I have a few questions these two slides on the topic of calculus on graphs:
What are the vertex fields defined here? My understanding is that it is a set of functions that takes in a vertex and gives a real number output. And because each vertex may need to undergo different transformation, each vertex $v_i$ has its corresponding function $f_i$. Is that right?
What does the inner product here means?
Why is there a square root of weight in gradient and divergence operator? Is it necessary? My understanding is that multiplying by weight and not square root of weight is sufficient.
What is $F$ in divergence operator?
These are all the slides that I have and I am having a lot of trouble understanding it. Is it that it is badly written? If not, can someone kindly explain to me please? Thanks.
A vertex field is a square-summable function from the set of vertices into $\mathbb{R}$. (If there are only finitely many vertices, saying "square-summable" is unnecessary.) Imagine a graph, say the triangle $K_3$. Put a number next to each vertex, say $3, 6, -2$. You have a vertex field.
The concept of inner product is explained on Wikipedia. Here the inner product of two vertex field $f,g$ means: multiply each value of $f$ by the corresponding value of $g$ and by the weight of that vertex. Add the results.
The reason for having a square root in $\sqrt{w_{ij}}$ will become apparent on a later slide, where the graph Laplacian is defined as the divergence of gradient. Since both the gradient and the divergence involve multiplying by $\sqrt{w_{ij}}$, the Laplacian will have $w_{ij}$. The author would rather have a simpler formula for Laplacian, because it will be used often in the future.
$F$ is a square-summable function defined on the edges. (If there are only finitely many edges, saying "square-summable" is unnecessary.) This is what the first line of definition "$\operatorname{div}:L^2(\mathcal E)\to L^2 (\mathcal V)$" is for, to state what are the domain and codomain of this map.