Some authors have written that derivation of an algebra is an important tools for studying its structure. Could you give me a specific example of how a derivation gives insight into an algebra's structure? More generally: Why should we study derivations of algebras? What is the advantage of knowing that an algebra has a non-trivial derivation?

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    $\begingroup$ I think this is a perfectly reasonable question, although it should probably be clarified if the OP is mainly interested in the functional-analytic versions or not. Certainly I have refereed far too many papers proving things (of varying depth and worth) about derivations on Banach algebras, which don't seem to pause to address the questions raised by the OP $\endgroup$ – Yemon Choi Jul 11 '18 at 14:31
  • $\begingroup$ Just as a side comment to the last question: I was surprised when I learned only much later than I would have liked to, that the algebra of all continuous functions (say on a manifold) has only the zero derivation, in striking contrast to its dense subalgebra of smooth functions. As for "a specific thing about importance" I suggest you to have a look at "Pursell-Shanks theorem". $\endgroup$ – Nicola Ciccoli Jul 11 '18 at 17:03
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    $\begingroup$ I am not sure what kind of answer you are looking for, but here is one basic remark. For an algebra, the composition of two derivations is not necessarily a derivation, but the set of derivations is closed under the commutator bracket and they form a Lie algebra. Derivations of Lie algebras are a fundamental concept in the study of Lie algebras and their representations. In some sense derivations of Lie algebras are analogous to automorphisms of groups (Lie algebras act on Lie algebras by derivations, while groups act on groups by automorphisms). $\endgroup$ – spin Jul 12 '18 at 3:08
  • $\begingroup$ @NicolaCiccoli Is this the original paper for said theorem? jstor.org/stable/2031961?seq=1#page_scan_tab_contents $\endgroup$ – rschwieb Jul 12 '18 at 17:14
  • $\begingroup$ Yes, that's it. $\endgroup$ – Nicola Ciccoli Jul 12 '18 at 19:59

I think you want an answer from a pure algebraist, and I would be very interested in seeing such an answer as well. However, as a functional analyst, there are some things I can say to the question of why derivations are important which I find very convincing.

The basic point is that differentiation is a derivation. Most simply, the map $D: f \mapsto f'$ from the smooth real-valued functions on $\mathbb{R}$ to itself. This map satisfies the derivation identity $D(fg) = fD(g) + D(f)g$. More generally, differentiation along a vector field is a derivation from $C^\infty(M)$ to itself, for any smooth manifold $M$. The exterior derivative is another example; this time the target space is not $C^\infty(M)$ itself but a bimodule over it.

If you're willing to consider derivatives which are not continuous but merely measurable, then a natural domain of the differentiation operator is the set of real-valued Lipschitz functions. These all have derivatives in $L^\infty$. Of course we need to assume a metric now, in order for the notion of being Lipschitz to make sense.

In fact there is a strong connection between metrics and derivations. Theorem: let $X$ be a $\sigma$-finite measure space and $\delta$ a derivation from a dense subalgebra of $L^\infty(X)$ to a bimodule over $L^\infty(X)$. Given a natural continuity condition and the right notion of "bimodule", we can conclude that there is a metric on $X$ such that the domain of $\delta$ is precisely the set of bounded Lipschitz functions on $X$, and the norm of $\delta(f)$ is the Lipschitz number of $f$.

Here's another connection between metrics and derivations. Let $X$ be any metric space equipped with a regular Borel measure. Define $\Omega(X)$ to be the set of derivations from ${\rm Lip}(X)$ to $L^\infty(X)$ which satisfy a natural continuity condition. If $X$ is a Riemannian manifold then $\Omega(X)$ will be the set of bounded measurable 1-forms. But this construction makes sense for arbitrary metric spaces! It trivializes in many cases, but there is a large array of non-manifold metric spaces for which this construction is nontrivial and interesting: sub-Riemannian manifolds, rectifiable sets, Hilbert cubes, various kinds of fractals, etc. We have a first order exterior derivative in all these cases.

I could go on, but I'll stop here. These topics are covered extensively in the second edition of my book Lipschitz Algebras, which is now in press.


Here are some algebraic aspects why one could be interested in derivations.

First, derivations actually appear pretty prominently in algebraic geometry. For a ring homomorphism $R\to S$, one can consider $R$-linear derivations on $S$, with image in some $S$-module $M$. There is a universal such module, the module of Kähler differentials $\Omega_{S/R}$, with a universal derivation $d:S\to\Omega_{S/R}$. This is a fairly important object to study in algebraic geometry, e.g., in the definition and study of smooth varieties, or in the algebraic versions of de Rham cohomology. (A chapter on derivations can be found in any algebraic geometry textbook.)

On the other hand, and a bit more specialized, there are also reasons to study locally nilpotent derivations on algebras. Since locally nilpotent derivations are related to actions of the additive group $\mathbb{G}_a$ on varieties, the locally nilpotent derivations also play a significant role in the study of quotients of $\mathbb{G}_a$-actions. In particular, (as found on the Wikipedia page) some counterexamples or Hilbert's 14th problem arise as kernels of locally nilpotent derivations. (survey paper here) Locally nilpotent derivations are also related to various open questions on the structure of affine space $\mathbb{A}^n$ (like the Zariski cancellation problem, see e.g. the Makar-Limanov invariant which can distinguish affine space from some affine varieties diffeomorphic to it). There is a whole book about the algebraic theory of locally nilpotent derivations:

  • G. Freudenburg. Algebraic theory of locally nilpotent derivations. Encyclopaedia of Mathematical Sciences, 136. Invariant Theory and Algebraic Transformation Groups, VII. Springer-Verlag, Berlin, 2006.

The two examples which I can give are the following. One is the algebra of octonions. The derivations of this algebra form a 14-dimensional Lie algebra. From this fact you can conclude that the automorphism group of the octonions is 14-dimensional. An automorphism is obtained from a given derivation by taking the $exp$ function. The exponential function doesn't work over finite fields. There are formulas for derivations of octonions e.g. $[L_x,L_y]+ [L_x,R_y]+[R_x,R_y]$ is a derivation for every pair of octonions $x,y$. In this formula $L_x$ and $R_x$ denote left and right multiplication by the octonion $x$. You can find more in Schafer's book or Barton-Sudbery https://arxiv.org/abs/math/0203010.

We can define the Jordan algebra $h_3(\mathbb O)$ of symmetric matrices of octonions. In this algebra $[L_x,L_y]$ is a derivation. The algebra of derivations is 52-dimensional in this case. It is the $f_4$ Lie algebra.

I wonder whether the $e_6$, $e_7$ and $e_8$ Lie algebras can be obtained as derivations of some algebra.

  • $\begingroup$ If the Killing form of a finite-dimensional Lie algebra $L$ is non-degenerate, then $L$ has zero center and all derivations of $L$ are inner (this works over any field). So if the Killing form is non-degenerate, then the Lie algebra of derivations of $L$ is isomorphic to $L$. Of course, this might be a bit unsatisfying since we rely on having constructed $L$ in the first place. $\endgroup$ – spin Jul 12 '18 at 8:22

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