The Lapacian on the 2-Torus

Visualization

Let us use this local parametrization to visualize the 2-torus for \(a=4\) and \(r=1\).

a <- 4
r <- 1

# Parameter defining how many points to take. 
frac_density <- 0.03
# Get parameter grid. 
u_seq <- seq(from = 0, to = 2*pi, by = frac_density)
v_seq <- seq(from = 0, to = 2*pi, by = frac_density)

# Generate surface from local coordinates. 
local_parametrization <- function(a, r, u, v) {
  
  x <- (a + r*cos(u))*cos(v)
  y <- (a + r*cos(u))*sin(v)
  z <- r*sin(u)
  
  local_chart <- list(x = x, y = y, z = z)
  local_chart_df <- data.frame(local_chart)
  
  return(local_chart_df)
}

coordinates_df <- expand.grid(... = u_seq, v_seq)
colnames(coordinates_df) <- c("u", "v")

local_parametrization_list <- apply(
  X = coordinates_df, 
  MARGIN = 1, 
  FUN = function(x) {local_parametrization(a = a, r = r, u = x[1] , v = x[2])}
)

local_parametrization_df <- do.call(what = rbind, args = local_parametrization_list)
rownames(local_parametrization_df) <- c(1:nrow(local_parametrization_df))

Now we use a 3d-scatter plot:

library(plotly)

p <- plot_ly(
  data = local_parametrization_df,
  x = ~ x, 
  y = ~ y, 
  z = ~ z, 
  color = ~ z
)

layout(p = p, scene = list(zaxis = list(range = c(-4,4))))

Euclidean Metric

The metric on \(T^2\) induced by the Euclidean metric of \(\mathbb{R}^3\) is given by \[ g^{T^2}=r^2du^2+(a+r\cos u)^2dv^2 \] Let us prove this. Given the local parametrization \(\phi(u, v) = (x(u, v), y(u, v), z(u, v))\) defined above, we need to compute \(\partial_u\phi\) and \(\partial_v\phi\). The components of the induced metric are then \[ g^{T^2} = E_{u}du^2 + E_{uv}du\otimes dv + E_{v}dv^2, \] where \[ E_{u} = \langle\partial_{u}\phi, \partial_{u}\phi\rangle_{\mathbb{R}^3}, \\ E_{uv} = \langle\partial_{u}\phi, \partial_{v}\phi\rangle_{\mathbb{R}^3}, \\ E_{v} = \langle\partial_{v}\phi, \partial_{v}\phi\rangle_{\mathbb{R}^3} \] We now calculate: \[ \frac{\partial x(u, v)}{\partial u} = -r\sin u\cos v, \quad \frac{\partial y(u, v)}{\partial u} = -r\sin u\sin v \quad \text{and} \quad \frac{\partial z(u, v)}{\partial u} = r \cos u, \] which implies: \[ E_u = ( -r\sin u\cos v)^2 + (-r\sin u\sin v )^2 + (r \cos u)^2 = r^2 \] Similarly, \[ \frac{\partial x(u, v)}{\partial v} = - (a+r\cos u)\sin v , \quad \frac{\partial y(u, v)}{\partial v} = (a+r\cos u)\cos v \quad \text{and} \quad \frac{\partial z(u, v)}{\partial v} = 0, \] and \[ E_v = (- (a+r\cos u)\sin v)^2 + (- (a+r\cos u)\cos v)^2 + 0^2 = (a+r\cos u)^2 \] Finally, we see that the crossed term vanishes: \[ E_{uv} = ( -r\sin u\cos v)(- (a+r\cos u)\sin v ) + (-r\sin u\sin v )(a+r\cos u)\cos v = 0 \]

Remark: In matrix form, with respect to the basis \(du, dv\), we can write the metric as \[ g^{T^2} = \left( \begin{array}{cc} r^2 & 0 \\ 0 & (a+r\cos u)^2 \end{array} \right) \]

Volume Form

The volume form \(\text{vol}_{T^2}\) on the 2-torus is defined by the relation \(\text{vol}_{T^2} = \sqrt{\det(g^{T^2})} du\wedge dv\) (the sign \(du\wedge dv\) is a choise orientation). Hence, \(\text{vol}_{T^2}=r(a+r\cos u)du\wedge dv\).

