I am reading a paper where they say the following is "easy" but I can't seem to see why.
Let $G$ be a finite undirected graph on an edge set $V$ and let $E$ be its set of oriented edges (i.e. each edge of $G$ appears twice in $E$; once in each direction). We call a function $f:E\to\mathbb{R}$ a flow on $G$ if $f(vw)=-f(wv)\forall v,w\in V$
For each edge $e=vw$ let $\chi^{e}:E\to\mathbb{R}$ by $\chi^{e}:=1_{vw}-1_{wv}$. $\chi^{e}$ is called the unit flow on $e$. Define now: \begin{eqnarray*} \mathbf{H} & = & \left\{ f:E\to\mathbb{R}:f(vw)=-f(wv)\forall v,w\right\} ,\text{ the set of all flows}\\ \bigstar & = & span\left\{ \sum_{w}\chi^{vw}:v\in V\right\} ,\text{ the span of "source" flows}\\ \Diamond & = & span\left\{ \sum_{i=1}^{n}\chi^{e}:e_{1},\ldots,e_{n}\text{ is an oriented cycle}\right\} ,\text{ the span of "cycle" flows} \end{eqnarray*}
Show that $\mathbf{H}=\bigstar\oplus\Diamond$
I can see that if $G$ is a tree, then $\bigstar=\mathbf{H}$. From here I think there might be a way to see the result by looking at a spanning tree of $G$, but I can't seem to get this to work and I think there is probably an easier way to see the result.
EDIT: I have been thinking about this a bit more, and now I suspect there might be a way to do it using Kirchoff's circuit laws someohow where you think of each edge as a resistor or something. For this reason, I'm tagging probability and mathematical physics where these comes up. (Probability is natural because this appears in a discussion about random uniform spanning trees of the graph too)
I found the answer to this in the book "Probability on Trees and Networks" by Russel Lyons and Yuval Peres in Chapter 2 in the Section 2.4 on energy. The proof does indeed come "easily" after a bit of overhead with some very useful definitions and results made in this book.