Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(..this is almost a continuation of my last question (which got closed!)...) Let me first rewrite one of the main results of this paper, http://calvino.polito.it/~camporesi/JMP94.pdf in a coordinate system where the metric on $\mathbb{H}_n$ looks like, $ds^2 = \frac{1}{z^2}(dz^2 + \sum_{i=1}^{n-1} dx_i^2)$.

Then the attached paper explains that on $\mathbb{H}_n$ the spin-s fields are modelled by symmetric transverse traceless tensors of rank-s and we denote such fields as $h_{i_1 i_2\dots i_s}^{\lambda l m \sigma}$. We further want them to satisfy the following equation,

\begin{align} (-\nabla^2 )h^{\lambda lm \sigma}_{i_1 i_2\dots i_s} = (\lambda ^2 + \rho^2 + s)h^{\lambda lm \sigma}_{i_1 i_2\dots i_s} \end{align}

where $\rho = \frac{n-1}{2}$. The index $\lambda \geq 0$ (the lower bound coming from the requirement of square integrability) can be thought to correspond to the non-compact direction of $\mathbb{H}_n$ thought of in the hyperboloid model and the indices $lm\sigma$ corespond to the $S^{n-1}$ cross-section of it. Here $s \leq l \leq \infty$, $m = 0,1,2,3...,s$ for $n \geq 4$ and $m=0,1$ for $n = 3$ and $\sigma$ is a further degeneracy with multiplicity of $g(s) = \frac{ (2s+n-3)(s+n-4)! }{(n-3)!s!}$ (...for the special case of $n=3$, $g(0)=1$ and $g(s\geq 1) =2$...)

  • Is my above understanding (from the paper) correct?

Further the asymptotic behaviour of these eigenfunctions in these coordinates can be derived to be such that,

\begin{align} h^{\lambda l m \sigma}_{i_1 i_2 \dots i_s } \sim \#z^{-s-\rho +i\lambda} + \#z^{s-\rho-i\lambda } \end{align}

(where $\#$ stands for the dependence on the coordinates orthogonal to the $z$ axis)

  • Now let us say that I move away from the eigenvalue equation above and ask as to what are the functions among the above which lie in the kernel of $(-\nabla^2 + a)$ (for some constant $a$) and which also goes as some $z^p$ near $z=0$.

    Is this above question a valid question to ask? If yes then how can that be answered? (hopefully by using the asymptotics just stated above!) (...in case my last question is not well-defined then kindly frame and answer what you think would be the closest meaningful question..)

share|improve this question
    
If you set $a = -(\lambda^2+\rho^2+s)$, you get $\lambda = \sqrt{-(a+\rho^2+s)}$ (recall that $\rho$ is fixed by $n$). It seems that you already have the formula for the $z$-asymptotic behavior of solutions in terms of $\lambda$ and hence in terms of $a$ as well. Is that not sufficient? –  Igor Khavkine Oct 7 '13 at 9:58
    
@IgorKhavkine I am not getting you :) This value of $\lambda$ that you state only ensures that the function, $h^{\lambda l m \sigma}_{i_1 i_2 \dots i_s}$ is a solution of $(-\nabla ^2 +a)$, but how do I ensure the $z^p$ behaviour near $z=0$? (...and can you also kindly confirm or correct the counting of the solutions as stated in my first point?..) –  user6818 Oct 8 '13 at 0:05
    
I'm sorry, I'm not sure I can confirm your counting, as it would take me quite some time to go through the details. Also, I may have misunderstood your question. The asymptotic formulas you gave are for $z\to \infty$, not $z\to 0$. Right? Then you can ignore my comment. Still, if you can follow the original argument that leads to the $z\to \infty$ asymptotics, the same method should work for $z\to 0$. I'm afraid I might not be more helpful beyond this suggestion. –  Igor Khavkine Oct 8 '13 at 13:42
    
@IgorKhavkine The asymptotics I wrote down are for $z \rightarrow 0$. Is there anyway that can be used to pick out which of these satisfy the new differential equation and also go as $z^p$ near $z=0$? –  user6818 Oct 8 '13 at 19:38
    
That's the thing. As far as I can tell, the 'new' differential equation is the same as the old one (up to identifying $a$ with $\lambda$ using the formula I gave). So I just don't see where the puzzle is. –  Igor Khavkine Oct 9 '13 at 7:26
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.