I am really struggling with understanding the Jacobi Identity. I am not struggling with verifying it or calculating commutators.. I just can't see through it. I can't see the motivation behind it (as an axiom for defining a Lie algebra). Could anyone give an intuitive explanation?
5 Answers
I don't think there is just one motivation. I will mention three.
First, the Jacobi identity says precisely that the bracket is a derivation with respect to itself, where a derivation of an algebra is a map $d$ with $d(a\cdot b)=d(a)\cdot b+a \cdot d(b)$. Thus, writing $\mathrm{ad}(a)$ for the map $b \mapsto [a,b]$, the Jacobi identity may be rewritten (using antisymmetry) as $$\mathrm{ad}(a)([b,c])=[\mathrm{ad}(a)(b),c]+[b,\mathrm{ad}(a)(c)].$$
Second, given an associative algebra $A$, one can produce a new, no longer associative, algebra $\mathrm{Lie}(A)$ by taking $A$ for the vector space underlying $\mathrm{Lie}(A)$, but with new product structure given by $(a,b) \mapsto abba$. This product, usually written as bracket (commutator) satisfies the Jacobi identity; this is probably the original motivation for the axiom.
Finally, start with a free associative algebra $T=T(V)$ on a vector space $V$. This turns out (Thm. 1 of §3 "Enveloping algebra of the free Lie algebra", chapter 1 of Bourbaki´s Lie groups and Lie algebras) to be the enveloping algebra of the free Lie algebra on $V$: thus just the Jacobi identity is enough to force all the other relations that the subLie algebra of $T(V)$ generated by $V$ with its bracket structure gets. So though one might be tempted to look for other axioms satisfied by commutator, it turns out that the only ones that exist in general are formal consequences of the Jacobi identity and antisymmetry. This last reason is the most sophisticated, and, at least to my mind, the most convincing demonstration that our axioms are the "right ones".
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14I was tempted to finish with, "...and uh, the third one is, um... let me think.... oops?" But I figured it was getting old by now. – Stephen Feb 26 '12 at 23:13

what's the reference for this quote? A quick search doesn't yield any results. – Lukas Juhrich Mar 07 '20 at 16:59

2@LukasJuhrich Rick Perry in the 2011 GOP primary debates. E.g. here: https://www.youtube.com/watch?v=YN8uFJz9gTk – Stephen Mar 08 '20 at 19:01
The group axioms are intended to abstract the properties of discrete symmetries (that is, bijections from a set to itself). That is, we may define a "concrete group" to be a group of permutations of some set with composition as the group operation. An abstract group is supposed to be a version of a concrete group that does not rely on a choice of group action, and Cayley's theorem shows that every abstract group is also a concrete group.
Okay, so what are the Lie algebra axioms intended to abstract? They are intended to abstract the properties of infinitesimal symmetries, which rather than permutations are derivations on algebras (such as the algebra $C^{\infty}(M)$ of smooth functions on a smooth manifold; derivations on such algebras are the same thing as vector fields on $M$). The commutator of two derivations is again a derivation, and the axioms of a Lie algebra are intended to be the abstract version of this. The analogue of Cayley's theorem here is hard; it is a corollary of the PoincaréBirkhoffWitt theorem.
So why should commutators satisfy the Jacobi identity? The reason is precisely that the bracket is supposed to be a derivation with respect to itself. Another way of saying this is that the bracket $y \mapsto [x, y]$ is a Lie algebra homomorphism. A thorough explanation of why we should expect this can be found in this blog post. (Roughly speaking this is the infinitesimal version of the fact that $x \mapsto gxg^{1}$ is a group homomorphism.)
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@Steve I agree with Qiaochu that manifolds are at the heart of the issue. The motivation for studying Lie algebras is that they are the infinitesimal elements of Lie groups. – Jim Belk Feb 27 '12 at 00:30

@Steve: I thought I was being pretty clear. It's clear that abstract groups are meant to capture concrete groups, but it's less clear what Lie algebras are meant to capture, and the best version of "concrete Lie algebra" I can come up with is "derivations acting on an algebra." Again, the Jacobi identity is special because it says that $y \mapsto [x, y]$ is a homomorphism, and again, this is the infinitesimal analogue of the statement that $x \mapsto g^{1} xg$ is a homomorphism. I mentioned manifolds because they are enormously important to applications of Lie algebras (e.g. to physics). – Qiaochu Yuan Feb 27 '12 at 00:37

