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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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Could you say something about the kinds of properties you are interested in – say, for example, connectedness? –  Zhen Lin Aug 20 '12 at 8:36
    
Connectedness, homotopy groups, homology groups, cohomology groups. These would all follow from general theory, if one could write down explicitly, what this space is, as for the case when one just imposes the second condition. –  Earthliŋ Aug 20 '12 at 9:32
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1 Answer

up vote 3 down vote accepted

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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Very nice! The first $=-1$ should be $\ne$. –  user31373 Aug 20 '12 at 11:25
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I'd use $Y=X+I/2$ instead of $X$, because then the condition becomes more symmetric: $\det(Y-I/2)\ne\det(Y+I/2)$ or $\operatorname{tr}Y \ne 0$. –  celtschk Aug 20 '12 at 14:24
    
@celschk: Good point. Will leave the original text, but add this explanation to the text. –  Einar Rødland Aug 21 '12 at 4:31
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