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(1) Let $M$ be a complex manifold of real dimension $2n$, and denote the line bundle of complex $(N,0)$-forms by $\Omega^{(N,0)}(M)$. When $M = CP^N$, the line bundles are indexed by the integers, and so, $\Omega^{(N,0)}(CP^N)$ must correspond to a integer. What is this integer? In the $N=1$ case, the corresponding integer is $2$. This suggests, that in general, $\Omega^{(N,0)}(CP^N)$ corresponds to $2N$. Is this true?

(2) For the anti-canonical spin$^c$ structure of $CP^N$, the spinor bundle is isomorphic to $$ S := (\Omega^{(0,0)}(CP^N) + \cdots + \Omega^{(0,N)}(CP^N)) \otimes S_N, $$ where $S_k$ is the square root of $\Omega^{(N,0)}(CP^N)$ (square root wrt tensoring as multiplication). What does the square root mean when when line bundle integer is odd? In the $N=2$ case, this is seen to reduce to $\cal{E}_{-1} \otimes \cal{E}_1$, where $\cal{E}_p$ is the line bundle corresponding to the integer $p$. Is this the $Z2$ grading on the spinor bundle? If so, what does this look like in higher dimensions?

(3) Finally, for a given spin connection $\nabla$, to define a Dirac operator we would need a Clifford representation, ie a map $$ C:(\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes S \to S. $$ For $N=2$, this should be given by a map $$ C: (\Omega^{(1,0)} \oplus \Omega^{(0,1)}) \otimes (\cal{E}_{-1} \oplus \cal{E}_1) \to \cal{E}_1 \oplus \cal{E}_1. $$ What is this rep? What does it look like for higher order $N$? Note: the first subindex in the second $\cal{E} \oplus \cal{E}$ above should be $-1$, I'm having tex problems when I try to write it as such though.

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1. If you mean the DEGREE of the line bundle, i.e., the first Chern class. Then that of the canonical bundle is equal to -(N+1). –  Guangbo Xu Nov 1 '10 at 20:24
    
What's the definition of the anti-canonical spin structure? –  Guangbo Xu Nov 1 '10 at 20:29
    
It's a little involved. See Section 3.4 in Friedrich's book on Dirac operators. –  Jean Delinez Nov 1 '10 at 20:35
    
What do you mean by "$\mathcal E_p$ is the line corresponding to the integer $p$"? Do you mean line bundle? In that case $\mathcal E_{-p}\otimes \mathcal E_{p}$ is the trivial line bundle... –  Sándor Kovács Nov 5 '10 at 17:23
    
I do mean the line bundle, and my $\otimes$ should, of course, be a $\oplus$. I have corrected it now. –  Jean Delinez Nov 5 '10 at 19:08

2 Answers 2

up vote 3 down vote accepted

ad 1.): No, the answer is that $\Omega^{1,0} CP^N$ corresponds to $N+1$. Proof sketch: As complex vector bundles, $TCP^N \oplus C= C^{N+1} \otimes L$, $L$ the tautological line bundle (the one with holomorphic sections). Thus $L^{\otimes (N+1)} = \Lambda^{N+1} (TCP^N \oplus C) = \Lambda^N TCP^N$. Dualizing gives the answer. For a more polished reasoning and more details, consult Griffith-Harris.

ad 2.) I am not sure whether your interpretation of $S_k$ is correct; Friedrich writes (bottom of page 77) that $S_0$ is the determinant line bundle and $S_N$ is trivial.

Here is how I would figure out the answer, switching perspective on $spin^c$ structures slightly (I do not have a handy reference for the following, I followed hints I found in Weyl's "The classical groups").

Let $V \to X$ be an oriented vector bundle of rank $2n$. A spin$^c$ structure is a bundle of complex $Cl(V)$-modules that is (fibrewise) irreducible. Given a line bundle $L$ on $X$, you can tensor a $CL(V)$-module with $L$ and obtain a new module. This gives an action of the group of line bundles on the set of spin^c-structures.

Note that since $2n$ is even, there is a unique irreducible representation of $Cl(R^{2n})$. Moreover, $Cl_{\mathbb{C}}(R^{2n})$ is the matrix algebra $\mathbb{C}(2^n)$, so the irrep has to be $2^n$-dimensional. However, this module is determined only up to isomorphism, which is responsible for the fact that not any vector bundle is spin^c.

