I am reading about lifts of graphs in relation to covering spaces. Before I pose my question I will explain some of the terminology.
Let $G$ and $H$ be two graphs. We say that a function $f: V(H) \rightarrow V(G)$ is a covering map if for every $v \in V(H)$, f maps a neighbor set $\Gamma_H(v)$ of $v$ one-to-one and onto $\Gamma_G(f(v))$. It is easy to see that if $h$ is an eigenfunction of $V(G)$ then $h \circ f$ is an eigenfunction of $V(H)$. We call the eigenfunctions inherited from $G$ old eigenfunctions and we call the eigenfunctions that are not inherited from $G$ new eigenfunctions.
I don't understand why the following claim holds.
Let $\phi$ be a new eigenfunction of $H$. Then $\sum\limits_{f(x)=v}\phi(x)=0$. This is suppose to follow from the following two facts.
1. Eigenfunctions of $G$ span the space of real functions on $V(G)$.
2. Distinct eigenfunctions can always be chosen to be mutually orthogonal.
I am not sure how these to facts imply the proposition. $\phi$ is a function on $V(H)$ and the two facts are about functions on $V(G)$.
View the adjacency matrix $A$ of the cover as acting on the vector space $V$ of real functions on the vertices of the cover. The subspace of function that are constant on the vertices in each fibre is $A$-invariant. Since $A$ is symmetric, the orthogonal complement to this subspace is $A$-invariant, it consists of the functions that sum to zero on each fibre. The old eigenvectors come from functions constant on fibres.