Currently I'm working through the exercises in Marsden's Introduction to Mechanics and Symmetry'-Book and got stuck at Exercise 1.3-1. I believe that there might be missing an antisymmetry-assumption in the problem statement, but maybe I'm just missing that one essential ingenious trick in the last of part (b) of the exercise. Anyway I've put here the problem statement and my complete solution attempt (all that I've got so far). Part (a) was rather easy and I had no problem there (I think) but in order to fully understand my attempt at part (b) I thought it best to put my solution to part (a) here as well.
My actual question concerns just the vanishing of the second line of some expression in the part (b) of the problem.
1. The problem statement (pp.15-16 in the book)
1.3-1. A linear operator $D$ on the space $\mathcal F(\mathbb R^n)$ of smooth functions on $\mathbb R^n$ is called a derivation if it satisfies the Leibniz identity: $D(FG) = (DF)G + F(DG)$. Accept the fact from the theory of manifolds that in local coordinates the expression of $DF$ takes the form $$(DF)(x) = \sum_{i=1}^n a^i(x) \frac{\partial F}{\partial x^i}(x)$$ for some smooth functions $a^1, \dots, a^n$.
(a) Use the fact just stated to prove that for any bilinear operation $\{,\}$ on $\mathcal F(\mathbb R^n)$ which is a derivation in each of its arguments, we have $$\{F,G\} = \sum_{i,j=1}^n \{x^i, x^j\} \frac{\partial F}{\partial x^i}\frac{\partial G}{\partial x^j}.$$
(b) Show that the Jacobi identity holds for any operation $\{,\}$ on $\mathcal F(\mathbb R^n)$ as in (a), if and only if it holds for the coordinate functions.
2. My solution attempt
Let $F, G, H \in \mathcal F(\mathbb R^n)$. I'm using the Einstein summation convention and the short-hand $F_i := \frac{\partial F}{\partial x^i}$ for partial derivatives.
Part (a)
Since $\{,\}$ is a derivation in both of its arguments it can be expressed in the two ways $$\{F, G\} = a_G^i F_i = b_F^j G_j,$$ where the $a_G^i$ and $b_F^j$ are smooth functions of the coordinates which depend on G and on F respectively. By choosing $F = x^i$ I obtain an expression for the $a_G^i$'s as follows: $\{x^i, G \} = a_G^j (x^i)_j = a_G^j \delta^i_j = a_G^i = b^{ij} G_j,$ where I use the shorthand $b^{ij} := b_{x^i}^j$. Hence the bracket can be expressed in the form $$\{F, G\} = a_G^i F_i = b^{ij} F_i G_j.$$ Finally the evaluation $$\{x^i, x^j\} = b^{kl} (x^i)_k (x^j)_l = b^{kl} \delta^i_k \delta^j_l = b^{ij}$$ proves the validity of the coordinate expression for $\{,\}$ that was given above. $\square$
Part (b)
The 'only if'-part is trivial and hence I concern myself now with the converse. I simplify the notation further by introducing the shorthand $i := x^i$. Observe that in this notation $\{i, j\} = \{x^i, x^j\}$ and $i_j := \frac {\partial x^i} {\partial x^j} = \delta^i_j.$
Using the result from (a) and the linearity and leibniz properties of $(\cdot)_i$ I first evaluate $$ \begin{align} \{\{F, G\}, H\} &= \{i, j\} \{F, G\}_i H_j\\ &= \{i, j\} (\{k, l\} F_k G_l)_i H_j\\ &= \{i, j\} (\{k, l\}_i F_k G_l + \{k, l\} (F_{ik} G_l + F_k G_{il})) H_j. \end{align} $$
Evaluating this expression two more times under the cyclic permutations $\{FGH \mapsto GHF,FGH \mapsto HFG\}$ and grouping the terms with first order and second order derivatives the expression for the (left-)Jacobi identity then becomes $$ \begin{align} &\{\{F, G\}, H\} + \{\{G, H\}, F\} + \{\{H, F\}, G\} =\\ \\ &\hphantom{+\ \ } \{i, j\} \{k, l\}_i (F_k G_l H_j + G_k H_l F_j + H_k F_l G_j)\\ &+ \{i, j\} \{k, l\} (F_{ik}G_lH_j + F_kG_{il}H_j + G_{ik}H_lF_j + G_kH_{il}F_j + H_{ik}F_lG_j + H_kF_{il}G_j). \end{align}$$
Since the first (result-)line in this expression contains only first order derivatives and the second line contains also second order derivatives they have to vanish separately (e.g. pick $F, G$ and $H$ as linear functions of the coordinates to 'extract' the first line):
line vanishes?
