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Comment by 082349872349872

1 month ago

OK, so if we have a distribution D (less nice than the average function) and a test function T (nicer than the average function), we have ⟨D,T⟩ = c: ℂ, so ⟨D,—⟩: test fn→ℂ and ⟨—,T⟩: distribution→ℂ ?

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082349872349872

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gradschoolfail  1 month ago

Wait i thought functions are predistributions..

[My bad, it was Matvei, not Manuel, no idea how i mixed that up..

Checkout his childrens books, as well as

https://archive.is/eaYRs

Note how the independent diagonals are what i consider interesting]

  • 082349872349872  1 month ago

    if there are no interiors (maybe edges but no faces nor volumes) then the vertices on the diagonals are truly independent: eg QM on small scales, GR on large ones.

    [I'm currently pondering how the "main diagonal" of a transition matrix provides objects, while all the off-diagonal elements are the arrows. This implies that by rotating into an eigenframe (diagonalising), we're reducing the diversion to -∞ (generalised eigenvectors have nothing to lose but their Jordan chains) and hence back in the world of classical boolean logic?]

    EDIT: https://mmozgovoy.dev/posts/solar-matter/