Euler Form of the Two-Torus

A torus has regions of positive and negative Gaussian curvature, but their total cancels:

$$\int_{T^2}e(TT^2)=\chi(T^2)=0.$$

This notebook computes the local Euler form from an embedded torus and makes that cancellation explicit.

Embedded torus and metric

Let $a$ be the tube radius and $b$ the distance from the tube center to the symmetry axis. The opaque quotes around $a\cos\theta+b$ preserve a compact symbolic atom during intermediate matrix computations.

declare symbol θ, φ, a, b : MathValue

def x : Vector MathValue := [| θ, φ |]
def X : Vector MathValue :=
  [| `(a * cos θ + b) * cos φ
   , `(a * cos θ + b) * sin φ
   , a * sin θ |]

def e_i_j : Matrix MathValue := ∂/∂ X_j x~i
def g_i_j : Matrix MathValue :=
  generateTensor (\[u, v] -> V.* e_u_# e_v_#) [2, 2]
def g~i~j : Matrix MathValue := M.inverse g_#_#
g_#_#
$\begin{pmatrix} a^{2} & 0 \\ 0 & (\cos(θ) a + b)^{2} \\ \end{pmatrix}_{\#\#}^{\;\;}$

Connection in an orthonormal frame

The vielbein removes the coordinate scale factors. As on the sphere, the inhomogeneous $A^{-1}dA$ term is essential when changing frames.

def Γ_i_j_k : Tensor MathValue :=
  (1 / 2) *
    (∂/∂ g_i_k x~j + ∂/∂ g_i_j x~k - ∂/∂ g_j_k x~i)

def Γ~i_j_k : Tensor MathValue := withSymbols [m]
  g~i~m . Γ_m_j_k

def A : Matrix MathValue :=
  [| [| 1 / a, 0 |]
   , [| 0, 1 / `(a * cos θ + b) |] |]

def d (t : Tensor MathValue) : Tensor MathValue :=
  !(flip ∂/∂) x t

def ω0~i_j : Matrix MathValue := Γ~i_j_#
def ω~i_j : Tensor MathValue := withSymbols [u, v]
  (M.inverse A)~i_u . ω0~u_v . A~v_j
  + (M.inverse A)~i_u . d A~u_j

The nonzero connection coefficient changes sign across the torus and drives the sign-changing curvature.

ω~1_2_2
$\sin(θ)$

Curvature and Euler form

We explicitly antisymmetrize the two form indices after applying Cartan's equation, then take the rank-two Pfaffian.

def Ω~i_j : Tensor MathValue := withSymbols [k]
  antisymmetrize (d ω~i_j + ω~i_k ∧ ω~k_j)

def eulerForm : Tensor MathValue :=
  (1 / (4 * π)) * withSymbols [t1, t2]
    (Ω~1_2_t1_t2 - Ω~2_1_t1_t2)
eulerForm
$\begin{pmatrix} 0 & \frac{1}{4} \cos(θ) π^{-1} \\ \frac{-1}{4} \cos(θ) π^{-1} & 0 \\ \end{pmatrix}$

The upper tensor component is $\cos\theta/(4\pi)$; including its antisymmetric partner gives the oriented density $\cos\theta\,d\theta\wedge d\phi/(2\pi)$. Its integral over one full meridian vanishes:

$$\int_0^{2\pi}\cos\theta\,d\theta=0.$$

Hence the positive outer curvature and negative inner curvature cancel, giving $\chi(T^2)=0$ independently of $a$ and $b$.

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