I am reading Bott and Tu.


On page 140 of that book, the exactness of co-boundary operator $\delta$ is used to prove that the double complex of forms with compact support has only one non zero column. "By the exactness of the rows, $H_{\delta}(K)$ is:" and then we are shown a figure with all but column zero filled with zeros. The zero column is $\Omega_c^i$ where $i=0, 1..$

When I look at the exactness and the definition of $H_{\delta}(k)$, I conclude that all columns have to be zero including the zero column. After all exactness means kernel=image and thus the cohomology should all be zero. What am I missing? Why is there a non-zero column at all?

ps: The answers in comments below have confused me even more :), so if anyone can give a comprehensive answer, I shall be grateful.

  • $\begingroup$ Clearly the row complexes are exact save in dimension zero, where the cohomology (just the cokernel of the final map) is the desired group $\Omega_c^i(M)$. I.e., the rows are resolutions of the $\Omega_c^i(M)$. $\endgroup$ – Lord Shark the Unknown Jul 29 '18 at 6:28
  • $\begingroup$ @LordSharktheUnknown but the kernel of the sum operator going to $\Omega_c^*(M)$ on the page 139 prop 12.12 is the image of the delta operator from previous step. ie. the cohomology is trivial. Take the very simple case of just two intersecting open sets, then the kernel of the sum are those forms which are zero outside the intersection and have opposite signs on the intersection which is exactly those forms which are in the image of delta. Delta is the extension by zero of the intersection with opposite signs on the intersection for the two sets. see the def. on page 139. $\endgroup$ – alireza Jul 29 '18 at 6:39
  • $\begingroup$ These are spaces of forms with compact support: for $U\subset V$ there is a map $\Omega_c^*(U)\to\Omega_c^*(V)$, not in the other dimension (so we are dealing with cokernels not kernels). $\endgroup$ – Lord Shark the Unknown Jul 29 '18 at 6:42
  • $\begingroup$ But on page 139, proof of exactness, it is shown that whenever $\delta w=0$ then $w$ is the image of the previous step. ie kernel/image is being argued as defining the cohomology. Also nowhere in the book so far we have defined the $\delta$ operator any other way. Would you see the text? $\endgroup$ – alireza Jul 29 '18 at 6:55
  • $\begingroup$ @Lord Shark the Unknown would you just write an answer with full definitions instead of comments so I understand better what you are saying about cokernel. $\endgroup$ – alireza Jul 29 '18 at 7:14

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