Is there an analogue of the module of differentials for "higher order derivations" in the Hochschild/cyclic senses?

Question

Note added by YC: the definition below of the cyclic sub-complex is incorrect; and the "higher order derivations" referred to here are traditionally known (since the 1940s) as n-cocycles.

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$\DeclareMathOperator{\Ker}{\mathrm{Ker}}\DeclareMathOperator{\Mod}{\mathrm{Mod}}\DeclareMathOperator{\Der}{\mathrm{Der}}\DeclareMathOperator{\Hom}{\mathrm{Hom}}$While reading about Hochschild cohomology, I learned that we could define derivations in terms of the Hochschild complex: writing \begin{align*} M &\xrightarrow{d^1} \Hom_{\Mod_R}(S,M)\\ &\xrightarrow{d^2} \Hom_{\Mod_R}(S\otimes_RS,M)\\ &\xrightarrow{d^3}\Hom_{\Mod_R}(S\otimes_RS\otimes_RS,M)\\ &\xrightarrow{d^4}\cdots. \end{align*} for the Hochschild cochain complex of an $R$-algebra $S$ with coefficients in an $S$-bimodule $M$, we have $$\Der_R(S,M)\cong\Ker(d^2).$$ Now, derivations play an important role in deformation theory, and we can build an universal object corepresenting them, the module of differentials $\Omega_{S/R}$ of $S$ over $R$, defined by $$\Hom_S(\Omega_{S/R},M)\cong\Der_R(S,M).$$ Naturally, this leads one to wonder about whether we have a similar universal object for the module $$\Der^{n}_R(S,M)\cong\Ker(d^{n+1})$$ of "$n$-order Hochschild derivations of $S$ into $M$". For example, here's what such a higher derivation looks like for $n=2$ and $n=3$ (where below we identify a map $D\colon S^{\otimes_R n}\to M$ with the unique $n$-multilinear map $D\colon S^{\times n}\to M$ it represents):

• A second order Hochschild derivation is a map $D\colon S\otimes_R S\to M$ satisfying the equation $$D(ab,c)-D(a,bc)=aD(b,c)-D(a,b)c$$ for each $a,b,c\in S$.
• A third order Hochschild derivation is a map $D\colon S\otimes_RS\otimes_RS\to M$ satisfying the equation $$D(ab,c,d)-D(a,bc,d)+D(a,b,cd)=aD(b,c,d)+D(a,b,c)d.$$ for each $a,b,c,d\in S$.

Lastly, we could also work with the cyclic complex of $S$ with coefficients with $R$, defining "higher cyclic derivations" in a similar manner. These satisfy one extra equation: $$D(a_1,\ldots,a_n)=(-1)^{n-1}D(a_n,a_1,\ldots,a_{n-1}).$$ So again, in the low degree cases, we have $D(a,b)=-D(b,a)$ and $D(a,b,c)=D(c,a,b)=D(b,c,a)$.

Now, write $\Der^{\mathrm{cycl},n}_R(S,M)$ for the set of "$n$-order cyclic derivations", and note that given an $S$-module morphism $f\colon M\to N$ and an $n$-order (cyclic) derivation $D$, the composition $f\circ D$ is still an $n$-order (cyclic) derivation. This gives us functors $\Der^{n}_R(S,-)$ and $\Der^{\mathrm{cycl},n}_R(S,-)$.

Question. The above two functors are corepresentable by $\Omega_{S/R}$ when $n=1$. Are they also corepresentable for $n\geq2$ (in the commutative case)?

Answer 1

The answer is yes and it is very simple. It helps to understand the case $n=1$ first in the way I explained in my thesis in Prop. 4.5.3. Namely, $\Omega^1_{S/R}$ can be constructed as the quotient of the right $S$-module $S \otimes_R S$ by the $S$-submodule generated by those $ab \otimes 1 - b \otimes a - a \otimes b$ with $a,b \in S$.

Similarly, a representing object of $\mathrm{Der}^2_R(S,-)$ can be constructed as the quotient of the right $S$-module $(S \otimes_R S) \otimes_R S$ modulo elements of the form $$(ab \otimes c) \otimes 1 - (a \otimes bc) \otimes 1 - (b \otimes c) \otimes a + (a \otimes b) \otimes c$$ with $a,b,c \in S$. So every time you use the $S$-module structure on $M$ in the definition of the derivation, you just put the scalar into the last tensor factor. This works by the very construction of the adjunction between scalar extension and scalar restriction.

The general definition is similar. For the cyclic variant you have to quotient out another relation.