My understanding of how derivations on commutative rings are like derivatives is that a derivation on $R$ is differentiation with respect to a vector field on $\text{Spec}(R)$. But derivations are supposed to be thought of as like derivatives in a wider context than commutative rings, and I don't really understand how.

Take anti-derivations on the exterior algebra of differential forms on a manifold, for instance. The exterior derivative and Lie derivatives both give you information about infinitesimal change in a differential form, but the interior derivative is defined pointwise, as an anti-derivation on the exterior algebra of the tangent space at each point, which ruins any attempt to think of anti-derivations on differential forms as capturing some information about infinitesimal change. So how can you think about interior differentiation as being like a derivative in any more concrete sense than that it obeys similar syntactic rules? More generally, how can you think about anti-derivations on exterior algebras (or more generally still, on anti-commutative graded rings) as being like a derivative?

There's also derivations on non-commutative rings. The adjoint action of an element of a ring $\text{ad}_x(y):=xy-yx$ is a derivation, but I don't see the significance of this. For example, the Pincherle derivative seems to act like a sort of "differentiation with respect to $d/dx$" insofar as it sends $d/dx$ to $1$, and the fact that it is a derivation forces certain other facts that this heutristic naively suggests to be true (for instance, that the shift operator $S_1=e^{d/dx}$ is its own Pincherle derivative). Is there some more precise way to describe the Pincherle derivative as differentiation with respect to $d/dx$? What about a way to characterize arbitrary derivations on non-commutative rings?

How about derivations on Lie algebras? The Jacobi identity can be interpreted as saying that adjoint actions are derivations, but as in the analogous fact I mentioned for derivations on non-commutative rings, I'm curious about what the significance of this is. And about how to think of arbitrary derivations on Lie algebras.


1 Answer 1


In all of these contexts, derivations are infinitesimal automorphisms, in the sense that $D$ is a derivation on $A$ (an algebra, a Lie algebra, etc.) iff $\exp(Dt)$ is an automorphism of $A \otimes k[t]/t^2$. On a commutative ring automorphisms correspond to automorphisms of the spectrum so derivations correspond to infinitesimal automorphisms of the spectrum, or vector fields.

On a noncommutative ring or a Lie algebra the commutator $[x, -]$ exponentiates to conjugation by $\exp(xt)$, which is an inner automorphism; that's why these derivations are called inner derivations. This observation leads to the sometimes useful identity

$$\exp(Xt) Y \exp(-Xt) = \sum_{n \ge 0} \frac{\text{ad}_X^n(Y)}{n!}.$$

Said another way, derivations always form a Lie algebra, and the "Lie group" that they're the Lie algebra of is, heuristically speaking, the automorphism group.

  • $\begingroup$ Thanks! Is something like this supposed to work for anti-derivations on graded rings too? It seems to work algebraically for homogeneous anti-derivations $D$ if you take $t$ to be homogeneous of degree $-|D|$ such that $t^2=0$, $Dt=0$, and $t$ (anti-)commutes with everything. But this feels harder to understand as an infinitesimal automorphism (and I'm also not sure how to generalize it to nonhomogeneous anti-derivations). For instance, for the exterior derivative, we'd need $|t|=-1$, which seems tricky to motivate, and for the interior derivative $i_X$, we'd need $t(X)=0$. $\endgroup$ Dec 28, 2017 at 6:58
  • 1
    $\begingroup$ @Alex: in the graded setting derivations of arbitrary degree form a graded Lie algebra rather than a Lie algebra and so the interpretation is a bit more complicated, and in particular elements in nonzero degree don't directly correspond to infinitesimal automorphisms; really we should be talking about dg Lie algebras and their homology / homotopy groups, I guess. $\endgroup$ Dec 28, 2017 at 9:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.