I am trying to determine the symplectic leaves of the singular distribution of $\mathfrak{su}(3)^*$ induced by the linear poisson bracket. Pick an element $x=(x^1,x^2,...,x^8)\in \mathfrak{su}(3)^*$. The coordinates are regarded with respect to the dual basis. Then, the subspace of the distribution at $x$ is generated by the following vectors:
$v_1=(0,0,0,0,x^5,-x^3,x^8,-x^7),v_2=(0,0,x^4,-x^3,0,0,0,x^7), v_3=(0,-x^4,0,2x^1,-x^7,-x^8,x^5,x^6) ,~~ v_4=(0,x^3,-2x^1,0,x^8,-x^7,x^6,x^5), v_5=(x^5,0,x^7,-x^8,0,-2x^1,x^3,x^4), v_6=(x^3,0,x^8,x^7,2x^1,0-x^4,-x^3) , v_7=(-x^8,0,-x^5,-x^6,x^3,x^4,0,2(x^1-x^2)) , v_8=(x^7,-x^7,x^6,x^5,-x^4,x^3,-2(x^1-x^2),0)$.
Even by taking some restrictions, for example $x$ to belong in the affine subspace $\{x^7=0,x^8=0\}$, it is complicated to find which combination of the above vectors is linearly independent and hence determine the dimension of $D_x$, let aside the symplectic leaf containing a given point. Any ideas and methods to simple down the calculations are deeply appreciated!
You really want to do this in coordinate form? Really, really?! Ok. We can do that. It's not as hard as you think; in fact quite workable and revealing in terms of the details of the structure uncovered. You just need to do it the right way, to make it easier. In fact, you should see it worked out in detail, because it forms a really nice - and symmetric - picture.
The first thing you can do is pull everything back to $U(3)$. The symplectic leaves of $SU(3)$ are a subset of those of $U(3)$, and you'll get a clearer view of matters if you broaden the scope of the problem.
For $U(3)$, in its natural matrix representation, denote the unit matrices by $$e^i_j: i, j = 1, 2, 3,$$ and note the product rule and corresponding Lie brackets: $$e^i_j e^k_l = δ^k_j e^i_l, \hspace 1em [e^i_j, e^k_l] = e^i_j e^k_l - e^k_l e^i_j = δ^k_j e^i_l - δ^i_l e^k_j \hspace 1em (i, j, k, l = 1, 2, 3).$$
This algebra is associated with the notion of "color". This is not a specious analogy. We can order and rename the off-diagonal elements like so: $$R = e^1_2, \hspace 1em O = e^1_3, \hspace 1em Y = e^2_3, \hspace 1em G = e^2_1, \hspace 1em B = e^3_1, \hspace 1em P = e^3_2,$$ thinking of them, respectively, as "Red", "Orange", "Yellow", "Green", "Blue" and "Purple". They form a hexagon of the color wheel. The representations $$ and $\bar{}$ lie at in-between angles on this hexagon, so we can think of them as {red, green, blue} and {cyan, amber, magenta}, respectively, though the "red"/"Red", "green"/"Green" and "blue"/"Blue" pairs are off from one another by 30 degrees, so they'd be shaded slightly differently.
The diagonal elements form a basis for a toroidal subalgebra. For $U(3)$, they are: $$L = e^1_1, \hspace 1em M = e^2_2, \hspace 1em N = e^3_3,$$ while, for $SU(3)$, they are restricted to the following linear combinations, $$U = L - M = e^1_1 - e^2_2, \hspace 1em V = M - N = e^2_2 - e^3_3, \hspace 1em W = N - L = e^3_3 - e^1_1.$$
For $SU(3)$, this presents a frame for a 2-dimensional spectrum which play a role analogous to the "chromaticity coordinates" in color theory; while the diagonal sum $$I = L + M + N,$$ plays a role analogous to "brightness".
For instance, in the Standard Model, the quarks can be treated as having colors red, green, blue, with brightness +⅓, the anti-quarks as having colors cyan, amber and magenta, with brightness -⅓, the leptons as being color-neutral, with brightness -1 and the anti-leptons also being color-neutral, but with brightness +1. Together, they form the vertices of a cube, analogous to the color cube. Brightness corresponds to (baryon minus lepton) number (or "B-L"), which has no (known) gauge force associated with it, though our state of knowledge is very likely going to change on this matter in the near future.
