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What I am asking here is a moral question. Mathematically moral, don't bother physics.

I mean, Euler's number is ubiquitous because, among all the exponentials, it alone is its own derivative with all the consequences we know.

I know that the Laplacian contains information about curvature, mean of function etc. so it is what you want in mean curvature flow, or geometrical like fields with the same flavor. It is not my field but it seems to me that it i pops out even in combinatoric arguments in Lie algebras. One can think billions of interesting facts, you got it.

So it seems that there is something fundamental that I can't reach and nobody has ever pointed out to me.

What are your thoughts?

Zappa
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  • I think it comes up mostly because its solutions, the harmonic functions, satisfy the mean value property. In combinatorics there are analogues of this property, and corresponding analogue of Laplacian comes up. I don't know about the specific Lie algebras context you mention. – Wojowu Jan 28 '19 at 15:51
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    I think you mean "philosophical" rather than "moral" question. – James Arathoon Jan 28 '19 at 16:14
  • Interestingly, some rather analytically minded geometers would say that curvatures are interesting because they involve the Laplacian, not the other way round. ;) – MaoWao Jan 28 '19 at 16:19
  • @JamesArathoon Nope. I asked about something I would feel to be the just way of looking at the Laplacian, even if there is no possible proof to support such a feeling. Thus a moral question. Philosophy is a much more serious quest. – Zappa Jan 28 '19 at 17:38
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    (Almost) the same question has also been asked on mathoverflow some time ago: https://mathoverflow.net/questions/54986/why-is-the-laplacian-ubiquitous – MaoWao Jan 29 '19 at 14:31
  • @MaoWao Thank you very much. And it seems it had better fortune than mine. – Zappa Jan 29 '19 at 15:13

1 Answers1

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The (negative) Laplacian is $-\text{div} \nabla$, and the integration by parts formula tells us that $-\text{div}$ is the adjoint of the gradient operator $\nabla$ (in settings where the boundary term vanishes). So the Laplacian has the familiar pattern $A^T A$, which recurs throughout linear algebra and math. This suggests that the (negative) Laplacian is a symmetric positive definite operator, so we would hope that there exists a basis of eigenfunctions for the Laplacian. This motivates the topic of eigenvalues of the Laplacian.

Gilbert Strang emphasizes the ubiquity of $A^T A$ in his linear algebra and applied math books.

littleO
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  • Till now, then, it seems that what is important is the fact that the Divergence theorem holds true, and that the Laplacian talks well with it. I am basing upon littleO and @Wojowu. – Zappa Jan 28 '19 at 17:47