3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.