Using $I$, we can express $L$, $M$ and $N$ in terms of $U$, $V$ and $W$ as: $$L = \frac{I + U - W}3, \hspace 1em M = \frac{I + V - U}3, \hspace 1em N = \frac{I + W - V}3.$$ The elements $U$, $V$ and $W$, themselves, satisfy the constraint: $$U + V + W = 0.$$
The Lie brackets take on a simple form, as follows: $$ [R,O] = [O,Y] = [Y,G] = [G,B] = [B,P] = [P,R] = 0,\\ [R,Y] = O, \hspace 1em [Y,B] = G, \hspace 1em [B,R] = P,\\ [G,O] = Y, \hspace 1em [O,P] = R, \hspace 1em [P,G] = B,\\ [R,G] = U, \hspace 1em [Y,P] = V, \hspace 1em [B,O] = W. $$
In addition, are the brackets of the form $[t,y] = t_y y$, where $t$ is a diagonal element and $y$ any of the off-diagonal elements. For $L$, $M$ and $N$ this includes the following: $$ (L_R, L_O, L_Y, L_G, L_B, L_P) = (+1, +1, 0, -1, -1, 0),\\ (M_R, M_O, M_Y, M_G, M_B, M_P) = (-1, 0, +1, +1, 0, -1),\\ (N_R, N_O, N_Y, N_G, N_B, N_P) = (0, -1, -1, 0, +1, +1). $$ For $U$, $V$ and $W$, the assignments are: $$ (U_R, U_O, U_Y, U_G, U_B, U_P) = (+2, +1, -1, -2, -1, +1),\\ (V_R, V_O, V_Y, V_G, V_B, V_P) = (-1, +1, +2, +1, -1, -2),\\ (W_R, W_O, W_Y, W_G, W_B, W_P) = (-1, -2, -1, +1, +2, +1). $$
Finally, all the toroidal basis elements have zero Lie brackets: $$[L,M] = [M,N] = [N,L] = 0, \hspace 1em [U,V] = [V,W] = [W,U] = 0,$$ as well as the mixed combinations, e.g. $[L,U] = 0$.
The Poisson manifold, which the symplectic leaves lie inside, is the function space over the dual ${u(3)}^*$ of the Lie algebra $u(3)$ of $U(3)$. Included in this are the linear functions over ${u(3)}^*$, which is just another way of describing the double-dual ${u(3)}^{**}$, which - in turn, is the equivalent of $u(3)$, itself. Therefore, the function space is a non-linear extension of the Lie algebra in which the basis plays the role of coordinates, and their respective Lie brackets the role of the fundamental Poisson brackets.
So, at this point we will revert to Poisson bracket notation, e.g. $\{R,G\} = U$.
The symplectic leaves are identified by which functions are invariant and by the values that these invariants take. So, the task at hand is to find the invariants. Given a function $Φ$ of the basis elements, write its total differential as $$dΦ = l dL + m dM + n dN + r dR + o dO + y dY + g dG + b dB + p dP.$$ The invariants are determined by the condition $\{Φ,⋯\} = 0$. Then, we have the following Poisson brackets and equations: $$ 0 = \{Φ,R\} = R(l - m) + Pb - Oy - Ug,\\ 0 = \{Φ,O\} = O(l - n) + Yg - Rp + Wb,\\ 0 = \{Φ,Y\} = Y(m - n) + Or - Gb - Vp,\\ 0 = \{Φ,G\} = G(m - l) + Bp - Yo + Ur,\\ 0 = \{Φ,B\} = B(n - l) + Gy - Pr - Wo,\\ 0 = \{Φ,P\} = P(n - m) + Ro - Bg + Vy,\\ 0 = \{Φ,L\} = Gg + Bb - Rr - Oo,\\ 0 = \{Φ,M\} = Pp + Rr - Yy - Gg,\\ 0 = \{Φ,N\} = Oo + Yy - Bb - Pp, $$ resulting in a linear system that has a rank of no larger than eight; since we already have $$\{Φ,I\} = \{Φ,L+M+N\} = 0,$$ with $I$ thus being a linear invariant.
In the following, we will write $F ≡ G$ to denote a "constraint": or an invariant condition that holds on a symplectic leaf. Each constraint may yield secondary constraints upon application of the Poisson brakcets, e.g. $\{F,R\} ≡ \{G,R\}$; and these, in turn, may yield further constraints down the line.
The symplectic leaves for $SU(3)$ are those of $U(3)$ where $I ≡ 0$.
Dimension-0 Symplectic Leaves: $(R,O,Y,G,B,P,U,V,W) ≡ (0,0,0,0,0,0,0,0,0)$.
First, we note that the constraint $R ≡ 0$ yields, as secondary constraints once removed: $O ≡ 0$, $U ≡ 0$, $P ≡ 0$, and then $Y ≡ 0$, $G ≡ 0$, $B ≡ 0$, $V ≡ 0$ and $W ≡ 0$ twice removed, thus resulting in $L ≡ M ≡ N ≡ ⅓I$. A similar conclusion applies if starting from any of these secondary constraints.
