TITLE: Motivation for equivariant sheaves?
QUESTION [9 upvotes]: Hello everyone;
i'm looking for a motivation for equivariant sheaves (see http://ncatlab.org/nlab/show/equivariant+sheaf) ~ Why are we interested in them?
More explicitely: Can I think of G-equivariant sheaves on a space X as a quotient of the category of sheaves (by some action? in a more general sense) by G?

REPLY [16 votes]: If you know that the sections of a vector bundle form a standard example of a sheaf, then the corresponding example of a $G$-equivariant sheaf on a space $X$ with $G$-action is a vector bundle $V$ over $X$ with a $G$-action compatible with the projection (i.e. making the projection $G$-equivariant, i.e. intertwining the actions). Such actions on vector bundles over homogeneous spaces were considered in representation theory by Borel, Weil, Bott and Kostant ("homogeneous vector bundles"; and later many generalizations to sheaves by Beilinson-Bernstein, Schmid, Miličić etc.). David Mumford introduced $G$-equivariant structures on sheaves under the name $G$-linearization for the purposes of geometric invariant theory. 
While for a function on a $G$-space the appropriate notion is the $G$-invariance, for sheaves the invariance is useful only up to a coherent isomorphism, what spelled out yields the definition of the $G$-equivariant sheaf. This is an example of a categorification (recall that functions form a set and sheaves form a category). Using the Yoneda embedding one can indeed consider the $G$-equivariant objects as objects in some fibered category of objects on $X$ with an action on each hom-space (see the lectures by Vistoli). 
While for a function to be invariant is a property, for a sheaf the $G$-equivariance entails the additional coherence data, so it is a structure. 
Category of $G$-equivariant sheaves on $X$ is not a quotient of the category of usual sheaves on $X$, but rather equivalent to the category of sheaves on the geometric quotient $G/X$, in the case when the action of $G$ on $X$ is principal; or in general if we replace the geometric quotient by the appropriate stack $[G/X]$.