Start with some standard stuff. Suppose we have a directed graph $\Gamma$. I'll write $e : v \to w \,$ when $e$ is an edge going from the vertex $v$ to the vertex $w$. We get a vector space of **0-chains** $C_0(\Gamma,\mathbb{R})$, which are formal linear combinations of vertices, and a vector space of **1-chains** $C_1(\Gamma,\mathbb{R})$, which are formal linear combinations of edges. We get a **boundary operator**

$$ \partial : C_1(\Gamma,\mathbb{Z}) \to C_0(\Gamma,\mathbb{Z}) $$

sending each edge $e: v \to w$ to the difference $w - v$. A **1-cycle** is a 1-chain $c$ with $\partial c = 0$. There is an inner product on 1-chains for which the edges form an orthonormal basis.

Any path in the graph gives a 1-chain. When is this 1-chain orthogonal to all 1-cycles?

To make this interesting, I need to rule out some obvious possibilities. If we have a graph consisting of two triangles connected by an edge, the path consisting of that one edge will be orthogonal to all 1-cycles:

To rule out this sort of situation, let's suppose $\Gamma$ has no **bridges**, meaning edges whose removal increases the number of connected components. The edge with the arrow on it in the picture above is a bridge.

**Question:** Suppose $\Gamma$ is a graph with no bridges. Any path in such an embedded graph gives a 1-chain. If this 1-chain is orthogonal to all 1-cycles, must it vanish?

To make this precise: I'm defining a **path** $\gamma$ to be a finite sequence of edges $e : v \to w$ and their formal 'inverses' $e^{-1}: w \to v$, like this:

$$ e_1^{\pm} : v_0 \to v_1, $$ $$ e_2^{\pm} : v_1 \to v_2 , $$ $$ \dots $$ $$ e_n^{\pm} : v_{n-1} \to v_n .$$

The corresponding chain is

$$ c(\gamma) = \pm e_1 \pm e_2 \pm \; \cdots \;\pm e_m. $$

**Question:** If $\Gamma$ is a graph with no bridges, and $\gamma$ is a path in $\Gamma$ such that the inner product of $c(\gamma)$ with every cycle vanishes, must we have $c(\gamma) = 0$?

I believe someone should have settled this by now, since it sounds easy, and the space of 1-chains orthogonal to all cycles has been studied quite a lot: it's called the **cut space** of the graph.

A **cut** is a partition of the vertices of a graph into two disjoint subsets. Any cut determines a **cut-set**, the set of edges that have one endpoint in each subset of the partition. If we take the sum of all those edges, we get a 1-chain orthogonal to all cycles. It's known that the cut space is spanned by 1-chains coming from cuts in this way. For example, in the graph I drew, the edge with the arrow on it spans the cut space.

A proof can be found here:

- Norman Biggs,
*Algebraic Graph Theory*.

but I suspect there's a lot more known about this subject!

[Note: I have edited my original question to simplify the hypotheses, and also called elements of $Z_1(\Gamma,\mathbb{R})$ **1-cycles**, to distinguish them from cycles in the sense of graph theory.]