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

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)?

• 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) Commented Nov 16, 2022 at 5:36
• @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) Commented Nov 16, 2022 at 23:40
• 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 Commented Nov 18, 2022 at 13:35
• @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 =/ Commented Apr 14 at 4:57
• 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. Commented Apr 14 at 4:57

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.
• 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" Commented Nov 18, 2022 at 13:36