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Dec 16, 2019 at 11:18 vote accept Jens Reinhold
Nov 19, 2019 at 18:39 comment added David E Speyer Basic observations: As in Neil Strickland's answer, I'll work in homology rather than cohomology and use his terms "basically toral" and "toral". We have a map $\lambda : H_n(X) \to \bigwedge^n H_1(X)$, dual to the cup product. If $\alpha \in H_n(X)$ is basically toral, then $\lambda(\alpha)$ is an elementary tensor so, if $\alpha$ is toral, then $\lambda(\alpha)$ is in the span of elementary tensors of $\lambda(H_n(X))$. In particular, if $\lambda$ is injective and its image is disjoint from the elementary tensors, there are no toral classes; this handles the case of curves of genus $\geq 2$.
Nov 19, 2019 at 18:04 history became hot network question
Nov 19, 2019 at 12:19 answer added Neil Strickland timeline score: 17
Nov 19, 2019 at 10:18 comment added Jens Reinhold Ah, thanks, I got this wrong. I have edited the question accordingly.
Nov 19, 2019 at 10:18 history edited Jens Reinhold CC BY-SA 4.0
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Nov 19, 2019 at 10:10 comment added Neil Strickland Using the structure of $H^*(T^n)$, it seems to be equivalent to say that $\alpha\in T^*(X)$ iff there is a map $f\colon T^n\to X$ for some $n$ with $f^*(\alpha)\neq 0$. This means that $T^*(X)$ is the complement of an ideal $U^*(X)\leq H^*(X)$, not a subring.
Nov 19, 2019 at 10:02 history edited Jens Reinhold CC BY-SA 4.0
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Nov 19, 2019 at 9:55 history asked Jens Reinhold CC BY-SA 4.0