Problem 8-5 (Completion of Local Frames)

Proposition 8.11 (Completion of Local Frames). Let M be a smooth n-manifold with or without boundary.

(a)
If ( X 1 , , X k ) is a linearly independent k -tuple of smooth vector fields on an open subset U M , with 1 k < n , then for each p U there exist smooth vector fields X k + 1 , , X n in a neighborhood V of p such that ( X 1 , , X n ) is a smooth local frame for M on U V .
(b)
If ( v 1 , , v k ) is a linearly independent k -tuple of vectors in T p M for some p M , with 1 k n , then there exists a smooth local frame ( X i ) on a neighborhood of p such that X i | p = v i for i = 1 , , k .
(c)
If ( X 1 , , X n ) is a linearly independent n -tuple of smooth vector fields along a closed subset A M , then there exists a smooth local frame ( X ~ 1 , , X ~ n ) on some neighborhood of A such that X ~ i | A = X i for i = 1 , , n .

Answers

Proof.

(a)
Fix p U and let ( v 1 , , v k ) be the derivations ( X 1 , , X k ) | p at the point p . Since v 1 , , v k T p M are k linearly independent elements of an n -dimensional vector space, we can find n k additional vectors v k + 1 , , v n such that v 1 , , v k , v k + 1 , , v n form a basis for T p M . By Proposition 8.7, we can find smooth global vector fields X k + 1 ( p ) , , X n ( p ) 𝔛 ( M ) on M such that X k + 1 ( p ) | p = v k + 1 , , X n ( p ) | p = v n . We argue that in a neighborhood V of p which is small enough, the resulting collection of derivations X 1 | q , , X k | q , X k + 1 | q , , X n | q

defines a basis for T q M for all q V .

By Proposition 8.1, the fact that our collection of vector fields is smooth means that the component functions (stacked together in a single matrix X ):

[ X 1 1 X k 1 X k + 1 1 X n 1 X 1 n X k n X k + 1 n X n n ]

are smooth. In particular, this matrix function is continuous. Since the determinant function det is continuous as well, their composition det X : M

det [ X 1 1 X k 1 X k + 1 1 X n 1 X 1 n X k n X k + 1 n X n n ]

is continuous too. We know that at p , its value is non-zero

det [ X 1 1 ( p ) X k 1 ( p ) X k + 1 1 ( p ) X n 1 ( p ) X 1 n ( p ) X k n ( p ) X k + 1 n ( p ) X n n ( p ) ] 0 .

By the definition of continuity, we can find a neighborhood V of p on which det X is non-zero. In other words, the columns ( X 1 , , X n ) | q form a basis for T q M at each point q V , i.e., ( X 1 , , X n ) is a local frame on V .

(b)
By Proposition 8.7, we can find smooth global vector fields X 1 , , X k 𝔛 ( M ) such that X 1 | p = v 1 , , X k | p = v k . Using a similar argumentation with determinant, one can demonstrate that the collection X 1 , , X k is linearly independent in some small neighborhood U of p . Using part (a), we can extend this linearly independent collection of vector fields to a local frame X 1 , , X k , X k + 1 , , X n in some neighborhood V U of p .
(c)
By Extension Lemma for Vector Fields (Problem 8-1), for each 1 j n , there exists a smooth global vector field X ~ j on M such that X ~ j | A = X j . By a similar argument with determinant, each point p A has an open neighborhood V p on which ( X ~ 1 , , X ~ n ) forms a local frame. Taking the union V : = p A V p of all these neighborhoods, we obtain the desired neighborhood of A .
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2023-09-03 08:21
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