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2. On the Fundamental Theorem for Curves in Space

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Supplementary Notes <strong>for</strong> MM08 Geometry I<br />

<strong>2.</strong> <strong>On</strong> <strong>the</strong> <strong>Fundamental</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Curves</strong><br />

<strong>in</strong> <strong>Space</strong><br />

Andrew Swann<br />

These notes are a supplement to <strong>Theorem</strong> <strong>2.</strong>3 <strong>in</strong> Pressley [1] which gives <strong>the</strong><br />

<strong>Fundamental</strong> <strong>Theorem</strong> <strong>for</strong> <strong>Curves</strong> <strong>in</strong> <strong>Space</strong>. The <strong>the</strong>orem has two parts. The rst<br />

states that two curves <strong>in</strong> R 3 parameterised by arclength with <strong>the</strong> same curvature and<br />

torsion , with > 0, are equivalent under rigid motions of R 3 . The second concerns<br />

existence of curves <strong>in</strong> R 3 with prescribed curvature and torsion. The proof of <strong>the</strong><br />

rst part is relatively direct and covered perfectly adequately by Pressley's text. The<br />

second part however relies on results from <strong>the</strong> <strong>the</strong>ory of l<strong>in</strong>ear dierential equations,<br />

<strong>in</strong> particular long-time existence of solutions, that are not covered by previous courses<br />

here <strong>in</strong> Odense. These notes will obta<strong>in</strong> <strong>the</strong> desired result us<strong>in</strong>g only <strong>the</strong> local existence<br />

result proved <strong>in</strong> MM04.<br />

We beg<strong>in</strong> by consider<strong>in</strong>g a smooth curve : (; ) ! R 3 , s 7! (s), which is<br />

parameterised by arclength and whose curvature is never zero. The latter<br />

assumption implies that <strong>the</strong> torsion is dened and we may write ft; n; bg <strong>for</strong> <strong>the</strong><br />

Frenet frame of . Then <strong>the</strong>se vectors satisfy <strong>the</strong> Serret-Frenet equations<br />

9>=<br />

t 0 = n<br />

n 0 >;<br />

= t +b<br />

(<strong>2.</strong>1)<br />

b 0 = n<br />

where 0 = d=ds. We wish to show that <strong>the</strong>se equations essentially determ<strong>in</strong>e . The<br />

rst part of Pressley's <strong>Theorem</strong> <strong>2.</strong>3 shows that is unique up to rigid motions. We<br />

are thus left with:<br />

<strong>Theorem</strong> <strong>2.</strong>1 (Existence part of <strong>the</strong> <strong>Fundamental</strong> <strong>Theorem</strong>). Let k; t : (; ) ! R be<br />

smooth functions with k(s) > 0 <strong>for</strong> all s 2 (; ). Then <strong>the</strong>re exists a smooth curve<br />

: (; ) ! R 3 which is parameterised by arclength, has curvature (s) = k(s) and<br />

has torsion (s) = t(s), <strong>for</strong> all s 2 (; ).<br />

<strong>2.</strong>1


To start <strong>the</strong> proof, let us look at <strong>the</strong> Serret-Frenet system (<strong>2.</strong>1) <strong>in</strong> matrix <strong>for</strong>m.<br />

Regard<strong>in</strong>g t, n and b<br />

0<br />

as row<br />

1<br />

vectors<br />

0<br />

we have<br />

1 1<br />

B@ t n<br />

b<br />

CA0<br />

We <strong>the</strong>re<strong>for</strong>e consider <strong>the</strong> equations<br />

<strong>for</strong> V : (; ) ! M 3 (R) with<br />

A(s) =<br />

=<br />

B@ 0 CA 0 B@ 0 t n<br />

0 <br />

0 0<br />

b<br />

CA : (<strong>2.</strong>2)<br />

V 0 = AV (<strong>2.</strong>3)<br />

0<br />

B@ 0 (s) 0<br />

(s) 0 (s)<br />

0 (s) 0<br />

1<br />

CA : (<strong>2.</strong>4)<br />

<strong>On</strong>e may get local existence of solutions to (<strong>2.</strong>3) with prescribed <strong>in</strong>itial data,<br />

by apply<strong>in</strong>g <strong>the</strong> follow<strong>in</strong>g <strong>the</strong>orem from MM04 (Supplerende noter til punktmængdetopologi,<br />

