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Thread: [SOLVED] Elliptic functions

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    MHF Contributor Bruno J.'s Avatar
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    [SOLVED] Elliptic functions

    Consider the function $\displaystyle w(z)=\int_1^z\frac{dt}{t}$ where the integral is along any rectifiable path from $\displaystyle 1$ to $\displaystyle z$. Clearly this is not a single-valued function of $\displaystyle z$, as the value of the integral will depend on the winding number of the path around $\displaystyle 0$. Thus $\displaystyle w(z)$ is defined up to an integral multiple of $\displaystyle 2\pi i$. Now consider the inverse function $\displaystyle z(w)$; we have $\displaystyle z(w+2\pi i)=z$ and therefore $\displaystyle z$ is a periodic function. (It's the exponential!)

    Now consider the function $\displaystyle w(z)=\int_1^z\left(\frac{1}{t}+\frac{i}{t-i}\right)dt$. Now $\displaystyle w$ is defined up to $\displaystyle 2\pi i m + 2\pi i n$ where $\displaystyle m,n \in \mathbb{Z}$. Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.
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    Quote Originally Posted by Bruno J. View Post
    Consider the function $\displaystyle w(z)=\int_1^z\frac{dt}{t}$ where the integral is along any rectifiable path from $\displaystyle 1$ to $\displaystyle z$. Clearly this is not a single-valued function of $\displaystyle z$, as the value of the integral will depend on the winding number of the path around $\displaystyle 0$. Thus $\displaystyle w(z)$ is defined up to an integral multiple of $\displaystyle 2\pi i$. Now consider the inverse function $\displaystyle z(w)$; we have $\displaystyle z(w+2\pi i)=z$ and therefore $\displaystyle z$ is a periodic function. (It's the exponential!)

    Now consider the function $\displaystyle w(z)=\int_1^z\left(\frac{1}{t}+\frac{i}{t-i}\right)dt$. Now $\displaystyle w$ is defined up to $\displaystyle 2\pi i m + 2\pi i n$ where $\displaystyle m,n \in \mathbb{Z}$. Explain why, mimicking the construction of the inverse function as above, we do not obtain an elliptic function.


    I may be off orbit by long miles, but let's give it a try:

    Spoiler:
    Indeed we get $\displaystyle z(w+2\pi i m+2\pi in)=z(w)$ , so this function is doubly periodic (...really? Read on), and thus, what's lacking to consider it an elliptic function? Well, the periods must be a basis for the real dimensional linear space $\displaystyle \mathbb{C}$ , and in this case we get $\displaystyle \frac{2\pi i}{2\pi i}=1\in\mathbb{R}$ , which screws the whole thing up.
    The complete, long and deep explanation may be well beyond what I'd be willing to explain by this means: We need a free abelian group in $\displaystyle \mathbb{C}$ which is also a maximal order there and etc.
    Making it short, though, we can simply say: that the above ratio is real and not complex non-real means the function $\displaystyle z(w)$ has actually just one single period (!) which, automatically, disqualifies it from being an elliptic function.


    Tonio
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    MHF Contributor Bruno J.'s Avatar
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    Oh, I'm sorry, my post should have read

    Now is defined up to $\displaystyle 2\pi in + 2\pi m$ where .
    That's why I used $\displaystyle \frac{i}{t-i}$ instead of $\displaystyle \frac{1}{t-i}$ : so that the residue at $\displaystyle i$ (or $\displaystyle 2\pi i$ times the residue, if you prefer) would be real.

    So the two "periods" are indeed linearly independent over $\displaystyle \mathbb{R}$.

    Sorry about the typo; corrected.
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    Quote Originally Posted by Bruno J. View Post
    Oh, I'm sorry, my post should have read



    That's why I used $\displaystyle \frac{i}{t-i}$ instead of $\displaystyle \frac{1}{t-i}$ : so that the residue at $\displaystyle i$ (or $\displaystyle 2\pi i$ times the residue, if you prefer) would be real.

    So the two "periods" are indeed linearly independent over $\displaystyle \mathbb{R}$.

    Sorry about the typo; corrected.

    Spoiler:
    Mind you, I did notice the $\displaystyle \frac{i}{t-i}$ thing and thought it is a little odd the second assumed period is pure imaginary, but of course I didn't check it (as I am not checking it, either ) .

    Well, then the two periods are a $\displaystyle \mathbb{R}-$basis for the complex, so then the only reason why that function wouldn't be an elliptic one I can think of right now is that the function isn't meromorphic on a fundamental parallelogram. Now, either the function is there holomorphic and thus bounded and thus constant (but STILL would be consider elliptic, imo), or else it has a singularity that it is not a pole....oh, I think I see now! Zero is there...hmmm.
    Anyway, it's late here so I shall check this closer tomorrow, perhaps.


    Tonio
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    MHF Contributor Bruno J.'s Avatar
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    Here's a hint : the universal cover of the twice punctured plane is the disc...
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    MHF Contributor Bruno J.'s Avatar
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    Here's the solution I know. I suspect there is a less technical one.

    The function $\displaystyle z \mapsto w$ is well defined $\displaystyle \mod 2\pi i \mathbb{Z} \oplus 2\pi \mathbb{Z}$, i.e. it is well defined on the elliptic curve $\displaystyle E=\mathbb{C}/(2\pi i \mathbb{Z} \oplus 2\pi \mathbb{Z})$. Now suppose there were an analytic inverse function $\displaystyle g : E \rightarrow \mathbb{C}-\{0,i\}$. Lifting $\displaystyle g$ to a map between the universal coverings (which are, respectively, $\displaystyle \mathbb{C}$ for $\displaystyle E$ and the unit disc $\displaystyle D$ for the twice-punctured plane $\displaystyle \mathbb{C}-\{0,i\}$), we obtain an analytic map $\displaystyle \tilde g : \mathbb{C} \rightarrow D$, which, by Liouville's theorem, must be constant...
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