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## Reps of $U(n)$ for the bundles of holomorphic and antiholomorphic forms of projective space

What are the representations of $U(n)$ that induce (see link text) the bundles of holomorphic $\Omega ^{(1,0)}$ and antiholomorphic $\Omega ^{(0,1)}$ forms of $\mathbb{CP}^n$ (recalling the well-known fact that $\mathbb{CP}^n = SU(n+1)/U(n)$).

Also, what are the representations that induce the line bundles?

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What is meant by "bundle of holomorphic forms" and "bundle of antiholomorphic forms"? – Kevin Lin Dec 11 2009 at 0:20
For the second question, the first paragraph of the section "Formulation" on this wikipedia page en.wikipedia.org/wiki/… might help? – Kevin Lin Dec 11 2009 at 1:58
I think that there might be some ambiguity in the notation in the original version of the question and this might be causing some consusion below. (I would edit, but I'm not sure if this is OK.) The projective space is $U(n)/(U(n-1) \times U(1))$ or, if you prefer to have $SU(n)$ on the top, $SU(n)/(S(U(n-1)\times U(1))$. The subgroup in the denominator consists of block diagonal $U(n)$ matrices of the form $(g,1/\det g)$, where $g$ is unitary of size $n-1$ embedded in $U(n)$ in the obvioua way. This specifies the $U(n-1)$ subgroup -- up to conjugation, of course. – José Figueroa-O'Farrill Dec 11 2009 at 9:47
Holomorphic forms appears to mean holomorphic 1-forms. At least, that's how Jean Delinez seemed to use this terminology in his previous question. mathoverflow.net/questions/8282/… It also seems that "antiholomorphic forms" is the dual vector bundle to "holomorphic forms", that is to say, it is the holomorphic tangent bundle. I find this terminology very strange, so I worry that I am misunderstanding it. If anyone has run across this terminology before, please leave a helpful comment. – David Speyer Dec 11 2009 at 13:15
I think that holomorphic and antiholomorphic refer, respectively, to (1,0) and (0,1) forms. This is a typical notation in the Physics literature, at least. – José Figueroa-O'Farrill Dec 11 2009 at 16:32

The group U(n-1) has an abelian factor U(1) and a semisimple factor SU(n-1).

I'll answer first the second part of the question: Since line bundles are induced from one dimensional character representations, and the semismple component has no non-trivial character representations, thus the line bundles are induced from character representations of the U(1) component. These representations are indexed by a single integer.

The answer to the first part of the question is that the bundle of holomorphic one forms is induced from the representation of U(n-1) which has a unit central charge in the U(1) part and equal to the contragredient fundamental n-1 dimensional representation of the SU(n-1) component. One of the ways to see that is say from the explicit expression of the U(n) action on CP(n-1) in the affine coordinates realized as a rational transformation. When restricted to U(n-1), one can see that the action of the U(1) is through its unit central charge representation and the action of SU(n-1) a multiplication of the affine coordinate vector by the reciprocal of the group element in the fundamental representation.

For higher holomorphic form bundles, the bundle is induced from the (irreducible) antisymmetric powers of the contragredient representations of SU(n-1) accomanied by the representation with the corresponding multiple central charge of U(1).

