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Let $A$ be a (not necessarily commutative, associative) $k$-algebra. The bimodule of non-commutative one-forms $\Omega^1_A$ is the free $A$-bimodule generated by symbols $da$, $a \in A$, subject to the relations $$ d(\lambda a + \mu b) = \lambda da + \mu db, ~~~~~~ d(ab) = a\,db + da \,b$$ for $a, b \in A$, $\lambda, \mu \in k$. Higher degree forms are then defined by $$\Omega^n_A = \Omega^1_A \otimes_A \cdots \otimes_A \Omega^1_A.$$ The direct sum of all forms $\Omega_A = \bigoplus_{n=0}^\infty \Omega_A^n$ becomes a differential graded algebra with the obvious differential.

Now Loday's book "cyclic cohomology", he defines the non-commutative de-Rham cohomology of $A$ to be the cohomology of the complex $((\Omega_A)_{\mathrm{ab}}, d)$, i.e. he takes the abelianization of $\Omega_A$.

Q: Why is that? What is wrong with defining the non-commutative de-Rham-cohomology just as the cohomology of $\Omega_A$?

/edit: To answer the question regarding the definition of $d$: Using the relations of $\Omega^1_A$, every element of $\Omega^n_A$ can be brought into the form $a_0 da_1\cdots da_n$. For these, $$d(a_0 da_1\cdots da_n) = da_0da_1 \cdots da_n.$$

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    $\begingroup$ What is the "obvious differential"? $\endgroup$ – abx May 15 '18 at 16:08
  • $\begingroup$ The generators are of the shape $a,da$ for $a\in A$. They are mapped by the "obvious differential" respectively to $da$, and $dda=0$. $\endgroup$ – dan_fulea May 15 '18 at 17:54
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    $\begingroup$ @ dan_fulea: OK, so what is the image of $a\,db$? $\endgroup$ – abx May 15 '18 at 20:20
  • $\begingroup$ It is $da\,db$. $\endgroup$ – Matthias Ludewig May 16 '18 at 20:54
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I do not really know the reason, but if I would guess I think it might be $\Omega(A)$ is no longer an abelian category. The exact sequences you written down are not just exact sequences of abelian groups, but may be interpreted as exact sequences of algebras (note that we have $d(ab)=da*b+a*db=0*b+a*0=0$ for $a,b\in \ker(d)$).

I am not entirely sure if the construction of abelian category carries through for non-commutative algebras. I do not know any counter-example, but it sounds very unlikely to be true as the construction is so general (consider some infinite dimensional non-commutative $\mathbb{C}$-Banach algebras, for example).

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  • $\begingroup$ Which exact sequences are you referring to? As far as I can see, I didn't write down any... $\endgroup$ – Matthias Ludewig Sep 24 '18 at 6:38
  • $\begingroup$ @MatthiasLudewig: I mean kernel and cokernel. Without abelian category this is not something well defined in general... $\endgroup$ – Bombyx mori Sep 24 '18 at 13:58

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