Show that $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$.











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Following is from Complex Analysis by Freitag :



enter image description here



My questions:



1- The text gives an example for $w_r(M,N)$ but it doesn't explain it or give a clear definition of it. For example where does $w_1(-S,-S) = dfrac{I_1(-E,i)}{I_1(-S,i)^2}$ come from? How $w_r(M,N)$ is defined exactly?



2- On the meaning of equation: For example suppose the second row of matrices M and N have (c,d) and (c',d') in the entries, respectively. Then the mentioned equation means $$((ca'+dc')z+(cb'+dd'))^{r/2} = w_r(M,N) (c frac{a'z+b'}{c'z+d'}+d)^{r/2} (c'z+d')^{r/2}$$ which no way to be correct for $w_r(M,N) in {{pm 1}}.$ What does the equation $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$ mean?



$M=begin{pmatrix}a&b\c&dend{pmatrix}$ and $N=begin{pmatrix}a'&b'\c'&d'end{pmatrix}$.










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  • What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
    – Zvi
    Nov 23 at 14:43












  • @Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
    – 72D
    Nov 23 at 14:55








  • 1




    Oh, I see your explanation at the end. sorry!
    – Zvi
    Nov 23 at 14:58






  • 1




    @Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
    – 72D
    Nov 23 at 14:59






  • 1




    See the edit i made to your question.
    – Zvi
    Nov 23 at 15:24















up vote
0
down vote

favorite












Following is from Complex Analysis by Freitag :



enter image description here



My questions:



1- The text gives an example for $w_r(M,N)$ but it doesn't explain it or give a clear definition of it. For example where does $w_1(-S,-S) = dfrac{I_1(-E,i)}{I_1(-S,i)^2}$ come from? How $w_r(M,N)$ is defined exactly?



2- On the meaning of equation: For example suppose the second row of matrices M and N have (c,d) and (c',d') in the entries, respectively. Then the mentioned equation means $$((ca'+dc')z+(cb'+dd'))^{r/2} = w_r(M,N) (c frac{a'z+b'}{c'z+d'}+d)^{r/2} (c'z+d')^{r/2}$$ which no way to be correct for $w_r(M,N) in {{pm 1}}.$ What does the equation $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$ mean?



$M=begin{pmatrix}a&b\c&dend{pmatrix}$ and $N=begin{pmatrix}a'&b'\c'&d'end{pmatrix}$.










share|cite|improve this question
























  • What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
    – Zvi
    Nov 23 at 14:43












  • @Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
    – 72D
    Nov 23 at 14:55








  • 1




    Oh, I see your explanation at the end. sorry!
    – Zvi
    Nov 23 at 14:58






  • 1




    @Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
    – 72D
    Nov 23 at 14:59






  • 1




    See the edit i made to your question.
    – Zvi
    Nov 23 at 15:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Following is from Complex Analysis by Freitag :



enter image description here



My questions:



1- The text gives an example for $w_r(M,N)$ but it doesn't explain it or give a clear definition of it. For example where does $w_1(-S,-S) = dfrac{I_1(-E,i)}{I_1(-S,i)^2}$ come from? How $w_r(M,N)$ is defined exactly?



2- On the meaning of equation: For example suppose the second row of matrices M and N have (c,d) and (c',d') in the entries, respectively. Then the mentioned equation means $$((ca'+dc')z+(cb'+dd'))^{r/2} = w_r(M,N) (c frac{a'z+b'}{c'z+d'}+d)^{r/2} (c'z+d')^{r/2}$$ which no way to be correct for $w_r(M,N) in {{pm 1}}.$ What does the equation $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$ mean?



$M=begin{pmatrix}a&b\c&dend{pmatrix}$ and $N=begin{pmatrix}a'&b'\c'&d'end{pmatrix}$.










share|cite|improve this question















Following is from Complex Analysis by Freitag :



enter image description here



My questions:



1- The text gives an example for $w_r(M,N)$ but it doesn't explain it or give a clear definition of it. For example where does $w_1(-S,-S) = dfrac{I_1(-E,i)}{I_1(-S,i)^2}$ come from? How $w_r(M,N)$ is defined exactly?



2- On the meaning of equation: For example suppose the second row of matrices M and N have (c,d) and (c',d') in the entries, respectively. Then the mentioned equation means $$((ca'+dc')z+(cb'+dd'))^{r/2} = w_r(M,N) (c frac{a'z+b'}{c'z+d'}+d)^{r/2} (c'z+d')^{r/2}$$ which no way to be correct for $w_r(M,N) in {{pm 1}}.$ What does the equation $I_r(MN, z) = w_r(M,N) I_r(M,Nz) I_r(N, z)$ mean?



