Signature is independent of the Riemannian metric chosen The signature of a 4-manifold is defined to be the dimension of harmonic self-dual two forms minus the dimension of harmonic anti-self-dual two forms, where the self or anti-self-dual is defined in terms of the eigenspaces of hodge star operator.
My question is, the hodge star operator is defined using the metric, i.e. we usually first define its operation on orthonormal basis then extend linearly, which makes clearly the point that this operator relies on the metric. But why is the dimension of the two eigenspaces independent of the metric?
 A: Quickly:
Assume you are dealing with closed oriented (simply-connected?) 4-manifold $M$.
There is an intersection form $Q_M$ on $H^2(M;\mathbb{Z})=H^2_{sing}(M;\mathbb{Z})$ given by $Q_M(\alpha,\beta)=(\alpha\smile\beta)([M])\in\mathbb{Z}$ which is symmetric for dimension reasons.  Under de Rham isomorphism, $Q_M(\alpha,\beta)=\int_M\alpha\wedge\beta$.  Since $\mathbb{Z}$ is torsion-free, $Q_M$ kills torsion elements so we can use $Q_M$ on $H^2_{sing}(M;\mathbb{R})\cong H^2_{dR}(M;\mathbb{R})$.  Under Hodge you get harmonic 2-forms $\mathcal{H}^2(M)$.
Diagonalizing the form $Q_M$ over $\mathbb{R}$ gives a set of positive and negative eigenvalues (counted with multiplicities, by the corresponding set of eigenvectors in $H^2_{dR}(M;\mathbb{R})$).  We define $b_2^+$ (resp. $b_2^-$) to be the number of positive (resp. negative) eigenvalues.  We can follow this construction onto $\mathcal{H}^2(M)$ and get $b^\pm$.  But since $Q_M$ is the same quadratic form on $H^2(M;\mathbb{R})$ and $\mathcal{H}^2(M)$, we have $b_i^\pm=b^\pm$, and so the signature $\sigma(M)=b^+-b^-$ is purely topological.
