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Let's say I have an elliptic differential $R(t,\sqrt{f(t)})$, where $f(t)$ is a fourth or third order polynomial. I want to prove it can be transformed by a Möbius transform $t\rightarrow\frac{at+b}{ct+d}$ into a form for which either $f(t)=t(t-1)(t-\lambda)$, $f(t)=t^3+at+b$ or$f(t)=(1-t^2)(1-k^2t^2)$).

EDIT: Reformulated question after remarks.

matti0006
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  • Your question is not entirely clear to me. If $\deg f=3$ then there exists Möbius transformations $\mu$ such that either $$\mu_1(f)=t(t-1)(t-\lambda)\qquad\text{or}\qquad\mu_2(f)=t^3+at+b,$$ and usually there will exist Möbius transformations for both cases. If $\deg f=4$ then there exists a Möbious transformation $\mu$ and a $\kappa\in\Bbb{C}$ such that $$\mu(f)=(1-t^2)(1-\kappa^2t^2),$$ though there is an exceptional case to be dealt with.

    But if $\deg f=3$ then $\deg\mu f=4$ is impossible, and conversely if $\deg f=4$ then $\deg\mu f=3$ is impossible if $\mu f\in\Bbb{C}[t]$.

     – Servaes Oct 04 '13 at 13:46
  • Ah yes, I see now you're right, the question asks me to prove it is always possible to write it in one of those forms. I will rephrase the question. Can you explain to me why if the degree of $f$ is $3$ you can write it in one of the forms you mentioned? Or why it works in the degree $4$ case, if you give me a hint on one I can probably figure out the other one, but I am not sure where to start. – matti0006 Oct 04 '13 at 13:55
  • But even then it seems that by a Mobius transformation the degree of the polynomial doesn't change. – matti0006 Oct 04 '13 at 15:46
  • @Servaes when a Möbius transformation moves the infinity, it will introduce extra poles to the "polynomial" $f$ at some finite value. The integration element will also pick up a double pole at that spot. Let says you are dealing with $\int \frac{1}{\sqrt{f(t)}} dt$ where $f$ is a polynomial of degree 3. The new poles from $f$ and $dt$ will combine togather and effectively give a new zero to the transformed $f$. – achille hui Oct 04 '13 at 16:50
  • @achille hui: You are right, I had the degree 4 case in mind. – Servaes Oct 04 '13 at 17:54
  • @matti0006: Note that the degree $3$ case can be solved by calculation: Pick an $f$, apply $t\ \mapsto\ \tfrac{at+b}{ct+d}$, and solve for $a$, $b$, $c$ and $d$ to get $\mu f=t(t-1)(t-\lambda)$ except if $f$ is a cube. The form $t^3+at+b$ is obtained by an affine transformation for any cubic $f$. Check that an appropriate rational function $R(x,y)\in\Bbb{C}(x,y)$ exists in each case. – Servaes Oct 04 '13 at 18:01
  • But I cannot "pick an $f$", I have to prove it in the general case. Also it seems like a lot of work to actually go and compute the whole thing. It seems the only thing you want to do is shift the zeros of $f$ to new positions (for example in case 1 to $0$, $1$ and $\lambda$, I just can't figure out how exactly the Mobius transformation accomplishes this. – matti0006 Oct 04 '13 at 18:20
  • Note that any $f\in\Bbb{C}[t]$ is a product of linear factors (and a nonzero constant), and that $f\left(\tfrac{at+b}{ct+d}\right)\in\Bbb{C}[t]$ if and only if $c=0$. Without loss of generality $d=1$, now solve for $a$ and $b$. – Servaes Oct 04 '13 at 20:05
  • I would do something like this: Assume $f(t)$ has zeros at $t_0$, $t_1$ and $t_2$, so it is of the form $(t-t_0)(t-t_1)(t-t_2)$ Now take $c=0$, $d=1$ as you said and replace $t$ with $at+b$: $(at+b-t_0)(at+b-t_1)(at+b-t_2)$. We want this to be equal to $t(t-1)(t-\lambda)$ for some $a,b$. But then $a=1$, and we obtain:

    $(t+b-t_0)(t+b-t_1)(t+b-t_2)=t(t-1)(t-\lambda)$. But the left side has a term $b^3$ which is constant, while the right side does not, hence this is not possible. Where did I go wrong?

     – matti0006 Oct 04 '13 at 20:19
  • Note that in general $f(t)=\alpha(t-t_0)(t-t_1)(t-t_2)$ with $\alpha\in\Bbb{C}^{\times}$. We do not need $f(\mu t)=t(t-1)(t-\lambda)$; we need $P\in\Bbb{C}(x,y)$ such that $$R\left(\mu t,\sqrt{f(\mu t)}\right)=P\left(t,\sqrt{t(t-1)(t-\lambda)}\right).$$ In particular, if we have $f(\mu t)=\beta t(t-1)(t-\lambda)$ for some $\beta\in\Bbb{C}^{\times}$, then $$P(x,y)=R\left(\mu^{-1}x,\sqrt{\beta}^{-1}\mu^{-1}y\right),$$ will do the job. So you prematurely conclude that $a=1$. – Servaes Oct 04 '13 at 20:35
  • This should be enough to figure the rest out! – matti0006 Oct 04 '13 at 20:47
  • If you have managed to solve the problem, consider answering your own question. Then the question will not need to be asked again (on this forum). – Servaes Oct 05 '13 at 11:57
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    Will do, not been working on it yet, but I need to hand in the question in latex anyway, so might as well put it up here – matti0006 Oct 05 '13 at 23:06

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