I know that the j-invariant is very important in mathematics and is related to several "almost integers" as well as monstrous moonshine. According to Wikipedia, the j-invariant is defined as the unique function on the upper half-plane of complex numbers such that it is holomorphic away from a simple pole at the cusp, j($e^{2\pi i/3}$) is $0$, and j($i$) is $1728$. Why is there an extra factor of $1728$ in the j-invariant definition?
Asked
Active
Viewed 307 times
1
-
1Maybe this helps: https://math.stackexchange.com/questions/531532/why-1728-in-j-invariant – garondal Oct 23 '22 at 16:39
-
I don't know much about elliptic curves so this won't help much. – mathlander Oct 23 '22 at 16:45
-
2With $j(i)=1728$ the pole at $\infty$ has residue $1$. The chosen normalizing constant doesn't matter much. – reuns Oct 23 '22 at 17:29
-
What's the residue of a pole? – mathlander Oct 23 '22 at 17:40
-
The residue of a pole is what you get when you integrate around it. If the pole is at $z=a$, then the residue is the coefficient of $(z-a)^{-1}$ in the Taylor expansion at $z=a$. https://en.wikipedia.org/wiki/Residue_(complex_analysis) – Gerry Myerson Oct 24 '22 at 01:45
-
But why specifically 1728 instead of some other multiple of 6? – mathlander Oct 24 '22 at 02:20
-
The Klein Absolute Invariant $J(\tau)$ is the same as $j(\tau)$ without the factor of $1728$. – Somos Jan 31 '23 at 11:56
2 Answers
1
With the definition of $j$ from Wikipedia, the $j$-function has a Fourier series $$j(τ) = q^{-1} + 744 + 196884q + 21493760q^2+\mathcal O(q^3) $$ at $\tau\to i \infty$, where $q=e^{2\pi i\tau}$. We can easily redefine $\tilde j=j/12^3$, such that $\tilde j(i)=1$, but from the above $q$-series you can see that $\tilde j$ does not have integer Fourier coefficients.
This singles out the overall factor in $j$. Another natural definition is $j-744$, which in most applications is completely equivalent.
El Rafu
- 608
0
COMMENT.- The $j$-invariant is a subtle notion and can be defined up to multiplicative constant according to the desired parameterization of certain modular forms. For example if you define $$j(z)=2^{12}\cdot3^3\cdot5^3\frac{G_4^3}{\Delta}(z)=1728\cdot 8000\frac{G_4^3}{\Delta}(z)$$ you get the basic fact that a meromorphic function $f$ defined over the complex upper half-plane is invariant by action of $SL(2,\mathbb Z)$ and meromorphic at $\infty$ if and only if $f$ is a rational function of $j$. While if you define, with Serre, $$j(z)=1728\frac{g_2^3}{\Delta}(z)$$ the basic fact above mentioned becomes stronger. There is not question here of to be clair because it would suppose many definitions and properties; only we say that it is basically a notion involving lattices and double periodic functions (when elliptic curves) and fastly fall on the function $\zeta$ de Riemann. So we bound to standard definitions on a lattice $\Gamma$
$$G_k(\Gamma)=\sum_{\gamma\in\Gamma}\frac{1}{\gamma^{2k}}\space\\g_2=60G_2,\space g_3=140G_3\\\Delta=g_2^3-27g_3^2$$
(Sorry for bad English).
Piquito
- 29,594
-
2How is this an answer to the OP's question? It doesn't distinguish $j$ from various multiples of $j$. – Qiaochu Yuan Oct 24 '22 at 05:35
-
Who said that I wanted to give an answer? You apparently, it would be said. – Piquito Oct 24 '22 at 11:19
-
Try to calculate $j(i)$ for both given formulations of $j$ Are the images equal? – Piquito Oct 24 '22 at 11:28