Relation between Jacobi identity, commutator, and derivation

Question

Here are some standard facts about these concept:

1. The commutator satisifies the Jacobi identity.
2. The commutator of two derivations is a derivation.
3. The Jacobi identity is equivalent to $\mathrm{ad}_x$ is a derivation (where $\mathrm{ad}_x(y)= [x,y]$).
4. The Jacobi identity is equivalent to $\mathrm{ad}_{[x,y]} = [\mathrm{ad}_x,\mathrm{ad}_y]$.

It is straightforward to prove each of them individually, but that does not explain why the concepts are so tightly linked. They even remain true in the graded context.

I can see the pathway from Jacobi identity to commutator and derivation, but I can't see how commutator and derivations relate to each other. I feel like there must be some kind of relation (other than commutator satisfies Jacobi), since the proof of statements 1 and 2 feel formally similar.

So the question is: Is there a way to relate derivation and commutator, so that the one can prove 1 or 2 from the other? And perhaps generalize this to the graded context.

Answer 1

Let $A$ be a finite-dimensional algebra over $\mathbb{R}$, not necessarily associative (so this includes both associative and Lie algebras), and let $\varphi_t : A \to A$ be a one-parameter family of automorphisms of $A$ (so a one-parameter subgroup of the Lie group $\text{Aut}(A)$). Then differentiating the identity

$$\varphi_t(ab) = \varphi_t(a) \varphi_t(b)$$

in the variable $t$ gives

$$D(ab) = D(a) b + a D(b)$$

where $D$ is the derivative $\frac{d \varphi}{dt}$ evaluated at $t = 0$. In other words, one-parameter automorphisms of $A$ have the form

$$\varphi_t(a) = e^{Dt} a$$

where $D$ is a derivation on $A$; said another way, the vector space $\text{Der}(A)$ of derivations on $A$ is the tangent space to the identity of $\text{Aut}(A)$. The slogan here is that "derivations are infinitesimal automorphisms."

This means $\text{Der}(A)$ is a Lie algebra! Moreover you can check that its Lie bracket is given by the commutator bracket. This is why the commutator of two derivations is another derivation, so this is where fact #2 comes from; this fact is the derivative of the fact that the product of two automorphisms is another automorphism.

Now, recall that if $G$ is a group then sitting inside its automorphism group $\text{Aut}(G)$ is the subgroup $\text{Inn}(G)$ of inner automorphisms given by conjugation $h \mapsto ghg^{-1}$. The analogue of this fact for Lie algebras $\mathfrak{g}$ is that sitting inside the Lie algebra $\text{Der}(\mathfrak{g})$ of derivations on $\mathfrak{g}$ is a Lie subalgebra of inner derivations, given by the adjoint action

$$\text{ad}_X(Y) = [X, Y].$$

If $\mathfrak{g}$ is the Lie algebra of a Lie group $G$ then inner derivations correspond precisely to the derivations obtained by starting from a one-parameter subgroup of inner automorphisms $h \mapsto g_t h g_t^{-1}$ of $G$, considering the induced one-parameter subgroup of automorphisms of $\mathfrak{g}$, and then differentiating the parameter. In particular, by the above calculation, inner derivations are derivations (with respect to the Lie bracket); this is where fact #3 comes from and why Lie algebras have to satisfy the Jacobi identity.

The second half of Fact #4 is the statement that the inner derivation map $\mathfrak{g} \to \text{Der}(\mathfrak{g})$ is a homomorphism of Lie algebras, which is the Lie algebra analogue of the statement that the inner automorphism map $G \to \text{Aut}(G)$ is a group homomorphism, although this isn't quite an explanation of why it's equivalent to the Jacobi identity.

For fact #1, take $A$ to be associative. In this context we can again consider one-parameter subgroups of inner automorphisms $x \mapsto a_t x a_t^{-1}$ of $A$, and differentiating these we again obtain a subalgebra of the Lie algebra $\text{Der}(A)$ of derivations on $A$ given by inner derivations

$$\text{ad}_a(x) = [a, x] = ax - xa.$$

So commutators are derivations! Since inner automorphisms of $A$ form a Lie subgroup of $\text{Aut}(A)$, inner derivations form a Lie subalgebra of $\text{Der}(A)$, which is where fact #1 comes from.

The assumption that $A$ is finite-dimensional and real can be removed by replacing $t$ with a nilpotent; more precisely we can consider "infinitesimal automorphisms to first order," namely automorphisms of $A[t]/t^2$ which reduce to the identity $\bmod t$, and a calculation shows that these have the form $1 + Dt = \exp(Dt)$ where $D$ is a derivation, in full generality ($A$ is an abelian group equipped with a bilinear multiplication, not necessarily associative). We can even consider "formal automorphisms" by working with formal power series $A[[t]]$ (at least if $A$ is a $\mathbb{Q}$-algebra, and this construction has to be modified a bit otherwise).

