On a smooth projective complex surface, if $C$ and $C'$ are distinct irreducible curves then their intersection is non-negative, $C \cdot C' \geq 0$. I am interested in cases where negative intersections occur because we drop either distinctness or irreducibility.

One side of this I am reasonably familiar with: it is common to consider self-intersections $C^2$ of irreducible curves. Non-negative self-intersections $C^2 \geq 0 $ still often occur: in these cases the curve can be deformed into another in the same linear equivalence class, and one takes the transverse intersection of this deformed curve with the original. In cases where $C^2 < 0$ the geometric picture is that no such deformation is possible.

However if instead of dropping distinctness we drop irreducibility but keep distinctness, one again finds examples of negative intersections, $C \cdot C ' < 0 $. I am wondering what is the geometric picture in this case?

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    $\begingroup$ Since intersection numbers are additive over irreducible components, the only way this can happen is if $C$ and $C'$ have an irreducible component $\Gamma$ in common, and $\Gamma^2$ is negative. $\endgroup$ – Art Mar 22 at 19:25
  • $\begingroup$ @Art Thanks for your comment. What you say is indeed true, but I perhaps should have stated in the question that this fact does not seem to me to give a geometric picture. It is a statement really about curve classes, but I think a geometric picture would deal in actual curves (the representatives of the classes). $\endgroup$ – diracula Mar 22 at 19:40

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