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hodge_star [2023/05/30 13:22] – created spencerhodge_star [2023/05/30 13:26] (current) – spencer
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 The Hodge star satisfies $\star^2 = (-1)^{k(n-k)}$ where $k$ is the degree of the form on which it acts. The Hodge star satisfies $\star^2 = (-1)^{k(n-k)}$ where $k$ is the degree of the form on which it acts.
 Thus $(-1)^{k(n-k)} \alpha \wedge \beta = (\alpha, \star \beta) \mathrm{vol}$. Thus $(-1)^{k(n-k)} \alpha \wedge \beta = (\alpha, \star \beta) \mathrm{vol}$.
-On the other hand, $(-1)^{k(n-k)} \alpha \wedge \beta = \beta \wedge \alpha = (-1)^{k(n-k)} (\star \alpha, \beta) \mathrm{vol}$, so the Hodge star is self-adjoint. +On the other hand, $(-1)^{k(n-k)} \alpha \wedge \beta = \beta \wedge \alpha = (-1)^{k(n-k)} (\star \alpha, \beta) \mathrm{vol}$, so the Hodge star is 'self-adjoint' in the sense that the adjoint to $\star$ on $k$-forms is $\star$ on $n-k$-forms
hodge_star.1685467350.txt.gz · Last modified: by spencer