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In the paper, page 5, Line 5, why the form $\eta_j$ is closed? We have to show that $d \eta_j=0$. Here $\eta_j = f(t)dt$ for some function $f$. Is it always true that $d\eta_j$ for any $f$? Thank you very much.

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2 Answers 2

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Your notation, though taken from the paper, does not make it clear that we are dealing with a complex paramter $t$ and complex valued $f$. In that setting, the form $f(t) dt$ is closed iff $f$ is holomorphic. To see this write $fdt = fdx+i fdy $ and apply $d$ to see that this is closed iff $\frac{\partial}{\partial y} f= i\frac{\partial}{\partial x} f $. In other words iff $$ 0 = \left(\frac{\partial}{\partial y} - i\frac{\partial}{\partial x}\right)f = -i\left( \frac{\partial}{\partial x} + i\frac{\partial}{\partial y}\right)f = -2i\frac{\partial}{\partial \overline{t}}f $$ (with $t=x+iy$, which is kind of against conventions). Now just check whether your $\eta_i$ is holomorphic.

(In general you will get a $dt \wedge d\overline{t}$ term).

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  • $\begingroup$ $\bar{t}=x-iy$. Why $\frac{\partial}{\partial \bar{t}} = \frac{\partial}{\partial x}+i\frac{\partial}{\partial y}$? $\endgroup$
    – LJR
    Jan 8, 2012 at 20:48
  • $\begingroup$ I see. $\frac{\partial}{\partial \bar{t}}=1/2(\frac{\partial}{\partial x} + i\frac{\partial}{\partial y})$ $\endgroup$
    – LJR
    Jan 8, 2012 at 21:49
  • $\begingroup$ @u Yes, sorry for not responding earlier, it seems you figured it out alone. $\endgroup$
    – user20266
    Jan 9, 2012 at 16:33
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This is just a definition. Any differential form $\eta $ such that $d\eta = 0$ is called a closed form. So in your case if $\eta _j=f(t)dt$ and you are able to show that $d\eta _j=0$, then it follows that $\eta _j$ is a closed form.

Since, $\eta _j=f(t)dt$ is a 1-form. By definition of the exterior differential operator you get $d\eta _j=f'(t)dt\wedge dt=0$, since $dt\wedge dt=0$.

This holds for the case where $f(t)$ is real valued.

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  • $\begingroup$ $d\eta_j=0$ is what we have to show. $\endgroup$
    – LJR
    Jan 4, 2012 at 16:32
  • $\begingroup$ By definition of differentiation of a 1-form, it seems that $d\eta_j=0$ for any $f$ since $f$ is only a function of $t$. So $\eta_j$ is a closed form. Is this true? $\endgroup$
    – LJR
    Jan 4, 2012 at 16:35
  • $\begingroup$ Yes. $d\eta_j=\frac{d\eta_j}{dt} dt\wedge dt=0$ since $dt\wedge dt=0$ (assuming $\eta_j$ is indeed only a function of $t$). $\endgroup$
    – SL2
    Jan 4, 2012 at 16:38
  • $\begingroup$ $f$ is complex valued. You have to use that $f$ is holomorphic for this, see my reply. $\endgroup$
    – user20266
    Jan 4, 2012 at 16:46
  • $\begingroup$ @SL2 this is not correct, if $f$ is complex valued and $t$ complex. $\endgroup$
    – user20266
    Jan 4, 2012 at 16:49

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