数学物理学报, 2026, 46(5): 1951-1961

双接触等价下分歧问题对的有限决定性

刘苏卉,

江汉大学人工智能学院数学与大数据系 武汉 430056

The Finite Determinacy for Pair of Bifurcation Problems Under Bi-Contact Equivalence

Liu Suhui,

Department of Mathematics and Big Data, School of Artificial Intelligence, Jianghan University, Wuhan 430056

收稿日期: 2025-01-20   修回日期: 2026-02-6  

基金资助: 国家自然科学基金(10671009)
国家自然科学基金(60534080)

Received: 2025-01-20   Revised: 2026-02-6  

Fund supported: NSFC(10671009)
NSFC(60534080)

作者简介 About authors

刘苏卉,E-mail:estellaliu@163.com

摘要

利用奇点理论方法, 通过引入分歧问题对的双接触等价概念, 给出了两个分歧问题对之间双接触等价的充要条件, 并借助代数条件, 得到分歧问题对的双接触有限决定性的若干判定准则.

关键词： 双接触等价; 双接触有限决定性; 切空间; 模

Abstract

In this paper, bi-contact equivalence about pairs of bifurcation problems is introduced by singularity-theoretic techniques. A sufficient and necessary condition for bi-contact equivalence between two pairs of bifurcation problems is given. Some criteria about bi-contact finite determination of pair of bifurcation problems are then obtained in terms of an algebraic condition.

Keywords： bi-contact equivalence; bi-contact finite determinacy; tangent space; module

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本文引用格式

刘苏卉. 双接触等价下分歧问题对的有限决定性[J]. 数学物理学报, 2026, 46(5): 1951-1961

Liu Suhui. The Finite Determinacy for Pair of Bifurcation Problems Under Bi-Contact Equivalence[J]. Acta Mathematica Scientia, 2026, 46(5): 1951-1961

1 引言

分歧现象普遍存在于各类非线性问题中, 这类问题可刻画为光滑芽

$ G(x, \lambda):(\mathbb{R}^{n}\times \mathbb{R}^{l},0) \longrightarrow (\mathbb{R}^{p},0) $

在 $(0,0)$ 处关于 $x$ 的导数 $(dG)_{(0,0)}$ 具有奇异性. 研究分歧问题时, 等价性是核心概念之一: 在等价性框架下, 一个关键问题是确定在 $0$ 的邻域内某点的泰勒展开中, 哪些项是可被舍弃而不改变由 $G$ 与分歧参数 $\lambda$ 共同决定的拓扑类型——即分歧问题的决定性研究, 而有限决定性则是分析分歧问题识别问题的核心工具.

分歧问题的决定性研究目前已有丰富成果 (例如文献 [1-4]): Golubisky 借助奇点理论与群论方法, 系统构建了分歧问题有限决定性的 $C^{\infty}$ 理论[1,2]; 文献 [4-6] 则通过代数几何工具, 展开了分歧问题有限决定性的 $C^{0}$ 理论研究.

本文将 Costa 与 Birbrair 提出的双 $K$-等价思想[7,8]拓展至分歧理论, 针对型为

$(f_{1}, f_{2}):(\mathbb{R}^{n}\times \mathbb{R}^{l},0)\longrightarrow (\mathbb{R}^{p}\times \mathbb{R}^{q},0)$

的分歧问题芽对, 建立了双接触等价关系 (此类分歧问题对可表示为发散图 $(\mathbb{R}^{q},0)\overset{f_2}{\leftarrow} (\mathbb{R}^{n}\times \mathbb{R}^{l},0)\overset{f_1}{\rightarrow} (\mathbb{R}^{p},0)$). 在此基础上, 本文进一步讨论了分歧问题对的识别问题, 给出了两个分歧问题对间双接触等价的充要条件, 并基于代数条件推导得到了分歧问题对双接触有限决定性的若干判定准则.

2 分歧问题芽对的等价关系

记状态变量为 $ x=(x_{1},\cdots,x_{n}) \in\mathbb{R}^{n} $, 分歧参数为 $ \lambda=(\lambda_{1},\cdots,\lambda_{l})\in\mathbb{R}^{l} $. 导数用下标表示, 例如 $ f_{x} $ 表示 $ \frac{\partial f}{\partial x} $; 上标 $ ^{\circ} $ 表示任一函数在原点处的值, 即 $ f^{\circ}=f(0) $, $ f_{x}^{\circ}=f_{x}(0) $ 等.

设 $ \mathcal{E}_{x} $ 为光滑芽 $ f:(\mathbb{R}^{n},0)\longrightarrow \mathbb{R} $ 构成的环, $ \mathcal{M}_{x} $ 为其极大理想. $ \overrightarrow{\mathcal{E}_{x}^{m}} $ 表示光滑芽 $ g: (\mathbb{R}^{n},0)\longrightarrow \mathbb{R}^{m} $ 构成的 $ \mathcal{E}_{x} $-模, $ \overrightarrow{\mathcal{M}_{x}^{m}} $ 为其在原点处消失的芽构成的子模. 令

$\mathcal{E}_{x,\lambda}=\{h:(\mathbb{R}^{n+l},0)\longrightarrow \mathbb{R}\}$

为光滑芽环, $ \mathcal{M}_{x,\lambda} $ 为其极大理想; 令

$\overrightarrow{\mathcal{E}_{x,\lambda}^{m}}=\{f:(\mathbb{R}^{n+l},0)\rightarrow \mathbb{R}^{m}\}$

为光滑芽构成的 $ \mathcal{E}_{x,\lambda} $-模. 设 $ GL(n) $ 为所有 $ n\times n $ 实可逆矩阵构成的群, $ \mathcal{L}^{\circ}(n) $ 表示 $ GL(n) $ 中恒等映射所在的连通分支. 令

$ \mathcal{L}^{\circ}_{p,q}(p+q)=\left\{ \left( \begin{array}{cc} S_{1}& 0\\ [6pt] 0 &S_{2}\\ [6pt] \end{array} \right)\in\mathcal{L}^{\circ}(p+q) \right\}, $

其中 $ S_{1} $ 为 $ p\times p $ 矩阵, $ S_{2} $ 为 $ q\times q $ 矩阵.

