# Homotopy type of specific space of matrices

I would like to determine topological properties of $\mathbb R^8$ minus the set determined by the equation $$\mathrm{det}\begin{pmatrix} a-a' & b-b'\\ c-c' & d-d' \end{pmatrix}=0$$ where $a,a',b,b',c,c',d,d'\in\mathbb R$.

How do I determine the homotopy type and how many connected components this space has? If this does not turn out to be a standard space, I would also like to determine (co)homology groups.

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$A$ such that $\mathrm{det}A\dots$? –  Olivier Bégassat Jul 11 '12 at 12:10
It should be $\mathbb R^8$ minus a closed subvariety determined by the equation in the 8 variables given by the determinant of the above matrix... So, for $a,b,c,d$ fixed, it is $\mathrm{GL}_2\mathbb R$. Maybe that makes it into a $\mathrm{GL}_2\mathbb R$ bundle over $\mathrm{GL}_2\mathbb R$, but I am not sure. –  Earthliŋ Jul 11 '12 at 12:20
Although there are $8$ parameters, the space of matrices has only 4 degrees of freedom, and the way you've described it, it is homeomorphic to $GL(2,R)$. Perhaps you want to ask what is the subset of $\mathbb R^8$ described by the condition that the above determinant is nonzero. –  Grumpy Parsnip Jul 11 '12 at 12:27
Assuming that's what you want, your space has a continuous map to GL(2,R), which you could try showing is a homotopy equivalence. (Just speculating, I haven't thought about it carefully.) –  Grumpy Parsnip Jul 11 '12 at 12:31
Thanks, I edited my question. –  Earthliŋ Jul 11 '12 at 13:04

Let's denote your subset of $\mathbb R^8$ by $X$. Then there is a surjective continuous map $X\to GL(2,\mathbb R)$. The homotopy type of $GL(2,R)$ is two copies of $SL(2,R)$. Anyway, from this you can already tell that $X$ has at least two connected components! Now, I claim that $X$ is actually homotopy equivalent to $GL(2,R)$. Given an $8$-tuple in $X$, perform a homotopy where $(a,a')\mapsto (a-t,a'-t)$ for $t\in[0,a]$. Similarly do this for the other coordinates. This deformation retracts $X$ onto the space where $a=b=c=d=0$. Which is exactly $GL(2,R)$. As mentioned by user8268, $SL(2,R)\simeq S^1$, so $X\simeq S^1\cup S^1$.
Change the coordinates: replace $a',\dots d'$ with $A=a-a',\dots,D=d-d'$. Then you see that the space is $GL_2(\mathbb{R})\times\mathbb{R}^4$, which is homotopy equivalent to $GL_2(\mathbb{R})$, and hence to $O_2(\mathbb{R})$, which is a disjoint union of two circles.
He said the complement of the set $\det=0$. –  Grumpy Parsnip Jul 11 '12 at 15:19