Part 22
Tangent Map of an Isometry
Printed version of the lecture Differential Geometry on 23. November 2009
Tommy R. Jensen, Department of Mathematics, KNU
22.1
Overview
Contents
1
Tangent Maps
1
2
Two Frames Determine a Unique Isometry
3
3
Conclusion
4
22.2
1
Tangent Maps
Tangent maps
Definition 7.4 (Chapter 1)
Let F : Rn → Rm be a mapping.
Let p be a point in Rn and v be a tangent vector to Rn at p.
The initial velocity vector β 0 (0) of the curve
β (t) = F(p + tv)
is denoted by F ∗ (v). It is a tangent vector to Rm at β (0) = F(p).
The function F∗ from tangent vectors to Rn to tangent vectors to Rm is called the tangent map of F.
Proposition 7.5 (Chapter 1)
Let F = ( f1 , f2 , . . . , fm ) : Rn → Rm be a mapping.
If v is a tangent vector to Rn at p, then
F ∗ (v) = (v[ f1 ], v[ f2 ], . . . , v[ fm ])F(p) .
22.3
1
The tangent map of an isometry
Theorem 2.1
Let F be an isometry of R3 with orthogonal part C.
Let v p be any tangent vector to R3 at some point p.
Then
F ∗ (v p ) = C(v)F(p) .
Proof of Theorem 2.1
Write F = T (C), where T is the translation by a, so
F(p) = T (C(p)) = a +C(p).
Then using that C is linear:
β (t) = F(p + tv) = a +C(p + tv) = a +C(p) + tC(v) = F(p) + tC(v).
This implies F ∗ (v p ) = β 0 (t)|t=0 = C(v)β (0) = C(v)F(p) .
22.4
Calculating Euclidean coordinates of the tangent map
Theorem 2.1 can be used to calculate F∗
Let F = T (C), where C is given by the orthogonal matrix


c11 c12 c13
 c21 c22 c23  .
c31 c32 c33
For any tangent vector v p we can write
v p = ∑ viUi (p).
Then using Theorem 2.1 we have
F ∗ (v p ) = F ∗ (∑ viUi (p)) = ∑ ci j v jUi (F(p)).
i, j
22.5
Tangent maps of isometries preserve dot products
Corollary 2.2
Let F be an isometry and v p and w p be tangent vectors to R3 at the same point p.
Then
F ∗ (v p ) • F ∗ (w p ) = v p • w p .
Proof of Corollary 2.2
Let C be the orthogonal part of F.
Then using Theorem 2.1, and that C preserves dot products:
F ∗ (v p ) • F ∗ (w p )
=
C(v)F(p) •C(w)F(p)
=
C(v) •C(w)
=
v•w
=
vp • wp.
22.6
2
Tangent maps of isometries preserve frames
Frames get mapped to frames
Let F be an isometry of R3 and let e1 , e2 , e3 be a frame at some point p.
Then
ei • e j = δi j for all i, j = 1, 2, 3.
From Corollary 2.2 follows that also
F ∗ (ei ) • F ∗ (e j ) = ei • e j = δi j for all i, j = 1, 2, 3.
Therefore F ∗ (e1 ), F ∗ (e2 ), F ∗ (e3 ) form a frame at point F(p).
22.7
2
Two Frames Determine a Unique Isometry
Two frames determine a unique isometry
Theorem 2.3
Let e1 , e2 , e3 be a frame at a point p, and let f1 , f2 , f3 be a frame at a point q, both in R3 .
Then there exists a unique isometry F of R3 such that
F ∗ (ei ) = fi for i = 1, 2, 3.
Proof of Theorem 2.3
There are two parts to the proof:
• Existence.
• Uniqueness.
22.8
Existence proof
Let eˆ1 , eˆ2 , eˆ3 be the vector parts of e1 , e2 , e3 . These are points in R3 .
Let fˆ1 , fˆ2 , fˆ3 be the vector parts of f1 , f2 , f3 .
Let C be the mapping
C : R3 → R3 , ∑ ai eˆi 7→ ∑ ai f̂i .
It is clear that C preserves dot products, so C is orthogonal.
Let T be the translation by q −C(p), and let F = T (C).
Then F(p) = T (C(p)) = C(p) + (q −C(p)) = q.
Using Theorem 2.1 to calculate F ∗ (ei ) gives:
F ∗ (ei ) = C(eˆi )F(p) = ( f̂i )q = fi for i = 1, 2, 3.
It follows that F is an isometry with a tangent map that maps the frame e1 , e2 , e3 to the frame f1 , f2 , f3 .
22.9
3
Uniqueness proof
Assume that F = T (C) has a tangent map that maps the frame e1 , e2 , e3 to the frame f1 , f2 , f3 .
Then it follows from Theorem 2.1 that C is a linear map that maps eˆi to f̂i for each i = 1, 2, 3.
Then C(∑ ai eˆi ) = ∑ aiC(eˆi ) = ∑ ai f̂i implies C = C.
It also follows from Theorem 2.1 that F maps p to q :
q = F(p) = T (C(p)) = T (C(p)) = T (T −1 (q)) ⇒ T = T.
So F = T (C) = F, which proves the Theorem.
22.10
How to calculate the isometry of Theorem 2.3
Assume that the attitude matrix of the frame e1 , e2 , e3 at the point p is A = (ai j ), and the attitude matrix
of the frame f1 , f2 , f3 at the point q is B = (bi j ).
This means that ei = (ai1 , ai2 , ai3 ) p , and fi = (bi1 , bi2 , bi3 )q , for i = 1, 2, 3.
Therefore A(eˆi ) = (e1 • ei , e2 • ei , e3 • ei ) = ui for i = 1, 2, 3.
It follows that
t
BA(eˆi ) = t B(ui ) = (bi1 , bi2 , bi3 ) = f̂i for i = 1, 2, 3.
If F = T (C) is the isometry which maps one frame to the other, then we know that C(eˆi ) = f̂i for
i = 1, 2, 3, and that C is uniquely determined.
So we have C = t BA.
It remains to calculate T as the translation by q −C(p).
22.11
3
Conclusion
The End
END OF THE LECTURE!
(Never give up. . . )
22.12
4
Next time:
Orientation
22.13
5