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Let $M$ be a manifold and let $\mathcal{U}:=\{U_\alpha\}_{\alpha\in I}$ be an open covering, $I$ be a totally ordered set. For every $p$ and for every $\alpha_0<\dots<\alpha_p$, $$ U_{\alpha_0\dots\alpha_p}:=U_{\alpha_0}\cap\dots\cap U_{\alpha_p} $$ and, for every $0\le i\le p$, consider the inclusion maps $$ \partial_i:U_{\alpha_0\dots\alpha_{i-1}\alpha_i\alpha_{i+1}\dots\alpha_p}\to U_{\alpha_0\dots\alpha_{i-1}\alpha_{i+1}\dots\alpha_p} $$ and the induced (via pull-back) restriction maps $$ \delta_i:\Omega^q(U_{\alpha_0\dots\alpha_{i-1}\alpha_{i+1}\dots\alpha_p})\to\Omega^q(U_{\alpha_0\dots\alpha_{i-1}\alpha_i\alpha_{i+1}\dots\alpha_p}) $$ Hence, we can map a form $\omega\in\Omega^q(U_{\alpha_0\dots\alpha_p-1})$ to $\Omega^q(U_{\alpha_0\dots\alpha_p-1\beta})$ for all $U_\beta$ by considering a suitable $\delta_i$ (depending on the position of $\beta$ w.r.t. $\alpha_j$ in $I$).

Consider the expression $$ \delta_0-\delta_1+\dots+(-1)^p\delta_p $$ For every $p,q$ and $\alpha_0<\dots <\alpha_{p-1}$ $$ \delta_0-\delta_1+\dots+(-1)^p\delta_p:\Omega^q(U_{\alpha_0\dots\alpha_p-1})\to\prod_{\beta_0<\dots<\beta_k}\Omega^q(U_{\beta_0\dots\beta_p}) $$ is defined setting zero the component of the image if $\{\alpha_i\}\nsubseteq\{\beta_j\}$ and using $(-1)^i\delta_i$ for the appropriate $i$ for the other components.

Gluing these maps together and defining

$$ C^p (\mathcal{U},\Omega^q):=\prod_{\alpha_0 < \dots < \alpha_p}\Omega^q(U_{\alpha_0 \dots \alpha_p}) $$

we get, for every $p,q$,

$$ \delta:C^{p-1}(\mathcal{U},\Omega^q)\to C^p(\mathcal{U},\Omega^q) $$

I have to prove that $\delta\circ\delta=0$.

I don't know how to proceed, since the definition of $\delta_0-\delta_1+\dots+(-1)^p\delta_p$ is not clear at all to me. Is anyone able to explain me that definition? Maybe in a simple case.

For example, if I consider $$ \delta:C^0(\mathcal{U},\Omega^q)\to C^1(\mathcal{U},\Omega^q) $$ then I can rewrite $$ \delta: \Omega^q (U_{\alpha_0})\to\prod_{\alpha_0 < \alpha_1}\Omega^q(U_{\alpha_0\alpha_1}) $$ i.e. it is the map which sends the $q$-forms on $U_{\alpha_0}$ on the direct product/direct sum (since it is finite) of the restriction of the form on the all possible intersection $U_{\alpha_0}\cap U_{\alpha_1}$, for every $\alpha_1\in I$ such that $\alpha_0<\alpha_1$. Right?

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2 Answers 2

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General idea: Take the set $U=U_{\alpha_0\ldots\alpha_p}$. By performing $\delta$ twice on some form on $U$, you will get the set $V=U_{\alpha_0\ldots\alpha_p\beta_0\beta_1}$ twice, and the two forms on $V$ will cancel each other out. This is due to the $(-1)^p$ in the definition of $\delta$.

To get the right feel, you may calculate $\delta$ explicitly for cases where $I$ is a small set.

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  • $\begingroup$ Which $\delta$ are you considering? I can't understand how it is defined $\endgroup$
    – avati91
    Aug 27, 2015 at 14:24
  • $\begingroup$ Is it $\delta_{p+1}$? $\endgroup$
    – avati91
    Aug 27, 2015 at 14:25
  • $\begingroup$ @avati91 I'm considering $\delta$ exactly how you defined it - the sum of all $\delta_i$'s. $\endgroup$ Aug 27, 2015 at 14:59
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If we denote inclusions $$I_{n_0 \dots n_p}^{n_i}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0\dots \widehat{n_i} \dots n_p}\hspace{5pt}\text{and}\hspace{5pt}I_{n_0 \dots n_p}^{n_in_j}:U_{n_0 \dots n_p}\hookrightarrow U_{n_0 \dots \widehat{n_i} \dots \widehat{n_j} \dots n_p},$$ then $$\delta:C^p(\mathfrak{U},\Omega^k)\rightarrow C^{p+1}(\mathfrak{U},\Omega^k)$$ is given by formula $$(\delta\omega)_{n_0 \dots n_{p+1}}=\sum_{i=0}^{p+1}(-1)^i(I_{n_0 \dots n_{p+1}}^{n_i})^*\omega_{{n_0 \dots \widehat{n_i} \dots n_{p+1}}}.$$ Hence $$(\delta\delta\omega)_{n_0\dots n_{p+2}}=\sum_{i=0}^{p+2}(-1)^i(I_{n_0\dots n_{p+2}}^{n_i})^*(\delta\omega)_{n_0\dots \widehat{n_i}\dots n_{p+2}}\\ =\sum_{i=0}^{p+2}(-1)^i(I_{n_0\dots n_{p+2}}^{n_i})^*\left(\sum_{j=0}^{i-1}(-1)^j(I_{n_0\dots \widehat{n_i}\dots n_{p+2}}^{n_j})^*\omega_{n_0\dots\widehat{n_j}\dots\widehat{n_i}\dots n_{p+2}}\right.\\ +\left.\sum_{j=i+1}^{p+2}(-1)^{j-1}(I_{n_0\dots \widehat{n_i}\dots n_{p+2}}^{n_j})^*\omega_{n_0\dots\widehat{n_i}\dots\widehat{n_j}\dots n_{p+2}}\right)\\ =\sum_{j<i}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_i})^*(I_{n_0\dots \widehat{n_i}\dots n_{p+2}}^{n_j})^*\omega_{n_0\dots\widehat{n_j}\dots\widehat{n_i}\dots n_{p+2}}\\ -\sum_{i<j}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_i})^*(I_{n_0\dots \widehat{n_i}\dots n_{p+2}}^{n_j})^*\omega_{n_0\dots\widehat{n_i}\dots\widehat{n_j}\dots n_{p+2}}\\ =\sum_{j<i}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_i n_j})^*\omega_{n_0\dots\widehat{n_j}\dots\widehat{n_i}\dots n_{p+2}}-\sum_{i<j}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_i n_j})^*\omega_{n_0\dots\widehat{n_i}\dots\widehat{n_j}\dots n_{p+2}}.$$ If you swap indeces $i$ nad $j$ in second term you end up with $$\sum_{j<i}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_i n_j})^*\omega_{n_0\dots\widehat{n_j}\dots\widehat{n_i}\dots n_{p+2}}-\sum_{j<i}(-1)^{i+j}(I_{n_0\dots n_{p+2}}^{n_j n_i})^*\omega_{n_0\dots\widehat{n_j}\dots\widehat{n_i}\dots n_{p+2}}.$$ Just form the definition $I_{n_0\dots n_{p+2}}^{n_j n_i}=I_{n_0\dots n_{p+2}}^{n_i n_j},$ so it vanishes.


How to look at Cech-de Rham complex

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