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Exam (elaborations) TEST BANK FOR Manifolds, Tensor and Forms An Intro
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, Solution Manualfor
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
1
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