Special values of $j$-invariant
Let $j(tau)$ be Klein's absolute invariant defined for $tau in mathbb{H}$ by
$$j(tau) = q^{-1} + 744 + 196884q + 21493760q^2 + 864299970q^3 + cdots$$
with $q := e^{2pi itau}$.
Are there any known special values of $j(tau)$ for which $textrm{Re}(tau)$ is irrational? Wikipedia has a list of special values but in all of those cases $textrm{Re}(tau)$ is rational.
modular-forms
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
add a comment |
Let $j(tau)$ be Klein's absolute invariant defined for $tau in mathbb{H}$ by
$$j(tau) = q^{-1} + 744 + 196884q + 21493760q^2 + 864299970q^3 + cdots$$
with $q := e^{2pi itau}$.
Are there any known special values of $j(tau)$ for which $textrm{Re}(tau)$ is irrational? Wikipedia has a list of special values but in all of those cases $textrm{Re}(tau)$ is rational.
modular-forms
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
add a comment |
Let $j(tau)$ be Klein's absolute invariant defined for $tau in mathbb{H}$ by
$$j(tau) = q^{-1} + 744 + 196884q + 21493760q^2 + 864299970q^3 + cdots$$
with $q := e^{2pi itau}$.
Are there any known special values of $j(tau)$ for which $textrm{Re}(tau)$ is irrational? Wikipedia has a list of special values but in all of those cases $textrm{Re}(tau)$ is rational.
modular-forms
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
Let $j(tau)$ be Klein's absolute invariant defined for $tau in mathbb{H}$ by
$$j(tau) = q^{-1} + 744 + 196884q + 21493760q^2 + 864299970q^3 + cdots$$
with $q := e^{2pi itau}$.
Are there any known special values of $j(tau)$ for which $textrm{Re}(tau)$ is irrational? Wikipedia has a list of special values but in all of those cases $textrm{Re}(tau)$ is rational.
modular-forms
modular-forms
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
asked Mar 3 '17 at 1:54
Ethan Alwaise
6,136517
6,136517
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Since it wasn't specified that $tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(tau)$ where $Re(tau)$ is irrational. However, $tau$ has a hypergeometric closed-form.
It is known that the equation,
$$j(tau) = n$$
can be solved for $tau$ in terms of the hypergeometric function. Using Method 4, we find,
$$tau = frac{_2F_1big(tfrac16,tfrac56,1,1-alphabig)}{_2F_1big(tfrac16,tfrac56,1,alphabig)}sqrt{-1}$$
where
$$alpha=frac{1+sqrt{1-frac{1728}n}}{2}$$
For example, negating the year $n=-2017$, then,
$$tau approx -0.273239 + 0.6868913,i$$
such that,
$$j(tau)=-2017$$
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
add a comment |
The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $phi$, meaning the golden ratio $phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).
Why integer values are special is a much more difficult question!
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since it wasn't specified that $tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(tau)$ where $Re(tau)$ is irrational. However, $tau$ has a hypergeometric closed-form.
It is known that the equation,
$$j(tau) = n$$
can be solved for $tau$ in terms of the hypergeometric function. Using Method 4, we find,
$$tau = frac{_2F_1big(tfrac16,tfrac56,1,1-alphabig)}{_2F_1big(tfrac16,tfrac56,1,alphabig)}sqrt{-1}$$
where
$$alpha=frac{1+sqrt{1-frac{1728}n}}{2}$$
For example, negating the year $n=-2017$, then,
$$tau approx -0.273239 + 0.6868913,i$$
such that,
$$j(tau)=-2017$$
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
add a comment |
Since it wasn't specified that $tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(tau)$ where $Re(tau)$ is irrational. However, $tau$ has a hypergeometric closed-form.
It is known that the equation,
$$j(tau) = n$$
can be solved for $tau$ in terms of the hypergeometric function. Using Method 4, we find,
$$tau = frac{_2F_1big(tfrac16,tfrac56,1,1-alphabig)}{_2F_1big(tfrac16,tfrac56,1,alphabig)}sqrt{-1}$$
where
$$alpha=frac{1+sqrt{1-frac{1728}n}}{2}$$
For example, negating the year $n=-2017$, then,
$$tau approx -0.273239 + 0.6868913,i$$
such that,
$$j(tau)=-2017$$
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
add a comment |
Since it wasn't specified that $tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(tau)$ where $Re(tau)$ is irrational. However, $tau$ has a hypergeometric closed-form.
It is known that the equation,
$$j(tau) = n$$
can be solved for $tau$ in terms of the hypergeometric function. Using Method 4, we find,
$$tau = frac{_2F_1big(tfrac16,tfrac56,1,1-alphabig)}{_2F_1big(tfrac16,tfrac56,1,alphabig)}sqrt{-1}$$
where
$$alpha=frac{1+sqrt{1-frac{1728}n}}{2}$$
For example, negating the year $n=-2017$, then,
$$tau approx -0.273239 + 0.6868913,i$$
such that,
$$j(tau)=-2017$$
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
Since it wasn't specified that $tau$ need to be an algebraic number, then yes, there are infinitely many special values of $j(tau)$ where $Re(tau)$ is irrational. However, $tau$ has a hypergeometric closed-form.
It is known that the equation,
$$j(tau) = n$$
can be solved for $tau$ in terms of the hypergeometric function. Using Method 4, we find,
$$tau = frac{_2F_1big(tfrac16,tfrac56,1,1-alphabig)}{_2F_1big(tfrac16,tfrac56,1,alphabig)}sqrt{-1}$$
where
$$alpha=frac{1+sqrt{1-frac{1728}n}}{2}$$
For example, negating the year $n=-2017$, then,
$$tau approx -0.273239 + 0.6868913,i$$
such that,
$$j(tau)=-2017$$
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
edited Dec 12 '17 at 9:43
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
answered Dec 12 '17 at 9:25
Tito Piezas III
26.8k365169
26.8k365169
add a comment |
add a comment |
The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $phi$, meaning the golden ratio $phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).
Why integer values are special is a much more difficult question!
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
add a comment |
The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $phi$, meaning the golden ratio $phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).
Why integer values are special is a much more difficult question!
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
add a comment |
The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $phi$, meaning the golden ratio $phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).
Why integer values are special is a much more difficult question!
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
The real values of the j-invariant are of interest in theoretical physics (for instance, Witten’s 3d gravity j-invariant is associated to black hole entropy). Now observe that the real number $phi$, meaning the golden ratio $phi$, is a special root of the (degree 10) polynomial that one obtains when writing $j$ as $A+iB/ C +iD$ and looking for real parts (using the usual ratio involving the modular discriminant). This is one of few integer values of $27 j$ (watch out for the normalisation factors), like the other special values on the ribbon graph (dessins of Grothendieck).
Why integer values are special is a much more difficult question!
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
edited Nov 28 '18 at 22:54
Marni Sheppeard
33
33
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
answered Oct 29 '17 at 22:37
Marni Dee Sheppeard
11
11
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
add a comment |
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
Hm, $tau$ must have a non-zero imaginary part...
– Tito Piezas III
Dec 12 '17 at 9:26
add a comment |
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