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 $\omega\in\Omega(X).$

  • $\begingroup$ 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. $\endgroup$ Commented Nov 2, 2019 at 17:54
  • $\begingroup$ @MoisheKohan thanks can you please hint how one can see it from Poincare Duality. $\endgroup$
    – Partha
    Commented Nov 2, 2019 at 18:21
  • $\begingroup$ 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. $\endgroup$ Commented Nov 2, 2019 at 18:56
  • $\begingroup$ @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. $\endgroup$
    – Partha
    Commented Nov 3, 2019 at 7:03

2 Answers 2

  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.

Any other answer is welcome as well.


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