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Alexander's trick, also known as the Alexander trick, is a basic result in geometric topology, named after J. W. Alexander.
## Statement

Two homeomorphisms of the n-dimensional ball $D^n$ which agree on the boundary sphere $S^\{n-1\}$, are isotopic.## Proof

Base case: every homeomorphism which fixes the boundary is isotopic to the identity relative to the boundary.## Radial extension

Some authors use the term Alexander trick for the statement that every homeomorphism of $S^\{n-1\}$ can be extended to a homeomorphism of the entire ball $D^n$.### Exotic spheres

The failure of smooth radial extension and the success of PL radial extension
yield exotic spheres via twisted spheres.

More generally, two homeomorphisms of D^{n} that are isotopic on the boundary, are isotopic.

If $fcolon\; D^n\; to\; D^n$ satisfies $f(x)\; =\; x\; mbox\{\; for\; all\; \}\; x\; in\; S^\{n-1\}$, then an isotopy connecting f to the identity is given by

- $J(x,t)\; =\; begin\{cases\}\; tf(x/t),\; \&\; mbox\{if\; \}\; 0\; leq\; ||x||\; <\; t,\; x,\; \&\; mbox\{if\; \}\; t\; leq\; ||x||\; leq\; 1.\; end\{cases\}$

Visually, you straighten it out from the boundary, squeezing $f$ down to the origin. William Thurston calls this "combing all the tangles to one point".

The subtlety is that at $t=0$, $f$ "disappears": the germ at the origin "jumps" from an infinitely stretched version of $f$ to the identity. Each of the steps in the homotopy could be smoothed (smooth the transition), but the homotopy (the overall map) has a singularity at $(x,t)=(0,0)$. This underlines that the Alexander trick is a PL construction, but not smooth.

General case: isotopic on boundary implies isotopic

Now if $f,gcolon\; D^n\; to\; D^n$ are two homeomorphisms that agree on $S^\{n-1\}$, then $g^\{-1\}f$ is the identity on $S^\{n-1\}$, so we have an isotopy $J$ from the identity to $g^\{-1\}f$. The map $gJ$ is then an isotopy from $g$ to $f$.

However, this is much easier to prove than the result discussed above: it is called radial extension (or coning) and is also true piecewise-linearly, but not smoothly.

Concretely, let $fcolon\; S^\{n-1\}\; to\; S^\{n-1\}$ be a homeomorphism, then

- $Fcolon\; D^n\; to\; D^n\; mbox\{\; with\; \}\; F(rx)\; =\; rf(x)\; mbox\{\; for\; all\; \}\; r\; in\; [0,1]\; mbox\{\; and\; \}\; x\; in\; S^\{n-1\}$

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Last updated on Thursday December 20, 2007 at 22:50:39 PST (GMT -0800)

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Last updated on Thursday December 20, 2007 at 22:50:39 PST (GMT -0800)

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