0
$\begingroup$

Let $X$ be the set of edges for an undirected connected graph with edge set $E(X)$ (So $e \mapsto e^{-1}$ is a bijection $E(X) \rightarrow E(X)$.)

A field on $X$ is a map $\phi : E(X) \rightarrow \Bbb R$ such that $\phi(e^{-1}) = - \phi(e)$. A function $f : X \rightarrow \Bbb R$ has a corresponding field $\Delta f(e):= f(e_+) - f(e_-)$ (if e_+ is the starting vertex of $e$ and $e_-$ the ending.) and a field $\phi$ has corresponding function $div \phi(x):= \sum_{e_- = x} \phi(e)$

There's a lemma in notes I'm reading that states:

$\frac{-1}{2} \sum_{e\in E(X)}\Delta f(e)\phi(e) = \sum_{x \in X} f(x)div \phi(x)$

We have

$\sum_{e \in E(X)} \Delta f(e) \phi(e) = \sum_{e} [f(e_+) - f(e_-)]\phi(e)=\sum_ef(e_+)\phi(e) + \sum_ef(e_-)\phi(e^{-1})$

but I don't see how to simplify anything else. Any help is appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

It's a bit weird that you write $ \Delta $ for that operator. I will write it as $ d $, $ df(e) = f(e_+) - f(e_-) $. Then we have: $$ \sum_e df(e) \phi(e) = \sum_e f(e_+) \phi(e) - \sum_e f(e_-) \phi(e)\\ = \sum_v \sum_{\substack{ e \\ e_+ = v }} f(e_+) \phi(e) - \sum_v \sum_{\substack{ e \\ e_+ = v }} f(e_-) \phi(e) \\ = \sum_v f(v) \sum_{\substack{ e \\ e_+ = v }} \phi(e) -\sum_v f(v) \sum_{\substack{ e \\ e_- = v }} \phi(e) \\ = \sum_v f(v) \; \text{div} \phi(v) -\sum_v f(v) (- \text{div} \phi(v) ) \\ = 2\sum_v f(v) \; \text{div} \phi(v) $$ Moving from the first to second line, I used the fact that every edge has a unique $ e_+ $. In moving from the third to fourth, I used the fact that $ \phi(e^{-1}) = - \phi(e) $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.