Principal bundles and associated vector bundles, the case of the complex projective space (1,0)-forms As can be guessed from some of my previous questions, I'm trying to understand, at the moment, the relationship between principal and their associated vector bundles. To this end I've been looking at $\mathbb{CP}^{n} = SU(n+1)/U(n)$ and trying to find the representation of $U(n)$ that gives $\Omega^{(1,0)}(\mathbb{CP}^n)$, for all $n$. For $n=1$, I worked it out using a transition function argument. But for $n>1$ this is proving very cumbersome. Can anyone point me in the direction of a more effective method. 
Sketch of transition function method (as requested): So $\Omega^{(1,0)}(\mathbb{CP}^n)$ is the dual of $T^{(0,1)}(\mathbb{CP}^n)$, which is in turn, by definition, the dual of $T^{(1,0)}(\mathbb{CP}^n)$. Now $T^{(1,0)}(\mathbb{CP}^n)$ is defined to be the bundle whose transition functions are the Jacobian of the change-of-coordinate maps. Dual complex bundles have conjugate trans fns, and so, the Jacobian also provides the functions of $\Omega^{(1,0)}(\mathbb{CP}^n)$. These functions $$\phi_{ij}:U_i \cap U_j \to GL(n,\mathbb{C})$$ are elementary to calculate.
Now let $$ \psi:U_i \cap U_j \to U(n)$$ be the transition functions of the principal bundle $U(n) \to SU(n+1) \to \mathbb{CP}^n$. For the right representation $\pi:U(n) \to GL(n,\mathbb{C})$, we will have
$$
\pi \circ \psi_{ij} = \phi_{ij}.
$$
In the $n=1$, case both sets of transition functions are uncomplicated and it's easy to spot what $\pi$ must be. For $n>1$, it's proving to be more messy, and that's why I'm wondering if there's a smarter way of doing things.
 A: In $SU(n+1)/U(n)$ there is a natural basepoint, the coset of the identity, which is fixed by
the action of $U(n)$ (thought of as acting on the quotient by virtue of being a subgroup
of $SU(n1)$).  Since $U(n)$ fixed this point, it acts on the (complexified) cotangent space to this point,
and the problem is then to understand this representation, and in particular, to decompose it
into two pieces, the $(1,0)$ piece and the $(0,1)$ piece.
Now the tangent space is ${\mathfrak su}(n+1)/{\mathfrak u(n)}$ (here I mean complexified Lie algebras), and so we have to decompose
this quotient under the adjoint action of $U(n)$.  It has dimension $(n+1)^2-1-n^2 = 2n,$
and in fact it will decompose as the sum of the standard representation of $U(n)$ direct sum
its dual (or equivalently, its complex conjugate).  One of these representations will 
give the $(1,0)$-subbundle of the tangent bundle, and the other the $(0,1)$-subbundle.
Dualizing (to pass from tangent to cotanget) will give your answer.  (I am not going
to actually stipulate which is which, just because I'm likely to blunder while tracing through the constructions and the duality; but it shouldn't
be hard to work out if you sit down with pen and paper.)
[The following is added in response to the comment below, asking for the movitation behind the above calculation; hopefully it is of some help:]
Since we are looking for an $SU(n+1)$-equivariant splitting of the (co)tangent bundle, it is enough to look for a $U(n)$-equivariant splitting of its fibre at the $U(n)$-fixed point.  (This is a manifestation of the very reason that $U(n)$-reps. give rise to equivariant bundles on the quotient.)   At that fixed point (the identity coset) the tangent space is ${\mathfrak su(n+1)}/{\mathfrak u(n)}$, just because it is the quotient of the tangent spaces at the identity of the corresponding groups.  Since this has to split into two complex conjugate halves under the $U(n)$ action (we know that is breaks up into
a $(1,0)$ and $(0,1)$ part), each of dimension $n$, it's not hard to guess what they must be.  A little computation
in the Lie algebra ${\mathfrak su(n+1)}$ confirms the guess.
Maybe a general lesson to be drawn is:  when trying to compute a $G$-equivariant
bundle on $G/H$, it is enough to compute the $H$-representation on the fibre at the
identity coset; indeed, passing from  $G$-equivariant bundles to this fibre is the quasi-inverse functor
to the one (implicitly) alluded to in the introduction, which associates a $G$-equivariant bundle to an $H$-representation.
[The following was added in response to the question about equivariance and transition functions in the comments:]
Dear Dyke, Let me hide behind Ben's answer, and leave the expression of equivariance in terms of transition functions as an exercise. (Note that, while it is in some sense routine, as Ben indicates, it may also be painful, because if $G$ acts transitively, as in your example, then you won't be able to choose the affine opens on which the bundle is trivialized to be $G$-invariant, and this will complicate things.) Instead, I'll note the following: when you describe a vector bundle by transition functions, you are giving a bunch of open sets that you glue together to get your space, and then the transition functions tell you
how to glue together the trivial bundle on these various opens into a bundle on the space under consideration. But the $G$-equivariant set-up gives a different way to think of the bundle: on $G/H$, since $G$ acts transitively, we just take the trivial bundle at the base point, and then move it around by $G$ to get a bundle over all of $G/H$. The only thing is that this is overdetermined (unless $H$ is trivial); there are lots of ways to go via an element in $G$ to a given point. This overdeterminacy is all encoded in the fact that $H$ stabilizes the base-point: so what we have to do is say, if we take our vector space
at the base-point, and then move it around by an element $h \in H$ (which doesn't actually
move the base-point at all), how we identify the ``moved space'' with the original space.
In other words (making this heuristic precise) we have to describe an $H$-action on
the vector space at the base-point.  Ergo, $H$-representations correspond to $G$-equivariant vector bundles on $G/H$.
So in this context, transition functions are not a very natural way to think about how
the vector bundle is constructed; a more representation-theoretic view-point is the way to go.