设$ \mathbf{M}_{x,\lambda}^{m}=\{T:(\mathbb{R}^{n+l},0)\longrightarrow M _{m}(\mathbb{R})\} $

为光滑矩阵值映射构成的 $\mathcal{E}_{x,\lambda} $-模. 我们还需要以下$ \mathcal{E}_{x,\lambda} $-模

$ \overrightarrow{\Theta} _{x,\lambda}=\{ X:(\mathbb{R}^{n+l},0)\rightarrow \mathbb{R}^{n}\}, $

及其子模

$ \overrightarrow{\Theta} _{\lambda}=\{ X:(\mathbb{R}^{l},0)\rightarrow \mathbb{R}^{l}\} $

以及它们的子模

$ \overrightarrow{\Theta} _{x,\lambda}^{\circ}=\{ X\in \overrightarrow{\Theta} _{x,\lambda}\mid X^{\circ}=0\} $

和$ \overrightarrow{\Theta} _{\lambda}^{\circ}=\{ \Lambda\in \overrightarrow{\Theta} _{\lambda}\mid \Lambda^{\circ}=0\}. $

设 $f_{1}:(\mathbb{R}^{n}\times \mathbb{R}^{l},0)\longrightarrow (\mathbb{R}^{p},0) $ 为一个分歧问题芽, $f_{2}:(\mathbb{R}^{n}\times \mathbb{R}^{l},0)\longrightarrow (\mathbb{R}^{q},0) $ 为另一个分歧问题芽, 则 $(f_{1}, f_{2}):(\mathbb{R}^{n}\times \mathbb{R}^{l},0)\longrightarrow (\mathbb{R}^{p}\times \mathbb{R}^{q},0) $ 称为分歧问题芽对. 该芽对可表示为发散图 $(\mathbb{R}^{q},0)\overset{f_2}{\leftarrow} (\mathbb{R}^{n}\times \mathbb{R}^{l},0)\overset{f_1}{\rightarrow} (\mathbb{R}^{p},0) $.

为研究分歧问题芽对的识别问题, 合适的坐标变换应保持零点集和分歧参数的特殊地位. 因此, 我们引入双切触群 $\mathcal{KK}_{\lambda} $, 定义如下

$ \mathcal{KK}_{\lambda}=\{(T, X, \Lambda)\in \mathbf{M}_{x,\lambda}^{p+q}\times \overrightarrow{\Theta} _{x,\lambda}^{\circ}\times \overrightarrow{\Theta} _{\lambda}^{\circ}\mid T^{\circ}\in \mathcal{L}^{\circ}_{p,q}(p+q), X_{x,\lambda}^{\circ}\in\mathcal{L}^{\circ}(n), \Lambda_{\lambda}^{\circ}\in\mathcal{L}^{\circ}(l)\}, $

该群以自然方式作用于分歧问题芽对 $(f_{1},f_{2}) $ (其中$f_{1} \in \overrightarrow{\mathcal{E}_{x, \lambda}^{p}} $,; $f_{2} \in \overrightarrow{\mathcal{E}_{x, \lambda}^{q}} $), 作用形式为

$ (T, X, \Lambda)\cdot (f_{1}(x, \lambda),f_{2}(x, \lambda))=T(x, \lambda)\left(f_{1}(X(x, \lambda),\Lambda(\lambda)),f_{2}(X(x, \lambda),\Lambda(\lambda))\right). $

定义 2.1 设两个光滑分歧问题芽对 $(f_{1},f_{2}) $, $ (g_{1},g_{2}) $ (其中 $f_{1}, g_{1} \in \overrightarrow{\mathcal{E}_{x, \lambda}^{p}} $,; $f_{2}, g_{2} \in \overrightarrow{\mathcal{E}_{x, \lambda}^{q}} $), 若存在 $(T, X, \Lambda)\in \mathcal{KK}_{\lambda} $, 使得

$ (g_{1}(x, \lambda),g_{2}(x, \lambda))=(T, X, \Lambda)\cdot (f_{1}(x, \lambda),f_{2}(x, \lambda)), $

则称 $(f_{1},f_{2}) $ 与 $(g_{1},g_{2}) $ 双切触等价. 若 $\Lambda(\lambda)=\lambda $, 则称 $(f_{1},f_{2}) $ 与 $(g_{1},g_{2}) $ 强双切触等价.

注 2.1 若分歧问题芽对 $(f_{1},f_{2}) $ 与 $(g_{1},g_{2}) $ 双切触等价, 则分歧问题 $(f_{1},f_{2}) $ 与 $(g_{1},g_{2}) $ 均切触等价 (参见文献[1]).

类似于经典分歧理论, 可将许多涉及切触等价的经典结果推广到双切触等价, 得到新的结论.

定义 2.2 分歧问题芽对 $(f_{1},f_{2})\in \overrightarrow{\mathcal{E}_{x, \lambda}^{p}}\times \overrightarrow{\mathcal{E}_{x, \lambda}^{q}} $ 的双切触切空间定义为

$ T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))=\left\{\left.\frac{\rm d}{{\rm d}t}(\Phi_{t}\cdot(f_{1},f_{2}))\right|_{t=0}\in\overrightarrow{\mathcal{E}_{x, \lambda}^{p}}\times \overrightarrow{\mathcal{E}_{x, \lambda}^{q}} \mid \Phi_{t}\in \mathcal{KK}_{\lambda},\Phi_{0} =Id\right\}, $

强双切触切空间定义为

$\begin{eqnarray*} RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) &=&\left\{\left.\frac{\rm d}{{\rm d}t}(\Phi_{t}\cdot(f_{1},f_{2}))\right|_{t=0}\in\overrightarrow{\mathcal{E}_{x, \lambda}^{p}}\times \overrightarrow{\mathcal{E}_{x, \lambda}^{q}} \mid \Phi_{t}\right.\\ &=&\left.(T(t, x, \lambda), X(t, x, \lambda), \lambda)\in \mathcal{KK}_{\lambda},\Phi_{0} =Id\right\}. \end{eqnarray*}$

下面给出分歧问题芽对 $(f_{1},f_{2}) $ 双切触切空间的代数构造.由于

$\Phi_{t}=(T(t, x, \lambda), X(t, x, \lambda), \Lambda)\in \mathcal{KK}_{\lambda}, $