Now assume that a complex structure has been given to $V$. Then the exterior algebra $\Lambda_{\bC} V$, together with the Clifford action defined by the formula $c(v)=v \wedge - \iota_v$ is a Clifford module. It has to be irreducible for dimension reasons!

This shows that any complex vector bundle is spin^c and gives an explicit desciption of the spin^c structures. Note that the complex structure was used here essentially, to write the exterior algebra $\lambda_{\bR} V$ (which is a Clifford module, but not irreducible) as a tensor product of $\Lambda_{\mathbb{C}}(V)$ with its conjugate. The result is pretty close to what you wrote and identical if you stick to Friedrichs notation. Now the clifford algebra is graded $Cl(V)_{\mathbb{C}}=Cl_{\mathbb{C}}(V)^0 _{\mathbb{C}}(V)^1$ and if you trace back the definition of the action, you see that $Cl_{\mathbb{C}}(V)^0$ preserves the even/odd-decomposition of the exterior algebra. This gives the grading of the spinor bundle, more or less as you suspected.

As I said before, spin^c structures can be twisted by line bundles (tensor products). The only line bundle available on a complex manifold is the determinant bundle of the tangent bundle (and its powers). By choosing different powers, you can switch between the anticanonical/canonical spin^c structures.

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Sorry, I meant to say spin$^c$ structure (I've changed it). You're right, $CP^N$ dosen't have a spin structure when $N$ is odd but it does have a spin^c structure. In this case (according to Friedrich anyway) the bundle $\Omega^{(k,0)}$ has a square root. Since it's a line bundle I would have thought this meant it would have had to have had an even Chern number. –  Jean Delinez Nov 1 '10 at 20:46
    
OK, then I deleted this part of the answer. –  Johannes Ebert Nov 1 '10 at 22:07
    
If you wish, I'll write more about why this notion of spin^c is the same as the one you use tomorrow –  Johannes Ebert Nov 10 '10 at 0:56

Here is a representation theoretical method for the computation of the (infinitesimal character of) the canonical bundle of any compact homogeneous Kahler manifold which include Grassmanians and complex projective spaces as particular cases. The computation for the complex projective space will be given explicitely. These spaces are coset spaces of a compact Lie group by an isotropy group which is a centralizer of a torus. The important step is to identify the generators of the torus in the Dynkin diagram of the compact Lie group. In the case of the $CP^N$, the torus is one dimensional and its generator corresponds to the simple root $\aplha_1$ or $\alpha_N$ of $A_N$. The other roots correspond to Grassmannians.

A nice explanation of this correspondence is given by John Baez here.

This method is sometimes called painted Dynkin diagrams is decribed in the following references by: by Alexeevsky and Bordemann, Forger and Romer).

Now the infinitesimal character corresponding to the canonical bundle is the sum of all roots of $A_N$ containing the generator of the torus which will be taken here as $\alpha_1$. In $A_N$ the roots are sums of continuous strings of simple roots, thus the roots containing $\alpha_1$ are: $\alpha_1, \alpha_1+\alpha_2, \alpha_1+\alpha_2+\alpha_3,. . .$, Their sum is therefore $\lambda = N\alpha_1+(N-1)\alpha_2+(N-2)\alpha_3,. . .$. The infinitesimal characters of the basic line bundles correspond to the fundamental weights, thus we need to transform the result to the weight basis. Using: $\alpha_1 = 2 w_1 - w_2, \alpha_i = -w_{i-1} + 2w_{i} + w_{i+1}$, for $i = 2, ..., N-1, \alpha_N = -w_{N-1}+ 2w_{N}$, we get: $\lambda = (N+1) \w_1$. The generalization to cases where the torus is multidimensional such as flag manifolds is obviousperformed by just duplication of the above procedure for evey torus generator

2) When N+1 is odd, tyen $CP^N$ does not admit a spin structure, however, one can tensor with the square root of a "line bundle" of odd infinitesimal character to obtain a $spin^c$ structure. Of course, neither the "square root of the canonical bundle" nor the square root of the line bundle exist separately, only their product is a line bundle over $CP^N$.

3) If we choose to work with a spin connection induced from the Levi-Civita connection, then the Dirac oprtaor '$D: \wedge(T^{0,1}) \otimes S\rightarrow \wedge(T^{0,1}) \otimes S$ is just the sum of the twisted Dolbeault operator and its adjoint as given in page 82 of Friedrich's book.

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