Timeline for How to understand the combinatorial Laplacian $\Delta$ which is defined on the graph?
Current License: CC BY-SA 4.0
9 events
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| Nov 15, 2023 at 18:56 | comment | added | ARG | @ThighCrush thanks! yes, switched the sign just after the sentence "extending by linearity" because it fits with what the OP was asking. But basically, if you define the divergence as the adjoint of the gradient (which is not what I did here) you get that $\nabla^* \nabla$ is positive definite. I could edit the post, but that might make your comment seem strange... | |
| Nov 14, 2023 at 12:33 | comment | added | ThighCrush | Very nice answer, that I am still digesting. However I think there's a minor mistake concerning a minus sign, in that MINUS the divergence is the adjoint of the gradient, and that this leads to the nice formula for the discrete divergence that you give... right? | |
| Aug 15, 2020 at 10:15 | comment | added | ARG | See this post for interpretation of the Laplacian | |
| Aug 15, 2020 at 10:14 | history | edited | ARG | CC BY-SA 4.0 |
added the fact that $c$ should be symmetric (was not properly written)
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| Jul 11, 2020 at 13:29 | comment | added | ARG | One can get rid of the factor of 2 (which appears in my computation). This factor comes in my construction just because I see $(x,y) $ and $(y,x)$ as two separate edges. To get rid of it, just set for every edge an orientation (that is, either $(x,y) \in E$ or $(y,x) \in E$ but not both). The notation is sometimes a bit more clumsy, but you get the exact same thing (without the factor of 2). | |
| Apr 20, 2020 at 22:36 | vote | accept | Hermi | ||
| Apr 20, 2020 at 18:08 | history | edited | Will Sawin | CC BY-SA 4.0 |
edited body
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| Apr 20, 2020 at 17:57 | history | edited | ARG | CC BY-SA 4.0 |
correcte typo, added an addendum
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| Apr 18, 2020 at 14:12 | history | answered | ARG | CC BY-SA 4.0 |