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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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    No. The complement of a circle in $\mathbb{R}^4$ is connected and the complement of $xy - zw = 1$ cannot be (see my answer).2012-06-20

2 Answers 2

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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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    I used Maple...2012-06-20