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_#_#
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
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
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 $b$.