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I am reading in Ralf Fröbergs notes on Koszul algebras. He writes that for any quadratic algebra $A$ there is a natural differential $$d:A_i\otimes (A^!_j)^*\to A_{i+1}\otimes (A^!_{j-1})^*$$ I think this differential $d$ can be defined as the composition, $$A_i\otimes (A^!_j)^*\to_{id\otimes d'}A_i\otimes(A_1^!\otimes A_{j-1}^!)^*\to A_i\otimes(A_1^!)^*\otimes (A_{j-1}^!)^*\to A_i\otimes A_1\otimes (A_{j-1}^!)^*\\ \to A_{i+1}\otimes (A^!_{j-1})^*$$ where $d'$ maps $f\in (A^!_j)^*$ to $d'f\in (A_1^!\otimes A_{j-1}^!)^*$ by defining $d'f(x\otimes y):=f(xy)$. The rest of the maps are natural.

Is this a correct understanding of the differential? If so, how do I show that $d^2=0$?

Much greatful for any response=)

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That is indeed the differential. As to why $d^2$ is zero, I suggest you do a concrete example, like $k\langle x,y\rangle/(xy-yx)$.

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  • $\begingroup$ OK will do, thanks! $\endgroup$
    – budwarrior
    Mar 7, 2017 at 15:47
  • $\begingroup$ A nice sourc of examples is the following: the quotient of a free algebra by the ideal generated by any set of quadratic monomials is Koszul. That gives you a lot of examples to try to see what is going on in $d^2$. $\endgroup$ Mar 7, 2017 at 15:48

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