kakarotyjn
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Let c be a singular k-cube and [tex]p:[0,1]^k \to [0,1]^k[/tex] a 1-1 function such that [tex]p([0,1]^k ) = [0,1]^k<br />
[/tex] and [tex]\operatorname{det} p'(x) \ge 0[/tex] for [tex]x \in [0,1]^k[/tex].If [tex]\omega[/tex] is a k-form,show that [tex]
<br />
\int\limits_c \omega = \int\limits_{c \circ p} \omega[/tex]
Note that
[tex]\int\limits_c \omega = \int\limits_{[0,1]^k } {c*\omega } = \int\limits_{[0,1]^k } {(f \circ c)(\det c')dx^1<br /> <br /> \wedge ... \wedge dx^k }[/tex]
[tex]\int\limits_{c \circ p} \omega = \int\limits_{[0,1]^k } {(c \circ p)*\omega } = \int\limits_{[0,1]^k } {(f \circ c<br /> <br /> \circ p)(\det (c \circ p)')dx^1 \wedge ... \wedge dx^k } = \int\limits_{[0,1]^k } {(f \circ c \circ p)((\det c') \cdot<br /> <br /> (\det p'))dx^1 \wedge ... \wedge dx^k }[/tex]
did I deduce it right?If it's right,how to prove
Note that
[tex]\int\limits_c \omega = \int\limits_{[0,1]^k } {c*\omega } = \int\limits_{[0,1]^k } {(f \circ c)(\det c')dx^1<br /> <br /> \wedge ... \wedge dx^k }[/tex]
[tex]\int\limits_{c \circ p} \omega = \int\limits_{[0,1]^k } {(c \circ p)*\omega } = \int\limits_{[0,1]^k } {(f \circ c<br /> <br /> \circ p)(\det (c \circ p)')dx^1 \wedge ... \wedge dx^k } = \int\limits_{[0,1]^k } {(f \circ c \circ p)((\det c') \cdot<br /> <br /> (\det p'))dx^1 \wedge ... \wedge dx^k }[/tex]
did I deduce it right?If it's right,how to prove
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