@Steve: I think you're taking my use of that example much too seriously. I just wanted to give nice geometric examples of derivations. I don't understand where you got the impression that the example was integral to my argument; it's in a parenthetical for a reason. – Qiaochu Yuan Feb 27 '12 at 02:43

@Steve: it's not a joke. It's just a parenthetical. I don't understand why you're interpreting it otherwise. – Qiaochu Yuan Feb 27 '12 at 16:45

My reflexive answer is: It makes ad a Liealgebra homomorphism, i.e. $\mathrm{ad}_{[x,y]}(z)=[\mathrm{ad}_x,\mathrm{ad}_y](z)$, so the adjoint representation $x \mapsto \mathrm{ad}_x$ really is a representation (well, that it's a homomorphism in combination with Steve's first point).
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3Note that one can think of the associativity axiom in a group as saying that the map $h \mapsto gh$ is a group homomorphism $G \to \text{Aut}_{\text{Set}}(G)$, so some people think of the Jacobi identity as the Lie algebra analogue of associativity. But this doesn't quite sit right with me; the group analogue is really that the map $h \mapsto ghg^{1}$ is a group homomorphism $G \to \text{Aut}_{\text{Grp}}(G)$. – Qiaochu Yuan Feb 27 '12 at 00:39
There is a famous identity in group theory due to HallWitt: if $x$, $y$, $z$ are elements of a group $G$, then $$[[x, y^{1}],z]^y\cdot[[y, z^{1}],x]^z\cdot[[z, x^{1}],y]^x=1.$$ It is trivial to prove; you can find a geometric interpretation (of a little variant of this) in terms of paths along a cube here.
Now suppose $X$, $Y$, $Z$ are three square matrices in $M_n(\mathbb C)$, and let $x(t)=\exp(t X)$, $y(t)=\exp(t Y)$, $z(t)=\exp(t Z)$ be the oneparameter subgroups of $\mathrm{GL}(n,\mathbb C)$ with initial tangent vectors $X$, $Y$ and $Z$, respectively. A little computation shows that, computing in the group $\mathrm{GL}(n,\mathbb C)$ we have $$[[x(t), y(t)^{1}],z(t)]^{y(t)}=1 + t^3 [[X,Y],Z]+ \text{terms of higher order in $t$.} $$ If follows that evaluating the HallWitt identity at $x=x(t)$, $y=y(t)$ and $z=z(t)$ we get $$ 1+t^3\bigl([[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]\bigr) + \cdots = 1 $$ so that in particular $$[[X,Y],Z]+[[Y,Z],X]+[[Z,X],Y]=0.$$
In a sense, then, the Jacobi identity is the infinitesimal version of the HallWitt identity.
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3(Of course, the actual history is backwards, and the HallWitt identity is really a group analog of the Jacobi identity... :) ) – Mariano SuárezÁlvarez Feb 27 '12 at 00:14

2@Steve: Well, the real motivation for that is the Jacobi identity :) – Mariano SuárezÁlvarez Mar 01 '12 at 03:51
For several reason one is interested in subspaces $L\subseteq End(V)$, with $V$ a vector space, which are closed under commutators —such things show up all over the place. In this situation, the commutator endows $L$ with an antisymmetric multiplication $L\times L\to L$.
Now suppose you start instead with a vector space $L$ endowed with an antisymmetric multiplication $L\times L\to L$ and you ask yourself: when is $L$ isomorphic to a subspace of some endomorphism algebra of a vector space closed under commutators?
The answer is, precisely, that
this happens when $L$ satisfies the Jacobi identity,
and in fact, in a sense which can be made precise, the identity is uniquely characterized by this condition.
Likewise, associativity is the precise condition on a vector space $A$ endowed with a multiplication $A\times A\to A$ which ensures that it is isomorphic to a subspace of an endomorphism algebra closed under composition.
Interestingly, such characterizations are not always possible: one famous example is that of Jordan algebras. A subspace $L$ of an endomorphism algebra $End(V)$ is called a special Jordan algebra if it is closed under the operation $$(x,y)\mapsto (x\circ y+y\circ x)/2.$$ It turns out that there is no set of identities which characterizes such things.
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