TITLE: Which paths in a graph are orthogonal to all cycles?
QUESTION [22 upvotes]: 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.]

REPLY [5 votes]: This answer is an attempt to slightly rephrase Ilya's argument; I would have written it as a comment, but there's not enough room.
Given a path $\gamma$ with endpoints $v_0$ and $v_n$, we have a 1-chain $c(\gamma)$, which is an integer-linear combination of edges of the graph.  Let $E_{c(\gamma)}$ be the set of edges with non-zero coefficients in $c(\gamma)$; this need not include every edge that appears in the path, since some might cancel to zero in the 1-chain.  If we think of these edges as decorated with the integer coefficients they inherit from the 1-chain $c(\gamma)$, we can associate a 1-chain with any subset of $E_{c(\gamma)}$.
It's possible that $E_{c(\gamma)}$ as a subgraph of $\Gamma$ contains several connected components that share no vertices with each other, but since the boundary of $c(\gamma)$ is only non-zero on $v_0$ and $v_n$, and it's impossible for a 1-chain to have a single-point boundary, these components would need to include exactly one whose 1-chain's boundary was non-zero on $v_0$ and $v_n$, and all the rest would have to give 1-cycles.  Since they're all disjoint, any one of the 1-cycles $\chi$ would satisfy $\langle c(\gamma), \chi \rangle = \langle \chi, \chi \rangle \gt 0$.
Assuming now that $E_{c(\gamma)}$ is connected as a subgraph, we can build up a 1-chain $\sigma$ with a positive inner product with $c(\gamma)$ as follows.  Starting at $v_0$, pick an edge $\epsilon_1$ incident on $v_0$, and put $\pm \epsilon_1$ in $\sigma$, with the sign chosen to be the same as the coefficient of $\epsilon_1$ in $c(\gamma)$.  Then advance to the other vertex of $\epsilon_1$, and choose an edge $\epsilon_2$ such that $\epsilon_2 \ne \epsilon_1$, and $\pm \epsilon_2$ with the same sign as the coefficient of $\epsilon_2$ in $c(\gamma)$ gives a boundary for $\pm \epsilon_1 \pm \epsilon_2$ that is zero at the current vertex.  This must be possible (assuming we haven't ended up at an endpoint of the path), since the boundary of $c(\gamma)$ is zero at the current vertex, so the signs of the edge coefficients in $c(\gamma)$ can't all give boundaries of the same sign here.
We continue this process until we reach either $v_n$, the endpoint of the path, or a vertex we've visited before.  If we reach a vertex we've visited before, we can drop any earlier edges from $\sigma$ and obtain a 1-cycle with a positive inner product with $c(\gamma)$.
If we reach $v_n$, there are two possibilities.  If $\sigma = c(\gamma)$ then $\sigma$ describes a simple path $\gamma'$ from $v_0$ to $v_n$ with the same 1-chain as our original path, and we can proceed to use that simple path in place of $\gamma$.  If $\sigma \ne c(\gamma)$, it nonethless has the same boundary.  Since $\sigma$ is supported on a subset of the same edges as $c(\gamma)$, and its coefficients are of the same sign but never greater in magnitude (and must be less on at least one edge), $\langle c(\gamma), \sigma \rangle$ will be strictly less than $\langle c(\gamma), c(\gamma) \rangle$.  So we'll have a non-trivial 1-cycle $\ell = c(\gamma) - \sigma$, and:
$$\langle c(\gamma), \ell \rangle  =  \langle c(\gamma), c(\gamma) \rangle - \langle c(\gamma), \sigma \rangle \gt 0$$
Finally, suppose we have a simple path $\gamma': v_0 \to v_1 \to \dots \to v_n$ with the same 1-chain as $\gamma$.  (The following refinement of Ilya's argument is something that John described to me in correspondence.)  Since the edge $e_n: v_{n-1} \to v_n$ cannot be a bridge, there must be a path joining $v_n$ to $v_0$ that does not include that $e_n$.  If we follow that path only as far as the first of the $v_i$ it reaches, we will have a path $\rho: v_n \to v_i$ which uses no edges of the simple path $\gamma'$.  We can then append the portion of $\gamma'$ that goes from $v_i$ to $v_n$, call it $\gamma'_i$, to obtain a 1-cycle $c(\rho) + c(\gamma'_i)$ that must have a positive inner product with $c(\gamma)$:
$$\langle c(\gamma), c(\rho) + c(\gamma'_i) \rangle = \langle c(\gamma'_i), c(\gamma'_i) \rangle \gt 0$$