These are the 0-dimensional symplectic leaves; indexed by $I$. For $SU(3)$, since $I ≡ 0$, then there's only one such leaf: the one corresponding to the zero representation.
Non-Trivial Symplectic Leaves
To find the other symplectic leaves, we will exclude those from consideration and proceed from there. With this in mind, we can rewrite the linear system in the following form: $$ Rr - Gg = Yy - Pp = Bb - Oo,\\ \frac{Oy - Pb + Ug}R = l - m = \frac{Bp - Yo + Ur}G,\\ \frac{Gb - Or + Vp}Y = m - n = \frac{Ro - Bg + Vy}P,\\ \frac{Pr - Gy + Wo}B = n - l = \frac{Yg - Rp + Wb}O. $$
Constrained Dimension-4 Symplectic Leaves: $RYB ≡ OPG$
The other case of a reduced-rank system occurs with the constraint $A = RYB - OPG ≡ 0$, which has the following secondary constraints: $$ \frac{BR}P - \frac{RY}O ≡ U ≡ \frac{GO}Y - \frac{PG}B,\\ \frac{RY}O - \frac{YB}G ≡ V ≡ \frac{PG}B - \frac{OP}R,\\ \frac{YB}G - \frac{BR}P ≡ W ≡ \frac{OP}R - \frac{GO}Y, $$ and, from this: $$ \frac{BR}P + Z ≡ L ≡ \frac{GO}Y + Z,\\ \frac{RY}O + Z ≡ M ≡ \frac{PG}B + Z,\\ \frac{YB}G + Z ≡ N ≡ \frac{OP}R + Z, $$ where $$Z = \frac{I - Φ_1}3, \hspace 1em \frac{BR}P + \frac{RY}O + \frac{YB}G ≡ Φ_1 ≡ \frac{GO}Y + \frac{PG}B + \frac{OP}R.$$ You may verify that $Φ_1$ is, itself, an invariant on the symplectic leaf, so that $Z$ is as well. The primary constraint $A ≡ 0$ and the secondary constraints on $U$, $V$ and $W$ together count as three independent constraints. Taken with the two invariants $I$ and $Φ_1$, that reduces the dimension from the 9 of $U(3)$, by 5, to 4.
We'll deal with the case $A ≡ 0$ here, first, and then the general case, later.
Let $$ γ = \frac{Rr - Gg}2 = \frac{Yy - Pp}2 = \frac{Bb - Oo}2,\\ λ = \frac{Bb + Oo + Rr + Gg}4, \hspace 1em μ = \frac{Rr + Gg + Yy + Pp}4, \hspace 1em ν = \frac{Yy + Pp + Bb + Oo}4. $$ Then we may write $$ r = \frac{+γ + λ + μ - ν}R, \hspace 1em y = \frac{+γ - λ + μ + ν}Y, \hspace 1em b = \frac{+γ + λ - μ + ν}B,\\ o = \frac{-γ + λ - μ + ν}O, \hspace 1em p = \frac{-γ - λ + μ + ν}P, \hspace 1em g = \frac{-γ + λ + μ - ν}G, $$ and find, upon substitution: $$2γU = 0, \hspace 1em 2γV = 0, \hspace 1em 2γW = 0 ⇒ γ = 0$$ and $$ l - m = - 2λ \frac Y{GO} + 2μ \frac B{PG},\\ m - n = - 2μ \frac B{PG} + 2ν \frac R{OP},\\ n - l = - 2ν \frac R{OP} + 2λ \frac Y{GO}. $$ from which we infer $$l = z - 2λ \frac Y{GO}, \hspace 1em m = z - 2μ \frac B{PG}, \hspace 1em n = z - 2ν \frac R{OP}.$$ Upon substitution into the differential equation for $Φ$, we get: $$\begin{align} dΦ &= z d(L + M + N)\\ &+ λ \left(+ \frac{dR}R - \frac{dY}Y + \frac{dB}B + \frac{dO}O - \frac{dP}P + \frac{dG}G - 2 \frac Y{GO} d\left(\frac{GO}Y + Z\right)\right)\\ &+ μ \left(+ \frac{dR}R + \frac{dY}Y - \frac{dB}B - \frac{dO}O + \frac{dP}P + \frac{dG}G - 2 \frac B{PG} d\left(\frac{PG}B + Z\right)\right)\\ &+ ν \left(- \frac{dR}R + \frac{dY}Y + \frac{dB}B + \frac{dO}O + \frac{dP}P - \frac{dG}G - 2 \frac R{OP} d\left(\frac{OP}R + Z\right)\right)\\ &= z dI - 2 \left(λ \frac Y{GO} + μ \frac B{PG} + ν \frac R{OP}\right) dZ. \end{align}$$ Thus $Φ$ reduces to a function $Φ(I, Z)$ of $I$ and $Z$ - or, equivalently: to a function of $I$ and $Φ_1$, thus showing that these are the only remaining independent invariants.