By relabelling the indices of the second and third summand in the first line its expression becomes $$\{i, j\} \{k, l\}_i (F_k G_l H_j + G_k H_l F_j + H_k F_l G_j) =\\(\{i, j\}\{k, l\}_i + \{i, l\}\{j, k\}_i + \{i, k\}\{l, j\}_i) F_k G_l H_j.$$ The first factor in this expression is just the evaluated form of the (left-)Jacobi identity $$\{\{k, l\}, j\} + \{\{j, k\}, l\} + \{\{l, j\}, k\} = 0 $$ which readily follows from the result in part (a): $\{\{k, l\}, j\} = \{i, p\} \{k, l\}_i j_p = \{i, p\} \{k, l\}_i \delta^j_p = \{i, j\} \{k, l\}_i$, and reevaluation of this expression under the cyclic permutations $\{klj \mapsto jkl, klj \mapsto ljk\}$.
line vanishes?
First I reorder the summands and factors and group them into three summand pairs $S$ with shared second order factors: $$\{i, j\} \{k, l\} (F_{ik} G_l H_j + F_{il} G_j H_k + G_{ik} H_l F_j + G_{il} H_j F_k + H_{ik} F_l G_j + H_{il} F_j G_k) = S(F, G, H) + S(G, H, F) + S(H, F, G),$$ where $S(F, G, H) := \{i, j\} \{k, l\} (F_{ik} G_l H_j + F_{il} G_j H_k)$. These have to vanish independently because $F, G, H$ can be varied independently from each other (e.g. to 'extract' $S(F, G, H)$ from this expression choose $G$ and $H$ as linear functions of the coordinates) . Hence it suffices to show that $S(F, G, H) = 0$:
By using the index relabelling $ljk \mapsto klj$ in the second summand of $S$ and using the fact the smoothness of $F$ implies that $F_{ik} = F_{ki}$ I get $$ \begin{align} S(F, G, H) &= (\{i, j\} \{k, l\} + \{i, l\} \{j, k\}) F_{ik} G_l H_j\\ &=\frac 1 2 (\{i, j\} \{k, l\} + \{i, l\} \{j, k\} + \{k, j\} \{i, l\} + \{k, l\} \{j, i\}) F_{ik} G_l H_j\\ &= \frac 1 2 ((\{i, j\} + \{j, i\})\{k, l\} + (\{j, k\} + \{k, j\}) \{i, l\}) F_{ik} G_l H_j, \end{align} $$ where the second line is obtained by $ik$-symmetrisation of the left factor on the first line, and the third line by subsequent reordering and grouping.
Varying $F, G$ and $H$ implies that I have to show that for all $i, j, k, l$ the factor $$(\{i, j\} + \{j, i\})\{k, l\} + (\{j, k\} + \{k, j\}) \{i, l\}$$ vanishes. And therein lies my problem. If $\{,\}$ were antisymmetric (i.e. $\{i, j\} = -\{j, i\}$) then I would be done. But this assumption was not made in the statement of the exercise. Specifically in the diagonal case were $i = k$ the expression even further reduces to $(\{i, j\} + \{j, i\}) \{i, l\}$ which (assuming a non-zero bracket) really can only be zero if the bracket is antisymmetric.
3. Question
Based on my experiences with part (b) of the exercise I conclude that ...
- ... either the problem statement needs to include the antisymmetry assumption?
- ... or I do have to show that bi-linearity, bi-leibniz-property and bi-jacobi-identity $\Rightarrow$ antisymmetry?
- ... or maybe I grouped the terms in the summation of the second line in a premature way which made it impossible for them to become zero? But I don't know where that would have happened. I even gave reasons why those terms can be 'isolated' from each other.
So far all the examples of brackets that I encountered in the book were antisymmetric which speaks for my first conclusion (or the last one). But of course I might be wrong.
Any comments, hints or other kinds of inspiration are are very welcome. Thanks in advance.