Hodge Star Operator

Recall that the Hodge star operator is definied by the relation \[ \alpha \wedge *\beta = \langle \alpha, \beta \rangle \text{vol}_{T^2}, \] for \(\alpha, \beta \in \bigwedge^*T^*(T^2)\) (or differential forms). The inner product is defined on the space of \(n\)-forms. In particular, \[ \langle du, du \rangle = \frac{1}{r^2} \quad\text{and}\quad \langle dv, dv \rangle = \frac{1}{(a+r\cos u)^2}. \] Note that \(*1 = \text{vol}_{T^2}\).

Let us now compute the Hodge star operator on the \(1\)-forms \(du\) and \(dv\):

\[ \langle du, du \rangle \text{vol}_{T^2}=r^{-2}r(a+r\cos u)du\wedge dv, \] so we see that \(*du=r^{-1}(a+r\cos u)dv\). Similarly we obtain \(*dv=-r(a+r\cos u)^{-1}du\).

Dirac Operator

The Dirac operator of the Clifford bundle of differential forms is given by \(D = d + d^{\dagger}\), where \(d^\dagger = -*d*\) (as the dimension is even). For example, for a smooth function \(f\in C^{\infty}(T^2)\), we have \[ D f = df = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv \] and for a \(1\)-form \(f du\) we obtain \[ D(fdu) = d(fdu) -f*d*du = - \frac{\partial f}{\partial dv} du\wedge dv-f*d\left(r^{-1}(a+r\cos u)dv\right) \] Hence, \[ D(fdu) = - \frac{\partial f}{\partial dv} du\wedge dv + f \sin u * (du\wedge dv) = - \frac{\partial f}{\partial dv} du\wedge dv + \frac{f \sin u}{r(a + r \cos u)} \]

Laplacian

The Laplacian is defined by \(\Delta = D^2\). Let us compute it explicitly for a smooth function \(f\in C^{\infty}(T^2)\). As we have seen \(Df = df\), hence \[ \Delta = D(df) = (d + d^\dagger) df = d^\dagger df \] On the one hand side, \[ df = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv \]

Let us study the first term: \[ d^{\dagger} \left(\frac{\partial f}{\partial u}du\right) = -*d*\left(\frac{\partial f}{\partial u}du\right) \\ = - * d\left(r^{-1}(a+r\cos u)\frac{\partial f}{\partial u} dv\right) \\ -*\left(-\sin u\frac{\partial f}{\partial u} + r^{-1}(a+r\cos u)\frac{\partial^2 f}{\partial u^2}\right)du\wedge dv\\ = \frac{1}{r(a + r\cos u)}\left(\sin u\frac{\partial f}{\partial u} - r^{-1}(a+r\cos u)\frac{\partial^2 f}{\partial u^2}\right) \] The second term can be computed similarly \[ d^{\dagger} \left(\frac{\partial f}{\partial v}dv\right) = -*d*\left(\frac{\partial f}{\partial v}dv\right) \\ = * d\left(r(a+r\cos u)^{-1}\frac{\partial f}{\partial v} du\right) \\ = - * \left(r(a+r\cos u)^{-1}\frac{\partial^2 f}{\partial v^2} du\wedge dv\right) \\ = - \frac{1}{r(a + r\cos u)}\left(r(a+r\cos u)^{-1}\frac{\partial^2 f}{\partial v^2}\right) \] Hence, \[ \Delta f =\frac{1}{r(a + r\cos u)}\left(\sin u\frac{\partial f}{\partial u} - r^{-1}(a+r\cos u)\frac{\partial^2 f}{\partial u^2}\right) - \frac{1}{r(a + r\cos u)}\left(r(a+r\cos u)^{-1}\frac{\partial^2 f}{\partial v^2}\right) \] which simplifies to \[ \Delta f = -\frac{1}{r^2}\frac{\partial^2 f}{\partial u^2} - \frac{1}{(a + r\cos u)^2}\frac{\partial^2 f}{\partial v^2} + \frac{\sin u}{r(a + r\cos u)}\frac{\partial f}{\partial u}. \]

For one and two forms the computation is very similar.