则

$\begin{align*} \left.\frac{\rm d}{{\rm d}t}(\Phi_{t}\cdot(f_{1},f_{2}))\right|_{t=0}=&\left.\frac{\rm d}{{\rm d}t}\left(T(t,x, \lambda)\left(f_{1}(X(t,x, \lambda),\Lambda(\lambda)),f_{2}(X(t,x, \lambda),\Lambda(\lambda))\right)\right)\right|_{t=0}\\ =&\left.\frac{{\rm d}}{{\rm d}t}\left( \begin{array}{cc} S_{1}& 0\\ [6pt] 0 &S_{2}\\ [6pt] \end{array} \right)\left(f_{1}(X(t, x, \lambda),\Lambda(\lambda)),f_{2}(X(t, x, \lambda),\Lambda(\lambda))\right)\right|_{t=0}. \end{align*}$

进一步展开得

$\begin{eqnarray*} &=&\left((\dot{S_{1}})_{0}f_{1}(x, \lambda), 0\right)+\left(0, (\dot{S_{2}})_{0}f_{2}(x, \lambda)\right)+\left((df_{1})_{x, \lambda}(\dot{X})_{0}(x, \lambda), 0\right)\\ &&+\left(0, (df_{2})_{x, \lambda}(\dot{X})_{0}(x, \lambda)\right)+\left((f_{1})_{ \lambda}(\dot{\Lambda})_{0}( \lambda), 0\right)+\left(0, (f_{2})_{ \lambda}(\dot{\Lambda})_{0}( \lambda)\right), \end{eqnarray*}$

其中

$ (\dot{S_{1}})_{0}\in\mathbf{M}_{x,\lambda}^{p}, \quad (\dot{S_{2}})_{0}\in\mathbf{M}_{x,\lambda}^{q}, \quad (\dot{X})_{0}(x, \lambda)\in\overrightarrow{\Theta} _{x,\lambda}^{\circ}, \quad (\dot{\Lambda})_{0}\in \overrightarrow{\Theta} _{\lambda}^{\circ}. $

特别地, 当 $\Lambda(t,\lambda)=\lambda $ 时, $(\dot{\Lambda})_{0}=0 $.

反之, 若

$ T=\left( \begin{array}{cc} S_{1}& 0\\ [6pt] 0 &S_{2}\\ [6pt] \end{array} \right),\quad S_{1}\in\mathbf{M}_{x,\lambda}^{p}, \quad S_{2}\in\mathbf{M}_{x,\lambda}^{q}, \quad X(x, \lambda)\in \overrightarrow{\Theta} _{x,\lambda}^{\circ}, \quad \Lambda\in \overrightarrow{\Theta} _{\lambda}^{\circ}, $

令

$ T(t, x, \lambda)=I+t\left( \begin{array}{cc} S_{1}& 0\\ [6pt] 0 &S_{2}\\ [6pt] \end{array} \right), $
$ X(t,x, \lambda)=x+tX(x, \lambda),\quad \Lambda(t,\lambda)=\lambda+\Lambda(\lambda). $

当 $t $ 充分小时, $(T(t, x, \lambda),X(t,x, \lambda),\Lambda(t,\lambda))\in \mathcal{KK}_{\lambda} $. 此外,

$\begin{align*} &\left.\frac{\rm d}{{\rm d}t}\left((T(t, x, \lambda), X(t, x, \lambda), \Lambda)\cdot(f_{1},f_{2})\right)\right|_{t=0}\\ =&\left.\frac{\rm d}{{\rm d}t}\left(T(t,x, \lambda)\left(f_{1}(X(t,x, \lambda),\Lambda(\lambda)),f_{2}(X(t,x, \lambda),\Lambda(\lambda))\right)\right)\right|_{t=0}\\ =&\left(S_{1}f_{1}(x, \lambda), 0\right)+\left(0, S_{2}f_{2}(x, \lambda)\right)+\left(({\rm d}f_{1})_{x, \lambda}(X)(x, \lambda), 0\right)+\left(0, ({\rm d}f_{2})_{x, \lambda}X(x, \lambda)\right)\\ =&\left(S_{1}f_{1}(x, \lambda), 0\right)+\left(0, S_{2}f_{2}(x, \lambda)\right)+\left(({\rm d}f_{1})_{x, \lambda}(X)(x, \lambda), 0\right)+\left(0, ({\rm d}f_{2})_{x, \lambda}X(x, \lambda)\right)\\ =&\left.\frac{\rm d}{{\rm d}t}\left(T(t,x, \lambda)\left(f_{1}(X(t,x, \lambda),\Lambda(\lambda)),f_{2}(X(t,x, \lambda),\Lambda(\lambda))\right)\right)\right|_{t=0}\\ &+\left((f_{1})_{ \lambda}\Lambda( \lambda), 0\right)+\left(0, (f_{2})_{\lambda}\Lambda( \lambda)\right). \end{align*}$

因此,

$\begin{eqnarray*} T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))&=&\Big\{\left(S_{1}f_{1}(x, \lambda),0\right)+\left(0, S_{2}f_{2}(x, \lambda)\right)+\left(({\rm d}f_{1})_{x, \lambda}X(x, \lambda), 0\right)\\ &&+\left(0, ({\rm d}f_{2})_{x, \lambda}X(x, \lambda)\right)+\left((f_{1})_{ \lambda}\Lambda(\lambda), 0\right)+\left(0, (f_{2})_{\lambda}\Lambda( \lambda)\right)\\ &&\mid S_{1}\in\mathbf{M}_{x,\lambda}^{p}, \, S_{2}\in\mathbf{M}_{x,\lambda}^{q}, \, X(x, \lambda)\in\overrightarrow{\Theta} _{x,\lambda}^{\circ}, \, \Lambda\in \overrightarrow{\Theta} _{\lambda}^{\circ}\Big\}\\ &=&\left\{I_{f_{1}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{p}}\times \{0\}+\{0\}\times I_{f_{2}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{q}}\right\}\\ &&+\mathcal{M}_{x,\lambda}\left\langle\frac{\partial(f_{1},f_{2})}{\partial x_{1}},\cdots,\frac{\partial(f_{1},f_{2})}{\partial x_{n}} \right\rangle+\mathcal{M}_{\lambda}\left\langle\frac{\partial(f_{1},f_{2})}{\partial \lambda_{1}},\cdots,\frac{\partial(f_{1},f_{2})}{\partial \lambda_{l}} \right\rangle. \end{eqnarray*}$

需注意, $T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 仅具有 $\mathcal{E}_{\lambda} $-模结构, 而非 $\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $ 的理想.

