[petsc-dev] proposed minor PetscPartitioner changes

[Jed Brown]

Matthew Knepley <knepley at gmail.com> writes:

>> >> A graph Laplacian certainly does transform under coordinate
>> >> transformation and indeed, we use that property to design effective
>> >> coarsening strategies.  That one basis strikes you as intrinsically
>> >> "more canonical" does not mean it isn't a linear operator.
>> >>
>> >
>> > That is one operator. This is argument by anecdote. An arbitrary graph
>> > is not a linear operator, but an arbitrary matrix definitely is (the
>> > coordinate representation of one).
>>
>> Dude, we solve linear systems and eigenproblems for arbitrary graphs.
>> It isn't an anecdote.
>>
>> Barry (rightly) objects to a 2D array representing a function on a grid
>> being considered a Matrix.  We don't "apply" it as a linear operator.
>> There is no "vector" on which it operates.
>>
>> But we absolutely do with a graph.  Our vectors are functions at the
>> vertices of the graph.  Applying the graph Laplacian tells us about
>> local compatibility of the field over the vertices.  It is entirely
>> analogous to fields over a grid.  You don't need a concept of "grid
>> refinement" to have matrices.
>>
>
> I don't think makes sense. You are saying, because a linear operator (the
> graph Laplacian) can be defined using the graph, then the graph is identical
> with this operator. 

It's a bijection; given an operator (with zero row sums - Laplacian, or
ignoring the diagonal, or allowing a "vertex weight"), you can draw the
graph.

> I do not agree. Whereas the matrix means nothing else but the linear
> operator which it represents.

I don't agree, but this is mindless pedantry.