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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?

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The bundle of holomorphic 1-forms (also known as the holomorphic cotangent bundle) on CP^1 is O(-2), not O(1). Its dual is the holomorphic tangent bundle to CP^1, which is O(2). I suspect you meant to ask about the holomorphic tangent and cotangent bundles, in which case Greg gave a very good description. You might want to edit your question to clarify, especially if you are looking for something else. – David Speyer Dec 9 '09 at 17:08
You're quite right that's exactly what I meant. – Jean Delinez Dec 10 '09 at 15:34
up vote 7 down vote accepted

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.

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.

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I turned to Maple because I overlooked what you noticed at a crucial point: The degree count lets you work with honest polynomials. Notice by the way that your polynomial f(x) has a cyclotomic-like factorizations. So you could ask, beyond the line bundle case, whether the tangent bundle has a topological splitting of the same type. – Greg Kuperberg Dec 10 '09 at 20:40
As you suspected, I am actually interested in finding a nice formulation of the tangent bundles. More specifically, I am looking for the representations of $U(n-1)$ that give the holomorphic and antiholomorphic forms (recalling the well-known fact that $\mathbb{CP}^n = SU(2)/U(n-1)$. I am going to pose this as separate question so that anyone who answers it will get credit. – Jean Delinez Dec 10 '09 at 20:59
... and, of course, I am also interested in the general Grassmannian answer to this question. – Jean Delinez Dec 10 '09 at 21:00
Yes! $CP^{2n-1}$ fibers over $HP^{n-1}$. The fiber is $CP^1$. There's your line bundle. – Greg Kuperberg Dec 10 '09 at 21:15
Very slick! Also, probably not holomorphic, which explains why I had so much trouble finding it. (HP^1 is a 4-sphere; since the 4-sphere has trivial H^2, it does not support a Kahler structure. This suggests but does not prove that the line bundle you built is not a holomorphic subbundle.) – David Speyer Dec 10 '09 at 21:38

I'll start with the second question first. Your question is to describe the tangent and cotangent bundles of projective spaces or a Grassmannian, and their exterior powers. The Grassmannian $\text{Gr}(k,n)$ is also the Grassmannian $\text{Gr}(n-k,n)$, so you could say that it has two tautological bundles $A$ and $B$, of dimension $k$ and $n-k$. Their direct sum is an $n$-dimensional trivial bundle. My intuition from the geometry is that the tangent bundle is $\text{Hom}(A,B)$. You can also write this as $A^* \otimes B$.

Now suppose that $k=1$ so that it is $\mathbb{C}P^n$. Then $A = \mathcal{O}(-1)$ (if I have my signs right) and $B$ is the quotient of $(n+1)\mathcal{O}(0)$ by $A$. So $A^* \otimes B$ is a quotient of $(n+1)\mathcal{O}(1)$ by $\mathcal{O}(0)$. This is thus a resolution of the tangent bundle by direct sums of line bundles. The cotangent bundle has a dual resolution, and you can explode a resolution like this to obtain a resolution of the exterior powers.

This realization of the tangent bundle of $\mathbb{C}P^n$ lets you compute its Chern classes. I think that the values of the Chern classes tell you that it isn't more directly a direct sum of line bundles. When $n=2$, Maple tells me that the Chern classes of the line bundles would be irrational. (Edit: As David Speyer explains, you can confirm with pen and paper that the Chern class doesn't factor completely, and doesn't even have a linear factor when $n$ is odd.)

I imagine that this is discussed better in some more recent textbook, but Google finds an old paper by Hangan with the same statement that the Grassmannian tangent bundle is a tensor product. The paper considers the geometric implications of having some tensor factorization of the tangent bundle of a manifold in general.

I would guess that you can't describe the tangent bundle of a general Grassmannian in terms of line bundles. The line bundles yield a certain subgroup of the semigroup of possible total Chern classes. I would think that this subgroup misses the Chern classes of the relevant bundles by a mile. (But I don't know a rigorous argument or a reference.)

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In the terminology of Bott and Tu in "Differential forms in algebraic topology", A manifold with a vector bundle splitting into a sum of line bundles is called a split manifold. Given a vector bundle, we can associate to it a (complete) flag bundle, and the pullback of the vector bundle to this flag bundle splits into a direct sum of line bundles. This is called the "splitting principle". If one considers a vector space as a vector bundle over a point, we obtain that the flag manifold itself is a split manifold and its tangent bundle splits ito a sum of line bundles, and its total Chern class is a product of Chern classes of line bundles. The CP1 example is the simplest case of this principle. The main point here is that the flag manifold needs to be complete. Thus, partial flag manifolds such as CPn, n>1 and the Grassmannian are not split manifolds.

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