Page 330 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 26: Lines of Curvature

In particular, Notes

 
 det k 2
det h ij det g ij ij
  
 
Note also that Equation (5) implies that the pair N , N has the same orientation as X , X if and
2
1
1
2
only if det k ij 0. Furthermore, since N(u) is also the outward normal to the unit sphere at
  
N(u), and since X , X , N is positively oriented in  , it follows that X (u), X (u) also gives the
3
2
1
2
1
det k
2
positive orientation on the tangent space to the  at N(u). Thus, we deduce that sign     .
ij
Consequently, we obtain:
  det h
sign det k ij  
 
ij
 det h ij  K det g ij
  
 
det g ij
The proposition follows by integrating over V.
We now turn to an interpretation of the mean curvature. Let X: U  be a parametric surface. A
3

variation of X is a smooth family F(u;t): U ( , )   such that F(u; 0) = X. Note that since

dF(u; 0) is non-singular, the same is true of dF(u; t ) for any fixed u , perhaps after shrinking the
0
0
interval (, ). Thus, all the maps F(u; t ) for t close enough to 0 are parametric surfaces. The
0
0
generator of the variation is the vector field dF/dt(u; 0). The variation is compactly supported if
F(u; t) = X(u) outside a compact subset of U. The smallest such compact set is called the support
of the variation F. Clearly, if a variation is compactly supported, then the support of its generator
is compact in U. We say that a variation is tangential if the generator is tangential; we say it is
normal if the generator is normal. Suppose now that the closure V is compact in U. We consider
the area AF (V) of F(u; t) as a function of t. The next proposition shows that the derivative of this
function depends only on the generator, and in fact is a linear functional in the generator.
Proposition 6. Let X : U   be a parametric surface, and let F(u; t) be a variation with generator
3
Y. Then:
dA (V)  g X Y dA
ij
F

dt t 0  V i j ...(6)

We first need the following lemma from linear algebra. We denote by S n×n the space of n × n
symmetric matrices, and by S n n the subset of those which are positive definite.



Lemma 2. Let B :(a,b)  S n n be continuously differentiable. Then we have:



log det B  tr B 1 B .   ...(7)
Proof. First note that (7) follows directly if we assume that B is diagonal. Next, suppose that B is
symmetric with distinct eigenvalues. Then there is a continuously differentiable orthogonal
matrix Q such that B = Q DQ, where D is diagonal. Note that dQ /dt = –Q (dQ/dt)Q, hence:
–1
–1
–1
B B = –Q D Q Q DQ + Q D D Q + Q Q,
–1
–1
–1
–1
–1
–1
–1
and in view of tr(AB) = tr(BA), we obtain:
1
tr B B   tr D D . 
1


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