$M=begin{pmatrix}a&b\c&dend{pmatrix}$ and $N=begin{pmatrix}a'&b'\c'&d'end{pmatrix}$.







complex-analysis proof-explanation modular-forms






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share|cite|improve this question













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edited Nov 23 at 15:24









Zvi

3,825328




3,825328










asked Nov 23 at 14:38









72D

53216




53216












  • What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
    – Zvi
    Nov 23 at 14:43












  • @Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
    – 72D
    Nov 23 at 14:55








  • 1




    Oh, I see your explanation at the end. sorry!
    – Zvi
    Nov 23 at 14:58






  • 1




    @Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
    – 72D
    Nov 23 at 14:59






  • 1




    See the edit i made to your question.
    – Zvi
    Nov 23 at 15:24


















  • What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
    – Zvi
    Nov 23 at 14:43












  • @Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
    – 72D
    Nov 23 at 14:55








  • 1




    Oh, I see your explanation at the end. sorry!
    – Zvi
    Nov 23 at 14:58






  • 1




    @Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
    – 72D
    Nov 23 at 14:59






  • 1




    See the edit i made to your question.
    – Zvi
    Nov 23 at 15:24
















What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
– Zvi
Nov 23 at 14:43






What are $M$ and $N$? How are they related to $c$ and $d$? What is $E$?
– Zvi
Nov 23 at 14:43














@Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
– 72D
Nov 23 at 14:55






@Zvi, $M$ and $N$ are elements of $SL_2(mathbb{Z})$ and e.g. $c$ and $d$ are M_21 and M_22 2nd row vector of $M$ (I don't know how to type $2 times 2$ matrix in latex). $E$ is $I_2$ the identity matrix.
– 72D
Nov 23 at 14:55






1




1




Oh, I see your explanation at the end. sorry!
– Zvi
Nov 23 at 14:58




Oh, I see your explanation at the end. sorry!
– Zvi
Nov 23 at 14:58




1




1




@Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
– 72D
Nov 23 at 14:59




@Zvi, $Nz$ is Mobius transformation $Mz=frac{M_{11}z+M_{12}}{M_{21}z+M_{22}}$
– 72D
Nov 23 at 14:59




1




1




See the edit i made to your question.
– Zvi
Nov 23 at 15:24




See the edit i made to your question.
– Zvi
Nov 23 at 15:24










1 Answer
1






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1
down vote



accepted











  1. The thing is the square root map $zmapsto sqrt{z}$ for a complex number $z$ does not satisfy $sqrt{z_1z_2}=sqrt{z_1}sqrt{z_2}$. This is due to the fact that we need a branch cut in $Bbb{C}$ to define $sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $arg z_1+arg z_2$ and $arg(z_1z_2)$ are unequal). So, the proper equality is $$sqrt{z_1z_2}=pmsqrt{z_1}sqrt{z_2},tag{1}$$ but what determine the choice of signs $pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality
    $$I_r(MN,z)=pm I_r(M,Nz) I_r(N,z).$$
    The number $w_r(M,N) =pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write
    $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$
    It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have begin{align}sqrt{-1}=I_1(-E,z)&=w_1(-S,-S) I_1(-S,-Sz) I_1(-S,z)\&=(-1) sqrt{frac{1}{z}} sqrt{-z}.tag{2}end{align}
    If $z=1$ with the usual convention $sqrt{1}=1$, we have $sqrt{-1}=-sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).


  2. I don't know why you think the expression is "in no way to be correct." Note from (1) that
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}&=left(frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}right)^{r/2}\&=left(pmfrac{sqrt{(ca'+dc')z+(cb'+dd')}}{sqrt{c'z+d'}}right)^{r}
    \&=pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}.end{align}

    Therefore,
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}&=left(pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}right)left(sqrt{c'z+d'}right)^{r}
    \&=pmleft(sqrt{(ca'+dc')z+(cb'+dd')}right)^r
    \&=pmbig((ca'+dc')z+(cb'+dd')big)^{r/2}.end{align}

    So, you see that
    $$big((ca'+dc')z+(cb'+dd')big)^{r/2}=pmleft(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}.$$
    This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)







share|cite|improve this answer



















  • 1




    In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
    – reuns
    Nov 23 at 18:46












  • @reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
    – Zvi
    Nov 23 at 18:52













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  1. The thing is the square root map $zmapsto sqrt{z}$ for a complex number $z$ does not satisfy $sqrt{z_1z_2}=sqrt{z_1}sqrt{z_2}$. This is due to the fact that we need a branch cut in $Bbb{C}$ to define $sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $arg z_1+arg z_2$ and $arg(z_1z_2)$ are unequal). So, the proper equality is $$sqrt{z_1z_2}=pmsqrt{z_1}sqrt{z_2},tag{1}$$ but what determine the choice of signs $pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality
    $$I_r(MN,z)=pm I_r(M,Nz) I_r(N,z).$$
    The number $w_r(M,N) =pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write
    $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$
    It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have begin{align}sqrt{-1}=I_1(-E,z)&=w_1(-S,-S) I_1(-S,-Sz) I_1(-S,z)\&=(-1) sqrt{frac{1}{z}} sqrt{-z}.tag{2}end{align}
    If $z=1$ with the usual convention $sqrt{1}=1$, we have $sqrt{-1}=-sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).