强双切触切空间的代数形式为

$\begin{align*} &RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\\ =&\left\{\left(S_{1}f_{1}(x, \lambda), S_{2}f_{2}(x, \lambda)\right)+\left(({\rm d}f_{1})_{x}X(x, \lambda), 0\right)+\left(0, ({\rm d}f_{2})_{x}X(x, \lambda)\right)\mid\right. \\ &\left.S_{1}\in\mathbf{M}_{x,\lambda}^{p}, \, S_{2}\in\mathbf{M}_{x,\lambda}^{q}, \, X(x, \lambda)\in\overrightarrow{\Theta} _{x,\lambda}^{\circ}\right\}. \end{align*}$

$ RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 是 $\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $ 的理想.

由于 $X(x, \lambda)\in\overrightarrow{\Theta} _{x,\lambda}^{\circ} $, 存在矩阵 $B_{1}=(b_{ij})_{n\times n} $ 和 $B_{2}=(b_{kl})_{n\times l} $ (其中 $b_{ij}, b_{kl} \in\mathcal{E}_{x,\lambda} $), 使得 $X(x, \lambda)=B_{1} x+B_{2}\lambda $. 因此,

$\begin{align*} RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) &=&\Big\{\left(S_{1}f_{1}(x, \lambda), S_{2}f_{2}(x, \lambda)\right)+\left(({\rm d}f_{1})_{x}(B_{1} x+B_{2}\lambda), 0\right)\\ &&+\left(0, ({\rm d}f_{2})_{x}(B_{1} x+B_{2}\lambda)\right)\Big\}. \nonumber \end{align*} $

定义 2.3 分歧问题芽对 $(f_{1},f_{2}) $ 的双切触余维数, 定义为 $T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 作为 $\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $ 的实向量子空间的余维数; 分歧问题芽对 $(f_{1},f_{2}) $ 的强双切触余维数, 定义为 $RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 作为 $\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $ 的实向量子空间的余维数.

定理 2.1 若两个分歧问题芽对 $(f_{1},f_{2}), (g_{1},g_{2})\in\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ} $ 强双切触等价, 则它们具有相同的强双切触余维数.

证 将集合 $\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ} $ 与直积 $\mathcal{E}_{x,\lambda}^{\circ}\times \cdots\times\mathcal{E}_{x,\lambda}^{\circ} $ (共 $p+q $ 个因子) 等同, 即将映射芽与其分量对应.

设

$ u= \left( \begin{array}{cc} a& 0\\ [6pt] 0 &b\\ [6pt] \end{array} \right) $

为可逆矩阵, 其中 $a=(a_{ij})_{p\times p} $ (元素 $a_{ij}\in\mathcal{E}_{x,\lambda} $), $b=(b_{ij})_{q\times q} $ (元素 $b_{ij}\in\mathcal{E}_{x,\lambda} $). 定义映射 $U:\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ}\rightarrow\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ} $ 为 $(f_{1},f_{2})\mapsto u\cdot(f_{1},f_{2}) $. 令$g_{1}=a\cdot f_{1} $,$g_{2}=b\cdot f_{2} $, 下面证明 $U: RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\rightarrow RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2})) $ 为同构.

首先证明

$\begin{equation} RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2}))\subset U\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right) \end{equation}$

和

$\begin{equation} RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\subset U^{-1}\left(RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2}))\right) \end{equation}$

由切空间的构造可知

$\begin{eqnarray*} RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))&=&\left\{I_{f_{1}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{p}}\times \{0\}+\{0\}\times I_{f_{2}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{q}}\right\}+\mathcal{M}_{x,\lambda}\left\langle\frac{\partial(f_{1},f_{2})}{\partial x_{1}},\cdots,\frac{\partial(f_{1},f_{2})}{\partial x_{n}} \right\rangle\\ &=&\left\{I_{f_{1}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{p}}\times \{0\}+\{0\}\times I_{f_{2}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{q}}\right\}+J_{(f_{1},f_{2})}, \end{eqnarray*}$

其中 $J_{(f_{1},f_{2})} $ 为 $(f_{1},f_{2}) $ 关于 $x $ 的雅可比理想.

结合映射 $U $ 的定义, 有

$\begin{eqnarray*} I_{g_{1}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{p}\times \{0\}+ \{0\}\times I_{g_{2}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{q}&\subset & U\left(I_{f_{1}}\cdot \overrightarrow{\mathcal{E}_{x,\lambda}^{p}}^{\circ}\times \{0\}+ \{0\}\times I_{f_{2}}\cdot \overrightarrow{\mathcal{E}_{x,\lambda}^{q}}^{\circ}\right)\\ &\subset & U\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right), \end{eqnarray*}$

且

$\begin{eqnarray*} J_{(g_{1},g_{2})} &\subset&U \left(I_{f_{1}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{p}}\times \{0\}+\{0\}\times I_{f_{2}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{q}}\right)+U\left(J_{(f_{1},f_{2})}\right)\\ &=& U \left(I_{f_{1}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{p}}\times \{0\}+\{0\}\times I_{f_{2}}\cdot\overrightarrow{\mathcal{E}_{x,\lambda}^{q}}+J_{(f_{1},f_{2})}\right)\\ &=& U\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right). \end{eqnarray*}$

再设 $\phi: (\mathbb{R}^{n}\times \mathbb{R}^{l},0 )\rightarrow (\mathbb{R}^{n}\times \mathbb{R}^{l},0 ) $ 为同构映射芽, 其诱导映射

$ \hat{\phi}:\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ}\rightarrow \overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ} $

定义为

$ (f_{1},f_{2})\mapsto (f_{1},f_{2})\circ \phi=(f_{1}\circ \phi, f_{2}\circ \phi)=(g_{1},g_{2}). $

则 $RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 与 $RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2})) $ 经 $\hat{\phi} $ 同构. 事实上,

$\begin{eqnarray*} \left\{I_{g_{1}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{p}\times \{0\}+ \{0\}\times I_{g_{2}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{q}\right\}&\subset &\hat{\phi}\left(I_{f_{1}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{p}\times \{0\}+ \{0\}\times I_{f_{2}}\cdot (\mathcal{E}_{x,\lambda}^{\circ})^{q}\right)\\ &\subset&\hat{\phi}\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right), \end{eqnarray*}$

