Consider the following question:
Let $X$ be a simply connected, symplectic 4-manifold. Does there exists a smoothly embedded Klein bottle $K\subset X$ such that the following conditions are both satisfied?


(1) The self intersection number of $K$ is 0 (which means that the normal bundle has a non-where vanishing section.)


(2) $[K]\neq 0\in H\_{2}(X;Z\_{2})$.


I am not aware of any kind of adjunction inequalities ruling out this possibility. But I also don't have a example of such embedded Klein bottle. 



Essential Klein bottle in simply connected symplectic 4 manifolds
Let $X$ be the algebraic variety over $\mathbb{R}$ obtained from the projective plain $\mathbb{P}^2$ by blowing up one point $P \in \mathbb{P}^2(\mathbb{R})$. Since $X$ is a smooth projective algebraic variety the space of complex points $X(\mathbb{C})$ admits a Kahler structure and hence the underlying $4$-dimensional real manifold admits a symplectic structure. Furthermore, $X(\mathbb{C}) \cong \mathbb{CP^2} \sharp \mathbb{CP}^2$ (**Correction**: $X(\mathbb{C}) \cong \mathbb{CP^2} \sharp \overline{\mathbb{CP}^2}$) is simply connected. Now the space $K = X(\mathbb{R})$ of real points can be identified with $\mathbb{RP}^2 \sharp \mathbb{RP}^2$, and is hence isomorphic to the Klein bottle. Let us check that $K$ satisfies Properties (1) and (2) of the question. Let $F$ be the restriction of the complex vector bundle $TX$ to $K$. Then the tangent bundle $TK$ can be identified with the real sub-bundle of $F$ and we have a decomposition $F \cong TK \oplus iTK$, identifying the normal bundle of $K$ with the imaginary sub-bundle of $F$. Since $TK$ admits a nowhere vanishing section it follows that $iTK$ admits a nowhere vanishing section. This shows (1). For (2), let $f: X \longrightarrow \mathbb{P}^2$ be the natural map. Then it will be enough to show that $f\_\*[K] \in H\_2(\mathbb{P}^2(\mathbb{C}),\mathbb{Z}/2)$ is non-trivial. Now note that the image of $K$ in $\mathbb{P}^2(\mathbb{C}) = \mathbb{CP}^2$ is given by $\mathbb{P}^2(\mathbb{R}) = \mathbb{RP}^2$ and that the corresponding map $K \longrightarrow \mathbb{RP}^2$ induces an isomorphism on $H\_2(-,\mathbb{Z}/2)$. It will now be enough to show that $[\mathbb{RP}^2] \in H\_2(\mathbb{CP}^2,\mathbb{Z}/2)$ is non-trivial. To see this, observe that if $\mathbb{CP}^1 \cong L \subseteq \mathbb{CP}^2$ is any complex line that is not defined over $\mathbb{R}$ then $L$ meets $\mathbb{RP}^2$ transversely at exactly one point (the intersection point of $L$ and $\overline{L}$), and that $[L] \in H\_2(\mathbb{CP}^2,\mathbb{Z}/2) \cong \mathbb{Z}/2$ is a generator.


