This may seem awkward, but I am interested in the topological properties of the space of pairs of real $2\times 2$ matrices $A$, $B$ satisfying the equations $$ \det A-\det B\neq0\\ \det(A-B)\neq0. $$ which should be $\mathbb R^8$ minus a closed subvariety. As a result of this question it should be an open subset of $\mathrm{GL}_2\mathbb R\times\mathbb R^4$, but I can't figure out how to determine what the first condition implies.
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There is a bijection $(A,B)\mapsto(U,X)$ where $U=A-B$ and $X=U^{-1}B$ since $U$ has to be invertible. The condition $\det A\not=\det B$ is then equivalent to $\det X\not=\det(X+I)$, which for a $2\times2$ matrix is equivalent to the trace $\mathop{\text{tr}} X\not=-1$. Thus, pairs $(A,B)$ matching the two clauses correspond to $(U,X)$ where $\det U\not=0$ and $\mathop{\text{tr}}X\not=-1$. Alternatively, as noted in a comment, to keep things symmetric, let $U=A-B$, $Y=U^{-1}(A+B)$. The conditions on $(U,Y)$ are then that $\det U\not=0$ and $\det(Y+I)\not=\det(Y-I)$. As mentioned in one of the comments, as well as indicated in the question, $U\in\text{GL}_2\mathbb{R}$, while the set of matrices $X$ corresponds to $(\mathbb{R}\setminus\{-1\})\times\mathbb{R}^3$ by the mapping $$X\mapsto(X_{11}+X_{22},X_{11}-X_{22},X_{12},X_{21}).$$ Thus, the space has two components (trace above or below $-1$) each of which is topologically equivalent to $\text{GL}_2\mathbb{R}\times\mathbb{R}^4$. For $Y$, the same map would map to $(\mathbb{R}\setminus\{0\})\times\mathbb{R}^3$. |
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