For $\mathbb{CP}^1$ the bundles of holomorphic and antiholomorphic forms are equal to the $\mathcal{O}(-2)$ and $\mathcal{O}(2)$ respectively. Do the holomorphic and antiholomorphic bundles of $\mathbb{CP}^2$ (or indeed) $\mathbb{CP}^n$) have a description in terms of line bundles. What happens in the Grassmannian setting?



Holomorphic and antiholomorphic forms of projective space
Greg's otherwise excellent answer gives the impression that computing Chern classes on projective space requires a computer algebra system. I'm writing to repell this impression. The cohomology ring of $\mathbb{P}^{n-1}$ is $\mathbb{Z}[h]/h^n$ where $h$ is Poincare dual to the class of a hyperplane. We have the short exact sequence


$$0 \to S \to \mathbb{C}^n \to Q \to 0$$,


where $S$ is the tautological line bundle, whose fiber over a point of $\mathbb{P}^{n-1}$ is the corresponding line in $\mathbb{C}^n$. The line bundle $S$ is also called $\mathcal{O}(-1)$, and has Chern class $1-h$. So $$c(Q) = 1/(1-h) = 1+h+h^2 + \cdots h^{n-1}.$$


As Greg explained, the tangent bundle is $\mathrm{Hom}(S, Q) = S^{\star} \otimes Q$. The formula for the Chern class of a general tensor product is painful to use in practice, but we can circumvent that here by tensoring the above exact sequence by $S^{\star}$.


$$0 \to \mathbb{C} \to (S^{\star})^{\oplus n} \to T\_{\mathbb{P}^{n-1}} \to 0$$,


So the Chern class of the tangent bundle to $\mathbb{P}^{n-1}$ is
$$(1+h)^n = 1 + n h + \binom{n}{2} h + \cdots + n h^{n-1}.$$


Yes, I deliberately ended that sum one term early. Remember that $h^n$ is $0$.


As Greg says, if this is to be expressible as a direct sum1 of line bundles, then this should be a product of $n-1$ linear forms, $c(T) = \prod\_{i=1}^{n-1} (1+ a\_i h)$. Since this is an equality of polynomials of degree $n-1$, we can forget that we are working modulo $h^n$ and just check whether the honest polynomial $f(x) := 1 + n x + \binom{n}{2} x + \cdots + n x^{n-1}$ factors in this way.


The answer is it does not, except when $n=2$ (the case of $\mathbb{P}^1$.) 
The unique factorization of $f(x)$ is 
$$\prod\_{\omega^n=1,\ \omega \neq 1} (1+x (1-\omega)).$$


Thus does raise an interesting question. When $n$ is even, $(1+2x)$ divides $f(x)$. This suggests that $\mathcal{O}(2)$ might be a subquotient of $T\_{\mathbb{P}^{n-1}}$. I can't figure out whether or not this happens.




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The case of Grassmannians is going to be worse for three reasons. The two minor ones are that (1) we may honestly have to use the formula for the chern class of a tensor product. and (2) the polynomials in question will be multivariate polynomials. The big problem will be that $H^{\star}(G(k,n))$ has relations in degree lower than $\dim G(k,n)+1$, so we can't pull the trick of forgetting that $h^n=0$ and working with honest polynomials.


I suspect that your question was more "give a nice description of the tangent bundle to the Grassmannian" than "can that tangent bundle be expressed as a direct sum of line bundles?". If you seriously care about the latter, I'll give it more thought.


1 Direct sum is the natural thing to ask for in the categories of smooth, or of topological, complex vector bundles. If you like the algebraic or holomorphic categories, as I do, it is more natural to ask for the weaker property that there is a filtration of the vector bundle, all of whose quotients are line bundles.


