# Question regarding the definition of linearization of line bundles

I'm reading Dolgachev's book 'Lectures on invariant theory'. In Chapter 7, the linearization of a group action is discussed. Let $$G$$ be a linear algebraic group acting on a quasi-projective variety $$X$$ over an algebraically closed field $$k$$ via $$\sigma: G\times X\rightarrow X$$. He then defines the notion of a $$G$$-linearized line bundle $$L$$ on $$X$$, and comments that for each $$g\in G$$ and $$x\in X$$, the induced map on the fibers $$L_x\rightarrow L_{gx}$$ is a linear isomorphism. We can view the set of such isomorphisms as of line bundles $$\overline{\sigma}(g): L\rightarrow g^*L\ ,$$ which satisfy certain cocycle conditions.

Upto here it's fine. Next, he makes the following remark: the collection of isomorphisms $$\overline{\sigma}(g)$$ can also be viewed as an isomorphism of line bundles $$\Phi : p_2^*L\rightarrow \sigma^*L\ ,$$ where $$p_2:G\times X\rightarrow X$$ is the second projection.

This last part is not clear to me. Surely, given such a $$\Phi$$, we can find the $$\overline{\sigma}(g)$$'s by restricting. But how to go the other way? Meaning, why do the collection of isomorphisms $$\overline{\sigma}(g)$$ given 'fiberwise' glue to give a global isomorphism? Is there a general theme like this, i.e. to define a morphism of sheaves, it is enough to define it on fibers in certain cases?

I think that you should regard the first definition as an imprecise version of the second definition. For example, suppose that $$X$$ is a point and so $$L$$ is simply a 1-dimensional vector space. Then the first definition is simply a map $$g : L \rightarrow L$$ for each group element $$g$$ satisfying multiplicativity; in other words a homorphism of groups $$G \rightarrow GL(L)$$ (i.e. a representation of $$G$$ as an abstract group). On the other hand, the second definition is equivalent to a homomorphism $$G \rightarrow GL(L)$$ as algebraic groups (i.e. a representation of $$G$$ as an algebraic group).