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I have a few questions these two slides on the topic of calculus on graphs:

  1. 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?

  2. What does the inner product here means?

  3. 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.

  4. 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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2 Answers 2

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  1. 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.

  2. 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.

  3. 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.

  4. $F$ 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.

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  • $\begingroup$ why must vertex field and F be square-summable function? $\endgroup$
    – user10024395
    Nov 12, 2017 at 14:54
  • $\begingroup$ with regards to 4, isn't the domain for gradient operator an edge and the codomain a real number? $\endgroup$
    – user10024395
    Nov 12, 2017 at 14:58
  • $\begingroup$ (a) because the slides tell you so. $L^2$ means square summable. The person who defines these concepts gets to make those calls. (b) No. The slides specify the domain of gradient operator as $L^2(\mathcal V)$ and the codomain as $L^2(\mathcal E)$. The domain consists of square-summable functions on vertices, the codomain is square-summable functions on edges. $\endgroup$
    – user357151
    Nov 12, 2017 at 15:52
  • $\begingroup$ The form of the gradient operator clearly shows that the result is a real number and since this function takes in an edge and spits out a real number, how is it true that the domain and codomain are $L^2(V)$ and $L^2(E)$?. The same applies to divergence. The domain and codomain defined doesnt seem right. $\endgroup$
    – user10024395
    Nov 13, 2017 at 2:09
  • $\begingroup$ For (3), how is Laplacian related to gradient and divergence? I understand the relationship in vector calculus, but the author's definition in this slides, they don't seem to be related $\endgroup$
    – user10024395
    Nov 13, 2017 at 2:12
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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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