Let $G$ be a connected graph and let $e$ be an edge in this graph. I would like to know if there are necessary and sufficient questions so that $e^{\vee}=0$ in $H^1(G)$? The question must be easy to experts but I did not find an explicit answer anywhere in literature and am not very much familiar with graph theory.
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2$\begingroup$ Is this equivalent to the edge being a bridge? $\endgroup$– Sam HopkinsCommented Dec 2, 2023 at 12:42
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$\begingroup$ @Sam Hopkins. I don't know. What do you mean by a bridge? $\endgroup$– divergentCommented Dec 2, 2023 at 13:55
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$\begingroup$ en.m.wikipedia.org/wiki/Bridge_(graph_theory) $\endgroup$– Phil TostesonCommented Dec 2, 2023 at 14:00
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1 Answer
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Sam Hopkins's suggestion is right, $e^{\vee}$ is trivial if and only if $e$ is a bridge. (I'm assuming that you are taking $e^{\vee}$ to be an element in the simplicial cochain complex, so technically its sign depends on a choice of orientation of $e$).
Here is one argument. We have that $e^{\vee} = 0$ if and only if it pairs to zero with every cycle $z \in H_1(G)$. If $e$ is a bridge, then $G-e$ is a union of two connected components $G_1, G_2$, and $H_1(G)$ is spanned by cycles contained in either $G_1$ or $G_2$. So $e$ pairs trivially with every element of $H_1(G)$. Conversely, if $e$ is not a bridge, then $G-e$ is connected. Choosing a cycle that crosses $e$ and then returns to its starting point through a path in $G-e$, we find an element $z \in H_1(G)$ with $e^{\vee}(z) = 1$.
answered Dec 2, 2023 at 17:47
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$\begingroup$ @ Phil Tosteson. Thank you very much for your nice answer. I have another question: How about the sum of two co-edges? When do we have $e^{\vee}_1+e^{\vee}_2=0$ in $H^1(G)$? Thank you very much! $\endgroup$ Commented Dec 2, 2023 at 18:01