(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.



Complex Projective Space Spin and Dirac: Part II
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.


