Jacobi Identity and Lie Algebra How can I demonstrate the Jacobi identity:
\begin{equation}
[S_{i}, [S_{j},S_{k}]] + [S_{j}, [S_{k},S_{i}]] + [S_{k}, [S_{i},S_{j}]] = 0 ~,
\end{equation}
using the infinitesimal generators $S_{\kappa}$ for a continuous group, where the generators satisfies the Lie algebra, such that:
$$[S_{\alpha},S_{\beta}] = \sum_{\gamma} f_{\alpha \beta \gamma}S_{\gamma}$$
where $f_{\alpha \beta \gamma}$ are the structure constants ?
I was doing the following:
$$\begin{align*}
[S_{i}, [S_{j},S_{k}]] &+ [S_{j}, [S_{k},S_{i}]] + [S_{k}, [S_{i},S_{j}]] = \\
 &= 
[S_{i}, \sum_{l} f_{jkl}S_{l}] + [S_{j}, \sum_{l} f_{kil}S_{l}] + [S_{k}, \sum_{l} f_{ijl}S_{l}]\\
 &=   \sum_{l} f_{jkl} [S_{i}, S_{l}] + \sum_{l} f_{kil}[S_{j}, S_{l}] + \sum_{l} f_{ijl}[S_{k}, S_{l}]\\ 
&= \sum_{l} f_{jkl} \sum_{m} f_{ilm}S_{m} + \sum_{l} f_{kil}\sum_{n} f_{jln}S_{n}\\
&\qquad + \sum_{l} f_{ijl}\sum_{p} f_{klp}S_{p}\\ 
&=  \sum_{l,~m} f_{jkl} f_{ilm}S_{m} + \sum_{l,~n} f_{kil}f_{jln}S_{n} + \sum_{l,~p} f_{ijl}f_{klp}S_{p}\\
 &=  f_{jk}^{l} f_{il}^{m}~S_{m} + f_{ki}^{l}f_{jl}^{n}~S_{n} + f_{ij}^{l}f_{kl}^{p}~S_{p}
\end{align*}$$
I know that these structure constants are antisymmetric:
\begin{equation}
f_{\alpha \beta}^{\gamma} = - f_{\beta \alpha}^{\gamma} ~~.
\end{equation}
Are there a way to go further and show that the expression will be equal to zero ?
 A: The Jacobi identity can be derives formally by expanding $[S_i,S_j]=S_iS_j-S_jS_i$. Define the associator of $S_i,S_j,S_k$ by
$$
(S_i,S_j,S_k)=(S_iS_j)S_k-S_i(S_jS_k)
$$
This is zero if the product is associative. However, a direct computation shows that we always have
$$
[[S_i,S_j],S_k]+[[S_j,S_k],S_i]+[[S_k,S_i],S_j]]=
$$
$$
(S_i,S_j,S_k)+(S_j,S_k,S_i)+(S_k,S_i,S_j)-(S_j,S_i,S_k)-(S_i,S_k,S_j)-(S_k,S_j,S_i).
$$
In particular, the Jacobi identity holds if the associator is always zero (but also, if it is, say, nonzero and left-symmetric, and so on).
A: Recapitulating @DietrichBurde's point, in different terms:
Again, emphatically, if we have a real or complex vector space $V$ with an anti-commutative binary (bilinear) operation $[,]$, this does not imply that $[,]$ satisfies the Jacobi identity. At this level of abstraction, the Jacobi identity must be required.
To see some sense in what the Jacobi asserts, rather than it just being "a formula", it asserts that the map $x\to \mathrm{ad}(x)$ is a Lie algebra homomorphism, where $\mathrm(ad)(x)(y)=[x,y]$. That is, it asserts/requires that $$
\mathrm{ad}(x)\circ\mathrm{ad}(y)-\mathrm{ad}(y)\circ\mathrm{ad}(x)
\;=\; \mathrm{ad}([x,y])
$$
The left-hand side is the natural Lie bracket $[A,B]=A\circ B-B\circ A$ inside the linear endomorphism algebra of $V$.
This also broaches the issue of possibly hoping to "open up" $[A,B]$ into $AB-BA$. This makes best sense if/when $A,B$ are elements of some associative algebra, such as square matrices. And, yes, if/when one has a vector subspace $V$ of square matrices of some size, closed under $[A,B]=AB-BA$, then it does make sense to "open up" the brackets, and, yes, one can prove by direct computation that the Jacobi identity holds.
The Jacobi identity also can be proven to hold when the vector space is the tangent space at the identity of a (real or complex) Lie group.
The two cases overlap because the space of square matrices is identifiable as the Lie algebra of the multiplicatively invertible matrices of that size.
EDIT: after a few minutes' foolling around, it's not so hard to make a three-dimensional algebra with a skew-symmetric operator $[,]$ which fails to satisfy the Jacobi identity. E.g., take basis $x,y,z$ and $[x,y]=x$, $[x,z]=y$, $[y,z]=0$. Then
$$
\Big(\mathrm{ad}(x)\circ\mathrm{ad}(y)-\mathrm{ad}(y)\circ\mathrm{ad}(x)\Big)(z)
\;=\; [x,[y,z]]-[y,[x,z]] \;=\; [x,0] - [y,y] \;=\; 0
$$
while
$$
[[x,y],z] \;=\; [x,z] \;=\; y \;\not=\; 0
$$