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If it's the defining representation on SU(n-1) or its dual, then it can't be trivial on the U(1) part. The U(1) part intersects SU(n-1) in a cyclic group that acts non-trivially. Surely you should use the defining representation on all of U(n-1). – Greg Kuperberg Dec 11 2009 at 7:18
The U(1) subgroup that I referred to is the U(1) subgroup of U(n) which intersects the SU(n-1) in the unit element only. According to Borel's theorem on Kaehlerian coset spaces of semisimple Lie groups (which include CPn), (A. Borel, Proc. Natl. Acad. Sci. USA 40, 1147 (1954)) The U(n-1) group in the denominator is a centralizer of a torus in U(n). The U(1) subgroup that I referred to is the torus itself, and its abelian and semisimple components commute. – David Bar Moshe Dec 11 2009 at 9:55
I corrected the error in my answer concerning the action of the abelian factor in the denominator on the holomorphic tanget bundle, thanks to Greg Kuperberg's remark. – David Bar Moshe Dec 11 2009 at 15:07
What does "contragradient" in "contragredient fundamental n-1 dimensional representation" mean? – Jean Delinez Dec 11 2009 at 19:39
As given in an earlier comment, a contragredient representation is the representation on the dual vector space. Let pi(g) be a representation, then its contragredient is pi(g^-1)^t. In the Wikipedia article on the dual representation, this terminology is mentioned. For SU(n), the contragredient of the fundamental representation is also the antisymmetric (n-1) power representation. In the following article by A. Borel there is a nice explanation: hkumath.hku.hk/~imr/records0001/borel.pdf – David Bar Moshe Dec 12 2009 at 5:44

In the comments above, I've been trying to figure out what the vector bundle of antiholomorphic forms is. José Figueroa-O'Farrill writes

I think that holomorphic and antiholomorphic refer, respectively, to (1,0) and (0,1) forms.

I write

It ... seems that "antiholomorphic forms" is the dual vector bundle to "holomorphic forms", that is to say, it is the holomorphic tangent bundle.

The point of this answer is to explain that we are both right.

First of all, I am used to distinguishing between two notions. For me, a "$(1,0)$-form" is a $\mathbb{C}$-valued $1$-form which locally looks like $\sum f_i(z) d z_i$, for some $C^{\infty}$ functions $f_i$. In a "holomorphic $1$-form" we also require the $f_i$ to be holomorphic functions. Apparently, not everyone makes this distinction. That's fine, but in this post I am going to use my language because it appears to be more precise.

Let $X$ be a complex manifold. The sheaf of $(1,0)$-forms on $X$ is a sheaf of modules for the sheaf of $C^{\infty}$ functions. Similarly, the sheaf of holomorphic $1$-forms is a sheaf of modules for the sheaf of holmorphic functions. Using the appropriate version of the Serre-Swan theorem in each case; we get a complex vector bundle over $X$, in the $C^{\infty}$ and holomorphic categories respectively. Taking the forgetful functor from holomorphic vector bundles to smooth vector bundles, we get naturally isomorphic vector bundles in either case, and this bundle is the cotangent bundle $T^*(X)$. So, no matter how you think about it, the "bundle of holomorphic $1$-forms" is the cotangent bundle.

That's what happens with $(1,0)$-forms. What happens with $(0,1)$-forms? In this case, I think it is best to generalize the $C^{\infty}$ approach above. That is to say, we consider the sheaf of all $1$-forms that locally look like $\sum f_i(z) d \overline{z}_i$, for some $C^{\infty}$ functions $f_i$. Again, Serre-Swan gives us a $C^{\infty}$ complex vector bundle which I'll call $A$. From this perspective, $A$ does not have an obvious holomorphic structure.

However, I claim that $A$ is noncanonically isomorphic to the tangent bundle to $X$. Here is the isomorphism. Choose a positive definite Hermitian structure on tangent bundle of $X$. This is always possible by a partition of unity argument. (My convention is that Hermitian forms are linear in the second argument and conjugate linear in the first.) For any $(1,0)$-vector field $v$, the $1$-form $\langle \ , v \rangle$ is a $(0,1)$-form.

The tangent bundle to $X$, of course, does have a natural holomorphic structure; the holomorphic sections are of the form $\sum f_i(z) \partial/\partial z_i$ where the $f_i$ are holomorphic. But this structure is very hidden in the presentation of $A$ as the bundle of $(0,1)$-forms, and that was the point that was confusing me.

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Right. In the Hermitian world, dual and complex conjugate are equivalent. And of course "contragredient" is another name for dual. – Greg Kuperberg Dec 11 2009 at 17:21