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Exam (elaborations) TEST BANK FOR Manifolds, Tensor and Forms An Intro

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, Solution Manual
for
Manifolds, Tensors, and Forms


Paul Renteln
Department of Physics
California State University
San Bernardino, CA 92407
and
Department of Mathematics
California Institute of Technology
Pasadena, CA 91125
prenteln@csusb. edu

, Contents




1 Linear algebra page 1
2 Multilinear algebra 20
3 Differentiation on manifolds 33
4 Homotopy and de Rham cohomology 65
5 Elementary homology theory 77
6 Integration on manifolds 84
7 Vector bundles 90
8 Geometric manifolds 97
9 The degree of a smooth map 151
Appendix D Riemann normal coordinates 154
Appendix F Frobenius’ theorem 156
Appendix G The topology of electrical circuits 157
Appendix H Intrinsic and extrinsic curvature 158




iii

, 1
Linear algebra




1.1 We have

0 = c1 (1, 1) + c2 (2, 1) = (c1 + 2c2 , c1 + c2 )
⇒ c2 = −c1 ⇒ c1 − 2c1 = 0 ⇒ c1 = 0 ⇒ c2 = 0,

so (1, 1) and (2, 1) are linearly independent. On the other hand,

0 = c1 (1, 1) + c2 (2, 2) = (c1 + 2c2 , c1 + 2c2 )

can be solved by choosing c1 = 2 and c1 = −1, so (1, 1) and (2, 2) are
linearly dependent (because c1 and c2 are not necessarily zero).
1.2 Subtracting gives
  
0= vi ei − vi ei = (vi − vi )ei .
i i i

But the ei ’s are a basis for V , so they are linearly independent, which implies
vi − vi = 0.
1.3 Let V = U ⊕ W , and let E := {ei }i=1 n
be a basis for U and F := { f j }mj=1 a
basis for W . Define a collection of vectors G := {gk }n+m k=1 where gi = ei for
1 ≤ i ≤ n and gn+i = f i for 1 ≤ i ≤ m. Then the claim follows if we can
show G is a basis for V . To that end, assume

n+m 
n 
m
0= ci gi = ci ei + ci f i .
i=1 i=1 i=1

The first sum in the rightmost expression lives in U and the second sum lives
in W , so by the uniqueness property of direct sums, each sum must vanish
by itself. But then by the linear independence of E and F, all the constants
ci must vanish. Therefore G is linearly independent. Moreover, every vector
v ∈ V is of the form v = u + w for some u ∈ U and w ∈ W , each of which

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