7/28/2019 Manifolds - Solutions 6

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1

(1)

(2) Let = i dxi, = i dx

i be 1-forms, then

= (i dxi) (j dx

j)

= (ij) dxi dxj

=1

2(ij ji) dx

i dxj

=i

7/28/2019 Manifolds - Solutions 6

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2

(4)

Fij dxi dxj =

1

2Fij (dx

i dxj dxj dxi)

=1

2(Fij dx

i dxj Fij dx

j dxi)

=1

2(Fij dx

i dxj Fji dx

i dxj)

by reindexing the second sum i j

=1

2(Fij dx

i dxj + Fij dx

i dxj)

using the anti-symmetry in the coefficients= Fij dx

i dxj

(5) (a) = 1 dx+ 2 dy + 2 dz and = 1 dydz+ 2 dzdx+ 1 dxdy

= 11 dx dy dz+ 2

2 dy dz dx + 33 dz dx dy

= (11 + 2

2 + 33) dx dy dz

(b) = 1 dx + 2 dy + 2 dz and = 1 dx + 2 dy + 2 dz

= (12 dx dy + 13 dx dz)+ (21 dy dx + 23 dy dz)

+ (31 dz dx + 32 dz dy)

= (12 21) dx dy + (23 32) dy dz+ (13 31) dx dz

using dxdx = dydy = dzdz = 0 and dxdy = dydx, etc.

(6)

(7)

7/28/2019 Manifolds - Solutions 6

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3

(8) Let f C(M), then

d df = d

f

xdx +

f

ydy

= d

f

x

dx +

f

xd2x + d

f

y

dy +

f

yd2y

=

2f

x2dx +

2f

yxdy

dx + 0 +

2f

xydx +

2f

y2dy

dy + 0

=2f

yxdy dx +

2f

xydx dy

= 0

Since 2f

yx=

2f

xyfor smooth functions and dx dy = dy dx.

(9) Let f(z) = u(x, y)+iv(x, y) be a holomorphic function, then the Cauchy-Riemann equations tell us

u

x=

v

yand

u

y=

v

x

so for A = f(z) dz we have

dA = df dz = (du + i dv) (dx + i dy)

=

u

xdx +

u

ydy

+ i

v

xdx +

v

ydy

(dx + i dy)

=

u

x+ i

v

x

dx +

u

y+ i

v

y

dy

(dx + i dy)

= i

u

x+ i

v

x

dx dy +

u

y+ i

v

y

dy dx

=

i

u

x

v

x

u

y i

v

y

dx dy

= 0

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4

(10) Let A = Ai dxi be a 1-form, where Ai C

(M) = 0(M) are smooth

functions (0-forms), then using properties (1) and (2) of exterior differ-ential we get

dA = dAi dxi + Ai d2xi = dAi dx

i

as d2xi = 0 (property (4) of exterior differential, see question 8), sosimilarly

d2A = d(dAi dxi) = d2Ai dx

i dAi d

2xi = 0

where again d2Ai = d2xi = 0.