When $J(tau)inmathbbR$? ($J$ is Klein's $j$-invariant.)

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I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.
The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.
modular-forms modular-function modular-group
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up vote
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I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.
The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.
modular-forms modular-function modular-group
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.
The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.
modular-forms modular-function modular-group
I would like to know the set $X=J^-1(mathbbR)$, where $J$ is Klein's $J$-invariant.
Since $J$ is a modular function, it suffices to know the intersection of $X$ and the fundamental domain $mathrmRe(tau)$.
The special values of $J$ suggest that $J(tau)>0$ for $tau=i y$ with $yge 1$, but I cannot prove that it holds generally.
modular-forms modular-function modular-group
asked Aug 23 at 3:25
user356126
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1106
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1 Answer
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The $j$-invariant is real on the fundamental domain, if and only if $tau$
lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).
To prove this, note that on the vertical lines with integer and half-integer real
values, the $h$-function is real, and the unit semicircle is equivalent uunder
$textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.
Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
lines, so takes every real value, by the IVT, on the stated set. Now since
the $j$-function is bijective on the "strict" fundamental region, then elsewhere
the $j$-function is non-real.
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The $j$-invariant is real on the fundamental domain, if and only if $tau$
lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).
To prove this, note that on the vertical lines with integer and half-integer real
values, the $h$-function is real, and the unit semicircle is equivalent uunder
$textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.
Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
lines, so takes every real value, by the IVT, on the stated set. Now since
the $j$-function is bijective on the "strict" fundamental region, then elsewhere
the $j$-function is non-real.
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
add a comment |Â
up vote
1
down vote
accepted
The $j$-invariant is real on the fundamental domain, if and only if $tau$
lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).
To prove this, note that on the vertical lines with integer and half-integer real
values, the $h$-function is real, and the unit semicircle is equivalent uunder
$textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.
Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
lines, so takes every real value, by the IVT, on the stated set. Now since
the $j$-function is bijective on the "strict" fundamental region, then elsewhere
the $j$-function is non-real.
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The $j$-invariant is real on the fundamental domain, if and only if $tau$
lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).
To prove this, note that on the vertical lines with integer and half-integer real
values, the $h$-function is real, and the unit semicircle is equivalent uunder
$textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.
Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
lines, so takes every real value, by the IVT, on the stated set. Now since
the $j$-function is bijective on the "strict" fundamental region, then elsewhere
the $j$-function is non-real.
The $j$-invariant is real on the fundamental domain, if and only if $tau$
lies on the boundary (both vertical lines and the circular arc) or if $tau$ has zero real part (lies on the imaginary axis).
To prove this, note that on the vertical lines with integer and half-integer real
values, the $h$-function is real, and the unit semicircle is equivalent uunder
$textSL_2(Bbb Z)$ to one of these vertical lines, so $j$ is real where I said.
Also $j$ tends to $infty$ and $-infty$ as $tautoinfty$ on these vertical
lines, so takes every real value, by the IVT, on the stated set. Now since
the $j$-function is bijective on the "strict" fundamental region, then elsewhere
the $j$-function is non-real.
answered Aug 23 at 6:21
Lord Shark the Unknown
88k954115
88k954115
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
add a comment |Â
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
Thank you for your answer. I understood the behavior of $J(tau)$ well.
â user356126
Aug 25 at 5:07
add a comment |Â
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