同时,

$ J_{(g_{1},g_{2})}\subset\hat{\phi}\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right). $

这是因为

$\begin{eqnarray*} \left(\frac{\partial g_{1}}{\partial x_{i}},\frac{\partial g_{2}}{\partial x_{i}}\right)&=&\left(\sum_{j=1}^{n}\frac{\partial \phi_{j}}{\partial x_{i}}\cdot \frac{\partial f_{1}}{\partial x_{j}}\circ \phi,\sum_{j=1}^{n}\frac{\partial \phi_{j}}{\partial x_{i}}\cdot \frac{\partial f_{2}}{\partial x_{j}}\circ \phi\right)\\ &=&\sum_{j=1}^{n}\frac{\partial \phi_{j}}{\partial x_{i}}\cdot \left(\frac{\partial f_{1}}{\partial x_{j}}, \frac{\partial f_{2}}{\partial x_{j}}\right)\circ \phi, \end{eqnarray*}$

故

$ J_{(g_{1},g_{2})}\subset\hat{\phi}\left(RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\right). $

因此, $RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $ 与 $RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2})) $ 经 $\hat{\phi} $ 同构.

当 $(T, X, \lambda)\in\mathcal{KK}_{\lambda} $ 时, 由上述结论可知 $RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\cong RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2})) $. 从而

$ \overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ}/RT(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))\cong \overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}}^{\circ}/RT(\mathcal{KK_{\lambda}}\cdot(g_{1},g_{2})). $

故二者的强双切触余维数相等.

例 2.1(参见文献 [2, 第 213 页) 考虑分歧问题芽对 $(h_{1},h_{2} )=(x^{2}-y^{2}+\lambda, 2xy) $, 其中 $\text{rank}((h_{1})_{x})|_{(0,0)}=0 $, $\text{rank}((h_{2})_{x})|_{(0,0)}=0 $.

若

$\begin{eqnarray*} p&=&(p_{1},p_{2} )\in (\mathcal{M}_{x,y,\lambda}^{3}+\mathcal{M}_{x,y,\lambda}\langle \lambda \rangle)\overrightarrow{\mathcal{E}_{x,y,\lambda}^{2}}\\ &=&(\mathcal{M}_{x,y,\lambda}^{3}+\mathcal{M}_{x,y,\lambda}\langle \lambda \rangle)(\mathcal{E}_{x,y,\lambda}\oplus \mathcal{E}_{x,y,\lambda}), \end{eqnarray*}$

则 $h=(h_{1},h_{2} ) $ 与 $h+p $ 强切触等价.

取 $p=(\lambda x, \lambda) $, 则 $h+p=(x^{2}-y^{2}+\lambda+\lambda x, 2xy+\lambda) $. 下面计算 $T(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) $ 与 $T(\mathcal{KK_{\lambda}}\cdot(h_{1}+p_{1},h_{2}+p_{2})) $.

首先计算 $T(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) $:

$\begin{eqnarray*} T(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) &=&\Big\{\left(S_{1}h_{1}(x, \lambda), 0\right)+\left(0, S_{2}h_{2}(x, \lambda)\right)+\left(({\rm d}h_{1})_{x, \lambda}(X)(x, \lambda), 0\right)\\ &&+\left(0, ({\rm d}h_{2})_{x, \lambda}X(x, \lambda)\right)+\left((h_{1})_{\lambda}\Lambda(\lambda), 0\right)+\left(0, (h_{2})_{\lambda}\Lambda(\lambda)\right)\\ &&\mid S_{1}\in\mathbf{M}_{x,y,\lambda}^{1}, \, S_{2}\in\mathbf{M}_{x,y,\lambda}^{1}, \, X(x, \lambda)=(a,b)\in\overrightarrow{\Theta} _{x,y,\lambda}^{\circ}, \, \Lambda(\lambda)=c\in\mathbf{M}_{\lambda}^{1}\Big\}\\ &=&I_{ h_{1}}\cdot \mathcal{E}_{x,y,\lambda}\times \{0\}+ \{0\}\times I_{ h_{2}}\cdot \mathcal{E}_{x,y,\lambda}+(2ax-2by, 2ay+2bx)+(c,0), \end{eqnarray*}$

其中 $a (0,0,0)=0 $, $b (0,0,0)=0 $, $c(0)=0 $, 故可设 $a=a_{1}x+a_{2}y+a_{3}\lambda $, $b=b_{1}x+b_{2}y+b_{3}\lambda $, $c=c_{1}\lambda $.

由此可得 $RT(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) $ 的第一个分量为 $\langle x^{2}-y^{2}+\lambda, x^{2},y^{2},xy,\lambda x, \lambda y,\lambda\rangle $, 第二个分量为 $\langle x^{2},y^{2}, xy,\lambda x, \lambda y \rangle $. 因此, $RT(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) $ 的强双切触余维数为

$ \dim \left( \mathcal{E}_{x,y,\lambda}\oplus \mathcal{E}_{x,y,\lambda}/RT(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) \right)=\infty. $

另一方面, 计算 $T(\mathcal{KK_{\lambda}}\cdot(h_{1}+p_{1},h_{2}+p_{2})) $:

$\begin{eqnarray*} T(\mathcal{KK_{\lambda}}\cdot(h_{1}+p_{1},h_{2}+p_{2})) &=&T(\mathcal{KK_{\lambda}}\cdot(x^{2}-y^{2}+\lambda+\lambda x, 2xy+\lambda))\\ &=&\left( \langle x^{2}-y^{2}+\lambda+\lambda x,x^{2},y^{2},xy,\lambda x, \lambda y,(1+x)\lambda\rangle, \right.\\ &&\left. \langle 2xy+\lambda,x^{2},y^{2},xy,\lambda x, \lambda y,\lambda\rangle \right). \end{eqnarray*}$

故 $T(\mathcal{KK_{\lambda}}\cdot(h_{1}+p_{1},h_{2}+p_{2})) $ 的双切触余维数为有限值.

3 分歧问题芽对双切触等价的充要条件及有限决定性

类似于经典分歧理论, 可将许多涉及切触等价的经典结果推广到双切触等价, 得到新的结论.

