Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
$A$ such that $\mathrm{det}A\dots$? – Olivier Bégassat Jul 11 '12 at 12:10
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
up vote 2 down vote accepted

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 $X\simeq S^1\cup S^1$.

share|cite|improve this answer

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.

share|cite|improve this answer
He said the complement of the set $\det=0$. – Grumpy Parsnip Jul 11 '12 at 15:19
@JimConant: oh I see; I edited out that discussion. – user8268 Jul 11 '12 at 16:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.