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Edit: According to essential comment of YCore I revise the question.

Let $A$ be a finite dimensional graded algebra which is a unital, super commutative and associative algebra. Is there a Lie group $G$ whose differential graded algebra of all $G$-left invariant differential forms be isomorphic to $A$? This is a differential form analogy of the classical fact that " Every finite dimensional Lie algebra is the Lie algebra of a Lie group".

I browsed this arXiv paper but I did not find a result for the graded algebra case.

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    $\begingroup$ First of all you certainly mean $A$ to be super-commutative, associative and unital, and also that $A_0$ is 1-dimensional. $\endgroup$
    – YCor
    Sep 14, 2019 at 7:44
  • $\begingroup$ @YCor yes. I should consider these extra conditions. I revise the question. Thank you for this point and your revision of my post. $\endgroup$
    – Ali Taghavi
    Sep 14, 2019 at 7:51
  • $\begingroup$ @YCor Is your email address the same as in your linked profile? Or you have another email too. I wish to send you a message. $\endgroup$
    – Ali Taghavi
    Sep 14, 2019 at 7:58
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    $\begingroup$ Restriction to the tangent space at the identity defined a cdga isomorphism between left-invariant forms and the Chevalley-Eilenberg cochains of the Lie algebra. In particular if you forget the differential you just get an exterior algebra on the dual of the Lie algebra, and all of these arise as CE cochains of abelian Lie algebras. More interestingly, the functor CE is part of a Quillen equivalence between suitably finite and connected dglas and augmented cdgas, but I don't know how to characterize the image of finite-dimensional Lie algebras. $\endgroup$
    – Bertram Arnold
    Sep 14, 2019 at 8:19
  • $\begingroup$ @YCor is it necessary to add the extra assumption $dim A^i =dim A^{n-i}$? $\endgroup$
    – Ali Taghavi
    Sep 14, 2019 at 8:52

1 Answer 1

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No. A commutative differential graded algebra A is isomorphic to the Chevalley-Eilenberg algebra of a finite-dimensional Lie algebra L if and only if its underlying graded algebra is the exterior algebra on A_1, which must be finite-dimensional. In this case we have L=(A_1)* and the Lie bracket on L is the dual of the differential A_1→A_2.

Accordingly, any commutative dga A whose underlying commutative graded algebra is not an exterior algebra on A_1 is a counterexample.

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  • $\begingroup$ Thank you very much for your answer. By commutarive do you mean graded commutative? So you are giving a necessary and sufficient condition for existence of a Lie group as in my question, yes? Does this necessary and sufficient condition automatically imply that $dim A^i=dim A^}n-i}$? the reason for the later question is thst for compact manifold $H^i\simeq H^{n-i}$. $\endgroup$
    – Ali Taghavi
    Sep 15, 2019 at 9:06
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    $\begingroup$ @AliTaghavi: Yes, commutative means "invariant under braiding", which in the old terminology was called "graded commutative". Hodge duality for exterior algebras immediately implies a positive answer to the question about dimensions. $\endgroup$
    – Dmitri Pavlov
    Sep 15, 2019 at 14:31

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