# Periods of a Compact Riemann Surface with genus $g$

Let $$X$$ be a compact Riemann Surface with genus $$g$$ and let $$\omega_1,\dots,\omega_g$$ be a basis of $$\Omega(X)$$, the space of holomorphic $$1$$-forms on $$X$$. As a smooth $$2$$ manifold we can think of $$X$$ as $$\Pi_{i=1}^{g}a_i b_i a_i^{-1}b_i^{-1}.$$ Then $$a_1,\dots,a_g,b_1\dots,b_g$$ forms a canonical basis of $$H_1(X,\mathbb{Z}).$$ Then we define periods of $$X$$:

$$P_i:=\big(\int_{a_i}\omega_1,\dots,\int_{a_i}\omega_g\big), P_{i+g}:= \big(\int_{b_i}\omega_1,\dots,\int_{b_i}\omega_g\big)$$ for $$i=1,\dots,g$$

And we have deRham-Hodge Theorem: $$H^1_{dR}(X,\mathbb{R})\cong\Omega(X)\oplus\bar{\Omega}(X)\cong\mathbb{C}^{2g}.$$ Using this how do I see that $$P_1,\dots,P_{2g}$$ are linearly independent pver $$\mathbb{R}$$?

So, if not then $$\exists \hspace{1 ex} c_1,\dots c_{2g}\in\mathbb{R}$$ not all zero, such that $$\sum_{i=1}^g c_i\int_{a_i} \omega_k+ \sum_{i=1}^g c_{i+g}\int_{b_i} \omega_k=0\hspace{1 ex}\forall k=1,\dots,g$$. But $$\because c_1,\dots,c_{2g}$$ are real numbers and not necessarily integers how exactly do I apply the theorem to prove the result?

$$\textbf{Edit:}$$ We can use Poincare Duality and Abel's Theorem which says the following: Suppose $$D$$ is a divisor on a compact Riemann surface $$X$$ with deg $$D=0.$$ Then $$D$$ has a solution iff $$\exists$$ a $$1$$-chain $$c\in C_1(X)$$ with $$\partial c=D$$ such that $$\int_c\omega=0$$ for every $$\omega\in\Omega(X).$$

• First of all, it is $H^1_{dR}(X, {\mathbb R})$. Secondly, you cannot derive linear independence from the Hodge-deRham theorem alone, what you need is the Poincare duality. Nov 2, 2019 at 17:54
• @MoisheKohan thanks can you please hint how one can see it from Poincare Duality. Nov 2, 2019 at 18:21
• Hint: The content of the PD in 2d is that for a compact oriented surface, the integration determines a nondegenerate pairing between the 1st de Rham cohomology and 1st singular homology group. Nov 2, 2019 at 18:56
• @MoisheKohan So, our condition says $\int_{\gamma}(\sum_{i=1}^{2g} c_i)\omega_k$ vanishes for all $k=1,\dots,g$ where $\gamma=a_1+\dots+a_g+b_1+\dots+b_g$. But I don't see any immediate contradiction. Nov 3, 2019 at 7:03

1. Hodge-deRham Theorem implies that whenever $$\omega_1,...,\omega_g$$ is a complex basis in $$\Omega^1(X)$$, the (necessarily closed) real forms $$Re(\omega_1), Im(\omega_1),...,Re(\omega_g), Im(\omega_g)$$ project to a basis of $$H^1_{dR}(X; {\mathbb R})$$. The Poincare Duality theorem for closed oriented connected 2-dimensional manifolds $$X$$ states that the pairing between singular homology $$H_1(X; {\mathbb R})$$ and deRham cohomology $$H^1_{dR}(X; {\mathbb R})$$, given by integration of 1-forms over 1-cycles is nondegenerate, i.e. defines an isomorphism to the dual vector spaces
$$H_1(X; {\mathbb R})\to H_1(X; {\mathbb R})^*, H^1_{dR}(X; {\mathbb R})\to H^1_{dR}(X; {\mathbb R})^*.$$
2. To complete the proof use the following general linear algebra fact: Let $$V, W$$ be $$n$$-dimensional real vector spaces and $$\langle \cdot, \cdot \rangle$$ is a nondegenerate bilinear pairing between $$V$$ and $$W$$. Then for any choice of bases $$\{v_1,...,v_n\}$$, $$\{w_1,...,w_n\}$$ in $$V, W$$, the "Gramian matrix" with the components $$\langle v_i, w_j \rangle$$ is nonsingular.
I didn't find any easy solution of this, anyway Riemann's Bilinear relations give us a way to see this. Interested readers should look up $$\textit{Compact Riemann Surfaces}$$ by Raghavan Narasimhan. There's a whole chapter on $$\textbf{Bilinear Relations}$$ and that solves it for me.