If we have an equation that looks like $$H=Y$$ and we want to multiply $H$ by either $ReM_{IJ}$ or $ImM_{IJ}$ where $M_{IJ}$ is a complex matrix. But the thing is that $$Y=\star(...)$$ where $\star$ is hodge star and (...) is set of complex functions and other numerical stuff, my question is technical here, say we decide to multiply H by $ReM_{IJ}$ can we move $ReM_{IJ}$ into the parenthesis and jump over the $\star$? That is to say $$ReM_{IJ}H=\star(ReM_{IJ} ...)$$ or this is absolutely wrong and we should keep $ReN_{IJ}$ outside the $\star$? That is to say $$ReM_{IJ}H=ReM_{IJ}\star( ...)$$
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What is your definition of the Hodge star? – Michael Albanese Apr 18 '15 at 21:26
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It is what takes, say, $\star dx$ to $dy\wedge dz$. Does this answer your question, if not may you elaborate? – PhilosophicalPhysics Apr 18 '15 at 21:44
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That's the Hodge star on $\mathbb{R}^3$. Here you're dealing with complex objects, so it matters whether $\ast$ is linear or conjugate linear; this will depend on your convention, hence my question. – Michael Albanese Apr 19 '15 at 02:33
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How do I know if it is linear or conjugate linear? And how does this relate to my question above? @MichaelAlbanese – PhilosophicalPhysics Apr 19 '15 at 11:03
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Also $ReN_{IJ}$ is not a complex object, it is a real thing. I still do not know whether I can leap it over the $\star$ or not? @MichaelAlbanese – PhilosophicalPhysics Apr 19 '15 at 11:08
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You are right. I misunderstood your question. – Michael Albanese Apr 19 '15 at 14:10
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So what do you say Michael? Can I jump with the $ReN_{IJ}$ over the hodge star like in the equation I wrote in the question just before the last line? – PhilosophicalPhysics Apr 19 '15 at 14:16
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Unless I am mistaken but as I remember that $\star 1 = dx \wedge dy \wedge dz$ and hence my question from the whole beginning. I thought even though $Re N_{IJ}$ is a real thing, maybe I should be careful leaping it over $\star$. So that's is exactly why I am asking. @MichaelAlbanese – PhilosophicalPhysics Apr 19 '15 at 14:30
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Are you asking whether $\star(cf) = c(\star f)$ for a real number $c$? – Michael Albanese Apr 19 '15 at 14:38
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If you agree that $ReN_{IJ}$ can be considered $c$ then yes that is pretty much my question. – PhilosophicalPhysics Apr 19 '15 at 14:43
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What is $H$ in your original question? Is it a function? – Michael Albanese Apr 19 '15 at 14:52
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That means nothing to me. Is it a function? If not, what type of mathematical object is it? – Michael Albanese Apr 19 '15 at 14:57
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Last question: $Re M_{IJ}$ is the $(I, J)^{\text{th}}$ entry of the matrix $Re M$. Is $M$ a matrix of numbers or functions? That is, is $Re M_{IJ}$ a real number or a real-valued function? – Michael Albanese Apr 19 '15 at 15:05
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After some discussion, it seems like your question is actually the following:
Let $\omega$ be a differential form and $c \in \mathbb{R}$. Is $\star(c\omega) = c(\star\omega)$?
The answer to this is yes. The Hodge dual is real-linear, in fact it is linear over real-valued functions, i.e. $\star(f\omega) = f(\star\omega)$.
If you want to bring out a complex number or complex-valued function, then it depends on your definition of $\star$; in particular, whether it is complex linear or conjugate linear.
Michael Albanese
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Aha, so no matter what $M$ is, given your last question in the discussion above, Re(M) is a real number and thus if Hodge dual is real-linear or linear over real-valued functions as you added in your answer then the answer is yes. Now since I am not very familiar with what you mean by complex linear or conjugate linear or even real-linear, how did you decide that mine was real-linear? – PhilosophicalPhysics Apr 19 '15 at 15:20
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Real linear means $\star(c\omega) = c(\star\omega)$ for any $c \in \mathbb{R}$, and $\star(\omega_1 + \omega_2) = (\star\omega_1) + (\star\omega_2)$. Complex linear is the same, except $c \in \mathbb{R}$, whereas conjugate linear requires $\star(c\omega) = \bar{c}(\star\omega)$.
The Hodge star is real linear, that follows from the definition. However, when defining the Hodge dual on a complex vector space or manifold, there are different conventions, one of which is complex linear, the other is conjugate linear. Regardless of which one is being used, the Hodge dual is still real linear.
– Michael Albanese Apr 19 '15 at 15:32 -
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I have a question now that I newly encountered your definition of real linear Hodge dual and this classification that I should be aware of. This is a quick Yes/No question: If we had instead of $ReN_{IJ}$, an $|\alpha|^2$ where $\alpha$ is a complex function, could we have said also that $\star(|\alpha|^2\omega)=|\alpha |^2(\star \omega)$? Where $\omega$ is what you defined up in your comment.
Please note that $\alpha \bar{\alpha} =|\alpha|^2$
Thanks in advance
– PhilosophicalPhysics Apr 19 '15 at 20:40 -
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1If you're talking about an inner product space, $\star$ is linear over real numbers. If you're talking about a manifold, $\star$ is linear over real functions. Given what you've said, it seems like you are in the latter case. – Michael Albanese Apr 19 '15 at 22:03