Let $X^4$ be a simply-connected closed smooth 4-manifold. Then every element $x \in H\_2(X; \mathbb{Z})$ can be represented by an embedded orientable surface and the minimal genus of such a surface is called the genus of $x$, denoted $g(x)$. An element $x \in H\_2(X; \mathbb{Z})$ is called characteristic if the reduction of $x$ to $H\_2(X; \mathbb{Z})$ is Poincar'{e} dual to $w\_2$.


Are there some methods for computing/lower-bounding the minimum genus amongst all characteristic homology classes in $H\_2(X; \mathbb{Z})$?


I saw in [this paper of Hamilton](https://arxiv.org/pdf/1301.3733.pdf) that the self-intersection number of characteristic homology classes of fixed genus can only be so small and my intuition is that the genus increases as the self intersection tends toward $+ \infty$ (and that this is even provable in some cases), so maybe that is somehow useful.



Minimal genus of characteristic surfaces?
If you want something specific to characteristic classes, the only thing I know you can leverage on is the fact that the complement of any surface representing a characteristic class is spin. Let's fix a 4-manifold $X$ and a characteristic class $x$.


The first result in this direction is probably the Kervaire-Milnor theorem: it says that if $x$ is represented by a sphere, then $x^2 \equiv \sigma(X) \pmod{16}$.


Hamilton (in the paper you link) uses the fact that the double cover of $X$ branched over a representative of $2x$ is spin, and then uses G-signature theorem to compute the signature and the 10/8-theorem to use the signature to bound b\_2. This in turn gives a lower bound on $g(x)$.


I guess that one could try to use the same tools that come into (some proofs of) the adjunction inequality and use the spin structure on the complement to say something more. Here I'm thinking of Pin(2)-equivariant Seiberg-Witten Floer homology (developed by Manolescu and then by Lin) and of involutive Floer homology (developed by Manolescu and Hendricks). I suspect that this strategy can only be (somewhat) effective if $b^+(X)$ is small.