Answer 2

Qiaochu's answer differentiates a 1-parameter family of automorphisms to show that they yield a derivation, that is, an operation on functions which satisfies Liebniz' rule $$ \tag{$\dagger$} D(a.b) = D(a).b+ a.D(b). $$

However, if you are willing to that operations satisfying $(\dagger)$ are interesting, then since the condition is purely algebraic, it makes sense to talk about derivations of any algebra $A$. Here, by "algebra" I mean an abelian group equipped with a bilinear map $*\colon A\times A \to A$ which need not be associative. Often $A$ is a vector space over a field $\mathsf k$, in which case $*$ is required to be a bilinear map when we view $A$ as a $\mathsf k$-vector space.

Now we set $$ \text{Der}_\mathsf k(A) = \{d\colon A \to A: d \text{ is } \mathsf k\text{-linear and } d(a*b) = d(a)*b+a*d(b), \forall a,b \in A\} $$ Now $\text{Der}_{\mathsf k}(A)$ is a $\mathsf k$-subspace of the vector space $\text{Hom}_\mathsf k(A,A)$, but it is easy to see in examples that it need not be closed under composition. A easy direct calculation, however, shows that if $d_1,d_2 \in \text{Der}_{\mathsf k}(A)$, then $[d_1,d_2]:=d_1\circ d_2 - d_2\circ d_1$ is again a derivation of $A$. Thus we see that the commutator bracket gives us a way of making the vector space $\text{Der}_{\mathsf k}(A)$ into an "algebra" (provided you are okay with non-associative algebras at least...).

Now suppose that $(A,*)$ is an associative algebra. Can we write down any interesting derivations on $A$? I claim that the commutator is one of the most natural examples. Indeed by definition $$ \tag{*} a*b =[a,b]+b*a $$ so that if I define $d_a\colon A \to A$ by setting $d_a(b) = [a,b]$ then $$ \begin{split} a*(b*c) = (a*b)*c = ([a,b]+b*a)*c &= [a,b]*c +b*(a*c)\\ &= [a,b]*c+ b*([a,c]+c*a) +\\ &= [a,b]*c+ b*[a,c] + (b*c)*a \end{split} $$ which rearranges to $d_a(b*c) = d_a(b)*c+ b*d_a(c)$, that is, $d_a$ is a derivation. (Essentially, every time we wish to move $a$ past an element $b$ we pick up a term $[a,b]$). Now if $A$ is an associative algebra, we can define $L(A)= (A,[-,-])$ to be the same $\mathsf k$-vector space equipped with the commutator bracket as its product instead of the associative product.

Fact: $\text{Der}_{\mathsf k}(A) \subseteq \text{Der}_{\mathsf k}(L(A))$.

Although this is completely trivial to check, it is surprisingly useful:

1. Applying this to $d= d_a$ we find that the commutator bracket obeys the Jacobi identity: Indeed we have $$ [a,[b,c]]=d_a([b,c]) = [d_a(b),c]+[b,d_a(c)] = [[a,b],c]+[b,[a,c]] $$ which rearranges (using the alternating property) to $$ [a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0. $$

2. We can also use the fact to see that the map $d\colon A \to \text{Der}_{\mathsf k}(A)$ has nice properties: if $\theta \in \text{Der}_{\mathsf k}(A)$ then $$ \begin{split} [\theta,d_b](c) = \theta\circ d_b(c) - d_b\circ \theta(c) &= \theta([b,c])- [b,\theta(c)] \\ &= [\theta(b),c]+[b,\theta(c)] -[b,\theta(c)]\\ & = d_{\theta(b)}(c) \end{split} $$ so that $[\theta,d_b] = d_{\theta(b}$. If we set $\theta= d_a$ for any $a \in A$ this shows that $d\colon L(A) \to \text{Der}_{\mathsf k}(A)$ is a Lie algebra homomorphism. Moreover, the identity proves that the image of $d$ is an ideal in $\text{Der}_{\mathsf k}(A)$.

Answer 3

Based on krm2233's answer, I just realised that Leibniz rule and Jacobi identity are secretly the same thing (or maybe not so secret). If $\theta$ is a linear endomorphism and we write $[\theta,b]:= \theta b$, then the Leibniz rule is just $$ [\theta,[b,c]] = [[\theta,b],c] + [b,[\theta,c]]. $$ Like Jacobi identity, it also has a commuator form $$ [\theta,d_b] = d_{[\theta,b]} = d_{\theta b}. $$ From this I can derive the fact that the commutator of derivation is a derivation from the Jacobi identity of commutator. If $\theta$ and $\theta'$ are derivations, then $$ [[\theta,\theta'],d_b] = [\theta, [\theta',d_b]] - [\theta', [\theta,d_b]] = d_{\theta(\theta'b)} - d_{\theta'(\theta b)} = d_{[\theta,\theta']b}. $$ To my surprise, the equation $$ [\theta,d_b] = d_{\theta b} $$ correctly characterizes the graded Leibniz rule so the above argument works in the graded case as well.