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LectureNotes11G

# LectureNotes11G - Oct 2 20061 Math 6455 Dierential Geometry...

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Math 6455 Oct 2, 2006 1 Differential Geometry I Fall 2006, Georgia Tech Lecture Notes 11 Orientability Any ordered basis ( b 1 , . . . , b n ) of R n may be viewed as a matrix B GL ( n ) whose i th column is b i . Thus the set of ordered basis of R n are in one-to-one correspondence with elements of GL ( n ), and so we may partition them into two subsets: those whose corresponding matrices have positive determinant, GL ( n ) + , and those whose corresponding matrices have negative determinant, GL ( n ) - . By an orientation for R n we mean the choice of GL ( n ) + or GL ( n ) - as the preferred source for picking a basis for R n . The standard orientation for R n is that determined by GL ( n ) + . Lemma 0.0.1. GL ( n ) has exactly two path-connected components: GL ( n ) + and GL ( n ) - . Proof. First note that since det: GL ( n ) R n - { 0 } is continuous, and R - { 0 } is not connected, then GL ( n ) is not connected. So it must have at least two compo- nents GL ( n ) + = det - 1 ((0 , )), and GL ( n ) - = det - 1 (( -∞ , 0)). Secondly note that multiplying the first row of any element of GL ( n ) + yields an element of GL ( n ) - . This implies that GL ( n ) + and GL ( n ) - are homeomorphic. So we just need to check that GL ( n ) + is path connected. This may be achieved in two steps: first we deform each element of GL ( n ) + to an element of SO ( n ) and then show that SO ( n ) is path connected. Step 1: This may be achieved with the aid of the Gram-Schmidt process. In particular recall that if ( b 1 , . . . , b n ) is any basis of R n and ( b 1 , . . . , b n ) is the corre- sponding orthonormalizaion of it, then b i is never antiparallel to b i and each pair b i and b i span a subspace which does not include any other of the elements of the basis. Thus tb i + (1 - t ) b i yields a continuous deformation of ( b 1 , . . . , b n ) to ( b 1 , . . . , b n ) through a family of basis. Step 2: Let B SO ( n + 1), then we may write B = ( b 1 , . . . , b n ) where b i are the columns of B . We may rotate B until b n coincides with (0 , . . . , 0 , 1). Then b 1 , . . . , b n - 1 lie in R n - 1 × { 0 } , and by a rotation in R n - 1 we may bring b n - 1 in coincidence with (0 , . . . , 0 , 1 , 0). Continuing this procedure we may continuously deform ( b 1 , . . . , b n ) through a family of orthonormal basis until b 2 , . . . , b n coincide with the last n - 1 elements of the standard basis of R n . At that point we must either have b 1 = (1 , 0 , . . . , 0) or b 1 = ( - 1 , 0 , . . . , 0). But the latter is impossible 1 Last revised: October 10, 2006 1

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because B has positive determinant, and continuous deformations of it through a family of basis preserve the sign of the determinant. So b 1 = (1 , 0 , . . . , 0) and we are done. This immediately yields that Corollary 0.0.2. Two basis of R n can be continuously transformed to each other, through a family of basis, if and only if they belong to the same orientation class. We may also define an orientation for an abstract finite dimensional vector space V over R : we say that a pair of ordered basis of V are equivalent provided that their image, under an isomorphism f : V R n belong to the same orientation class.
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