Sætn<strong>in</strong>g <strong>2.</strong>2, side 6):<br />

<strong>Theorem</strong> <strong>2.</strong><strong>2.</strong> Let B R n+1 be an open set. Suppose f : B ! R n , (s; x) 7! f (s; x)<br />

is a cont<strong>in</strong>uous function whose partial derivatives @f=@x j , j = 1; : : : ; n, exist and<br />

are cont<strong>in</strong>uous <strong>in</strong> B.<br />

Fix a po<strong>in</strong>t (s 0 ; x 0 ) <strong>in</strong> B. Then <strong>the</strong> dierential equation<br />

x 0 (s) = f (s; x(s))<br />

with <strong>in</strong>itial condition<br />

x(s 0 ) = x 0 ;<br />

has a unique solution x(s) with s <strong>in</strong> an <strong>in</strong>terval (s 0 "; s 0 + ") <strong>for</strong> some " > 0.<br />

To apply this to (<strong>2.</strong>3), note that M 3 (R) = R 9 , so take n = 9 and put<br />

f (s; V ) = A(s)V;<br />

B = (; ) ¢ M 3 (R):<br />

As <strong>the</strong> entries of A are smooth <strong>in</strong> s, <strong>the</strong> required partial derivatives of f exist and are<br />

smooth. Thus, we have local solutions to (<strong>2.</strong>3) around any given po<strong>in</strong>t s 0 2 (; )<br />

with any given <strong>in</strong>itial data V 0 2 M 3 (R).<br />

<strong>2.</strong>2


F<br />

or<br />

two solutions V 1 and V 2 to (<strong>2.</strong>3) on some sub<strong>in</strong>terval of ( 0 ; 0 ) (; ),<br />

consider <strong>the</strong> matrix V T<br />

2<br />

V 1<br />

obta<strong>in</strong>ed by multiply<strong>in</strong>g <strong>the</strong> transpose of V 2 onto V 1 .<br />

Dierentiat<strong>in</strong>g we have<br />

(V T<br />

2<br />

V 1<br />

) 0 = (V 0<br />

2 )T V 1<br />

+ V T<br />

2<br />

V 0<br />

1 = (AV 2) T V 1<br />

+ V T<br />

2 (AV 1 ) = V T<br />

2 (AT + A)V 1<br />

= 0;<br />

s<strong>in</strong>ce A <strong>in</strong> (<strong>2.</strong>4) skew-symmetric. We have thus proved<br />

Lemma <strong>2.</strong>3. Suppose V 1 and V 2 satisfy V 0<br />

i<br />

= AV i<br />

, i = 1; 2 on ( 0 ; 0 ) with A as<br />

<strong>in</strong> (<strong>2.</strong>4). Then V T<br />

2<br />

V 1<br />

is constant on ( 0 ; 0 ).<br />

Fix a po<strong>in</strong>t s 0 2 (; ) and let V be <strong>the</strong> maximal solution to (<strong>2.</strong>3) with <strong>in</strong>itial<br />

condition V (s 0 ) = Id, where Id is <strong>the</strong> (3 ¢ 3) identity matrix. Here maximal<br />

means that V is dened on some sub<strong>in</strong>terval ( 0 ; 0 ) of (; ) and that <strong>the</strong>re is no<br />

such solution dened on a larger sub<strong>in</strong>terval. Our aim is to prove that ( 0 ; 0 ) = (; ).<br />

Lemma <strong>2.</strong>4. V (s) is an orthogonal matrix <strong>for</strong> each s 2 ( 0 ; 0 ), i.e.,<br />

V T (s)V (s) = Id :<br />

Proof. By Lemma <strong>2.</strong>3, V T V is constant on ( 0 ; 0 ). At s 0 it has value Id T Id = Id.<br />

