\omega}(z + \lambda) = \tilde{\omega}(z)$ for all $z \in \mathbb{C}$ and $\lambda \in L$. In simpler terms, this lifting $\tilde{\omega}$ is periodic with respect to the lattice $L$.

Now, since $\omega$ is closed, so is $\tilde{\omega}$. In $\mathbb{C}$, every closed 1-form is exact (this is a consequence of the Poincaré lemma). Therefore, there exists a smooth function $f: \mathbb{C} \to \mathbb{R}$ such that $\tilde{\omega} = df$. However, $f$ is not necessarily periodic, and that is perfectly fine!

Consider a basis {$ \lambda_1, \lambda_2 $} for the lattice $L$, where $ \lambda_1 = a + bi $ and $ \lambda_2 = c + di $, with $a, b, c, d \in \mathbb{R}$. Since $ \tilde{\omega} $ is periodic, we have $ df(z + \lambda_1) = df(z) $ and $ df(z + \lambda_2) = df(z) $. Integrating these, we get

$\qquad f(z + \lambda_1) = f(z) + A$ $\qquad f(z + \lambda_2) = f(z) + B$

for some constants $A, B \in \mathbb{R}$. Now, define a new function

$\qquad g(z) = f(z) - \frac{A}{a^2 + b^2}(ax + by) - \frac{B}{c^2 + d^2}(cx + dy)$

where $z = x + iy$. We want to show that $dg$ is periodic, i.e., $g(z + \lambda_1) = g(z)$ and $g(z + \lambda_2) = g(z)$. Let's compute:

$\qquad g(z + \lambda_1) = f(z + \lambda_1) - \frac{A}{a^2 + b^2}(a(x+a) + b(y+b)) - \frac{B}{c^2 + d^2}(c(x+a) + d(y+b))$ $\qquad = f(z) + A - \frac{A}{a^2 + b^2}(ax + a^2 + by + b^2) - \frac{B}{c^2 + d^2}(cx + ca + dy + db)$ $\qquad = f(z) + A - \frac{A}{a^2 + b^2}(ax + by) - A - \frac{B}{c^2 + d^2}(cx + dy) - \frac{B}{c^2 + d^2}(ca + db)$ $\qquad = f(z) - \frac{A}{a^2 + b^2}(ax + by) - \frac{B}{c^2 + d^2}(cx + dy) + A - A - \frac{B}{c^2 + d^2}(ca + db)$

This simplifies to

$\qquad g(z + \lambda_1) = g(z) - \frac{B}{c^2 + d^2}(ca + db)$

Similarly,

$\qquad g(z + \lambda_2) = g(z) - \frac{A}{a^2 + b^2}(ac + bd)$

Since {$ \lambda_1, \lambda_2 $} form a lattice, we must have $ad - bc \neq 0$, which means that $ca + db$ and $ac + bd$ cannot both be zero unless $A = B = 0$. If $A = B = 0$, then $\tilde{\omega}$ is already periodic, meaning it descends to a well-defined 1-form on the torus. Otherwise, we need to tweak our function $g$ a bit more.

However, there's a more direct approach. Since $ \tilde{\omega} = df $, we can write $ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy $. We want to show that $ \omega $ is cohomologous to a linear combination of $dx$ and $dy$. Consider the integrals

$\qquad A = \int_{\gamma_1} \omega = \int_0^1 \tilde{\omega}(\gamma_1(t)) \gamma_1'(t) dt$ $\qquad B = \int_{\gamma_2} \omega = \int_0^1 \tilde{\omega}(\gamma_2(t)) \gamma_2'(t) dt$

where $ \gamma_1 $ and $ \gamma_2 $ are loops on the torus corresponding to the generators of the lattice. Now, consider the 1-form

$\qquad \eta = A dx + B dy$

Then $ \int_{\gamma_1} \eta = A $ and $ \int_{\gamma_2} \eta = B $. Thus, the 1-form $ \omega - \eta $ has zero periods, meaning it's exact. This implies that $ \omega $ is cohomologous to $A dx + B dy$, so $dx$ and $dy$ generate $H^1(T, \mathbb{R})$.

ii) Expressing a Basis of $H^1(T, \mathbb{Z})$ using $dx$ and $dy$

Now, let's find a basis for $H^1(T, \mathbb{Z})$. This is a bit more subtle because we're now dealing with cohomology classes that have integer periods. A 1-form $\omega$ represents a class in $H^1(T, \mathbb{Z})$ if its integral over any closed loop on $T$ is an integer.

Consider the 1-forms $dx$ and $dy$. Integrating $dx$ along a loop that goes once around the torus in the $x$-direction gives 1, and integrating it along a loop in the $y$-direction gives 0. Similarly, integrating $dy$ along the $y$-direction gives 1, and along the $x$-direction gives 0. These are integers, so $dx$ and $dy$ represent classes in $H^1(T, \mathbb{Z})$.

Moreover, any element in $H^1(T, \mathbb{Z})$ can be written as $m dx + n dy$, where $m$ and $n$ are integers. This is because the integral of $m dx + n dy$ over any loop will be an integer combination of the integrals of $dx$ and $dy$ over that loop, which are integers by assumption. Therefore, {$dx, dy$} forms a basis for $H^1(T, \mathbb{Z})$.

iii) Relating Period Integrals to De Rham Cohomology

Finally, let's connect period integrals to De Rham cohomology. The period integrals of a 1-form $\omega$ over the torus are simply the integrals of $\omega$ over a basis of loops that generate the first homology group $H_1(T, \mathbb{Z})$. In our case, these loops are the ones we used before, going around the torus in the $x$ and $y$ directions.

The magic happens because of De Rham's theorem. This theorem states that there's a natural isomorphism between De Rham cohomology and singular cohomology with real coefficients. In other words, $H^1(T, \mathbb{R})$ is isomorphic to $Hom(H_1(T, \mathbb{Z}), \mathbb{R})$. This means that a cohomology class in $H^1(T, \mathbb{R})$ is completely determined by its values on the homology classes in $H_1(T, \mathbb{Z})$, which are precisely the period integrals.

So, the period integrals capture all the information about the cohomology class of a 1-form. If two 1-forms have the same period integrals, then they represent the same cohomology class, meaning their difference is an exact form. This is a powerful connection that allows us to study the topology of the torus using analytic tools like differential forms and integration.

Wrapping Up

Alright, guys, that was a whirlwind tour of tori, lattices, cohomology, and period integrals! We've shown that $dx$ and $dy$ generate $H^1(T, \mathbb{R})$, found a basis for $H^1(T, \mathbb{Z})$ using $dx$ and $dy$, and related period integrals to De Rham cohomology. Hopefully, this has shed some light on these concepts and made them a bit less intimidating.

Keep exploring, keep questioning, and never stop diving deeper into the beautiful world of Riemann surfaces!