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We know that in this $$\star {f(...)}$$ the $\star$ represents the Hodge dual.

But in this: $\star_3 f(...)$ what does specifically the $\star_3$ symbol mean?

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    $\begingroup$ Is $f$ a 3-form by any chance? $\endgroup$ – Phoenix87 Feb 1 '15 at 20:16
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    $\begingroup$ Can you give an example of its use? $\endgroup$ – Timaeus Feb 1 '15 at 20:17
  • $\begingroup$ Soryy for not being specific, okay so $f(...) = dlog(\beta\bar{\alpha})$ where $\alpha$ and $\beta$ are complex functions. @Phoenix87 $\endgroup$ – beyondtheory Feb 1 '15 at 20:19
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    $\begingroup$ It may mean a Hodge dual wrt. 3-space to distinguish it from a 4-spacetime. $\endgroup$ – Qmechanic Feb 1 '15 at 20:19
  • $\begingroup$ @Timaeus I did in my last comment but coul not tag two users at the same time! $\endgroup$ – beyondtheory Feb 1 '15 at 20:21
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This notation arises often in supergravity. Suppose one has a $d$-dimensional theory. The Hodge $\star$ operator has the usual definition, and $\star_p$ is the Hodge star operation defined on a $p$-dimensional sub-manifold. The question of which sub-manifold is often either explicitly stated or obvious from the context. For example, in the Klebanov-Strassler solution of ten-dimensional Type IIB supergravity (http://arxiv.org/abs/hep-th/0007191), the complex 3-form $G_3$ is imaginary self-dual, $\star_6 G_3 = i G_3$. The six-dimensional space here is the deformed conifold--the full ten dimensional solution is a warped product of this 6-dimensional space with 4-dimensional Minkowski space.

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  • $\begingroup$ Oh does it mean that if we had (t, x, y, z) in the context as you were referring to. And then we bumped into this notation, does it mean that as @Qmechanic said that $\star_3$ means wrt space and not spacetime? $\endgroup$ – beyondtheory Feb 1 '15 at 21:00
  • $\begingroup$ Well Qmechanic is right, my answer is confirming, clarifying, and generalizing what he said. In regards to your question, the subspace does not need to be Euclidean--you could just as easily defined the $\star_3$ operator to act on the Minkowski space. $\endgroup$ – Surgical Commander Feb 1 '15 at 21:05
  • $\begingroup$ Then it would act on the whole Minkowski space and not on the "Sub"-space you referred to. Or did I not understand your last comment? $\endgroup$ – beyondtheory Feb 1 '15 at 21:10
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    $\begingroup$ Ok, I'll try to clarify with the example I mentioned above, the Klebanov-Strassler solution. The metric looks like $ds_{10}^2 = H^{-1/2}dx_{\mu}dx^{\mu} + H^{1/2} ds_6^2$, where $dx_{\mu}dx^{\mu}$ is 4-dimensional Minkowski space and $ds_6^2$ is the deformed conifold. $H$ is a function of the deformed conifold coordinates. This is a warped product space, the $H$ provides the warping. $\star_6$ as is used in this context is w.r.t. the $ds_6^2$. You could also have defined a $\star_4$ w.r.t. the Minkowski space. $\endgroup$ – Surgical Commander Feb 1 '15 at 21:14

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