  2. I don't know why you think the expression is "in no way to be correct." Note from (1) that
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}&=left(frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}right)^{r/2}\&=left(pmfrac{sqrt{(ca'+dc')z+(cb'+dd')}}{sqrt{c'z+d'}}right)^{r}
    \&=pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}.end{align}

    Therefore,
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}&=left(pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}right)left(sqrt{c'z+d'}right)^{r}
    \&=pmleft(sqrt{(ca'+dc')z+(cb'+dd')}right)^r
    \&=pmbig((ca'+dc')z+(cb'+dd')big)^{r/2}.end{align}

    So, you see that
    $$big((ca'+dc')z+(cb'+dd')big)^{r/2}=pmleft(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}.$$
    This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)







share|cite|improve this answer



















  • 1




    In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
    – reuns
    Nov 23 at 18:46












  • @reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
    – Zvi
    Nov 23 at 18:52

















up vote
1
down vote



accepted











  1. The thing is the square root map $zmapsto sqrt{z}$ for a complex number $z$ does not satisfy $sqrt{z_1z_2}=sqrt{z_1}sqrt{z_2}$. This is due to the fact that we need a branch cut in $Bbb{C}$ to define $sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $arg z_1+arg z_2$ and $arg(z_1z_2)$ are unequal). So, the proper equality is $$sqrt{z_1z_2}=pmsqrt{z_1}sqrt{z_2},tag{1}$$ but what determine the choice of signs $pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality
    $$I_r(MN,z)=pm I_r(M,Nz) I_r(N,z).$$
    The number $w_r(M,N) =pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write
    $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$
    It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have begin{align}sqrt{-1}=I_1(-E,z)&=w_1(-S,-S) I_1(-S,-Sz) I_1(-S,z)\&=(-1) sqrt{frac{1}{z}} sqrt{-z}.tag{2}end{align}
    If $z=1$ with the usual convention $sqrt{1}=1$, we have $sqrt{-1}=-sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).


  2. I don't know why you think the expression is "in no way to be correct." Note from (1) that
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}&=left(frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}right)^{r/2}\&=left(pmfrac{sqrt{(ca'+dc')z+(cb'+dd')}}{sqrt{c'z+d'}}right)^{r}
    \&=pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}.end{align}

    Therefore,
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}&=left(pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}right)left(sqrt{c'z+d'}right)^{r}
    \&=pmleft(sqrt{(ca'+dc')z+(cb'+dd')}right)^r
    \&=pmbig((ca'+dc')z+(cb'+dd')big)^{r/2}.end{align}

    So, you see that
    $$big((ca'+dc')z+(cb'+dd')big)^{r/2}=pmleft(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}.$$
    This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)







share|cite|improve this answer



















  • 1




    In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
    – reuns
    Nov 23 at 18:46












  • @reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
    – Zvi
    Nov 23 at 18:52















up vote
1
down vote



accepted







up vote
1
down vote



accepted







  1. The thing is the square root map $zmapsto sqrt{z}$ for a complex number $z$ does not satisfy $sqrt{z_1z_2}=sqrt{z_1}sqrt{z_2}$. This is due to the fact that we need a branch cut in $Bbb{C}$ to define $sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $arg z_1+arg z_2$ and $arg(z_1z_2)$ are unequal). So, the proper equality is $$sqrt{z_1z_2}=pmsqrt{z_1}sqrt{z_2},tag{1}$$ but what determine the choice of signs $pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality
    $$I_r(MN,z)=pm I_r(M,Nz) I_r(N,z).$$
    The number $w_r(M,N) =pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write
    $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$
    It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have begin{align}sqrt{-1}=I_1(-E,z)&=w_1(-S,-S) I_1(-S,-Sz) I_1(-S,z)\&=(-1) sqrt{frac{1}{z}} sqrt{-z}.tag{2}end{align}
    If $z=1$ with the usual convention $sqrt{1}=1$, we have $sqrt{-1}=-sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).


