Frobenius の定理 - yoshitake-hのブログ

定理
$\theta =\left[ \begin{array}{c} \theta^{(1)} \\ \vdots \\ \theta^{(k)}\end{array}\right],\quad \theta^{(a)} =\sum_{i=1}^n \theta^{(a)}_i\mathrm{d}x_i +\sum_{j=1}^k \theta^{(a)}_{n+j}\mathrm{d}y_j$
を $\mathbb{R}^n\times \mathbb{R}^k$ 上の $C^\infty$ 1-form を成分とする列ベクトルとし，
$k\times k$ 行列 $\Theta =(\theta^{(a)}_{n+j})$ は可逆であるとする．
このときつぎは同値．
(1) 任意の $(x^0,\,y^0)\in \mathbb{R}^n\times \mathbb{R}^k$ に対し，
$y^0=f(x^0),$
$\theta^{(a)}_i(x,\,f(x)) +\sum_{j=1}^k \theta^{(a)}_{n+j}(x,\,f(x)) \frac{\partial f_j}{\partial x_i} =0,\quad i\in \{1,\,\dots,\,n\},\quad a\in \{1,\,\dots,\,k\}$
をみたす $C^\infty$ 写像 $f=(f_j)_{j=1}^k:\mathbb{R}^n\to \mathbb{R}^k$ が存在する．
(2) $\mathbb{R}^n\times \mathbb{R}^k$ 上の $C^\infty$ 1-form を成分とする $k\times k$ 行列
$\omega =(\omega^{(a,\,b)}),\quad \omega^{(a,\,b)} =\sum_{i=1}^n \omega^{(a,\,b)}_i\mathrm{d}x_i +\sum_{j=1}^k \omega^{(a,\,b)}_{n+j}\mathrm{d}y_j$
が存在して，
$\mathrm{d}\theta = \omega \wedge \theta.$

証明
$k\times k$ 可逆行列に値をとる関数 $P(x,\,y)$ に対し，
$\theta \longmapsto P\theta$
とおきかえても，条件 (1)，(2) は変わらない．
したがって，
$\theta^{(a)} =\sum_{i=1}^n \theta^{(a)}_i\mathrm{d}x_i -\mathrm{d}y_a$
の場合に示せばよい．このとき (1) は，
$\frac{\partial f_a}{\partial x_i} =\theta^{(a)}_i(x,\,f(x)).$

(1) ⇒ (2)
(1) より，
$\theta^{(a)}(x,\,f(x)) =\mathrm{d}(f_a(x)-y_a).$
よって
$\mathrm{d}\theta^{(a)} =0.$

(2) ⇒ (1)
$\theta^{(a)} =\sum_{i=1}^n\theta^{(a)}_i\mathrm{d}x_i -\mathrm{d}y_a,$
$\mathrm{d}\theta^{(a)} =\sum_{i<i'} (\frac{\partial \theta^{(a)}_{i'}}{\partial x_i} -\frac{\partial \theta^{(a)}_i}{\partial x_{i'}}) \mathrm{d}x_i\wedge \mathrm{d}x_{i'} -\sum_{i=1}^n \sum_{j=1}^k \frac{\partial \theta^{(a)}_i}{\partial y_j} \mathrm{d}x_i\wedge \mathrm{d}y_j,$
$\sum_{b=1}^k \omega^{(a,\,b)}\wedge \theta^{(b)} =\sum_{i<i'} \sum_b (\omega^{(a,\,b)}_i \theta^{(b)}_{i'} -\omega^{(a,\,b)}_{i'} \theta^{(b)}_i) \mathrm{d}x_i\wedge \mathrm{d}x_{i'}$
$-\sum_i\sum_b (\omega^{(a,\,b)}_i-\sum_c \omega^{(a,\,c)}_{n+b} \theta^{(c)}_i) \mathrm{d}x_i\wedge \mathrm{d}y_b$
$+\sum_{j<b} (\omega^{(a,\,b)}_{n+j} -\omega^{(a,\,j)}_{n+b}) \mathrm{d}y_j\wedge \mathrm{d}y_b.$
したがって，
$\frac{\partial \theta^{(a)}_{i'}}{\partial x_i} -\frac{\partial \theta^{(a)}_i}{\partial x_{i'}} =\sum_{b=1}^k (\omega^{(a,\,b)}_i \theta^{(b)}_{i'} -\omega^{(a,\,b)}_{i'} \theta^{(b)}_i),$
$-\frac{\partial \theta^{(a)}_i}{\partial y_b} =-\omega^{(a,\,b)}_i+\sum_c \omega^{(a,\,c)}_{n+b} \theta^{(c)}_i,$
$\omega^{(a,\,b)}_{n+j} -\omega^{(a,\,j)}_{n+b}=0.$

