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   "source": [
    "# 2次元トーラスのオイラー形式\n",
    "\n",
    "トーラスにはガウス曲率が正の領域と負の領域がありますが、その総和は\n",
    "相殺されます。\n",
    "\n",
    "$$\\int_{T^2}e(TT^2)=\\chi(T^2)=0.$$\n",
    "\n",
    "このNotebookでは、埋め込まれたトーラスから局所的なオイラー形式を計算し、\n",
    "この相殺を明示します。\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "bb3dad5f",
   "metadata": {},
   "source": [
    "## 埋め込まれたトーラスと計量\n",
    "\n",
    "$a$ を管の半径、$b$ を管の中心から対称軸までの距離とします。\n",
    "$a\\cos\\theta+b$ を囲むopaque quote（不透明引用）により、行列計算の途中でも、\n",
    "この式をひとまとまりのコンパクトな記号式として保持します。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "id": "441f6f87",
   "metadata": {},
   "outputs": [],
   "source": [
    "declare symbol θ, φ, a, b : MathValue\n",
    "\n",
    "def x : Vector MathValue := [| θ, φ |]\n",
    "def X : Vector MathValue :=\n",
    "  [| `(a * cos θ + b) * cos φ\n",
    "   , `(a * cos θ + b) * sin φ\n",
    "   , a * sin θ |]\n",
    "\n",
    "def e_i_j : Matrix MathValue := ∂/∂ X_j x~i\n",
    "def g_i_j : Matrix MathValue :=\n",
    "  generateTensor (\\[u, v] -> V.* e_u_# e_v_#) [2, 2]\n",
    "def g~i~j : Matrix MathValue := M.inverse g_#_#\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "008a9076",
   "metadata": {},
   "outputs": [
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     "data": {
      "text/html": [
       "$\\begin{pmatrix} a^{2} & 0 \\\\ 0 & (\\cos(θ) a + b)^{2} \\\\ \\end{pmatrix}_{\\#\\#}^{\\;\\;}$"
      ]
     },
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     "output_type": "display_data"
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   "source": [
    "g_#_#\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "0a8ae65f",
   "metadata": {},
   "source": [
    "## 正規直交枠での接続\n",
    "\n",
    "多脚場（vielbein）によって座標のスケール因子を取り除きます。球面の場合と同様に、\n",
    "枠を変換するときには非斉次項 $A^{-1}dA$ が不可欠です。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "1d78635e",
   "metadata": {},
   "outputs": [],
   "source": [
    "def Γ_i_j_k : Tensor MathValue :=\n",
    "  (1 / 2) *\n",
    "    (∂/∂ g_i_k x~j + ∂/∂ g_i_j x~k - ∂/∂ g_j_k x~i)\n",
    "\n",
    "def Γ~i_j_k : Tensor MathValue := withSymbols [m]\n",
    "  g~i~m . Γ_m_j_k\n",
    "\n",
    "def A : Matrix MathValue :=\n",
    "  [| [| 1 / a, 0 |]\n",
    "   , [| 0, 1 / `(a * cos θ + b) |] |]\n",
    "\n",
    "def d (t : Tensor MathValue) : Tensor MathValue :=\n",
    "  !(flip ∂/∂) x t\n",
    "\n",
    "def ω0~i_j : Matrix MathValue := Γ~i_j_#\n",
    "def ω~i_j : Tensor MathValue := withSymbols [u, v]\n",
    "  (M.inverse A)~i_u . ω0~u_v . A~v_j\n",
    "  + (M.inverse A)~i_u . d A~u_j\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "2e214987",
   "metadata": {},
   "source": [
    "ゼロでない接続係数はトーラス上で符号を変え、それに伴って曲率の符号も\n",
    "変化します。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "4b78f197",
   "metadata": {},
   "outputs": [
    {
     "data": {
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       "$\\sin(θ)$"
      ]
     },
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    }
   ],
   "source": [
    "ω~1_2_2\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f92fa3c1",
   "metadata": {},
   "source": [
    "## 曲率とオイラー形式\n",
    "\n",
    "カルタンの方程式を適用した後、2つの形式添字を明示的に反対称化し、\n",
    "階数2のパフィアン（Pfaffian）を取ります。\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "id": "d808dc93",
   "metadata": {},
   "outputs": [],
   "source": [
    "def Ω~i_j : Tensor MathValue := withSymbols [k]\n",
    "  antisymmetrize (d ω~i_j + ω~i_k ∧ ω~k_j)\n",
    "\n",
    "def eulerForm : Tensor MathValue :=\n",
    "  (1 / (4 * π)) * withSymbols [t1, t2]\n",
    "    (Ω~1_2_t1_t2 - Ω~2_1_t1_t2)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "88b639ca",
   "metadata": {},
   "outputs": [
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     "data": {
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       "$\\begin{pmatrix} 0 & \\frac{1}{4} \\cos(θ) π^{-1} \\\\ \\frac{-1}{4} \\cos(θ) π^{-1} & 0 \\\\ \\end{pmatrix}$"
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   "source": [
    "eulerForm\n"
   ]
  },
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   "cell_type": "markdown",
   "id": "a1005146",
   "metadata": {},
   "source": [
    "上三角側のテンソル成分は $\\cos\\theta/(4\\pi)$ です。その反対称な相方も\n",
    "含めると、向き付き密度\n",
    "$\\cos\\theta\\,d\\theta\\wedge d\\phi/(2\\pi)$ が得られます。子午線方向を\n",
    "一周する積分はゼロになります。\n",
    "\n",
    "$$\\int_0^{2\\pi}\\cos\\theta\\,d\\theta=0.$$\n",
    "\n",
    "したがって、外側の正の曲率と内側の負の曲率が相殺され、\n",
    "$\\chi(T^2)=0$ が得られます。この結果は $a$ と $b$ に依存しません。\n"
   ]
  }
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