# Clifford Action for Kahler Manifolds

I'm working with Kahler manifolds at the moment and looking at their spin$^c$ structure. I really don't know much about spin$^c$ structures in general and don't have enough time to learn it all at present, so I was hoping someone might be able to show me a shortcut in this case.

As far as I understand, for the usual spin$^c$ structure on an $N$-dimensional Kahler manifold $M$, the spinor bundle $S$ is given by the the direct sum of all the anti-holomorphic, that is $S = \bigoplus_{i=1}^N \Omega^{(0,i)}(M)$. Let $$\nabla^s:S \to S \otimes \Omega^{(1,1)}, ~~~~~~~~~ s \mapsto \sum s_i \otimes \omega_i$$ be the spin$^c$ connection.

Is there a simple direct algebraic description of the Clifford action $$c:S \otimes \Omega^{(1,1)} \to S$$ in this case? For example, an initial stupid guess might be, for $\omega_i = \omega_i^h + \omega_i^{ah}$, with $\omega_i^{h} \in \Omega^{(1,0)}$ and $\omega_i^{ah} \in \Omega^{(0,1)}$, we would have $$c(\sum_i s_i \otimes \omega_i) = \sum_i s_i\omega^{ah}_i.$$ I am quite sure this is complete rubbish, but it illustrates the kind of result-from-heaven I'm hoping exists

-
I think you probably mean Spin-c structure, not Spin structure. The complex projective plane has $w_2\neq 0$. – Tim Perutz Feb 5 '11 at 20:02
Yes, of course. I've edited accordingly. – Ago Szekeres Feb 6 '11 at 17:12
@Ago: I think your summation $\oplus_{i=1}^N$ should read $\oplus_{i=0}^N$. See, for example Friedrich's book. – Jean Delinez Feb 11 '11 at 18:22

You ask for a short-cut, and I'm going to interpret this as asking "How can I see that on a Kaehler manifold $X$ the forms $\Omega^{0,\bullet}_X$ are a complex Clifford module, without first going through all that stuff about spin groups and their representations?"

Warning: My scalar factors are probably wrong.

I'm going to start with the Hodge Laplacian $$\Delta = 2(\overline{\partial}\;\overline{\partial}^\ast+\overline{\partial}^\ast\overline{\partial})$$ acting on $S:=\Omega^{0,\bullet}_X$.

What makes $\Delta$ a Laplacian is that it's a second order differential operator whose symbol $\sigma \colon T^\ast X \to \mathrm{End}(S)$ is the quadratic map $\lambda \mapsto - \| \lambda\|^2 \mathrm{Id}_S$. (The symbol is just the coefficient matrix of the leading (i.e. second) derivative terms in $\Delta$, written invariantly.) The symbol of $\overline{\partial}$ is $\lambda\mapsto \lambda^{0,1}\wedge \cdot$, that of $\overline{\partial}^\ast$ is $\lambda\mapsto \iota(g(\lambda))$ where $g\colon T^\ast X \to TX$ is the metric tensor, and from this you can check the symbol of $\Delta$.

$\Delta$ has an evident square root, $D=\sqrt{2}(\overline{\partial}+\overline{\partial}^\ast)$, and this is a Dirac operator: a first order differential operator whose square is a Laplacian. The (complexified) symbol $\rho\colon T^*X\otimes \mathbb{C}\to \mathrm{End}(S)$ of a Dirac operator $D\colon \Gamma(S)\to \Gamma(S)$ makes $S$ into a Clifford module. The natural Clifford multiplication on $(0,\bullet)$-forms is the symbol of $D$, that is $$\rho(\lambda) = \sqrt{2}(\iota(g(\lambda))+\lambda^{0,1}\wedge \cdot ) \in \mathrm{End}(S).$$ In fact, this Clifford module is a $Spin^c$-structure, which means that pointwise on $X$ it's a isomorphic to a standard Clifford module.

-
Great, this seems to be what I'm looking for. Just some questions about notation though: What do you mean by $\lambda^{(1,0)}$? Also, I assume that by $g(\lambda)$ you mean $g(\lambda, \cdot )$, but what does $\iota$ stand for? – Ago Szekeres Feb 7 '11 at 19:25
Also, what happens when we tensor $S$ with the square root of the canonical bundle $\Omega^{(N,0)}(X)$. And in this case is the spin connection the Levi-Civita connection? – Ago Szekeres Feb 7 '11 at 19:34
By $\lambda^{0,1}$ I mean the $(0,1)$-component of $\lambda\in T^\ast X$ under the decomposition $T^\ast X\otimes \mathbb{C}=T^\ast_{1,0}X \oplus T^\ast_{0,1}X$ into $dz$'s and $d\overline{z}$'s. By $\iota(v)$ I mean the contraction of forms by the vector field $v$. – Tim Perutz Feb 7 '11 at 19:59
Other spin-c structures differ from the canonical one $s_{can}$ by tensoring with line bundles (the Clifford multiplication is given by the same formula tensored with the identity map on the line bundle). And if we have a square root $L$ of $K_X$ then $L\otimes s_{can}$ arises from a spin structure. Not quite sure what you're asking about spin connections... Bear in mind that there's not a unique spin connection in the spinor bundle of a spin-c structure. – Tim Perutz Feb 8 '11 at 0:46
Thanks a lot for your help. Just one last question: So there's no unique spin connection, but will the Levi-Civita connection a spin connection? – Ago Szekeres Feb 8 '11 at 14:45