Talk:Whitehead torsion (Ex)
Markus Land (Talk | contribs) (Created page with "We can compute that <wikitex>;$(1-t-t^{-1})\cdot(1-t^2-t^3) = 1$ </wikitex> so the element is a unit. The map <wikitex> $\mathrm{Wh}(\Z/5\Z) \to \Rr$</wikitex> can be describ...") |
Markus Land (Talk | contribs) |
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$$\xymatrix{ \Zz\lbrack \Zz/5\Zz \rbrack \ar[r]^-{\alpha} & \Zz\lbrack \Zz/5\Zz \rbrack}$$ </wikitex> | $$\xymatrix{ \Zz\lbrack \Zz/5\Zz \rbrack \ar[r]^-{\alpha} & \Zz\lbrack \Zz/5\Zz \rbrack}$$ </wikitex> | ||
and inducing it up to the complex numbers gives the map <wikitex>; | and inducing it up to the complex numbers gives the map <wikitex>; | ||
− | + | $$\xymatrix{\Cc \ar[r]^-{e^{\frac{2\pi i\alpha}{5}}} & \Cc}$$ </wikitex> | |
given by multiplication by <wikitex> $e^{\frac{2\pi i\alpha}{5}}$ </wikitex>. On <wikitex> $\mathrm{GL}_1(\Cc)$ </wikitex> the determinant is of course the identity map, so we need to compute <wikitex>; | given by multiplication by <wikitex> $e^{\frac{2\pi i\alpha}{5}}$ </wikitex>. On <wikitex> $\mathrm{GL}_1(\Cc)$ </wikitex> the determinant is of course the identity map, so we need to compute <wikitex>; | ||
$$ \ln(| e^{\frac{2\pi i\alpha}{5}} |) = 0 $$</wikitex> | $$ \ln(| e^{\frac{2\pi i\alpha}{5}} |) = 0 $$</wikitex> |
Latest revision as of 18:10, 28 August 2013
We can compute that
- so the element is a unit. The map can be described as follows.
There is a ring homomorphism sending a generator to . Now we consider the composite
By definition it takes the class of some automorphism to the real number . The above composite is a well-defined group homomorphism as all single maps are well-defined group homomorphisms, so it remains to check whether it factors over the Whitehead group. So let . Then by left multiplication it defines a map ;
and inducing it up to the complex numbers gives the map ;
given by multiplication by . On the determinant is of course the identity map, so we need to compute ;
which follows since for all real numbers , in particular for .
Now to show that generates an infinite cyclic subgroup of it suffices to show that because then is not a torsion element.
For this we simply calculate that ;
hence it follows that ;
since . In particular applying the logarithm we get ;
which we wanted to show.