# homologically trivial $1$-cycles and surfaces

Let $X$ be a smooth (complex) threefold and $\gamma\in {\rm CH}_1(X)$ a homologically trivial $1$-cycle. Is there a way to construct a (singular) surface $S\subset X$ supporting $\gamma$ such that, denoting $\tau:\widetilde S\rightarrow S$ a desingularization, there is a homologically trivial $\widetilde\gamma\in Pic^0(\widetilde S)$ satisfaying $$i_*\tau_*\widetilde \gamma =\gamma \ \ {\rm in\ CH}_1(X)$$ where $i:S\hookrightarrow X$ is the inclusion?

• You mean $\gamma$ in $\mathrm{CH}^1(X)$ (codimension $1$)? Mar 12, 2018 at 12:26
• Thank you. No, I do mean "curves".
– pi_1
Mar 12, 2018 at 12:38
• If you take any surface containing the curves in the support of $\gamma$, and take a resolution of it, you get a class in $Pic(\widetilde S)$ that pushes forward to $\gamma$ like you say. Is your question whether we can take this to be homologically trivial?
– byu
Mar 12, 2018 at 14:19
• I suspect that this fails already for Hironaka's example, as described in Appendix B of Hartshorne's "Algebraic geometry". Mar 12, 2018 at 15:49
• For a projective example, simply take any cycle which is homologically equivalent to zero but not algebraically equivalent to zero. (On a surface, any 1-cycle which is homologically trivial is algebraically trivial.)
– naf
Mar 13, 2018 at 5:36