Consider $\Bbb R^4$ with standard symplectic form $dx_1\wedge dy_1+dx_2\wedge dy_2$. This determines an orientation of $\Bbb R^4$ so that $(\partial_{x_1},\partial_{y_1},\partial_{x_2},\partial_{y_2})$ is a positive basis, and we can define intersection products of two 2-dimensional submanifolds intersecting transversally. How can we find two symplectically-embedded 2-planes that intersect negatively? (This question https://mathoverflow.net/questions/50474/negative-intersection-of-symplectic-submanifolds in overflow is related, but doesn't give an answer.)
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This doesn't appear to be possible. Naively, don't we expect that if the planes are transverse their intersection is $+1$, and if they coincide their intersection is $0$? The reason we get compact examples (e.g., the exceptional divisor of the blow-up of $\mathbf{C}^{2}$ at the origin) has to do with global structure. – Andrew D. Hwang Jun 14 '21 at 21:40