As is well known, the line bundles over *CP*$^1$ are indexed by the integers. My question is how are the line bundles over *CP*$^n$, $n > 1$, and *Gr*$(n,k)$ indexed? Moreover, do there exist any other interesting classifications of line bundles over spaces (I remember something about Atiyah and elliptic curves)?

$\begingroup$ Tagging this question with both algebraicgeometry and differentialgeometry is potentially confusing, since it isn't clear which category you want to work in. Fortunately for the Grassmannian, every smooth complex line bundle is algebraisable, but for more general complex varieties, this is not the case. $\endgroup$ – Peter McNamara Dec 3 '09 at 0:35
Algebraic line bundles on a smooth variety $X$ are classified by the Picard group $Pic(X) = H^1(X, \mathbf O_X^*)$. This is an exercise in Hartshorne's book, basically every line bundle is mapped to it's gluing cocycle. The group $Pic(X)$ is also equal to the group $CH^1(X)$ of divisors modulo rational equivalence. The map sends a line bundle to it's first Chern class.
Now for a projective space Picard group is $\mathbf Z$ generated by $\mathbf O(1)$. Picard group of a Grassmannian is also $\mathbf Z$. I believe that the generator is a pullback of $\mathbf O(1)$ for a Plucker embedding $Gr(n,k) \subset \mathbf P^N$.
One can prove it as follows: both $\mathbf P^n$ and $Gr(n,k)$ are algebraically cellular, meaning that they consist of pieces isomorphic to affine spaces: for $\mathbf P^n$ this is obvious, and for $Gr(n,k)$ the cells are Schubert cells.
For such varieties Chow groups coincide with cohomology groups and are generated by cells (it's like computing cohomology of CW complex without cells of odd dimension). Since there is only one cell of complex codimension 1 for projective spaces and Grassmannians, we get $Pic(X) = H^2(X, \mathbf Z) = \mathbf Z$.
EDIT:
Actually, a projective space IS a Grassmannian. :)
As other people commented, the generator of the $Pic(Gr(n,k))$ is the n'th wedge power of the canonical vector bundle of rank n.

1$\begingroup$ The nth wedge of the tautological nbundle is the pullback of $\mathcal{O}(1)$ by a (some might say the) Plucker embedding. $\endgroup$ – Fran Burstall Dec 4 '09 at 21:49

$\begingroup$ Yes, thank you for this comment. (Also someone told me that "Russian English" is immediately recognized by weird use of articles a/the. We just don't have them in Russian.) $\endgroup$ – Evgeny Shinder Dec 5 '09 at 4:57

$\begingroup$ generator of the picard group is not the pullback of $\mathcal{O}(1)$. For example take Gr(2,4). its degree 2 hypersurface in $\mathbb{P}^5$. So the pulled back bundle is actually twice the generator. $\endgroup$ – user100841 Aug 7 '17 at 10:28
Algebraic line bundles over generalized flag varieties of semisimple groups should be classifiable by characters of the Cartan subgroup of the group modulo the Cartan subgroup of the semisimple part of the parabolic subgroup. This is because all line bundles over a generalized flag variety are equivariant with respect to the semisimple group action, so they correspond to the characters of the parabolic subgroup (see David's answer). For a Grassmannian Gr(n,k), my guess is that line bundles are indexed by the integers, the generator line bundle being the kth exterior power of the tautological kdimensional bundle. I hope better educated people would come up with references and/or corrections to the above.
As a compact homogeneous Kahler manifold the Grassmannian is a coset space of the general linear group by a maximal parabolic subgroup. Homogeneous line bundles over the Grassmannian are in a one to one correspondence with the character representations of the maximal parabolic, which are indexed by one integer. According to the BottBorelWeil theorem, the space of holomorphic sections of the line bundle carries an irreducible representation of the special unitary group SU(n). In the case of the Grassmannian Gr(n,k), these representations correpond to Young tableaux with k rows. The integer characterizing the line bundle is just the number of columns in this tableaux.
If your question is about complex line bundles up to an algebraic isomorphism, such bundles are classified by [the group of divisors](http://en.wikipedia.org/wiki/Divisor_(algebraic_geometry) for any algebraic variety. There is also only one definition of the group of divisors for smooth varieties, as in your case.
It's not hard to see that group of divisors of $\mathbb P^n $ is generated by a hyperplane $H$. The key observation is that a variety defined by a homogenous polynomial of degree $n$ is rationally equivalent to $n$ copies of $H$ (because you can deform the coefficients of polynomial so that it becomes just $x_1^n = 0$).
Therefore the complex line bundles on any $\mathbb P^n$ are classified by integer numbers: $\mathcal O,\ \mathcal O(1),\ \mathcal O(2)$ etc.
Update: see also the answer by Evgeny who explains the same things as me, but better.
Grassmannians have a cell decomposition into Schubert_cells where each cell is just an affine space $\mathbb A^l$. This particularly simple structure means that divisors up to rational equivalence will be the same as algebraic cycles up to rational equivalence and the same as topological $H^2$. Moreover, it will be generated by cells in complex codimension one.
Now it's an exercise to find out from the definition of Schubert cells that there's only 1 cell with this property. To see it, fix a flag $(F_i), F_i\in V$. Then the cell is given by the conditions $$ \text{dim}\\, V\cap F_{nk1+i} = i \quad \text{for}\ 0\le i\le k$$ for subspaces $V$ of dimension $k$.
Therefore the bundles on $Gr(n, k)$ are also numbered by integers.
First of all: this is from the "differentialgeometric point of view"
If you want to "classify" vector bundles chern classes are a very helpful tool. Well, it can happen that two bundles are "different" but the chern classes coincide, but if the Chern classes of a pair of vector bundles do not agree, then the vector bundles are different.
In the case of line bundles, the only nontrivial chern class is the first chern class and it turns out that the first chern class is a complete invariant which classifies complex line bundles. (In other words: there is a bijection between the isomorphism classes of line bundles over a topological space $X$ and the elements of $H^2(X;\mathbb{Z})$)
But note: line bundles can have equivalent first Chern class, but different holomorphic structures. In this case, see the posts of Evgeny,Ilya...
Smooth real line bundles are classified (in the same way) by the first StiefelWhitney class. As in the complex case, the only nonvanishing StiefelWhitney class is $w_1$.
For references see: Wikipedia: Line bundle
Wikipedia: Chern class
Wikipedia: StiefelWhitney class
Milnor, Stasheff, Characteristic classes (google books link)