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This should be an easy question, but I found it ungooglable and not obvious to visualize...

What geometric object is defined by the equation $xy-zw=1$ in $\mathbb R^4$? And what is the homotopy type of the complement?

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By trivial identification with the subset of the space of $2\times2$ matrices, it is just $SL(2)$. I know not much on it, but this may help you google it. –  sos440 Jun 20 '12 at 4:28
    
You mean $\text{SL}_2(\mathbb{R})$. –  Qiaochu Yuan Jun 20 '12 at 4:28
    
Think before asking. Yes, sorry, it is just $\mathrm{SL}_2\ \mathbb R$, which is homotopy equivalent to a circle. Thus the complement has fundamental group $\mathbb Z^2$ and second homotopy group $\mathbb Z$. Thanks. –  Earthliŋ Jun 20 '12 at 4:35
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Taking complements does not respect homotopy equivalence in general. For example, $\mathbb{R}^2$ admits both a point and $\mathbb{R}$ (as a coordinate axis) as contractible (hence homotopy-equivalent) subsets; for the former the complement is homotopy equivalent to $S^1$ but for the latter the complement is homotopy equivalent to two points. –  Qiaochu Yuan Jun 20 '12 at 4:39
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No. The complement of a circle in $\mathbb{R}^4$ is connected and the complement of $xy - zw = 1$ cannot be (see my answer). –  Qiaochu Yuan Jun 20 '12 at 4:44
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2 Answers 2

up vote 5 down vote accepted

It's $\text{SL}_2(\mathbb{R})$, of course! As a "geometric object" it may equivalently be realized as the unit sphere $x^2 + y^2 - z^2 - w^2 = 1$ in $\mathbb{R}^{2,2}$, which exhibits it (or maybe one of its connected components?) as a homogeneous space for the orthogonal group $\text{O}(2, 2)$. In particular it can be given the structure of a pseudo-Riemannian manifold.

The complement of the unit sphere in $\mathbb{R}^{2, 2}$ has two connected components $$X = \{ (x, y, z, w) : x^2 + y^2 - z^2 - w^2 > 1 \}$$ $$Y = \{ (x, y, z, w) : x^2 + y^2 - z^2 - w^2 < 1 \}.$$

$X$ deformation retracts via the straight-line homotopy $(x, y, (1-t)z, (1-t)w)$ to $\{ (x, y) : x^2 + y^2 > 1 \}$, which is homotopy equivalent to $S^1$.

$Y$ deformation retracts via the straight-line homotopy $((1-t)x, (1-t)y, z, w)$ to $\{ (z, w) : z^2 + w^2 > -1 \}$, which is contractible.

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It's hard to visualize a 3-dimensional hypersurface in 4 dimensions. But perhaps this animation may help, showing cross-sections at different values of $w$. Note that these are hyperbolic paraboloids except at $w=0$ where you have a hyperbolic cylinder.

http://www.math.ubc.ca/~israel/problems/surf2.gif

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Great animation. How ever did you manage to come up with that in such a short time? –  Earthliŋ Jun 20 '12 at 4:37
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I used Maple... –  Robert Israel Jun 20 '12 at 6:26
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