$\mathrm{End}(\mathbb{R}^2)$的商集、商空间分析及Hausdorff性判定问询
Let's break down this problem step by step, starting with the quotient set structure, then moving to the quotient topology, and finally addressing the Hausdorff property.
1. Quotient Set $X/\sim$: Similarity Classes
Two 2x2 real matrices are similar if and only if they have identical real Jordan canonical forms. This gives us four distinct types of similarity classes:
Distinct real eigenvalues $\lambda \neq \mu$:
Represented by the diagonal matrix $\begin{pmatrix} \lambda & 0 \ 0 & \mu \end{pmatrix}$. Classes here are unordered pairs of distinct real numbers (since swapping $\lambda$ and $\mu$ gives a similar matrix via $\begin{pmatrix}0&1\1&0\end{pmatrix}$).Repeated real eigenvalue $\lambda$, non-diagonalizable:
Represented by the Jordan block $\begin{pmatrix} \lambda & 1 \ 0 & \lambda \end{pmatrix}$. These are matrices with trace $2\lambda$, determinant $\lambda^2$, and cannot be diagonalized.Repeated real eigenvalue $\lambda$, diagonalizable:
Represented by the scalar matrix $\begin{pmatrix} \lambda & 0 \ 0 & \lambda \end{pmatrix}$. Each scalar matrix is only similar to itself.Complex conjugate eigenvalues $\alpha \pm i\beta$ ($\beta \neq 0$):
Represented by the real Jordan form $\begin{pmatrix} \alpha & \beta \ -\beta & \alpha \end{pmatrix}$. These matrices have no real eigenvalues, and their characteristic polynomial is irreducible over $\mathbb{R}$.
In short, the quotient set is a union of these four families of classes, parameterized by real numbers (or pairs of real numbers) corresponding to the invariants above.
2. Quotient Space Topology
Since $X$ is homeomorphic to $\mathbb{R}^4$ (via mapping a matrix $\begin{pmatrix}a&b\c&d\end{pmatrix}$ to the tuple $(a,b,c,d)$), we use the standard Euclidean topology on $X$. The quotient topology on $X/\sim$ is defined as follows:
A subset $U \subseteq X/\sim$ is open if and only if its preimage under the quotient map $\pi: X \to X/\sim$ (i.e., all matrices in $X$ that are similar to any matrix in $U$) is open in $X$.
This is the finest topology that makes $\pi$ continuous. Convergence in the quotient space means: a sequence of classes $[A_n]$ converges to $[A]$ if there exists a sequence of matrices $B_n \sim A_n$ such that $B_n$ converges to $A$ in $X$.
3. Is the Quotient Space Hausdorff?
No, the quotient space $X/\sim$ is not Hausdorff. Here's why:
Consider the scalar matrix $A = \begin{pmatrix}0&0\0&0\end{pmatrix}$ (its own similarity class) and the Jordan block $B = \begin{pmatrix}0&1\0&0\end{pmatrix}$ (non-diagonalizable, repeated eigenvalue 0). Suppose we try to separate these two classes with disjoint open neighborhoods:
- Any open set containing $[A]$ must include all matrices in $X$ sufficiently close to $A$ (by continuity of $\pi$). Take the sequence of matrices $C_t = \begin{pmatrix}t&t\0&t\end{pmatrix}$ for small $t \neq 0$. Each $C_t$ is similar to the Jordan block $\begin{pmatrix}t&1\0&t\end{pmatrix}$ (conjugate by $\begin{pmatrix}1/t&0\0&1\end{pmatrix}$), so $[C_t]$ is in the same class as the Jordan block with eigenvalue $t$.
- As $t \to 0$, $C_t$ converges to $A$ in $X$, so $[C_t]$ converges to $[A]$ in the quotient space.
- At the same time, $[C_t]$ converges to $[B]$: the Jordan block $\begin{pmatrix}t&1\0&t\end{pmatrix}$ converges to $B$ as $t \to 0$, so $\pi(\begin{pmatrix}t&1\0&t\end{pmatrix}) = [C_t]$ converges to $[B]$.
This means the sequence $[C_t]$ has two distinct limits ($[A]$ and $[B]$), which violates the Hausdorff property (in Hausdorff spaces, sequences can have at most one limit). Thus, no disjoint open neighborhoods exist for $[A]$ and $[B]$.
Summary
- Quotient Set: $X/\sim$ is the collection of similarity classes of 2x2 real matrices, each represented by one of the four real Jordan canonical forms.
- Quotient Topology: Open sets are unions of similarity classes whose preimage in $\mathbb{R}^4$ is open.
- Hausdorff Property: The quotient space is non-Hausdorff, as shown by sequences of classes converging to two distinct limits.
内容的提问来源于stack exchange，提问作者Metso