$g_j(t,\,u) =f_j(x^0+tu)$ とおくと，
$\frac{\partial g_j}{\partial t} =\sum_iu_i\frac{\partial f_j}{\partial x_i}(x^0+tu),\qquad \frac{\partial g_j}{\partial u_i} =t\frac{\partial f_j}{\partial x_i}(x^0+tu).$

常微分方程式
$\frac{\partial g_a}{\partial t}(t,\,u) =\sum_iu_i\theta^{(a)}_i(x^0+tu,\,g(t,\,u)),\qquad g_a(0,\,u)=y^0_a$
の解に対し，
$\frac{\partial}{\partial u_i}\frac{\partial g_a}{\partial t}$
$=\frac{\partial}{\partial u_i}\sum_{i'}u_{i'}\theta^{(a)}_{i'}(x^0+tu,\,g(t,\,u))$
$=\theta^{(a)}_i +\sum_{i'}u_{i'}(t\frac{\partial \theta^{(a)}_{i'}}{\partial x_i} +\sum_b\frac{\partial \theta^{(a)}_{i'}}{\partial y_b} \frac{\partial g_b}{\partial u_i}).$
$\frac{\partial}{\partial t}t\theta^{(a)}_i(x^0+tu,\,g(t,\,u))$
$=\theta^{(a)}_i +\sum_{i'}tu_{i'}\frac{\partial \theta^{(a)}_i}{\partial x_{i'}}+\sum_b t\frac{\partial \theta^{(a)}_i}{\partial y_b} \frac{\partial g_b}{\partial t}$
$=\theta^{(a)}_i +\sum_{i'}tu_{i'}\frac{\partial \theta^{(a)}_i}{\partial x_{i'}} +\sum_b t\frac{\partial \theta^{(a)}_i}{\partial y_b} \sum_{i'}u_{i'}\theta^{(b)}_{i'}.$
したがって，
$\frac{\partial}{\partial t} (\frac{\partial g_a}{\partial u_i}-t\theta^{(a)}_i(x^0+tu,\,g(t,\,u)))$
$=\sum_{i'}tu_{i'}(\frac{\partial \theta^{(a)}_{i'}}{\partial x_i} -\frac{\partial \theta^{(a)}_i}{\partial x_{i'}})$
$+\sum_{i',\,b}u_{i'}(\frac{\partial \theta^{(a)}_{i'}}{\partial y_b} \frac{\partial g_b}{\partial u_i} -\frac{\partial \theta^{(a)}_i}{\partial y_b} t\theta^{(b)}_{i'})$
$=\sum_{i',\,b}tu_{i'}(\omega^{(a,\,b)}_i\theta^{(b)}_{i'}-\omega^{(a,\,b)}_{i'}\theta^{(b)}_i)$
$+\sum_{i',\,b}u_{i'} ((\omega^{(a,\,b)}_{i'}-\sum_c\omega^{(a,\,c)}_{n+b}\theta^{(c)}_{i'})\frac{\partial g_b}{\partial u_i} -(\omega^{(a,\,b)}_i-\sum_c\omega^{(a,\,c)}_{n+b}\theta^{(c)}_i)t\theta^{(b)}_{i'})$
$=\sum_{i',\,b}u_{i'} (\omega^{(a,\,b)}_{i'}-\sum_c\omega^{(a,\,c)}_{n+b}\theta^{(c)}_{i'})\frac{\partial g_b}{\partial u_i} -\sum_{i',\,b}tu_{i'}\omega^{(a,\,b)}_{i'}\theta^{(b)}_i +\sum_{i',\,b,\,c}u_{i'}\omega^{(a,\,c)}_{n+b}\theta^{(c)}_it\theta^{(b)}_{i'}$
$=\sum_{i',\,b}u_{i'} (\omega^{(a,\,b)}_{i'}-\sum_c\omega^{(a,\,c)}_{n+b}\theta^{(c)}_{i'})(\frac{\partial g_b}{\partial u_i} -t\theta^{(b)}_i).$
最後に， $\omega^{(a,\,c)}_{n+b}=\omega^{(a,\,b)}_{n+c}$ を用いた．

$t=0$ のとき
$\frac{\partial g_a}{\partial u_i} -t\theta^{(a)}_i=0$
なので，常微分方程式の解の一意性より，
$\frac{\partial g_a}{\partial u_i} -t\theta^{(a)}_i=0.$
よって， $f_a(x)=g_a(1,\,x-x^0)$ とおくと，
これは (1) の解になる．//