Explanation for Lurie's SAG Remark 25.1.3.7

Question

I am trying to understand the theory of simplicial commutative rings or animated rings. I just find a remark in Lurie's book Spectral Algebraic Geometry:

Remark 25.3.1.7. Let $f : R[x_1,\ldots ,x_n]\to R[y_1,\ldots,y_m]$ be a homomorphism of polynomial rings over $R$, given by $x_i\mapsto f_i(y_1,\ldots,y_m)$. For every object $\DeclareMathOperator{\calg}{CAlg^\Delta_\textit{k}} A\in \calg$, composition with $f$ induces a map of spaces $$\DeclareMathOperator{\mapcalg}{Map_{CAlg^\Delta_\textit{R}}} \mapcalg (R[y_1,\ldots,y_m],A) \to \mapcalg R([x_1,\ldots,x_n],A). $$ Passing to homotopy groups at some point $\eta\in\mapcalg R([y_1,\ldots,y_m],A)$, we get a map $\pi_\ast(A)^m\to\pi_\ast (A)^n$. For $\ast =0$, this map is given by $$ (a_1,\ldots,a_m)\mapsto (f_1(a_1,\ldots,a_m),\ldots,f_n(a_1,\ldots,a_m)). $$ For $\ast >0$, it is given instead by the action of the Jacobian matrix $[\frac{\partial f_i}{\partial y_j} ]$ (which we regard as a matrix taking values in $\pi_0(A)$ using the morphism $\eta$).

I do not understand why the induced map of higher homotopy groups is given by the Jacobian. Also I cannot see how the map of polynomial rings influences the homotopy groups. It seems too mysterious for me.

Answer 1

We have a ring map $\phi\colon R[x_1,\dotsc,x_n]\to R[y_1,\dotsc,y_m]$ with $\phi(x_i)=f_i(y_1,\dotsc,y_m)$ say. The question calls the ring homomorphism $f$ rather than $\phi$, but I think that that creates confusion. The list of polynomials gives a map $f\colon R^m\to R^n$, or more generally $A^m\to A^n$ for any $R$-algebra $A$, but that is not a ring map and is not the same as $\phi$. By applying $\pi_0$ we get a map $\pi_0(A)^m\to\pi_0(A)^n$ which is natural for $R$-algebras $A$ and sends $(a_1,\dotsc,a_m)$ to $f_1(a),\dotsc,f_n(a))$.

If $A$ is an $R$-algebra, then so is the mapping spectrum $B=F(S^k_+,A)$. There are maps $S^0\to S^k_+\to S_0$ whose composite is the identity, and these give maps $A\xrightarrow{\eta}B\xrightarrow{\theta}A$ with $\theta\eta=1$. We have $\pi_0(B)=\pi_0(A)\oplus\pi_k(A)\epsilon$ with $\epsilon^2=0$. For a polynomial $f(t)=\sum_ic_it^i$ in one variable and an element $a+b\epsilon\in\pi_0(B)$ we have $$ f(a+b\epsilon) = \sum_ic_i(a+b\epsilon)^i = \sum_ic_ia^i + \sum_ii c_i a^{i-1}b\epsilon = f(a) + f'(a)b\epsilon $$ Essentially the same argument shows that for a homomorphism $\phi$ corresponding to an $n$-tuple of polynomials $f_i$, and an element $(a_1+\epsilon b_1,\dotsc,a_m+\epsilon b_m)\in\pi_0(B)^m$, we have $$ f(a+\epsilon b) = f(a) + (Jf)(a).b\epsilon $$ This is the sense in which $\phi$ acts on higher homotopy groups via the Jacobian.

Answer 2

First, we may reduce to the case $n=1$. Then, we may reduce to the case that $f_1$ is a monomial. The statement can be directly checked when $\deg f_1\le1$. It then suffices to check that the action is zero when $\deg f_1>1$, which reduces to the following statement:

Proposition. Let $A$ be a homotopy associative ring in spectra. Then the multiplication map $\Omega^\infty A\times\Omega^\infty A\to\Omega^\infty A$ induces zero maps on higher homotopy groups.

I learned this from some notes by Akhil Mathew. Indeed, for $n\in\mathbb N_{>0}$, the multiplication map $\Omega^\infty A\times\Omega^\infty A\to\Omega^\infty A$ induces a map $\pi_n(A)\oplus\pi_n(A)=\pi_n(\Omega^\infty A\times\Omega^\infty A)\to\pi_n(A)$ of abelian groups. To see that this map is zero, it suffices to show that two composites $$\pi_n(A)\xrightarrow{\iota_j}\pi_n(A)\oplus\pi_n(A)\longrightarrow\pi_n(A)$$ for $j=1,2$ are zero, where $\iota_j\colon\pi_n(A)\to\pi_n(A)\oplus\pi_n(A)$ are two canonical inclusions. We show for the first inclusion $\iota_1$, and the second is similar. Note that the map $\iota_1$ is induced by the map $\alpha_1\colon\Omega^\infty A\to\Omega^\infty A\times\Omega^\infty A,x\mapsto(x,0)$, and the composite $$\Omega^\infty A\xrightarrow{\alpha_1}\Omega^\infty A\times\Omega^\infty A\to\Omega^\infty A$$ is zero, and the result follows.