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Let $R$ be a discrete commutative ring. The Proposistion 25.1.2.4 in Lurie's Spectral Algebraic Geometry says that the natural functor from the $\infty$-category of animated $R$-algebras to the $\infty$-category of connective $\mathbb E_{\infty}$-algebras over $R$ admits left and right adjoints. I am wondering if both adjoints are identity on discrete objects?

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  • $\begingroup$ The left adjoint certainly won't be, because it has to preserve the cotangent complex. $\endgroup$ Commented May 1 at 8:33
  • $\begingroup$ @JonPridham Thanks for the comment. This is a good point. $\endgroup$
    – Y.M
    Commented May 1 at 10:17

1 Answer 1

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Let $\DeclareMathOperator\CAlg{CAlg}\DeclareMathOperator\an{an}\CAlg_R^{\an}$ denote the $\infty$-category of animated commutative $R$-algebras, and $\DeclareMathOperator\cn{cn}\CAlg_R^{E_\infty,\cn}$ denote the $\infty$-category of connective $E_\infty$-$R$-algebras.

Proposition 1. The left adjoint to the forgetful functor $\CAlg_R^{\an}\to\CAlg_R^{E_\infty,\cn}$ is identity on usual commutative $R$-algebras if and only if $R$ is a commutative $\mathbb Q$-algebra.

Proof. The "if" part holds since in the char 0 case, the forgetful functor in question is the identity functor. For the "only if" part, note that the left adjoint carries polynomial $R$-algebras to free $E_\infty$-$R$-algebras, Q.E.D.

Proposition 2. The right adjoint to the forgetful functor $\CAlg_R^{\an}\to\CAlg_R^{E_\infty,\cn}$ is identity on usual commutative $R$-algebras.

Proof. The composite functor $$\CAlg_R^{\an}\longrightarrow\CAlg_R^{E_\infty,\cn}\xrightarrow{\tau_{\le0}}\CAlg_R^{\heartsuit}$$ coincides with the truncation $\tau_{\le0}$. Taking right adjoints, we get what we want, Q.E.D.

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  • $\begingroup$ Thanks for the answer. In the proof of Prop 1, what do you mean by saying the left adjoint carries poly $R$-algebras to free $E_{\infty}$-$R$-algebras? My original concern is how to compare $Map(A,B)$ and $Map(A,\Theta(B))$ with $A$ a discrete $R$-algebra and $B$ an animated $R$-algebra. So I guess there is no general way to compare them? $\endgroup$
    – Y.M
    Commented May 1 at 10:16
  • $\begingroup$ @Y.M The left adjoint carries free objects to free objects: consider taking left adjoints of the forgetful functor from animated rings to connective $E_\infty$-rings and then to anima. $\endgroup$
    – Z. M
    Commented May 1 at 10:34
  • $\begingroup$ So it should be the left adjoint takes free $E_{\infty}$-$R$-algebras to poly $R$-algebras? $\endgroup$
    – Y.M
    Commented May 1 at 10:41
  • $\begingroup$ No, the left adjoint is $\DeclareMathOperator\An{An}\An\DeclareMathOperator\CAlg{CAlg}\DeclareMathOperator\cn{cn}\DeclareMathOperator\an{an}\to\CAlg_R^{\an}\to\CAlg_R^{E_\infty,\cn}$ which carries a finite set to the polynomial $R$-algebra generated by this finite set and then to the free $E_\infty$-$R$-algebra generated by this finite set. $\endgroup$
    – Z. M
    Commented May 1 at 10:44

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