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Let $M$ be a an oriented riemannian manifold. I have seen the following definition for the Hodge-star operator acting on a differential form. Starting with $\beta\in \Omega^p(M)$ we have $$\alpha \wedge \star \beta = \left<\alpha,\beta\right>\text{vol} ~~\forall \alpha \in \Omega^p(M)$$

where $ \left<\alpha,\beta\right> = \left<e^1\wedge\ldots\wedge e^p, f^1\wedge \ldots \wedge f^p\right> = \det[\left<e^i,f^j\right>]$.

My question is if we have a Lie algebra (of a matrix group) valued form $F\in \Omega^p(M,g)$. How do we now define the Hodge star operator ?. I am asking this question because I want to understand expressions such as $$ F \wedge \star F$$ in the context of Yang-Mills theory. Thank you very much.

EDIT: I think my confusion really comes from the fact wheter I should use the commutator or the matrix product as the first answer kindly mentioned. In the case of Yang-Mills functional is it meant as: $$ L_{YM}= \int_{M}Tr(F\wedge\star F), ~~F = \sum\limits_{j}\omega_j \otimes g_j\in \Omega^2(M,g) \\Tr(F\wedge \star F) = \sum\limits_{j,k}\left<\omega_j, \omega_k\right>\text{vol}~\text{Tr}([g_j,g_k])$$ or is it meant as $$F\in \Omega^2(M,g) \\Tr(F\wedge \star F) = \sum\limits_{j,k}\left<\omega_j, \omega_k\right>\text{vol}~\text{Tr}(g_jg_k).$$ Thank you for you help.

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2 Answers 2

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The Hodge star is not any different. But, depending on the notation you're using, if you have Lie-algebra valued forms, you use wedge on the forms and Lie bracket on the Lie-algebra part. To be explicit, if you have $F= \sum\limits_j \omega_j\otimes v_j$ and $G = \sum\limits_k\eta_k\otimes w_k$, with $\omega_j,\eta_k$ differential forms and $v_j,w_k\in\mathfrak g$, then $F\wedge G = \sum\limits_{j,k} (\omega_j\wedge\eta_k) \otimes [v_j,w_k]$.

Some authors will use regular matrix product rather than matrix commutator, when working with matrix-valued forms, so you need to check your particular author.

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  • $\begingroup$ Thank you. I have added a small question at the end of my original post. $\endgroup$ Sep 16, 2017 at 14:50
  • $\begingroup$ Well, trace of a commutator is $0$ :) $\endgroup$ Sep 16, 2017 at 17:10
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If I'm not misteaking, the first expression you propose vanishes as the trace is conjugaison invariant. Moreover YM action is not suppose to measure only an error of commutation (with G=U(1), YM should be different from 0).

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