The symplectic leaves, thus have 9 dimensions, reduced by the constraints to 6 and by the invariants to 4.
General 6-Dimensional Symplectic Leaves
Finally, for the general case, we assume that we're away from the constraint $A ≡ 0$. Setting $$ x = \frac{Rr - Gg}A = \frac{Yy - Pp}A = \frac{Oo - Bb}A,\\ λ = \frac{Yy + Pp - x (RYB + OPG)}{2YP},\\ μ = \frac{Bb + Oo - x (RYB + OPG)}{2BO},\\ ν = \frac{Rr + Gg - x (RYB + OPG)}{2RG}, $$ (and recycling $λ$, $μ$ and $ν$ to different use) we can write: $$\begin{align} y = λP + xBR, \hspace 1em p = λY + xGO,\\ b = μO + xRY, \hspace 1em o = μB + xPG,\\ r = νG + xYB, \hspace 1em g = νR + xOP. \end{align}$$ Upon substitution into half of the remaining equations, we get: $$\begin{align} 0 = \left(\frac{OP}R - \frac{YB}G\right)(λ - μ + xU) = A\frac{(λ + xL) - (μ + xM)}{RG}\\ 0 = \left(\frac{GO}Y - \frac{BR}P\right)(μ - ν + xV) = A\frac{(μ + xM) - (ν + xN)}{YP}\\ 0 = \left(\frac{PG}B - \frac{RY}O\right)(ν - λ + xW) = A\frac{(ν + xN) - (λ + xL)}{BO} \end{align}$$ and, since we're away from $A ≡ 0$, defining $$γ = λ + xL = μ + xM = ν + xN,$$ we have $$(λ, μ, ν) = (γ - xL, γ - xM, γ - xN)$$ Upon substitution into the other half of the remaining equations, we get: $$\begin{align} Uν + x(BO - YP) = l - m,\\ Vλ + x(RG - BO) = m - n,\\ Wμ + x(YP - RG) = n - l. \end{align}$$ Thus, setting $$z = ⅓ (l + m + n - γ(L + M + N) + x(RG + YP + BO) - x(LM + MN + NL)).$$ we have: $$(l, m, n) = (z + γL + x(MN - YP), z + γM + x(NL - BO), z + γN + x(LM - RG)).$$
Upon substitution into the differential equation for $Φ$, we get: $$\begin{align} dΦ &= (G(γ - xN) + xYB) dR + (R(γ - xN) + xOP) dG\\ &+ (P(γ - xL) + xBR) dY + (Y(γ - xL) + xGO) dP\\ &+ (O(γ - xM) + xRY) dB + (B(γ - xM) + xPG) dO\\ &+ (z + γL + x(MN - YP)) dL\\ &+ (z + γM + x(NL - BO)) dM\\ &+ (z + γN + x(LM - RG)) dN\\ &= z dI + ½ γ dΦ_2 + x dΦ_3 \end{align}$$ where $$ Φ_2 = RG + OB + YP + GR + BO + PY + L^2 + M^2 + N^2,\\ Φ_3 = RYB + OPG - RGN - YPL - BOM + LMN. $$ Thus $Φ = Φ\left(I, Φ_2, Φ_3\right)$ is a function of the three invariants $I$, $Φ_2$ and $Φ_3$.
Subtracting the 3 invariants from the 9 coordinates of $U(3)$ leaves 6 free coordinates for the highest dimension symplectic leaves of $U(3)$. Those which belong also to $SU(3)$ are obtained by setting $I ≡ 0$. Upon substitution of $L$, $M$ and $N$, for $I$, $U$, $V$ and $W$, the quadratic and cubic invariants reduce to: $$\begin{align} Φ_2 &= RG + OB + YP + GR + BO + PY + ⅓ (U^2 + V^2 + W^2),\\ Φ_3 &= RYB + OPG\\ &+ \frac{RG(V - W) + YP(W - U) + BO(U - V)}3 \\ &- \frac{(U - V)(V - W)(W - U)}{27}. \end{align}$$
When subject to the constraint $A ≡ 0$, the invariants reduce to: $$ I ≡ 3Z + Φ_1,\\ Φ_2 = 3Z^2 + 2ZΦ_1 + Φ_1^2 ≡ ⅓ I^2 + ⅔ Φ_1^2,\\ Φ_3 = Z^2 (Z + Φ_1) ≡ \frac{I^3 - 3IΦ_1^2 + 2Φ_1^3}{27}, $$ and, for $SU(3)$, where $I ≡ 0$, to: $$Φ_2 = \frac{2}{3} Φ_1^2, \hspace 1em Φ_3 = \frac{2}{27} Φ_1^3.$$