# $\mathbb{R}$ -Linear independence of periods in Riemann surfaces.

Let $X$ be a compact Riemann Surfaces of genus $g \geq 0$. Let $a_1,b_1,...,a_g,b_g$ a basis for the homology group $H_1(X)$, and let $w_1,...,w_g$ a basis for the space of holomorphic differentials. Now we define the period vectors by $$A_i = \left(\int_{a_i}w_1, ..., \int_{a_i}w_g\right)$$ $$B_i = \left(\int_{b_i}w_1, ..., \int_{b_i}w_g\right)$$ for each $1 \leq i \leq g$. Then I want to prove that the family of $2g$ period vectors $$\{A_i, B_i: 1\leq i \leq g \}$$ is linearly independent over $\mathbb{R}$.

If we try by a contradiction: Asumme that the family of periods vectors is linearly dependent, then there exists $c_1, .. ,c_{2g} \in \mathbb{R}$ such that $$\sum_{i = 1}^g c_i\int_{a_i}\omega _j + \sum_{i = g+1}^{2g}c_i\int_{b_i}\omega _j = 0$$ for each $1 \leq j\leq g$. I think that, this maybe is equivalent to $$\int_{c_1a_1 + \cdots + c_{2g}b_g} \omega_j = 0$$ but only happens when $c_i \in \mathbb{Z}$, then not works.

Some help? Or this is false?

Sorry by my question, but in most of literature about this topic the author assume that the basis for $H_1(X, \mathbb{Z})$ is canonical and this clean the proof in an easy but not trivial way, however I needed the proof when the basis is arbitrary. I found it!!
$\text{P. Griffiths, } \textit{Introduction to algebraic curves}, \text{Translations of Mathematical Monographs, vol 76, American Math. Society, 1989.}$