I'm interested in $c_1^2(L)$ where $L$ is the determinant line bundle associated to a $Spin^c$ structure on a compact oriented four-manifold $X^4$.

Specifically I'd like to know the numerical value of this characteristic number for some simple examples; let's say $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}$ and $\mathbb{CP}^2\#\overline{\mathbb{CP}^2}\#\overline{\mathbb{CP}^2}$. I've seen this number show up while reading about Seiberg-Witten theory but I can never find any sources that give it's value on an actual manifold. Any reference to a source which has calculated some of this would be excellent.

As an alternative or add-on, what methods are/would be used to calculate such a thing? My knowledge about characteristic classes is pretty weak. (Speaking of which, one misc. question: Is this class sensitive to orientation? i.e. does it change sign if one changes the orientation of $X$?)

Thank you for any information.


Last question first. Recall that a $\text{Spin}^c$-structure on a 4-manifold is a lift of the structure group of the tangent bundle from $O(4)$ to $\text{Spin}^c(4)$. The map $\text{Spin}^c(4) \to O(4)$ factors through $SO(4)$, so a $\text{Spin}^c$-structure comes equipped with an orientation. When we talk about $\text{Spin}^c$ structures, we're usually doing so on manifolds that are oriented in the first place, in a way compatible with the $\text{Spin}^c$-structure; if these two orientations disagreed, there's no reason to believe stuff coming from the orientation (eg the Hodge star) would play well with stuff coming from the $\text{Spin}^c$-structure (eg Clifford algebras).

Onto your actual question. Any Kahler manifold carries a $\text{Spin}^c$-structure; often this is defined in terms of spinor bundles (eg in Morgan's book), but I'll keep with the language of structure groups above. Because in 4D $\text{Spin}^c(4) = (SU(2) \times SU(2)) \times_{\Bbb Z/2} U(1)$, writing an element of $U(2)$ as $\lambda g$, where $\lambda \in U(1)$ and $g \in SU(2)$, we have a homomorphism $f: U(2) \to \text{Spin}^c(4)$ given by $f(\lambda g) = (g, \lambda, \lambda)$. You can verify that this is well-defined. The map $\text{Spin}^c(4) \to SO(4)$ is given by $(g,h,\lambda) \mapsto \pm(g,h) \in (SU(2) \times SU(2))/\pm I = SO(4)$. This means that our maps $U(2) \to \text{Spin}^c(4) \to SO(4)$ are compatible, the composition being the natural homomorphism $U(2) \to SO(4)$; so given a Kahler structure, we inherit a $\text{Spin}^c$-structure.

The map $U(2) \to \text{Spin}^c(4)$ is compatible with the determinant map (of both) to $U(1)$. This tells me that the determinant line bundle of the Kahler manifold is the same as the determinant line bundle of the $\text{Spin}^c$-structure, and hence they have the same Chern class. So it suffices to see what $c_1(L)^2$ is for the determinant line bundle of the complex structure; but it is a general fact of complex vector bundles that $c_1(E) = c_1(\Lambda^{top}(E))$, so that $c_1(L) = c_1(TM)$ here.

To calculate it, I'm just going to invoke some theorems. First, we have the relationship $c_1(E)^2 - 2c_2(E) = p_1(E)$ for any complex vector bundle $E$; second, we have the fact that for a 4-manifold, $p_1(M) = 3\sigma(M)$; and last, for a complex manifold of complex dimension $n$, $c_2(M) = \chi(M)$. Putting this all together, we get $c_1(E)^2 = 2\chi(M) + 3\sigma(M)$. If you know the index formula for the linearized Seiberg-Witten operator, you'll realize this gives us an exciting fact: the Seiberg-Witten moduli space has expected dimension zero on a Kahler manifold!

If I twisted the Spin-c structure from the "standard one", I would get $c_1(\mathfrak s \otimes L) = c_1(\mathfrak s) + 2c_1(L)$ (the point being that the determinant map is really a $z \mapsto z^2$-thing, because of that $_{\Bbb Z/2}$ in the definition of $\text{Spin}^c$-structure), so we can in principle then calculate $c_1^2$ for every $\text{Spin}^c$ structure on a Kahler manifold.

As for your specific examples: $\chi(\Bbb{CP}^2 \# n \overline{\Bbb{CP}^2}) = 3+n$, and $\sigma(\Bbb{CP}^2 \# n \overline{\Bbb{CP}^2}) = 1-n$, so we get $c_1(\Bbb{CP}^2 \# n \overline{\Bbb{CP}^2})^2 = 9-n$ from the previous computation.

  • $\begingroup$ Thank you very much for your answer. I am quite ignorant of complex/Kähler geometry. Is it the case that $\mathbb{CP}^2\#n\overline{\mathbb{CP}^2}$ is Kähler for every n? What happens if we consider $k\mathbb{CP}^2\#l\overline{\mathbb{CP}^2}$ for arbitrary $k$ and $l$? These aren't all Kähler manifolds, are they? $\endgroup$ – Brian Klatt Aug 2 '16 at 17:26
  • $\begingroup$ @BrianKlatt Yeah, they're Kahler manifolds for $k=1$ - you took the Kahler manifold $\Bbb{CP}^2$ and "blew it up" at n points, which gives you another Kahler manifold. With $k>1$, those are never Kahler (or even symplectic) manifolds, though that's actually a theorem of Seiberg-Witten theory. Actually, everything in this answer is valid for almost complex manifolds, with the same results, though you have to be a little more careful with your phrasing. It's true that if $k$ is odd, your manifolds admit almost complex structures, which then have naturally associated spin-c structures. $\endgroup$ – user98602 Aug 2 '16 at 17:30
  • $\begingroup$ So if $k$ is odd, is the basic point that $c_1^2=2\chi + 3\sigma$ still holds? $\endgroup$ – Brian Klatt Aug 2 '16 at 17:35
  • $\begingroup$ There's a more fundamental point to be made - manifolds do not come equipped with a $\text{Spin}^c$ structure; you have to give them one. If you equip your manifold with an almost complex structure, then the induced $\text{Spin}^c$ structure indeed satisfied the relation you gave. If you don't, it doesn't make sense to ask for a calculation of $c_1(\mathfrak s)^2$. In the above cases, where you say "$c_1$ of..." etc, I assume you're equipping these manifolds with the $\text{Spin}^c$ structures coming from their (almost) complex structures. $\endgroup$ – user98602 Aug 2 '16 at 17:38
  • $\begingroup$ Ah, thanks for pointing that out. I think I was assuming there was a canonical choice if the manifold was simply-connected (as in the spin case). $\endgroup$ – Brian Klatt Aug 2 '16 at 17:44

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