Lemma <strong>2.</strong>5. The set<br />

O(3) = f X 2 M 3 (R) : X T X = Id g<br />

of orthogonal matrices is a compact subset of M 3 (R).<br />

Proof. If X is orthogonal, <strong>the</strong>n each column x i of X is a unit vector <strong>in</strong> R 3 . Thus as<br />

vector <strong>in</strong> R 9 , X has length kx 1 k 2 + kx 2 k 2 + kx 3 k 2 = 3. So O(3) is bounded <strong>in</strong> M 3 (R).<br />

The function F : M 3 (R) ! M 3 (R) given by F (X) = X T X is cont<strong>in</strong>uous, so O(3) =<br />

F 1 (fIdg) is closed. There<strong>for</strong>e, O(3) is closed and bounded and hence compact.<br />

Suppose 0 < . Let (~s i ) 1 i=1 be a sequence of po<strong>in</strong>ts <strong>in</strong> ( 0; 0 ) converg<strong>in</strong>g to 0 .<br />

Then as V (~s i ) lie <strong>in</strong> <strong>the</strong> compact set O(3), <strong>the</strong>re is a subsequence (s n ) 1 n=0<br />

and a po<strong>in</strong>t<br />

~V 2 O(3) such that<br />

V (s n ) ! ~ V as n ! 1:<br />

Let W be a solution of <strong>the</strong> system (<strong>2.</strong>3) with <strong>in</strong>itial condition W ( 0 ) = V ~ . Then W (s)<br />

is dened on some <strong>in</strong>terval ( 1 ; 1 ) around 0 . We may assume 1 > 0 . Then on<br />

( 1 ; 0 ) we have by Lemma <strong>2.</strong>3 that W T V is constant. However, (W T V )(s n ) ! V ~ T ~V =<br />

Id as n ! 1, so W T V = Id on ( 1 ; 0 ). Now V is orthogonal, so W = (V T ) 1 = V<br />

on this <strong>in</strong>terval. Putt<strong>in</strong>g<br />

^V (s) =(<br />

V (s) 0 < s < 0 ;<br />

W (s) 0 6 s < 1 ;<br />

<strong>2.</strong>3


we thus obta<strong>in</strong> a solution to <strong>the</strong> equations (<strong>2.</strong>3), with <strong>in</strong>itial value Id at s = s 0 , on <strong>the</strong><br />

larger <strong>in</strong>terval ( 0 ; 1 ). This contradicts <strong>the</strong> maximality of V and we conclude that<br />

0 = . A similar argument shows 0 = and thus our maximal solution V is dened<br />

on <strong>the</strong> whole of (; ).<br />

To conclude <strong>the</strong> proof of <strong>Theorem</strong> <strong>2.</strong>1,<br />

0<br />

write<br />

1<br />

our maximal solution V as<br />

B@ t CA V = n :<br />

b<br />

Put<br />

(s) =<br />

Z<br />

s<br />

s 0<br />

t(u) du:<br />

Then : (; ) ! R 3 is a curve with 0 = t. As V is orthogonal, we have V 1 = V T ,<br />

so V V T = Id show<strong>in</strong>g that ft; n; bg is orthonormal. In particular, t is a unit vector<br />

and is parameterised by arclength. We now have t 0 = kn, so = k and n is <strong>the</strong><br />

normal to . Now b is proportional to t ¢ n with ht ¢ n; bi = det(V ). For V 2 O(3),<br />

1 = det(Id) = det(V T V ) = det(V T ) det(V ) = det(V ) 2 , so det V = ¦1. As det V (s)<br />

is cont<strong>in</strong>uous on (; ) and det V (s 0 ) = det Id = 1, we have det V (s) = 1 <strong>for</strong> all s,<br />

b = t ¢ n, and b is <strong>the</strong> b<strong>in</strong>ormal. F<strong>in</strong>ally, b 0 = tn shows = t, as required.<br />

References<br />

[1] A. Pressley, Elementary dierential geometry, Spr<strong>in</strong>ger Undergraduate Ma<strong>the</strong>matics<br />

Series, Spr<strong>in</strong>ger-Verlag London Ltd., London, 2001.<br />

<strong>2.</strong>4

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