# A technical question that has to do with Hodge Star

1. Apr 19, 2015

### PhyAmateur

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 Hodge star? That is to say $$ReM_{IJ}H=\star(ReM_{IJ} ...)$$ or this is absolutely wrong and we should keep $$ReN_{IJ}$$ outside the Hodge star? That is to say $$ReM_{IJ}H=ReM_{IJ}\star( ...)$$

EDIT: M_{IJ} is a complex matrix

Last edited: Apr 19, 2015
2. Apr 19, 2015

### robphy

What's your definition of the Hodge Star in this context?

3. Apr 19, 2015

### PhyAmateur

Hodge star in R^3.

4. Apr 20, 2015

### Ben Niehoff

The Hodge star is a linear map. Careful though, in complex geometry sometimes the Hodge star is a conjugate-linear map. Hence the reason for the question.

P.S. Your posts would be way easier to read if you use in-line Latex where appropriate.

5. Apr 20, 2015

### PhyAmateur

What do you mean here by conjugate-linear map? Let me be more specific with my question, do you mean if "c" is a real number or real function then
$$\star(c\omega)=(c\star \omega)$$
If so, then what will happen if c were a complex function instead?

Note: I am trying to using a single "\$" for in-line Latex with no results :(!

6. Apr 20, 2015

### Ben Niehoff

Actually, I think the conjugate-linear version is typically written $\bar \star$. It means the combination of taking the Hodge dual and the complex conjugate. Hence for a complex number $c$,

$$\bar \star \; c \omega = \bar c \; \bar \star \; \omega.$$

As for why you would want to do this: it allows you to define a natural Hermitian product on compact complex manifolds:

$$\langle \alpha, \, \beta \rangle = \int_M \alpha \wedge \bar \star \, \beta.$$

On PF, use $for in-line Latex. 7. Apr 20, 2015 ### PhyAmateur Oh great, I understand now, then for a linear map$\star$and if$c$is a complex function then it will not affect the complex function upon leaping it over the$\star##

That is to say, is the following correct if the c is a complex function:

$$\star(c\omega)=c\star\omega$$

Thank you for the in-line tip, much neater! And if you may suggest for me a good read about those mappings (linear and complex conjugate) I'd be grateful!

8. Apr 20, 2015