设 $f=(f_{1},f_{2})\in\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $. 若 $p=(p_{1},p_{2})\in\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $ 满足 $(f_{1},f_{2})+t(p_{1},p_{2}) $ 与 $(f_{1},f_{2}) $ 双切触等价, 则 $p=(p_{1},p_{2})\in T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})) $.

反之, 设 $f=(f_{1},f_{2}) $, $p=(p_{1},p_{2})\in\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $, 且满足

$ T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))=T(\mathcal{KK_{\lambda}}\cdot((f_{1},f_{2})+t(p_{1},p_{2}))), $

那么 $(f_{1},f_{2})+t(p_{1},p_{2}) $ 与 $(f_{1},f_{2}) $ 是否双切触等价? 为此, 我们有如下定理.

定理 3.1 设 $f=(f_{1},f_{2}) $, $p=(p_{1},p_{2})\in\overrightarrow{\mathcal{E}_{x,\lambda}^{p+q}} $, 且

$ T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))=T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})+t(p_{1},p_{2})), $

则 $(f_{1},f_{2})+t(p_{1},p_{2}) $ 与 $(f_{1},f_{2}) $ 双切触等价.

证 令 $g(x,\lambda,t)=(g_{1}(x,\lambda,t),g_{2}(x,\lambda,t))=(f_{1},f_{2})+t(p_{1},p_{2}) $, 我们需要找到

$(T(x,\lambda,t), X(x,\lambda,t), \Lambda(\lambda,t))\in\mathcal{KK}_{\lambda}, $

使得

$\begin{align*} (f_{1}(x, \lambda),f_{2}(x, \lambda))&=&(T, X, \Lambda)\cdot (g_{1}(x, \lambda,t),g_{2}(x, \lambda,t))\\ &=&T(x, \lambda,t)\left(g_{1}(X(x, \lambda,t),\Lambda(\lambda,t),t),g_{2}(X(x, \lambda,t),\Lambda(\lambda,t),t)\right), \end{align*}$

其中

$ T( x, \lambda,t)=\left( \begin{array}{cc} S_{1}(x, \lambda,t)& 0\\ [6pt] 0 &S_{2}(x, \lambda,t)\\ [6pt] \end{array} \right). $

对式 (3.1) 关于 $t $ 求导, 得

$ 0= \left( \begin{array}{cc} (S_{1})_{t}(x, \lambda,t)& 0\\ [6pt] 0 &(S_{2})_{t}(x, \lambda,t)\\ [6pt] \end{array} \right) g(X(x, \lambda,t),\Lambda(\lambda,t))$
$ + \left( \begin{array}{cc} S_{1}(x, \lambda,t)& 0\\ [6pt] 0 &S_{2}(x, \lambda,t)\\ [6pt] \end{array} \right) g_{t}(X(x, \lambda,t),\Lambda(\lambda,t)) $
$\begin{eqnarray*} &=&\left((S_{1})_{t}(x, \lambda,t)g_{1}(x, \lambda,t), (S_{2})_{t}(x, \lambda,t)g_{2}(x, \lambda,t)\right)\\ &&+\left(S_{1}(x, \lambda,t)(f_{1})_{x}(X(x, \lambda,t), \Lambda(\lambda,t))X_{t}(x, \lambda,t), S_{2}(x, \lambda,t)(f_{2})_{x}(X(x, \lambda,t), \Lambda(t,\lambda))X_{t}(t,x, \lambda)\right)\\ &&+\left(S_{1}(x, \lambda,t)(f_{1})_{\lambda}\Lambda_{t}(x, \lambda), S_{2}(x, \lambda,t)(f_{2})_{\lambda}\Lambda_{t}(x, \lambda)\right)\\ &&+\left(S_{1}(x, \lambda,t)t(p_{1})_{x}(X(x, \lambda,t),\Lambda( \lambda,t))X_{t}(x, \lambda,t), S_{2}(x, \lambda,t)t(p_{2})_{x}(X(x, \lambda,t),\Lambda( \lambda,t))X_{t}(x, \lambda,t)\right)\\ &&+\left(S_{1}(x, \lambda,t)t(p_{1})_{\lambda}(X(x, \lambda,t),\Lambda( \lambda,t))\Lambda_{t}( \lambda,t), S_{2}(x, \lambda,t)t(p_{2})_{\lambda}(X(x, \lambda,t),\Lambda( \lambda,t))\Lambda_{t}( \lambda,t)\right)\\ &&+\left(S_{1}(x, \lambda,t)p_{1}(X(x, \lambda,t),\Lambda( \lambda,t)), S_{2}(x, \lambda,t)p_{2}(X(x, \lambda,t),\Lambda( \lambda,t))\right). \end{eqnarray*}$

求解上述方程关于 $p=(p_{1},p_{2}) $ 的表达式, 得

$\begin{align*} p(X(x, \lambda,t),\Lambda( \lambda,t))=&-T^{-1}(x, \lambda,t)T_{t}(x, \lambda,t)g(X(x, \lambda,t),\Lambda( \lambda,t),t)\\ &-(g)_{x}(X(x, \lambda,t), \Lambda(\lambda,t),t)X_{t}(x, \lambda,t)-(g_{\lambda})(X(x, \lambda,t),\Lambda(\lambda,t),t)\Lambda_{t}( \lambda,t). \end{align*}$

假设 $p $ 可表示为

$\begin{equation} p(x, \lambda)=-a(x, \lambda,t)g(x, \lambda,t)-(g)_{x}(x, \lambda,t)b(x, \lambda,t)-(g)_{\lambda}(x, \lambda,t)c(x, \lambda,t), \end{equation}$

其中

$ a(x, \lambda,t)= \left( \begin{array}{cc} a_{1}(x, \lambda,t)& 0\\ [6pt] 0 &a_{2}(x, \lambda,t)\\ [6pt] \end{array} \right), $

且 $a_{1}(x, \lambda,t)\in\mathbf{M}_{x,\lambda}^{p} $,$a_{2}(x, \lambda,t)\in\mathbf{M}_{x,\lambda}^{q} $,$b(x, \lambda,t)\in \overrightarrow{\mathcal{E}}_{(x, \lambda,t)}^{n} $,$c(x, \lambda,t)\in \overrightarrow{\mathcal{E}}_{(x, \lambda,t)}^{l} $. 接下来, 我们需要构造满足式 (3.2) 的 $a(x, \lambda,t) $、$ b(x, \lambda,t) $ 和 $c(x, \lambda,t) $. 事实上, 模

