Involution on variety lifts to vector bundle Let $X$ be a smooth projective variety. Let $i$ be an involution on $X$, which induces a $G=\mathbb{Z}\setminus (2)$ action on $X$.
Consider a vector bundle  $E$ on $X$. It is mentioned that $E$ has an $i$-action if there is a $G$-action on $E$ which lifts the $i$ action from $X$ to $E$. Then $E$ decomposes into $E^+\oplus E^-$, the Eigen bundles.
1) what exactly is the action on $E$? Is $E$ the locally free sheaf, or the space of sections?
2) what are the Eigen bundles? How do we get them?
3) are the Eigen bundles vector bundles or just torsion -free?
4) Do we have such a decomposition for any finite group action on a variety which lifts to $E$?
I will be grateful for any clarification. Thank you.
 A: Item 1 may be obviated by the edit: The action on $E$ is assumed to exist. (In practice, the lift of $i$ to $E$ might be defined in a variety (no pun) of ways.) As for the remaining points:
If $f:X \to X$ is a map and $p:E \to X$ is a vector bundle, a vector bundle map $\tilde{f}:E \to E$ is said to cover $f$ if $p \circ \tilde{f} = f \circ p:E \to X$. (In words, $\tilde{f}$ maps the fibre $E_{x}$ over $x$ linearly to the fibre $E_{f(x)}$ over $f(x)$.)
Every involution of a vector space (finite-dimensional or not) induces a canonical splitting into eigenspaces:
$$
v = \tfrac{1}{2}(v + i(v)) + \tfrac{1}{2}(v - i(v)).
$$
(The summands are eigenvectors of $i$ with eigenvalue $1$ and $-1$, respectively.)
Applied fibrewise, if $i$ is an involution of a vector bundle $E$ that fixes fibres (i.e., covers the identity map on the base, so that $i$ induces an involution on each fibre of $E$), then $E$ splits canonically into eigenbundles. These are vector subbundles, not just torsion-free sheaves.
If $i:X \to X$ is an involution that lifts to $\tilde{\imath}:E \to E$, there is a push-forward vector bundle $i_{*}E \to X/G$ whose fibre at each point $[x] = \{x, i(x)\}$ in $X/G$ is $E_{x} \oplus E_{i(x)}$, the direct sum of the fibres of $E$ over the two preimage points of $[x]$. The involution $\tilde{\imath}:E \to E$ induces an involution of $i_{*}{E}$ that covers the identity on the quotient $X/G$, so $i_{*}E$ decomposes into $\pm1$-eigenbundles as above.
Explicitly, the involution of $i_{*}E$ sends $(v, w)$ to $(\tilde{\imath}(w), \tilde{\imath}(v))$. The respective eigenbundles comprise elements of the form $(v, \tilde{\imath}(v))$ of eigenvalue $1$, and $(v, -\tilde{\imath}(v))$ of eigenvalue $-1$.
If instead one wants to decompose $E \to X$, the issue is that $\tilde{\imath}:E_{x} \to E_{i(x)}$ isn't an involution of the fibres of $E$ individually, only an involution of the total space $E$.
