A = norm-preserving linear map (+other conditions) => A = lin isometry

In summary: The next step is$$\langle v,w\rangle = \frac12 \left[\langle A(v+w),A(v+w)\rangle-\langle v+w,v+w\rangle \right]$$If you expand ##\langle A(v+w),A(v+w)\rangle## you get four terms, each of which is of the form ##\langle a,a\rangle##, which is what you want.In summary, Theorem 4.4.4 in "Semi-Riemannian Geometry: The Mathematical Language of General Relativity" by Stephen Newman states that if a linear map A:V→V is a linear isometry, then the norm of Av is equal to the norm of v for all
  • #1
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I'm studying "Semi-Riemannian Geometry: The Mathematical Langauge of General Relativity" by Stephen Newman. Theorem 4.4.4 in that book:

Let ##(V,g)## be a scalar product space, and let ##A:V\to V## be a linear map. Then:
  1. If ##A## is a linear isometry, then ##\|A(v)\|=\|v\|\ \forall\ v\in V##
  2. If ##\|A(v)\|=\|v\|\ \forall\ v\in V##, and if ##A## maps spacelike (resp. timelike and lightlike) vectors to spacelike (resp. timelike and lightlike) , then ##A## is a linear isometry.

The proof of part 2 is given like this:

Since ##\|A(v)\|=\|v\|## is equivalent to ##|\langle A(v),A(v)\rangle|=|\langle v,v\rangle|##, the assumption regarding the way ##A## maps vectors yields ##\langle A(v),A(v)\rangle=\langle v,v\rangle##.

Seems a bit incomplete. I'd like to know if my approach is correct:

$$\langle A(v+tw),A(v+tw)\rangle=\langle A(v)+tA(w),A(v)+tA(w)\rangle$$
$$=\langle A(v),A(v)\rangle+t^2\langle A(w),A(w)\rangle+2t\langle A(v),A(w)\rangle=\langle v,v\rangle+t^2\langle w,w\rangle+2t\langle A(v),A(w)\rangle$$

But $$\langle A(v+tw),A(v+tw)\rangle=\langle v+tw,v+tw\rangle=\langle v,v\rangle+t^2\langle w,w\rangle+2t\langle v,w\rangle$$

(I'm not even sure if the ##t## coefficient matters) The above shows that ##\langle v,w\rangle=\langle A(v),A(w)\rangle##.

Is this fine? Secondly, wouldn't this same theorem more generally hold for a linear map ##A:V\to W##, where ##V,W## are both scalar product spaces (regardless of dimensions of ##V## and ##W##)?
 
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  • #2
The book proof is just a sketch. The point is that without the second condition you could have$$\langle Av, Aw\rangle = -\langle v,w\rangle$$Your proof needs to take this into account.
 
  • #3
PeroK said:
The book proof is just a sketch. The point is that without the second condition you could have$$\langle Av, Aw\rangle = -\langle v,w\rangle$$Your proof needs to take this into account.
Yep that I understood already (I mean about why the additional conditions are needed apart from norm preservation), but I was still wondering if the whole theorem holds in more general conditions
 
  • #4
Shirish said:
Yep that I understood already (I mean about why the additional conditions are needed apart from norm preservation), but I was still wondering if the whole theorem holds in more general conditions
Yes, it must. The proof is not dependent on the image space being ##V##.
 
  • #5
From
##\langle v+w,v+w\rangle=\langle v,v\rangle+\langle w,w\rangle+2\langle v,w\rangle##

you have

##\langle v,w\rangle = \frac12 \left[\langle v+w,v+w\rangle-\langle v,v\rangle-\langle w,w\rangle \right]##

Thus the inner product of any pair of vectors is determined by inner products of the form ##\langle a,a\rangle##. That is why they didn't write any more.
 
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