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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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    $\begingroup$ 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) $\endgroup$ Commented Nov 16, 2022 at 5:36
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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
    Commented Nov 16, 2022 at 23:40
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    $\begingroup$ Thanks for the explanation. BTW, you have defined cyclic cocycles incorrectly: they are (special kinds of) n-cocycles which take values in Hom_R(S,R), and not in a general coefficient module M as you claim. So a cyclic derivation can be viewed as a function $S \times S \to R$, a cyclic 2-cocycle as a function $S\times S\times S \to R$, and so on. With the definition you have tried to make above, I don't think you get a subcomplex - the cyclic symmetry has to involve the coefficient module if you look at the original definition of Connes $\endgroup$
    – Yemon Choi
    Commented Nov 18, 2022 at 13:35
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    $\begingroup$ @YemonChoi Ah, I fear too much time has passed since I last tried understanding Hochschild co/homology, and now I don't remember things well enough to the point that I'm comfortable in editing the question at the moment (to be fair, I got the definition of cyclic cocycles wrong the first time; so it isn't like I ever understood it too well). I'll make a note to revisit this question once I get to Hochschild stuff again. Sorry for taking so long to address these things =/ $\endgroup$
    – Emily
    Commented Apr 14 at 4:57
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    $\begingroup$ In the meantime, though, feel free to freely edit the question as you wish, doing any and all changes you see fit, as if it were a community wiki one. $\endgroup$
    – Emily
    Commented Apr 14 at 4:57

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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
    Commented Nov 18, 2022 at 13:36

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