# Green's function on torus

I have a question about the Green's function $$G(z,w)$$ on torus which takes the form (for example the first equation in the paper https://annals.math.princeton.edu/wp-content/uploads/annals-v172-n2-p03-p.pdf) $$$$-\nabla^2_z G(z,w)=\delta_w(z)-1/|T|$$$$ where the first term on the RHS is a delta function, and the second term on the RHS $$-1/|T|$$ is a constant term. This Green's function may have some relevance in CFT on a torus. My question is why this constant term come into play? What is the meaning of the constant term?

• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Jan 5 at 15:35

## 2 Answers

When both "space" ($$x$$) and "time" ($$y$$) directions are periodic, the Laplacian on torus with coodinate $$z=x+iy$$ has a normalized zero mode $$\varphi_0(z) = \frac 1 {\sqrt{{\rm Im}(\tau)}}$$ (Here $$\tau$$ is the modular parameter defining the torus.)

As $$-\nabla^2 \varphi_0=0.$$ the zero mode means that the Laplace operator is not 1-1 and so prevents the Laplacian with periodic boundary conditions from having an inverse. There is therefore no actual Green function. Instead we must therefore resort to a modified Green function. We can make use of a theta function with characteristics defined by $$\theta\left[ \matrix{a\cr b}\right] (z|\tau)= \sum_{m=-\infty}^\infty \exp\{i\pi \tau(m+a)^2 +2\pi i (m+a)(z+b)\}, \quad {\rm Im}(\tau)>0\quad a,b \in {\mathbb R}.$$

Observe that $$F(x,y) \equiv e^{-\pi y^2/ {\rm Im}(\tau)} \theta\left[ \matrix{\textstyle{\frac 12}\cr \textstyle{\frac 12}}\right] (z|\tau)$$ obeys $$F(x+1,y) =-F(x,y), \quad F(x+ {\rm Re}(\tau), y+{\rm Im}(\tau)) =({\rm phase}) F(x,y)$$ so $$G_0(x,y)) = -\frac{1}{2\pi} \ln |F| = -\frac{1}{2\pi} \ln\left| \theta\left[ \matrix{\textstyle{\frac 12}\cr \textstyle{\frac 12}}\right] (z|\tau)\right| + \frac 12 y^2 / {\rm Im}(\tau)\\ = -\frac 1{2\pi} \ln |E(z)| + \frac 12 y^2 / {\rm Im}(\tau)+const.$$ is both periodic on the torus and obeys $$-\nabla^2 G_0(x,y) = \delta^2(x,y) - 1/{\rm Im}(\tau)\\ = \sum_n \varphi_n(z) \varphi^*_n(0)- \varphi_0(z)\varphi_0(0)$$ Here the sum is over all $$n$$ including $$n=0$$. The $$\varphi_n(z)$$, $$n>0$$ are the eigenfuctions of the Laplacian with non-zero eigenvalues.

The modified Green function can be used to solve $$-\nabla^2 \phi= f(x,y)$$ as $$\phi(x,y) = \int_{\rm torus} G_0(x-x', y-y'))f(x',y') dx' dy'=0.$$ provided that $$f$$ is perpendicular to the the zero mode, i.e. $$\int_{\rm torus} f(x,y) dx dy=0.$$

This is consistent with the Fredholm alternative for linear operators.

• Thank you very much! This paper sciencedirect.com/science/article/pii/… "Conformal and current algebras on a general Riemann surface" have considered similar Green function for higher genus case in eq(8) which is $\nabla^z G^z(z,w)=\delta^{(2)}(z-w)-\sum_{j=1}^{3g-3}g^{z\bar{z}}\eta^z_{\bar{z},j}(z,\bar{z})h_{ww}^j(w)$, where the second term should correspond zero modes. So here what the zero modes is $\eta$ or $h$?($\eta$ is called Beltrami form and $h$ is holomophic quadratic differential) – phys_student Jan 5 at 16:14
• If $h$ or $\eta$ corresponds to zero mode, then the second term should be of the form with like $h^* h$ or $\eta^*\eta$, as $\phi^*_0\phi_0$ in the answer. However, the second term take the form $\eta h$. (Where I omit the indices $z,\bar{z}$) – phys_student Jan 5 at 16:37
• I don't have access to the Eguchi paper here, although I have read it a long time ago. I'll have a look at what they do when I get into the office tomorrow. – mike stone Jan 5 at 17:07
• h is the zero mode(s) and eta specifies a gauge slice – Wakabaloola Jan 6 at 14:40
• @Wakabaloola Yes, the Beltrami form is tangent to the gauge slice. – phys_student Jan 13 at 2:26

The constant term is needed because on a compact manifold with periodic boundary conditions there is a zero mode in the spectrum of the laplacian. This is easier to see on the circle, where $$\frac{d^2}{dt^2}f = \lambda f$$ has the periodic constant solution $$f(t) =c$$. This makes the operator not-inveritble.

One should instead work in the space orthogonal to the zero modes. Let's look at the spectral decomposition of the Green function: $$G(t, t') = \sum_\lambda u_\lambda^*(t) u_\lambda(t') \times \lambda^{-1},$$ where $$u_\lambda(t)$$are the eigenfunctions of the Operator. Note that $$\nabla^2 G = \sum_\lambda u_\lambda^*(t) u_\lambda(t') = \delta(t - t')$$ by completeness.

Now you see in the expression for the Green function why $$\lambda = 0$$ would be problematic. We remove this from the sum to work on the subspace orthogonal to this function, getting $$G'(t, t') = \sum_{\lambda \neq 0} u_\lambda^*(t) u_\lambda(t') \times \lambda^{-1}$$ Following which we get a modified Green equation $$\nabla^2 G = \sum_{\lambda \neq 0} u_\lambda^*(t) u_\lambda(t') = \delta(t - t') - u_0^*(t)u(t').$$ This last term corresponds to the constant piece you're subtracting in the case of periodic boundary conditions, because we saw that the zero eigenfunctions is constant. The factor of T is simply from some normalisation condition.

The other way of seeing it is to think in terms of electrostatics. On a compact manifold the periodicity is inconsistent with the Green function that represents the response to a point charge placed at some point: $$\int_{M} \delta(t, t') = \int_{M} \nabla^{2} G = \int_{\partial M}\nabla G \cdot dn = 0$$ since the Green function will be periodic. Yet the left hand side does not integrate to 0. To fix this we must add a constant negative "background charge" which modifies the LHS to $$\int_{M}\delta(t, t') - |c|^{2}$$ with $$c$$ chosen so that the integral vanishes. In other words this background charge is needed to ensure compatibility with the periodic boundary conditions.

• Thank you! Another question why the integral of Green function over the Torus equal zero, as discussed below the first eqn in the paper I cited. – phys_student Jan 5 at 16:25
• The Green function will inherit the boundary conditions of the operator and if periodic over a compact manifold its derivative will also be so, integrating around the boundary will then give you zero. – lux Jan 5 at 16:26
• If you're asking about the Green function itself the reasoning is pretty much the same – lux Jan 5 at 16:27
• P.s. Please don't forget to upvote if the answer helped – lux Jan 5 at 17:08
• Um, there's no inherent reason why the integral of the Green function itself would be zero, is there? This just arises because all these Green functions are, by construction, orthogonal to $u_0$ – and since that's a constant, the $L^2$ scalar product is just a simple integral. – leftaroundabout Jan 6 at 9:25