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Jun 19 at 22:17 history edited LSpice CC BY-SA 4.0
Typo
Jun 19 at 20:34 comment added Quarto Bendir When you replace $K$ by $\lambda$ you get (up to normalization) the integral of $\omega^m$, which is (up to normalization) the volume. (I think this is sometimes called Wirtinger theorem.) The particular constants $r$ and $(m-1)!$ are just normalizations depending on some conventions of how the various quantities are defined.
Jun 19 at 17:14 comment added Nikolai Thank you for the answer, but I'm not entirely sure how you go from $K_A = \lambda(E)\text{id}$ to the obtained expression for $\lambda(E)$ by the steps you mentioned? Wedging with $\omega^{m-1}$ gives $$\frac{i}{2\pi} K \wedge \omega^{m-1}$$ and integrating over $X$ gives the expression $$\int_X \frac{i}{2\pi} K \wedge \omega^{m-1}$$. @quarto-bendir
Jun 19 at 15:19 history answered Quarto Bendir CC BY-SA 4.0