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I have a very basic question about the exterior derivative of differential forms and de Rham complexes. It is very basic, I know that the exterior derivative satisfies $d^2=0$. Knowing that, how is a de Rham complex even possible ?

My definition of a de Rham complex is the following : Let $M$ be a manifold of dimension $n$, the de Rham complex is the following chain : $$ 0 \xrightarrow[]{d} \Omega ^{0}(M) \xrightarrow[]{d} \Omega ^{1}(M) \xrightarrow[]{d} ... \xrightarrow[]{d} \Omega ^{m}(M) \xrightarrow[]{d} 0 $$

My question : since $d^2 =0$, shouldn't we always have that $\Omega ^2 (M) = 0$ ?

Thank you very much for your help

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    $\begingroup$ Why do you think that if $f : V \to W$ is the constant homomorphism $x \mapsto 0$ between two vector spaces $V$ and $W$, then $W$ cannot have any non-zero elements? (Take $f = d^2$, $V = \Omega^0(M)$ and $W = \Omega^2(M)$ to relate this to your question.) $\endgroup$ – Rob Arthan Aug 26 '18 at 23:13
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$\Omega^2(M)$ is the space of all the differential $2$-forms on $M$. You have $d(\Omega^1(M))\subseteq\Omega^2(M)$ and $d^2(\Omega^0(M))\subseteq d(\Omega^1(M))\subseteq\Omega^2(M)$. As you said, $d^2(\Omega^0(M))=\{0\}$, but there's no reason why $d^2(\Omega^0(M))$ would be equal to $\Omega^2(M)$.

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  • $\begingroup$ Thank you, I understand my mistake now. $\endgroup$ – Alain Aug 27 '18 at 11:58
  • $\begingroup$ @Kevinklbrmt You're welcome :) Glad it helped. Since that's the case, it'd be nice to mark the answer as accepted by checking the green mark. $\endgroup$ – Scientifica Aug 27 '18 at 13:32

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