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.
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$\begingroup$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.
I think @user357151 basically answered correctly all your questions, and you should probably accept this one.
However, regarding the slides, while trying to implement the divergence operator numerically, I noticed that there may be an error (or I did an error implementing it) because the way div was written defines the exact conjugate of the gradient operator, instead of the opposite of the conjugate of the gradient. In order to verify $grad^\star = -div $ I prefer to use the following notation:
$(div F)_{i} = \sum_{(k,l) \in \mathcal{E}} \sqrt{a_{kl}}(\delta_{i}(l)-\delta_{i}(k)) F_{kl}$
Please correct me if I am wrong
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