If I have understood things correctly, the Koszul dual coalgebra of a quadratic algebra $A=A(V,R)$ looks as follows $$A^{¡}=C(sV,s^2R)=\mathbb{K}1\oplus sV\oplus s^2R \oplus ... \oplus\big(\bigcap_{i+2+j=n}sV^{\otimes i}\otimes s^2R\otimes sV^{\otimes j}\big)\oplus...\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ } (\star)$$ where the $s'$s just modify the degrees.

In Algebraic Operads (Loday & Vallette)the Koszul dual algebra is defined as follows $$(A^!)^{(n)}:=s^n(A^{¡*})^{(n)}$$ My question is, when taking the linear dual of $A^{¡}$ are we dualizing each (finite dimensional) summand in the direct sum $(\star)$ separetaly and then taking direct sum or are we dualizing the vector space $A^¡$ all at once?

If we dualize it all at once, how do the graded parts of $(A^{¡*})$ look like?

Much grateful for any response=)

Edit: I just realized $(A^{¡})$ does not have to be infinite dimensional as a vector space.



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