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Let $X$ be a quasi-projective $k$-variety. In this case the symmetric power $S^d(X)$ is well-defined. If $S^\bullet(X)=\bigsqcup_{n>0}S^d(X)$, where we suppose $S^0(X)=\operatorname{spec}(k)$, then $S^\bullet(X)$ is the universal monoidal $k$-scheme. Now if $G$ is a commutative quasi-projective $k$-group, then $S^\bullet(G)$ is defined and, moreover, there is a map of algebraic groups $S^d(G)\to G$ given by $d$-fold summation map $G^d\to G$.

If $\mathcal{C}$ is the category of quasi-protective $k$-varieties and $\mathcal{D}$ is the category of monoidal quasi-projective $k$-varieties, then $S^\bullet$ is left adjoint to a forgetful functor $U$. This defines a comonad on $\mathcal{D}$ by taking $\bot=(S^\bullet\circ U)\colon\mathcal{D}\to \mathcal{D}$. Thus one has an augmented simplicial monoidal (commutative) $k$-scheme $\bot_*(G)\to G$ such that $U(\bot_*(G))\to U(G)$ is contractible (Proposition 8.6.10 of Weibel's book). This gives extra degeneracies for simplicial scheme $S^\bullet_*(G)\to G$, its right part is given by $S^\bullet(S^\bullet(G))\rightrightarrows S^\bullet(G)\to G$. Here by $S^\bullet_n(G)$ I mean applying $S^\bullet(-)$ to $G$ iteratively $n$ times.

If $V$ is any smooth $k$-variety, I could apply $\operatorname{Mor}_{\operatorname{Sch}_k}(V, -)$ to this simplicial scheme and get an augmented simplicial set $\operatorname{Mor}(V, S^\bullet_*(G))\to \operatorname{Mor}(V, G)$. As far as I understand the previous shows that this is a contractible simplicial set but all the maps (except of extra degeneracies) are also the maps of monoids. A priori, extra degeneracies are only maps of sets, but I guess they also should be maps of monoids. Is there any simple way to see this?

Why do I care about this? If $\operatorname{char}(k)=0$ then the groupification of each commutative monoid $\operatorname{Mor}(V, S^\bullet_*(G))$ is the group of correspondences $\operatorname{Cor}(V, S^\bullet_{*-1}(G))$. Thus if the above simplicial guy is contractible (as commutative monoid), I think that I could take the gropification and get contractible augmented simplicial abelian group. After applying Dold-Kan we see that the complex $$\ldots\to \operatorname{Cor}(V, S^\bullet (S^\bullet(G))) \to \operatorname{Cor}(V, S^\bullet(G))\to \operatorname{Cor}(V, G)\to \operatorname{Mor}(V, G)$$ is exact and thus we can conclude that there is an exact sequence of presheaves with transfers given by $$\ldots\to\mathbb{Z}_{\operatorname{tr}}(S^\bullet(S^\bullet(G)))\to \mathbb{Z}_{\operatorname{tr}}(S^\bullet(G))\to\mathbb{Z}_{\operatorname{tr}}(G)\to G^{\operatorname{tr}}.$$

Does anyone has any suggestion how to modify the above construction or how to make it clearer? Anything could be helpful.

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    $\begingroup$ Why is $S^d(G)$ a group? For example, if $d=2$, how do you multiply the unordered pair $(x,y)$ by itself when $x \neq y$? $\endgroup$
    – S. Carnahan
    Commented Oct 6, 2023 at 0:15
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    $\begingroup$ Regarding the extra degeneracy: in any adjunction of that form, it's very rare for that morphism in $\mathcal{C}$ to lift to a morphism in $\mathcal{D}$; that's effectively asking for the generators of a free algebra to be a subalgebra. $\endgroup$ Commented Oct 6, 2023 at 8:40
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    $\begingroup$ @S.Carnahan +1. For example, symmetric power of an elliptic curve cannot ever be an algebraic group by cohomological reasons; its cohomology is not an exterior algebra. $\endgroup$
    – Denis T
    Commented Oct 7, 2023 at 0:03
  • $\begingroup$ Sorry, this is my fault. $S^n(G)$ is not an algebraic group and one has only the map of schemes $S^n(G)\to G$. Nevertheless, $S^\bullet(G)$ is a commutative monoidal scheme and the question is still valid $\endgroup$
    – Sam
    Commented Oct 8, 2023 at 0:02

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