Return to Question

bug workarounds

Source Link

edited Mar 6, 2013 at 0:08

45.7k
6
114
220

Hi all

I am currently trying to understand Borcherd's lifts for a part of my thesis work. I am dreadfully confused about some comments made in the example given in Bruinier's book mentioned in the title. I understand the example itself however I am interested in applying some of the statements he makes in a similar case but things are not adding up for me.

The example begins by identifying the Shimura variety associated to the orthogonal group for the lattice L= \mathbb{H} \perp \mathbb{H} \perp <-1>$L= \mathbb{H} \perp \mathbb{H} \perp \langle -1\rangle$ with the one associated to the symplectic group of genus 2.

He then states ``the"the space M^!{-1/2, L} of nearly holomorphic modular forms of weight -1/2 can be identified with the space \tilde{J}{0,$M^!_{-1/2, L}$ of nearly holomorphic modular forms of weight -1}/2 can be identified with the space $\tilde{J}_{0,1}$ of weakly holomorphic Jacobi forms of weight 0 and index 1.''" I feel comfortable with this statement as Eichler-Zagier explain this identification for holomorphic Jacobi forms and it certainly looks like the proof immediately generalizes to nearly holomorphic ones.

Next he says ``The"The obstruction space S_{5/2, L}$S_{5/2, L}$ can be identified in the same way with the space of skew-holomorphic Jacobi forms of weight 3 and index 1.''" This is the first thing I do not understand. Why is it skew-holomorphic and why not cusp form? It seems to me like it should be regular Jacobi cusp forms just looking at generalizing the proof in E-Z mentioned above.

In any case, he then says ``This"This space equals 0. Therefore every Fourier polynomial is the principal part of a weak Jacobi form in \tilde{J}{0,1}.'' This is the most important issue for me which I cannot reconcile with my understanding of Borcherd's products (which is tenuous at best). According to E-Z, \tilde{J}{0$\tilde{J}_{0,1}$." This is the most important issue for me which I cannot reconcile with my understanding of Borcherd's products (which is tenuous at best). According to E-Z,1} $\tilde{J}_{0,1}$ is 1 dimensional, spanned by \phi_{0,1}$\phi_{0,1}$ which is given in the example. However if this is so then M^!{-1/2, L} must also be 1-dimensional. But if the obstruction space is trivial as stated, then by Bruinier's Theorem 1.17, every possible principal part of a Fourier expansion should be realized for some element of M^!{$M^!_{-1/2, L}$ must also be 1-1/2dimensional. But if the obstruction space is trivial as stated, L}then by Bruinier's Theorem 1.17, every possible principal part of a Fourier expansion should be realized for some element of $M^!_{-1/2, L}$ (as Bruinier was alluding to in his statement), meaning in particular that the space is infinite dimensional.

I would really appreciate any insight into this, particularly the last part regarding the dimension of M^!_{-1/2, L}$M^!_{-1/2, L}$, because I am curious as to whether other Humbert surfaces can be realized as the divisor of a Borcherd's lift.

Thanks for your time.

edit: The compiler wasn't displaying the math tex correctly even though it was compiling fine in WinEdt so I removed all the dollar signs.

Hi all

The example begins by identifying the Shimura variety associated to the orthogonal group for the lattice L= \mathbb{H} \perp \mathbb{H} \perp <-1> with the one associated to the symplectic group of genus 2.

He then states ``the space M^!{-1/2, L} of nearly holomorphic modular forms of weight -1/2 can be identified with the space \tilde{J}{0,1} of weakly holomorphic Jacobi forms of weight 0 and index 1.'' I feel comfortable with this statement as Eichler-Zagier explain this identification for holomorphic Jacobi forms and it certainly looks like the proof immediately generalizes to nearly holomorphic ones.

Next he says ``The obstruction space S_{5/2, L} can be identified in the same way with the space of skew-holomorphic Jacobi forms of weight 3 and index 1.'' This is the first thing I do not understand. Why is it skew-holomorphic and why not cusp form? It seems to me like it should be regular Jacobi cusp forms just looking at generalizing the proof in E-Z mentioned above.

In any case, he then says ``This space equals 0. Therefore every Fourier polynomial is the principal part of a weak Jacobi form in \tilde{J}{0,1}.'' This is the most important issue for me which I cannot reconcile with my understanding of Borcherd's products (which is tenuous at best). According to E-Z, \tilde{J}{0,1} is 1 dimensional, spanned by \phi_{0,1} which is given in the example. However if this is so then M^!{-1/2, L} must also be 1-dimensional. But if the obstruction space is trivial as stated, then by Bruinier's Theorem 1.17, every possible principal part of a Fourier expansion should be realized for some element of M^!{-1/2, L} (as Bruinier was alluding to in his statement), meaning in particular that the space is infinite dimensional.

I would really appreciate any insight into this, particularly the last part regarding the dimension of M^!_{-1/2, L}, because I am curious as to whether other Humbert surfaces can be realized as the divisor of a Borcherd's lift.

Thanks for your time.

edit: The compiler wasn't displaying the math tex correctly even though it was compiling fine in WinEdt so I removed all the dollar signs.

Hi all

The example begins by identifying the Shimura variety associated to the orthogonal group for the lattice $L= \mathbb{H} \perp \mathbb{H} \perp \langle -1\rangle$ with the one associated to the symplectic group of genus 2.

He then states "the space $M^!_{-1/2, L}$ of nearly holomorphic modular forms of weight -1/2 can be identified with the space $\tilde{J}_{0,1}$ of weakly holomorphic Jacobi forms of weight 0 and index 1." I feel comfortable with this statement as Eichler-Zagier explain this identification for holomorphic Jacobi forms and it certainly looks like the proof immediately generalizes to nearly holomorphic ones.

Next he says "The obstruction space $S_{5/2, L}$ can be identified in the same way with the space of skew-holomorphic Jacobi forms of weight 3 and index 1." This is the first thing I do not understand. Why is it skew-holomorphic and why not cusp form? It seems to me like it should be regular Jacobi cusp forms just looking at generalizing the proof in E-Z mentioned above.

In any case, he then says "This space equals 0. Therefore every Fourier polynomial is the principal part of a weak Jacobi form in $\tilde{J}_{0,1}$." This is the most important issue for me which I cannot reconcile with my understanding of Borcherd's products (which is tenuous at best). According to E-Z, $\tilde{J}_{0,1}$ is 1 dimensional, spanned by $\phi_{0,1}$ which is given in the example. However if this is so then $M^!_{-1/2, L}$ must also be 1-dimensional. But if the obstruction space is trivial as stated, then by Bruinier's Theorem 1.17, every possible principal part of a Fourier expansion should be realized for some element of $M^!_{-1/2, L}$ (as Bruinier was alluding to in his statement), meaning in particular that the space is infinite dimensional.

I would really appreciate any insight into this, particularly the last part regarding the dimension of $M^!_{-1/2, L}$, because I am curious as to whether other Humbert surfaces can be realized as the divisor of a Borcherd's lift.

Thanks for your time.