Let $c$ be an orientation preserving k-cube in $M$( k dimensional manifold with boundary with orientation $\mu$) such that $c_{(k,0)} $ lies in $\partial M$ and is the only face that has any interior points in $\partial M$.
$c_{(i,\alpha)}=c\circ (I^n_{(i,\alpha)})$ and $I^n_{(i,\alpha)}=(x^1,\cdots , x^{i-1},\alpha,x^i,\cdots,x^{n-1})$
$\omega$ is a k-1 form on $M$ which is $0$ outside of $c([0,1]^k)$.

1) In $\int_{c_{(k,0)}} \omega =(-1)^k\int _{\partial M}\omega$ How did we get $(-1)^k$ ?
2) In $ \int _{\partial c}\omega=\int_{(-1)^kc_{(k,0)}}\omega=(-1)^k\int_{c_{(k,0)}} \omega =\int _{\partial M}\omega $ I am concerned with only the first equality and included the others just to give sense of what he is trying to show. Shouldn't it be $\sum_{i,\alpha} (-1)^k\int_{c_{(i,\alpha)}}\omega$ ? Why other faces not appear in the integration as a sum?

  • $\begingroup$ You should give more context, else people need a copy of the book to answer your question. $\endgroup$ – Pedro Tamaroff Jul 1 '18 at 11:46
  • $\begingroup$ Aren't a chains defined so that the following is valid? $$\int_{\lambda c} \omega = \lambda \int_c \omega$$ $$\int_{c_1+c_2} \omega = \int_{c_1} \omega + \int_{c_2} \omega$$ $\endgroup$ – md2perpe Jul 1 '18 at 20:32
  • $\begingroup$ Yes but how does that imply the result $\endgroup$ – mathemather Jul 2 '18 at 4:42
  • $\begingroup$ What is $c_{(k,0)$? $\endgroup$ – klirk Jul 6 '18 at 12:55

The answer to (2) is based on the fact that a continuous function that is zero outside a set must also be zero on the boundary of the set. This applies since $c$ is orientation preserving (see Spivak page 122) so that $c$ takes boundary points to boundary points; thus $\omega$ is zero on all the faces other than the $k$ face.

  • $\begingroup$ So it is implicit that $\omega$ is $0$ outside $c[0,1]^k$ right ? I dont see why orientation preserving map should take boundary point to boundary point. Even if so, how come then $c[0,1]^k$ is not $0$ ? $\endgroup$ – mathemather Jul 11 '18 at 16:57

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