Let \(T\) be a tree with exactly two vertices of degree at least three. It looks like two spiders joined by a path:
$$ T=T(A,c,B), $$
where \(A\) and \(B\) are the two multisets of leg lengths and \(c\ge1\) is the length of the trunk between the branching vertices.
The whole tree is determined by one symmetric-function derivative:
$$ \boxed{ \frac{\partial X_T}{\partial p_1} = \frac{\partial X_{T'}}{\partial p_1} \quad\Longrightarrow\quad T\cong T'. } \tag{1} $$
This is smaller than the full chromatic symmetric function. Wang, Yu, and Zhang already proved that the full \(X_T\) distinguishes this class. The point here is the compression: one unspecialized marked-singleton derivative still remembers every leg and the trunk joining the two bouquets.
I find that slightly absurd. A derivative that looks like a sum of cards turns out to preserve the complete two-center geometry.
Turn the derivative into a marked deck#
Write
$$ U_F(\mathbf x) = \sum_{E_0\subseteq E(F)} \prod_{C\in\operatorname{Comp}(V(F),E_0)}x_{|C|} $$
for the connected-partition polynomial of a forest. Coefficientwise sign conversion turns
$$ D(T)=\frac{\partial X_T}{\partial p_1} $$
into
$$ M_T=\frac{\partial U_T}{\partial x_1} = \sum_{v\in V(T)}U_{T-v}. \tag{2} $$
So \(D(T)\) is exactly a marked-singleton deck polynomial. Its specialized image already recovers the degree sequence. Any competitor \(T'\) in (1) therefore has the same two branching degrees and is itself a double spider.
The reconstruction problem is now concrete: recover \(A\), \(B\), and \(c\) from \(M_T\).
The giant component exposes the smaller bouquet#
Put
$$ \alpha=\sum A,\qquad \beta=\sum B, $$
and orient the tree so that \(\alpha<\beta\). The smaller terminal spider has order
$$ h=\alpha+1, $$
while deleting its branching vertex leaves a unique larger component of order
$$ N=\beta+c>\frac{|T|}{2}. $$
That strict half-order inequality is the first lever. In a monomial of \(M_T\), a component of order \(N\) is unique. Extracting its coefficient gives
$$ [x_N]M_T = M_{S(A)} + \text{marked suffix-path contributions}. \tag{3} $$
The second term comes from connected \(N\)-sets containing both branching vertices. Their complements are disjoint terminal suffixes of the pendant legs. Earlier marked-cut coefficients recover exactly the truncated global leg capacities needed to calculate and subtract that contamination.
What remains is the complete marked-singleton polynomial \(M_{S(A)}\) of the smaller terminal spider.
If \(A\) has at least three legs, its branching vertex is intrinsic. The spider reconstruction theorem recovers \(A\), and a finite-difference transform of the connected-subtree polynomial then forces both the trunk length and the larger side. Equal side totals are handled separately by the specialized deck layer.
That leaves one genuinely annoying boundary:
$$ |A|=2. $$
Then \(S(A)\) is only a path. The unrooted path remembers \(\alpha=a_1+a_2\), but not where its hidden branching attachment sat. This is exactly where the giant-component argument stops.
The folded two-edge response unfolds#
Take the part of \(M_T\) coming from an ordinary degree-two marked singleton, and map
$$ x_ax_b\longmapsto y^a+y^b \qquad(a+b=|T|-1). $$
Call the result \(H_T(y)\). Every pendant leg of length \(d\) contributes
$$ (L+1)\sum_{j=1}^{d-1} \left(y^j+y^{|T|-1-j}\right) + (L-1)\left(y^d+y^{|T|-1-d}\right), \tag{4} $$
where \(L\) is the total number of legs. The trunk contributes one more explicit interval with endpoint coefficients determined by the two branching degrees.
After the already-known short legs are removed, the unknown long legs and trunk form a polynomial \(\mathcal A_{C,c}(y)\), but the invariant shows it only through the reciprocal fold
$$ \overline H_T(y) = \mathcal A_{C,c}(y) + y^{|T|-1}\mathcal A_{C,c}(y^{-1}). \tag{5} $$
This looks lossy. It is not lossy on the realized double-spider domain.
If two candidates had the same fold, their difference \(Q\) would satisfy
$$ Q(y)=-y^{|T|-1}Q(y^{-1}). \tag{6} $$
Hence the first and last nonzero exponents of \(Q\) would have to sum to \(|T|-1\). Literal leg and trunk coefficients force those two exponents to sum to at most \(|T|-2\), even in the one cancellation pattern where the first trunk discrepancy is swallowed by a long-leg discrepancy.
Contradiction.
So the folded marked-cut response uniquely recovers
$$ \boxed{\text{the trunk length and the complete global leg multiset}.} \tag{7} $$
That anti-reciprocal endpoint argument is my favorite move in the proof. The fold is globally noninjective, but the geometry of actual legs leaves no room for an anti-reciprocal difference to exist.
Put the legs back on the correct side#
Knowing the global leg multiset is not yet knowing its partition into \(A\) and \(B\). Subtract the completely known sector in which the trunk is never cut. The remaining crossed-trunk polynomial factors into two side responses joined by a path response.
Three cases close the grouping.
- If the centers are adjacent, a known path factor multiplies the unknown larger spider. A nonzero product of positive \(x_1\)-degree cannot have zero \(x_1\)-derivative, so the larger side is unique.
- If the side-total gap is at least two, one giant coefficient exposes the three-leg spider obtained by adding a unit leg to the smaller pair.
- In the sole unit-gap ambiguity, the two candidate groupings differ in the connected-subtree transform by
$$ z(1-z^c)(J_{A'}-J_A). \tag{8} $$
Equality forces \(J_A=J_{A'}\), and descending cyclotomic inversion recovers the leg multiset.
That exhausts every finite double spider.
Why the full derivative matters#
Several smaller projections fail before the theorem succeeds.
The degree sequence plus every marked two-edge-cut count first collides at order nine:
$$ ((1,2),1,(1,3)) \quad\text{and}\quad ((1,1),2,(2,2)). $$
The next three-part marginal repairs that pair, but degree plus the complete two- and three-part blocks collide again at order eighteen. A four-part block repairs that witness.
So this is not a theorem saying that one cute low-order statistic happens to work. The lower marginals keep producing exact counterfeits. What survives is the complete unspecialized derivative, used structurally: giant components, marked cuts, reciprocal unfolding, and the crossed-trunk factorization all carry different pieces of the reconstruction.
The result stops at trees with two branching vertices. It does not prove that the specialized deck layer distinguishes every double spider, and it does not reconstruct arbitrary trees from \(\partial X_T/\partial p_1\). Beyond two centers, terminal spiders arrive in orbit sums and centroid products can collide before their equatorial continuation repairs them.
Still, one derivative remembers every double spider. That is already a beautifully unreasonable amount of information.
Notebook references: Q-0029, R-0511, R-0512, R-0513, R-0515, R-0530, R-0532, R-0541, R-0543