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As indicated by the title, I am looking to find the Green's function for the Laplacian on $S^1 \times S^2$. Is such a function known? If not, does anyone have an approach to constructing such a function? My first idea isto combine the Green's function on $S^2$ in a nice way with some $1$-periodic function on $\mathbb{R}$, but I haven't had much luck.

GiantTortoise1729
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    Is it possible to do the following: an element in $S^1 \times S^2$ may be written $(x_1, x_2, x_3, x_4, x_5)$ with $x_1^2+x_2^2 =1$, $x_3^2+x_4^2+x_5^2 = 1$. Embed into $\mathbb{R}^5$ via the identity map, say $\varphi$. Then just compose with the Green's function on $\mathbb{R}^5$, i.e. take $G(\varphi(x),\varphi(x'))$. It just seems like a rather naive approach and I'm unsure of myself. – GiantTortoise1729 Jul 15 '16 at 14:35
  • What are you mean by construction? Would a representation as some unwieldy series or an integral do? – Andrew Jul 18 '16 at 15:02
  • Yes, that would suffice – GiantTortoise1729 Jul 18 '16 at 18:02

1 Answers1

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Here's a way to get a representation as series. It works in a more general situation. Let $M$ and $N$ be smooth closed Riemannian manifolds. Denote by $\{\varphi_i\}_{i=1}^\infty$, $\{\psi_j\}_{i=j}^\infty$ the eigenfunctions with eigenvalues $\lambda_i$ and $\mu_j$ for the Laplace operators on $M$ and $N$ respectively. Let $L_2$ norms of eigenfunctions be equal to one so delta-functions can be expanded as $$ \delta_M(x-x')=\sum_{i=1}^\infty \varphi_i(x) \varphi_i(x'), $$ $$ \delta_N(y-y')=\sum_{i=1}^\infty \psi_i(y) \psi_i(y'). $$ Then Green's function for $M$ is given by series $$ G_M(x,x')=\sum_{i=1}^\infty \frac{\varphi_i(x) \varphi_i(x')}{\lambda_i} $$ and analogously for $N$. The Green's function for $M\times N$ is $$ G_{M\times N}(x,y,x',y')= \sum_{i,j=1}^\infty \frac{\varphi_i(x)\varphi_i(x') \psi_j(y)\psi_j(y')}{\lambda_i+\mu_j} $$ since $$ \Delta_{x,y} G_{M\times N}(x,y,x',y')= \sum_{i,j=1}^\infty \varphi_i(x)\varphi_i(x') \psi_j(y)\psi_j(y')= $$ $$ \delta_M(x-x')\delta_N(y-y')=\delta_{M\times N}(x-x',y-y'). $$ So one has to combine eigenfunctions rather than Green's functions themselves.

For your case $\varphi_i$ are cosines (up to a constant) and $\psi_j$ are spherical harmonics. A possibility to get a closed form for this series seems rather thin to me.

Andrew
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  • Yes, getting a closed form does seem dubious. But thank you very much! – GiantTortoise1729 Jul 19 '16 at 15:14
  • I will wait to award the bounty. – GiantTortoise1729 Jul 19 '16 at 15:14
  • do you know a nice way of writing down these eigenfunctions? – GiantTortoise1729 Jul 21 '16 at 15:33
  • Nothing more nice that a standard definition https://en.wikipedia.org/wiki/Spherical_harmonics – Andrew Jul 21 '16 at 19:55
  • +1 very nice, any good books on this subject? – nik Jul 22 '16 at 14:32
  • @nik Such an expansion for $G_M$ is well known and is called the bilinear expansion of the Green function. Search gives, for example https://books.google.ru/books?id=WLEVEY-3gIAC&pg=PA294&dq=bilinear+expansion+Green+function&hl=ru&sa=X&ved=0ahUKEwi46OPWionOAhWFjSwKHbhBAGEQ6AEIHTAA#v=onepage&q=bilinear%20expansion%20Green%20function&f=false and http://www2.ph.ed.ac.uk/~mevans/amm/section11.pdf – Andrew Jul 23 '16 at 07:51
  • I haven't a reference for $G_{M\times N}$, put up it thinking about the OP question. But it's the same bililnear formula if to note that $\varphi_i(x)\psi_j(y)$ are eigenfunctions for the Laplacian on $M\times N$ with eigenvalues $\lambda_i+\mu_j$. – Andrew Jul 23 '16 at 08:26
  • Sorry to bring up an old question...but are we allowed to pass the differential operator through the infinite sum and change the order of integration without first proving absolute convergence? Will absolute convergence always hold? – GiantTortoise1729 Aug 02 '16 at 19:54
  • @GiantTortoise1729 delta-functions are distributions and the series for them does not converge pointwise. E.g. for $S^1$ it is $\delta(x- x')=\frac1{2\pi}+\frac1\pi\sum_{n=1}^\infty \cos nx \cos nx'$. Convergence is understood here in the sense of distributions. And yes in the sense of distribution termwise differentiation is ok. – Andrew Aug 03 '16 at 06:11