$ \left\{ \left( \begin{array}{cc} a_{1}(x, \lambda,t)& 0\\ [6pt] 0 &a_{2}(x, \lambda,t)\\ [6pt] \end{array} \right) \mid a_{1}(x, \lambda,t)\in\mathbf{M}_{x,\lambda}^{p}, a_{2}(x, \lambda,t)\in\mathbf{M}_{x,\lambda}^{q} \right\} $

由 $\{ S_{0}, S_{ij}, S_{kl} \mid 1\leq i,j\leq p; p+1\leq k,l\leq p+q\} $ 生成, 其中 $S_{0}=I_{p+q} $,

$ S_{ij}= \left( \begin{array}{cc} E_{ij}& 0\\ [6pt] 0 &0\\ [6pt] \end{array} \right),\quad S_{kl}=\left( \begin{array}{cc} 0& 0\\ [6pt] 0 &E_{kl}\\ [6pt] \end{array} \right), $

且 $\overrightarrow{\mathcal{M}}_{x,\lambda}=\langle x_{1}e_{1},\cdots,x_{n}e_{n},\lambda_{1}e_{n+1},\cdots,\lambda_{l}e_{n+l} \rangle $.

若 $h=(h_{1},h_{2})\in \overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q} $, 则 $T(\mathcal{KK_{\lambda}}\cdot(h_{1},h_{2})) $ 可由生成元集

$ \{ J_{ij}h,J_{kl}h,J_{s}h, J_{t}h \mid 1\leq i,j\leq p; p+1\leq k,l\leq p+q;$
$ p^{2}+q^{2}+1\leq s\leq p^{2}+q^{2}+n; p^{2}+q^{2}+n+1\leq t\leq p^{2}+q^{2}+n+l\} $

生成, 其中

$\begin{align*} J_{ij}h=&S_{ij}h\ (1\leq i,j\leq p);\quad J_{kl}h=S_{kl}h\ (p+1\leq k,l\leq p+q);\\ J_{s}h=&(h_{x}) x_{s-(p^{2}+q^{2})}e_{s-(p^{2}+q^{2})}\ (p^{2}+q^{2}+1\leq s\leq p^{2}+q^{2}+n);\\ J_{t}h=&(h_{\lambda}) \lambda_{t-(p^{2}+q^{2}+n)}e_{t-(p^{2}+q^{2}+n)}\ (p^{2}+q^{2}+n+1\leq t\leq p^{2}+q^{2}+n+l). \end{align*}$

令

$\begin{align*} J=&\Big(J_{11},\cdots,J_{1p},J_{21},\cdots,J_{2p},\cdots,J_{p1},\cdots,J_{pp},\\ &J_{(p+1)(p+1)},\cdots,J_{(p+1)(p+q)},\cdots,J_{(p+q)(p+1)},\cdots,J_{(p+q)(p+q)},\\ & J_{s}\ (p^{2}+q^{2}+1\leq s\leq p^{2}+q^{2}+n), J_{t}\ (p^{2}+q^{2}+n+1\leq t\leq p^{2}+q^{2}+n+l)\Big)^{T}. \end{align*}$

由条件 $T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2}))=T(\mathcal{KK_{\lambda}}\cdot(f_{1},f_{2})+t(p_{1},p_{2})) $, 存在矩阵

$ A=(a_{ij})_{(p^{2}+q^{2}+n+l)\times(p^{2}+q^{2}+n+l)}, $

其中 $a_{ij}\in\mathcal{E}_{x,\lambda}\ (1\leq j\leq p^{2}+q^{2}+n) $, $a_{ij}\in\mathcal{E}_{\lambda}\ (p^{2}+q^{2}+n+1\leq j\leq p^{2}+q^{2}+n+l) $, 使得

$\begin{equation} J(f+p)=AJ(f). \end{equation}$

由于 $J $ 是线性算子, 结合式 (3.3) 可得

$ J(p)=(A-I)J(f)=(A-I)J(g-tp). $

令 $B=A-I $, 则 $(I+tB)J(p)=BJ(g) $. 当 $t $ 充分小时,

$\begin{equation} J(p)=(I+tB)^{-1}BJ(g). \end{equation}$

又因为 $p=\sum_{1\leq s\leq p+q}J_{ss}p $, 结合式 (3.4) 有

$\begin{eqnarray*} J_{ss}p&=&\sum_{1\leq i,j\leq p}c_{i,j}^{s}J_{ij}g+\sum_{p+1\leq k,l\leq p+q}c_{k,l}^{s}J_{kl}g\\ &&+\sum_{p^{2}+q^{2}+1\leq v\leq p^{2}+q^{2}+n}c_{v}^{s}J_{v}g+\sum_{p^{2}+q^{2}+n+1\leq w\leq p^{2}+q^{2}+n+l}c_{w}^{s}J_{w}g, \end{eqnarray*}$

其中

$\begin{align*} &c_{i,j}^{s}\in\mathcal{E}_{x,\lambda,t}\ (1\leq i,j\leq p;1\leq s\leq p+q);\\ &c_{k,l}^{s}\in\mathcal{E}_{x,\lambda,t}\ (p+1\leq k,l\leq p+q;1\leq s\leq p+q);\\ & c_{v}^{s}\in\mathcal{E}_{x,\lambda,t}\ (p^{2}+q^{2}+1\leq v\leq p^{2}+q^{2}+n;1\leq s\leq p+q);\\ & c_{w}^{s}\in\mathcal{E}_{\lambda,t}\ (p^{2}+q^{2}+n+1\leq w\leq p^{2}+q^{2}+n+l;1\leq s\leq p+q). \end{align*}$

定义

$\begin{align*} a^{s}=& \sum_{1\leq i,j\leq p}c_{i,j}^{s}S_{ij}+\sum_{p+1\leq k,l\leq p+q}c_{k,l}^{s}S_{kl},\\ b^{s}= &\sum_{p^{2}+q^{2}+1\leq v\leq p^{2}+q^{2}+n}c_{v}^{s}x_{v-(p^{2}+q^{2})}e_{v-(p^{2}+q^{2})},\\ c^{s}=&\sum_{p^{2}+q^{2}+n+1\leq w\leq p^{2}+q^{2}+n+l}c_{w}^{s}\lambda_{w-(p^{2}+q^{2}+n)}e_{w-(p^{2}+q^{2}+n)}. \end{align*}$

最终构造

$ a=\sum_{1\leq s\leq p+q}a^{s},\quad b=\sum_{1\leq s\leq p+q}b^{s},\quad c=\sum_{1\leq s\leq p+q}c^{s}, $

则 $a,b,c $ 满足式 (3.2).