  2. I don't know why you think the expression is "in no way to be correct." Note from (1) that
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}&=left(frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}right)^{r/2}\&=left(pmfrac{sqrt{(ca'+dc')z+(cb'+dd')}}{sqrt{c'z+d'}}right)^{r}
    \&=pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}.end{align}

    Therefore,
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}&=left(pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}right)left(sqrt{c'z+d'}right)^{r}
    \&=pmleft(sqrt{(ca'+dc')z+(cb'+dd')}right)^r
    \&=pmbig((ca'+dc')z+(cb'+dd')big)^{r/2}.end{align}

    So, you see that
    $$big((ca'+dc')z+(cb'+dd')big)^{r/2}=pmleft(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}.$$
    This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)







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  1. The thing is the square root map $zmapsto sqrt{z}$ for a complex number $z$ does not satisfy $sqrt{z_1z_2}=sqrt{z_1}sqrt{z_2}$. This is due to the fact that we need a branch cut in $Bbb{C}$ to define $sqrt{z}$, and the product $z_1z_2$ may take the product across the boundary (i.e., when $arg z_1+arg z_2$ and $arg(z_1z_2)$ are unequal). So, the proper equality is $$sqrt{z_1z_2}=pmsqrt{z_1}sqrt{z_2},tag{1}$$ but what determine the choice of signs $pm$ depends on the values of $z_1,z_2$ and on how you select your branch cut, and I don't know which branch cut your book is using. However, that is precisely why you have the equality
    $$I_r(MN,z)=pm I_r(M,Nz) I_r(N,z).$$
    The number $w_r(M,N) =pm1$ is just something to handle the signs so that the reader can do more definite operations (i.e., no guessing which sign it is going to be, as the signs are already encoded in $w_r(M,N)$), and so the author can write
    $$I_r(MN,z)=w_r(M,N) I_r(M,Nz)I_r(N,z).$$
    It has to be verified, though, that $w_r(M,N)$ does not depend on $z$, and I disagree, as the text seems to be suggesting, that $w_r(M,N)$ does not depend on $z$. In the book's example, we have begin{align}sqrt{-1}=I_1(-E,z)&=w_1(-S,-S) I_1(-S,-Sz) I_1(-S,z)\&=(-1) sqrt{frac{1}{z}} sqrt{-z}.tag{2}end{align}
    If $z=1$ with the usual convention $sqrt{1}=1$, we have $sqrt{-1}=-sqrt{-1}$, which is false. So, I would write $w_r(M,N,z)$, rather than $w_r(M,N)$. I am certain that no matter which branch cut you use, there are always $z_1,z_2$ that violate (2). However, I should like to mention that $w_r(M,N)$ only depends weakly on $z$ (that is, for fixed $M,N$, and for almost all $z$, $w_r(M,N)$ is constant in an open neighbourhood of $z$).


  2. I don't know why you think the expression is "in no way to be correct." Note from (1) that
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}&=left(frac{(ca'+dc')z+(cb'+dd')}{c'z+d'}right)^{r/2}\&=left(pmfrac{sqrt{(ca'+dc')z+(cb'+dd')}}{sqrt{c'z+d'}}right)^{r}
    \&=pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}.end{align}

    Therefore,
    begin{align}left(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}&=left(pmfrac{left(sqrt{(ca'+dc')z+(cb'+dd')}right)^r}{left(sqrt{c'z+d'}right)^r}right)left(sqrt{c'z+d'}right)^{r}
    \&=pmleft(sqrt{(ca'+dc')z+(cb'+dd')}right)^r
    \&=pmbig((ca'+dc')z+(cb'+dd')big)^{r/2}.end{align}

    So, you see that
    $$big((ca'+dc')z+(cb'+dd')big)^{r/2}=pmleft(c frac{a'z+b'}{c'z+d'}+dright)^{r/2}left(c'z+d'right)^{r/2}.$$
    This equation simply means that you have to worry about the branch cut, and taking $(r/2)$th power is not a multiplicative operation, but it is multiplicative up to sign change. (Taking $(r/2)$th power is multiplicative only when $r$ is even.)








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edited Nov 23 at 18:38

























answered Nov 23 at 16:59









Zvi

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  • 1




    In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
    – reuns
    Nov 23 at 18:46












  • @reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
    – Zvi
    Nov 23 at 18:52
















  • 1




    In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
    – reuns
    Nov 23 at 18:46












  • @reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
    – Zvi
    Nov 23 at 18:52










1




1




In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
– reuns
Nov 23 at 18:46






In the context of modular forms we look only at $Im(z) > 0$. Replace it by $z = ix, Re(x) > 0$, assume $c ge 0$, pick the principal branch of $sqrt{.}$, replace $sqrt{cz+d}$ by $sqrt{cx-id}$. Then the $pm $ sign doesn't depend on $x, Re(x) > 0$.
– reuns
Nov 23 at 18:46














@reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
– Zvi
Nov 23 at 18:52






@reuns Then that makes sense for $w_r(M,N)$ to not depend on $z$. Thank you very much for the information.
– Zvi
Nov 23 at 18:52




















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