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.


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 '17 at 6:58
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    $\begingroup$ @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 '17 at 9:43

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