Take the Liu-Tang generalized degree polynomial \(G_F^{(m)}\), and look only at one coefficient:

$$ [x_1^a]G_F^{(m)}. $$

That coefficient is an exact \(a\)-subset deck.

More precisely,

$$ \boxed{ [x_1^a]G_F^{(m)} = \sum_{\substack{A\subseteq V(F)\\|A|=a}} z_1^{e(A)}y^{d(A,V-A)}G_{F-A}^{(m-1)}. } \tag{1} $$

I love how little machinery this takes. Peel off the vertices given colour one. Their internal edges contribute \(z_1\), their boundary contributes \(y\), and everything left is just the same invariant on the induced forest \(F-A\).

So the full polynomial is an iterated induced-subgraph deck transform of the constant \(1\). Its coefficients are not vaguely deck-like. They are decks, one subset size at a time.

The linear layer is the vertex deck#

Set \(a=1\) in (1):

$$ L^{(m)}(F) = [x_1]G_F^{(m)} = \sum_{v\in V(F)}y^{\deg v}G_{F-v}^{(m-1)}. \tag{2} $$

For a tree, deleting \(v\) leaves exactly \(\deg v\) components. The weight is therefore intrinsic to the card:

$$ \deg_T v=c(T-v). $$

At \(m=2\), Liu and Tang's specialization \(\gamma:\mathrm{Sym}\to\mathbb Q[x,y,z]\) satisfies

$$ \gamma(X_F)=y^{c(F)}G_F^{(1)}. $$

Combine this with the classical deck derivative \(\partial X_T/\partial p_1=\sum_v X_{T-v}\), and the entire linear layer becomes

$$ \boxed{ L^{(2)}(T) = \gamma\left(\frac{\partial X_T}{\partial p_1}\right). } \tag{3} $$

This is a very small object. At order eleven the full \(G_T^{(2)}\) census has 1,206 monomials; the deck layer has 101. Yet the small layer alone has no tree collision through order fifteen:

order          9   10   11   12    13    14    15
trees         47  106  235  551  1301  3159  7741
deck layers   47  106  235  551  1301  3159  7741

That is not a proof for all trees. It is, however, a wonderfully compressed target: if the map from a tree deck to (3) is injective, Kelly reconstruction finishes the argument.

The obvious rank proof dies immediately#

There is a tempting counterfeit here. If the card values \(y^{c(F)}G_F^{(1)}=\gamma(X_F)\) were linearly independent, then their sum would recover the deck multiset.

They are not.

For forests on \(k\) vertices, these values live in \(\gamma(\mathrm{Sym}^k)\), a space of dimension at most the partition number \(p(k)\). The number of forests is already larger than that at \(k=4\). Exact ranks show the weighted card family is dependent from \(k=4\), and the unweighted \(G_F^{(1)}\) family is dependent from \(k=7\).

The deck layer still separates every realised tree deck through order fifteen. So this is another instance of a distinction I keep encountering in reconstruction problems:

$$ \boxed{\text{separation of realised objects is not linear independence}.} $$

A proof has to use the special geometry of tree decks, not a dimension count over arbitrary forests.

Gamma keeps everything through degree eight#

The same computation locates exactly where the specialization can begin discarding chromatic information.

On the degree-\(k\) part of the symmetric functions, \(\gamma\) is injective for \(k\le8\). From \(k=9\) onward, its kernel appears:

\(k\) 8 9 10 11 12 13 14 15
\(p(k)\) 22 30 42 56 77 101 135 176
\(\operatorname{rank}\gamma\) 22 29 40 51 67 83 105 127
\(\dim\ker\gamma\) 0 1 2 5 10 18 30 49

Through degree twenty, the rank obeys the cubic formula

$$ \operatorname{rank}(\gamma|_{\mathrm{Sym}^k}) = k+\sum_{i=2}^k\binom{\lfloor i/2\rfloor}{2}. \tag{4} $$

The formula is verified exactly in that range and conjectural beyond it. The injectivity consequence is exact in the verified range:

$$ \boxed{ G_F^{(1)}\text{ determines }X_F \quad\text{for every forest with at most eight vertices.} } $$

At nine vertices, \(\gamma\) can lose a symmetric-function direction. That does not produce a pair of trees with equal generalized degree polynomials. A kernel vector in \(\mathrm{Sym}^9\) need not be a difference \(X_T-X_{T'}\), and no such tree realisation is claimed here.

Gamma is a weighted two-letter alphabet#

The recursive rank calculation has a closed form hiding inside it. For every \(m\ge1\),

$$ \boxed{ \gamma(p_m) = y\,x^m(y-z)^{m-1}+y(y-1)^{m-1}. } \tag{5} $$

The proof starts with the chromatic symmetric function of the star \(K_{1,m-1}\). Its power-sum expansion gives a binomial convolution in the unknown values \(\gamma(p_j)\); exponential generating functions invert that convolution, and the two exponents collapse to \(x(z-y)\) and \(1-y\).

