Definition of the map $M \to \operatorname{Alb}(M)$

Question

I'm reading about the Albanese variety in Griffiths' and Harris' Principles of algebraic geometry. For a complex manifold $M$, the Albanese variety is defined as $$\operatorname{Alb(M)} := H^0(M, \Omega_M^1)^* / H_1(M, \mathbb{Z}).$$ There is also a map $\mu: M \to \operatorname{Alb}(M)$, defined by taking a base point $p_0 \in M$ and a basis $\omega_1, \dotsc, \omega_q \in H^0(M, \Omega_M)$, and setting $$\mu(p) := \left(\int_{p_0}^p \omega_1, \dotsc, \int_{p_0}^p \omega_q\right).$$

My questions:

1. What does that tuple mean? I mean if I choose a basis of $H^0(M, \Omega_M^1)^*$, the tuple could be coefficients of the basis elements. But what basis to choose? One dual to $\omega_1, \dotsc, \omega_q$?
2. Is it clear that the definition does not depend on the choice of $\omega_1, \dotsc, \omega_q$?

Answer 1

Basis is irrelevant. Fix a base point $p_0$ and $p$ any point. Take a path $\gamma$ from $p_0$ to $p$. Then, we have for any 1-form $\omega$ the integral $\int_{\gamma} \omega$ which gives a map $(p,\gamma)\to H^0(\Omega^1)^*$ and thus a map to $H^0(\Omega^1)^*/H_1(\mathbb{Z})$. Check that this does not depend on $\gamma$ and thus we get a map from $M\to H^1(\Omega^1)^*/H^1(\mathbb{Z})$.

Answer 2

A better definition of $\mu$ is given in the answer by Mohan:

$$ \mu(p) = \int_{p_0}^p\cdot, \quad \text{i.e.} \quad \mu(p)(\omega) = \int_{p_0}^p \omega.$$

To see that this coincides with the definition of Griffiths and Harris, choose a basis $\omega_1, \dotsc, \omega_q \in H^0(\Omega^1)$, and a dual basis $\delta_1, \dotsc, \delta_q$. Write $\omega = \sum_i c_i \omega_i$. Then $$\mu(p)(\omega) = \int_{p_0}^p \omega = \sum_i c_i \int_{p_0}^p \omega_i = \sum_i \delta_i(\omega) \int_{p_0}^p \omega_i,$$ i.e. $\mu(p) = \sum_i \delta_i \int_{p_0}^p \omega_i$.