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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.


$\DeclareMathOperator{\Ker}{\mathrm{Ker}}\DeclareMathOperator{\Mod}{\mathrm{Mod}}\DeclareMathOperator{\Der}{\mathrm{Der}}\DeclareMathOperator{\Hom}{\mathrm{Hom}}$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)?

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  • $\begingroup$ (Also, if yes; how are these related to the usual "universal objects" found in deformation theory, like the de Rham complex or the cotangent complex?) $\endgroup$
    – Emily
    Nov 16, 2022 at 4:59
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    $\begingroup$ I think people usually consider rather higher derivations or Hasse-Schmidt derivations, and for these there is a module of differentials. Googling should turn up both old papers of Nakai and such people, on one hand, and much more recent developments ("Hasse-Schmidt algebra" is a good keyword to find these) $\endgroup$ Nov 16, 2022 at 5:36
  • $\begingroup$ What is your definition of a derivation with values in a module? All definitions known to me require the target to be a bimodule. $\endgroup$
    – Denis T
    Nov 16, 2022 at 13:10
  • $\begingroup$ In general there's a thing called (Connes-)Tsygan calculus, which is a most reasonable known generalisation of interplay between de Rham complex and polyvector fields on a smooth/algebraic manifold to an arbitrary ring. If you want higher derivations for some other reason, probably it's better to formulate that reason first, at least vaguely. $\endgroup$
    – Denis T
    Nov 16, 2022 at 13:14
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    $\begingroup$ @YemonChoi It's to motivate the question, as they look like a natural analogue of derivations but in higher degrees. (In any case I can't change the terminology/notation above anymore (apart from adding the qualifier "Hochschild" to disambiguate with the usual ones), as Martin's answer refers to it) $\endgroup$
    – Emily
    Nov 16, 2022 at 23:40

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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.

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    $\begingroup$ Since the OP has chosen not to update the terminology: the traditional notation in Hochschild cohomology for ${\rm Der}^n_R(S, -)$ is $Z^n_R(S,-)$, and elements of this space have been known for 80 years as cocycles not "higher derivations" $\endgroup$
    – Yemon Choi
    Nov 18, 2022 at 13:36

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