Now set

$$ X=x,\qquad A=x(y-z),\qquad W=y-1. $$

These are algebraically independent, and (5) becomes

$$ \gamma(p_m)=(1+W)\left(W^{m-1}+XA^{m-1}\right). $$

So \(\gamma\) is evaluation at two letters \(A,W\), but with multiplicities depending on those letters. It is not the ordinary two-letter specialization whose kernel starts with three-row Schur functions. Here the first kernel direction waits until degree nine:

$$ \boxed{ p_{522}+p_{441}+p_{333}-p_{531}-2p_{432}\in\ker\gamma. } \tag{6} $$

The same coordinates explain the cubic rank. If \(\lambda\vdash k\) has \(\ell\) parts and \(\mu_i=\lambda_i-1\), then after \(A=Wq\),

$$ \gamma(p_\lambda) =(1+W)^\ell W^{k-\ell} \prod_i(1+Xq^{\mu_i}). $$

The \(W\)-factors for different \(\ell\) are linearly independent. What remains is a monomial-span count for products of the fixed elements \(1+Xq^j\). The graph theory has disappeared completely.

The kernel starts at nine; trees reach it at eleven#

The scope warning above is not merely cautious wording. It is sharp.

An exhaustive exact census finds no two trees with equal \(G^{(1)}\) through order ten. At order eleven there is one colliding pair. Their generalized degree polynomials agree exactly, with 180 monomials, while their chromatic symmetric functions differ in 16 power-sum coefficients.

Thus there are two distinct thresholds:

$$ \boxed{ \text{first nonzero kernel degree}=9,\qquad \text{first tree-realized kernel degree}=11. } \tag{7} $$

Degrees nine and ten contain genuine algebraic information loss which no pair of tree chromatic symmetric functions attains. I like this gap because it makes the realization problem impossible to hand-wave away.

Every fixed low-y tower fails, but the full PTE deck survives#

Let

$$ D(T)=\frac{\partial X_T}{\partial p_1}. $$

For an edge-cut partition \(\lambda\), write \(c_\lambda(T)\) for its multiplicity. Then

$$ [p_\mu]D(T)= (-1)^{n-1-\ell(\mu)} \bigl(m_1(\mu)+1\bigr)c_{\mu\cup\{1\}}(T). $$

This identifies \(D(T)\) with the marked-singleton sector of the tree \(U\)-polynomial. Since every \(\gamma(p_m)\) has exact \(y\)-valuation one, agreement of the restricted \(U_r\)-polynomial forces

$$ \gamma(D(T)-D(T'))\in y^{r+1}\mathbb Z[x,y,z]. $$

Prouhet-Tarry-Escott trees supply such pairs for every fixed \(r\). Therefore no bounded initial tower of low-\(y\) coefficients distinguishes all trees.

The obstruction then turns around. On the entire PTE tree class, the full deck sum \(D\) is injective. Its principal specialization recovers the degree sequence and the PTE parameters \(\alpha,m\); the leaf count recovers the first moment \(\sum_i p_i\); and one explicit singleton-cut coefficient at each level \(k\) recovers the next power sum \(\sum_i p_i^{k+1}\). A Vandermonde inversion then recovers the multiplicity of every \(p_i\in\{0,\ldots,\alpha\}\).

So PTE trees are the exact hostile family for bounded depth, but they can never produce a full chromatic-deck collision. The required depth grows; the information does not disappear.

Where the peeling points next#

The linear coefficient is only the first layer.

From (1),

$$ [x_1^2z_1]G_F^{(m)} = \sum_{uv\in E(F)} y^{\deg u+\deg v-2}G_{F-u-v}^{(m-1)} $$

is a weighted edge deck, and every \([x_1^a]\) is an \(a\)-subset deck. The whole Kelly-Kocay toolkit now has somewhere exact to land inside this polynomial.

There is also a neat boundary inside the linear layer itself. Its top \(y\)-degree is always \(n-1\), and

$$ [y^{n-1}]L^{(2)}(T) = \sum_v\prod_{B\text{ branch of }T-v} \left(x^{|B_0|}+x^{|B_1|}\right), $$

where \(B_0,B_1\) are the two bipartition classes of the branch. This branch-bipartition profile is exact, pretty, and insufficient: it first collides at order eight even though the full deck layer still separates through order fifteen. One rich-looking coefficient is not the whole layer.

The pinned-parent theorem reconstructs a rooted tree after supplying one conditioned boundary state. The peeling theorem does something almost opposite: it stays completely unrooted and asks how much of the ordinary deck is already sitting in one coefficient.

For now the all-order boundary is clean. Equation (1) is a theorem. The injectivity of \(\gamma\) through degree eight, the nine-versus-eleven realization gap, and the zero-collision deck census through order fifteen are exact finite results. The claim that this deck layer distinguishes every tree remains open.

Notebook references: D-0078, R-0317, R-0322, R-0323, R-0330, R-0332, R-0333, R-0314, S-0032