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Sean McIlroy science forum beginner
Joined: 05 May 2005
Posts: 24

Posted: Sat Jul 15, 2006 6:39 pm Post subject:
originfixing isometry



hi all
is an isometry necessarily linear if the origin is a fixed point? why?
peace,
stm 

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William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Sun Jul 16, 2006 5:56 am Post subject:
Re: originfixing isometry



On Sat, 15 Jul 2006, Sean McIlroy wrote:
Quote:  is an isometry necessarily linear if the origin is a fixed point? why?
No, you might have an obnoxious metric space that's not a linear space or 
a metric space that forgot to include certificate of origin. 

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Ulysse from CH science forum beginner
Joined: 30 May 2006
Posts: 16

Posted: Fri Jul 21, 2006 12:44 pm Post subject:
Re: originfixing isometry



On Sat, 15 Jul 2006 22:56:39 0700, William Elliot
<marsh@hevanet.remove.com> wrote:
Quote:  On Sat, 15 Jul 2006, Sean McIlroy wrote:
is an isometry necessarily linear if the origin is a fixed point? why?
No, you might have an obnoxious metric space that's not a linear space or
a metric space that forgot to include certificate of origin.
OK, but Sean probably had the case in mind, where the metric space 
is a normed linear space or even an ordinary euclidian vector space
(isomorphic to R^n with its standard positive definite
symmetric bilinear form  the scalar product s:
( (x_1, ... , x_n) , (y_1, ... , y_n) ) > x_1*y_1 + ... + x_n*y_n,
which leads to the usual "L_2" norm defined by x = sqrt(s(x,x)) ).
At least in this special case, the answer to the question is Yes.
This can be proven along the following lines:
By definition the distance d(x,y) = yx and so x = d(0,x).
This shows that an isometric with 0 fixed keeps norms unchanged.
Also yx = y'x' if x',y' are the images of x,y; taking
squares gives s(yx,yx) = s(y'x',y'x'); this, the special cases
s(y,y) = s(y',y') , s(x,x) = s(x',x') and bilinearity of s show that
s(x,y) = s(x',y'). Therefore the images of the vectors in
a orthonormal basis have the same property. To conclude
that the isometry is the unique linear map sending a given
orthonormal basis to the corresponding basis (which is of
course an isometry) it is enough to prove that if an isometry
fixes an orthonormal basis, then it is the identity: this follows
from the shown invariance of s by an isometry and the fact
that the components of a vector in an orthonormal basis are
the scalar products of this vector and the vectors of the basis. 

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