利用区间 $[0,1] $ 的紧性, 可得到对所有 $t\in [0,1] $, 存在 $a(x,\lambda,t) $、$ b(x,\lambda,t) $、$ c(x,\lambda,t) $ 满足式 (3.2).

定义 3.1 对任意映射 $f=(f_{1},f_{2}) $, 记 $j^{k}f $ 为 $f $ 的 $k $ 阶泰勒多项式 (或 $k $-芽). 若对所有满足 $j^{k}g=j^{k}f $ 的芽 $ g=(g_{1},g_{2})\in\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q} $, 均有 $g $ 与 $f $ 双切触 (强双切触) 等价, 则称芽 $ f=(f_{1},f_{2})\in\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q} $ 是 $k $-双切触决定的 ($ k $-强双切触决定的). 若存在整数 $k $ 使得 $f $ 是 $k $-双切触决定的 ($ k $-强双切触决定的), 则称 $f $ 是有限双切触决定的 (有限强双切触决定的).

通常, 有限双切触决定性与有限余维数之间存在密切联系.

引理 3.1(Nakayama 引理) 设 $A $ 为局部环, $m $ 为其极大理想, $P $ 为 $A $-模. 若 $M\subset P $ 是有限生成子模, $N\subset P $ 是子模, 且满足 $M\subset N+mM $, 则 $M\subset N $.

引理 3.2 分歧问题芽对 $f=(f_{1},f_{2})\in\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q} $ 的强双切触余维数有限, 当且仅当存在整数 $l\geq 1 $, 使得

$ \mathcal{M}_{x,\lambda}^{l}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}\subset RT\mathcal{KK_{\lambda}}(f_{1},f_{2}). $

证 考虑如下包含序列

$ RT\mathcal{KK_{\lambda}}(f_{1},f_{2})\supset RT\mathcal{KK_{\lambda}}(f_{1},f_{2})+\mathcal{M}_{x,\lambda}^{0}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ} \supset RT\mathcal{KK_{\lambda}}(f_{1},f_{2})+\mathcal{M}_{x,\lambda}^{1}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}\supset. $

由$\textbf{引理 3.1}$, 该序列在第 $l $ 步终止 (即稳定) 的充要条件为

$ RT\mathcal{KK_{\lambda}}(f_{1},f_{2})\supset \mathcal{M}_{x,\lambda}^{l}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}. $

定理 3.2 若分歧问题芽对 $f=(f_{1},f_{2})\in\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q} $ 的强双切触余维数有限, 则 $f $ 是有限强双切触决定的.

证 令 $p=j^{l}(f)-f\in \mathcal{M}_{x,\lambda}^{l+1}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ} $. 对任意 $t\in [0,1] $, 有

$ tp, \ t(dp)_{x}B_{1}x, \ t(dp)_{x}B_{2}\lambda \in \mathcal{M}_{x,\lambda}^{l}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}, $

其中 $B_{1}=(b_{ij})_{n\times n} $($ b_{ij}\in\mathcal{E}_{x,\lambda} $) 为 $n\times n $ 矩阵, $B_{2}=(b_{kl})_{n\times l} $($ b_{kl}\in\mathcal{E}_{x,\lambda} $) 为 $n\times l $ 矩阵.

由 $\mathcal{M}_{x,\lambda}^{l+1}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}\subset RT\mathcal{KK_{\lambda}}(f_{1},f_{2}) $ 及式 (2.2), 可得

$ RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2})\subset RT\mathcal{KK_{\lambda}}(f_{1},f_{2}). $

反之, 由于

$ f=(f_{1}+tp_{1},f_{2}+tp_{2})-t(p_{1},p_{2})\in RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2}) +\mathcal{M}_{x,\lambda}^{l}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}, $

结合 $\mathcal{M}_{x,\lambda}^{l}\overrightarrow{\mathcal{E}}_{x,\lambda}^{p+q,\circ}\subset RT\mathcal{KK_{\lambda}}(f_{1},f_{2}) $, 有

$ f\in RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2})+\mathcal{M}_{x,\lambda} RT\mathcal{KK_{\lambda}}(f_{1},f_{2}). $

同理可证, 对矩阵 $B_{1}=(b_{ij})_{n\times n} $ ($ b_{ij}\in\mathcal{E}_{x,\lambda} $) 和 $B_{2}=(b_{kl})_{n\times l} $($ b_{kl}\in\mathcal{E}_{x,\lambda} $), 有

$ t(df)_{x}B_{1}x, \ t(df)_{x}B_{2}\lambda \in RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2})+\mathcal{M}_{x,\lambda} RT\mathcal{KK_{\lambda}}(f_{1},f_{2}). $

因此

$ RT\mathcal{KK_{\lambda}}(f_{1},f_{2})\subset RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2})+\mathcal{M}_{x,\lambda} RT\mathcal{KK_{\lambda}}(f_{1},f_{2}). $

由$\textbf{引理 3.1}$, 得

$ RT\mathcal{KK_{\lambda}}(f_{1},f_{2})\subset RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2}). $

综上,

$ RT\mathcal{KK_{\lambda}}(f_{1},f_{2})= RT\mathcal{KK_{\lambda}}(f_{1}+tp_{1},f_{2}+tp_{2}). $

根据$\textbf{定理 3.1}$, 芽对 $f=(f_{1},f_{2}) $ 与 $f+tp=(f_{1}+tp_{1},f_{2}+tp_{2}) $ 强双切触等价. 特别地, 芽对 $f=(f_{1},f_{2}) $ 与 $j^{l}(f)=(j^{l}f_{1},j^{l}f_{2}) $ 强双切触等价, 故 $f $ 是有限强双切触决定的.

致谢

作者感谢审稿人仔细审阅本文并提出富有建设性的意见, 这些意见为完善论文终稿提供了重要帮助.

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