# When is a (co)edge trivial in graph cohomology?

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.

• Is this equivalent to the edge being a bridge? Dec 2, 2023 at 12:42
• @Sam Hopkins. I don't know. What do you mean by a bridge? Dec 2, 2023 at 13:55
• en.m.wikipedia.org/wiki/Bridge_(graph_theory) Dec 2, 2023 at 14:00

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 $$e^{\vee}(z) = 1$$.
• @ 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! Dec 2, 2023 at 18:01