For a tree \(T\), one derivative of its chromatic symmetric function is already a sum over vertex-deleted cards:
$$ D(T)=\frac{\partial X_T}{\partial p_1}. $$
After the usual coefficientwise sign conversion, this becomes
$$ M_T=\frac{\partial U_T}{\partial x_1} =\sum_{v\in V(T)}U_{T-v}. $$
The large-component half of \(M_T\) has an exact factorization:
$$ \boxed{ [x_{n-k}]M_T =[w^k]\frac{\partial}{\partial x_1} \prod_i\Pi_{B_i,r_i}(w), \qquad 1\le k<\frac n2, } $$
where the rooted trees \((B_i,r_i)\) are the branches obtained by deleting the centroid core of \(T\).
So the giant half of one unrooted derivative is a derivative of one rooted product at the centroid. I did not expect the geometry to line up this cleanly.
Connected deletions are already derivatives#
The identity is part of a larger hierarchy. Stanley's power-sum expansion gives, for every \(k\ge1\),
$$ \boxed{ (-1)^{k-1}\frac{\partial X_T}{\partial p_k} = \sum_{\substack{S\subseteq V(T)\\|S|=k\\T[S]\text{ connected}}} X_{T-S}. } $$
Differentiating by \(p_k\) marks one component of order \(k\). In a tree, fixing its vertex set forces all \(k-1\) internal edges, forces every boundary edge absent, and leaves an arbitrary edge subset of \(T-S\). That is the whole proof.
The same layer is a coefficient of the order-two generalized-degree polynomial:
$$ \sum_{\substack{|S|=k\\T[S]\text{ connected}}} y^{d_T(S)}G^{(1)}_{T-S} = [x_1^kz_1^{k-1}]G_T^{(2)}. $$
Thus the vertex deck, weighted edge deck, and every higher connected-subtree deletion layer belong to one fixed CSF-determined object.
Reverse the rooted factors at the centroid#
Let \(K(T)\) be the one- or two-vertex centroid core. Deleting it leaves rooted components \((B_1,r_1),\ldots,(B_d,r_d)\). For one rooted branch, define
$$ \Pi_{B,r}(w) = \sum_{\substack{ Q=\varnothing\text{ or}\\ B[Q]\text{ connected},\ r\in Q }} w^{|B|-|Q|}U_{B-Q}. $$
If \(F_{B,r}=R_{B,r}+U_B\) is the genuine rooted factor, then
$$ \boxed{ \Pi_{B,r}(w)=w^{|B|}F_{B,r}(w^{-1}). } $$
Every connected set larger than half the tree contains the centroid core. Its complement therefore consists of independent pruning choices inside the centroid branches, and those choices multiply:
$$ \mathcal Z_T(w) = \prod_i\Pi_{B_i,r_i}(w). $$
Marking a singleton component now gives the displayed giant-half identity. The derivative has converted an unrooted deletion sum into a low-order jet of a product of reversed rooted factors.
The strict half really can forget#
This factorization is not automatically injective.
At order seven, the spider \(S(1,1,1,3)\) and an adjacent double spider with centroid branches
$$ P_1,\qquad P_2\text{ rooted at an endpoint},\qquad P_3\text{ rooted at its center} $$
have different degree sequences but identical centroid products through the entire strict half. Writing
$$ L=1+wx_1,\qquad C=x_1^3+2x_1x_2+x_3, $$
their products satisfy
$$ \boxed{ P_-(w)-P_+(w)=w^4LC(x_1-wx_2). } $$
Both trees have order seven, so the available product modulo \(w^4\) is identical. This is a genuine tree collision, not an ambient polynomial counterfeit.
The first missing coefficient is repaired exactly by the equatorial layer. For a tree of order \(2h+1\),
$$ \partial_{x_h}M_T = [w^{h+1}]\partial_{x_1}\mathcal Z_T + \sum_{i:|B_i|=h}M_{T-B_i}. $$
In the seven-vertex pair, the residual cards are \(K_{1,3}\) and \(P_4\), whose marked-singleton difference is \(x_3\). That is precisely the missing term.
I like this boundary very much. The strict giant half loses information, but the exact middle layer knows the card that repairs it.
Three equal terminal spiders are rigid in the visible half#
The next obstruction is not a single branch but a terminal orbit. Suppose a hub has three equal-order terminal spiders with leg multisets
$$ \mathcal A=\{A_1,A_2,A_3\}, \qquad \sum_{a\in A_i}a=h-1, \qquad |A_i|\ge3. $$
Let
$$ C_a(t)=\sum_{q=0}^at^qU_{P_q} $$
and form the terminal derivative collar
$$ \Psi_{\mathcal A,\Lambda}(t) = \partial_{x_1} \sum_iU_{S(A_i)} \prod_{b\in\Lambda\setminus A_i}C_b(t), $$
where \(\Lambda=A_1\uplus A_2\uplus A_3\).
Only the first half of this collar is visible from the unique-giant sector. It is enough:
$$ \boxed{ \Psi_{\mathcal A,\Lambda} \equiv \Psi_{\mathcal B,\Lambda} \pmod {t^{\lfloor h/2\rfloor+1}} \Longrightarrow \mathcal A=\mathcal B. } $$
Repeated blocks and mixed twig counts are included. The proof first recovers clean inverse sums, weighted rows, the exact block sum, and the block product. After those peel, every remaining three-versus-three transfer has rank-one barycentric form. Exact contact bounds and a convexity argument for three positive factor values rule out the transfer.
This closes the visible three-terminal collar problem. It does not reconstruct an arbitrary tree from \(D(T)\): the remaining task is to expose and transport these rigid collars through the descending centroid equations, and the bicentroid endpoint grouping is still open.
But the architecture is no longer foggy. One derivative gives connected deletion layers, the giant layers multiply around the centroid, the equator repairs the first strict-half collision, and a complete visible terminal collar is rigid. That is a surprisingly coherent amount of tree geometry inside one marked-singleton derivative.
Notebook references: D-0078, Q-0029, Q-0042, R-0353, R-0516, R-0544, R-0547, R-0548, R-0550, R-0559