Kernel of a matrix and the Catalan numbers

Question

Let $B_n$ denote the Boolean lattice of a set with $n \geq 2$ elements and $C_n$ the matrix with entries $c_{i,j}=1$ if $i \leq j$ and $c_{i,j}=0$ else, where $i,j\in B_n$.

Let $M_n:=C_n+C_n^T$ (this is also the Cartan matrix of a certain Frobenius algebra associated to $B_n$), which is a symmetric matrix. Thus geometric and algebraic multiplicity should coincide. Let $I$ denote the identity matrix.

Question: Is there a bijective proof that for $n$ even (odd) we have that the basis of the kernel of $M_n - 2 I$ ($M_n-3 I$) is enumerated by the Catalan numbers?

This is true for $n \leq 15$.

(see also Factorisation of a polynomial from the Boolean algebra )

Here a bijective proof asks for a bijection of a basis of the kernel to known combinatorial objects that are enumerated by the Catalan numbers.

For example for $n=6$, the kernel has dimension 5 and basis vectors are given by

(0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, -1, 1, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, -1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, -1, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 1, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0),

(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)

I was not able to see a pattern so far for a nice basis of the kernel as the vectors are so big for larger n.

For example for $n=2$, $C_2$ is given by $\begin{bmatrix} 1 & 1 & 1 &1 \\ 0 & 1 & 0 &1\\ 0 & 0 & 1 &1\\ 0 & 0 & 0 &1\\ \end{bmatrix}$ and $M_2$ is given by $\begin{bmatrix} 2 & 1 & 1 &1 \\ 1 & 2 & 0 &1\\ 1 & 0 & 2 &1\\ 1 & 1 & 1 &2\\ \end{bmatrix}$.

Answer 1

Another suggestion: let $A_{2n}=M_{2n}-2I$ ($I$=identity matrix), so we are interested in $\ker A_{2n}$. Let $V$ be the vector space on which $A_{2n}$ acts, so we can regard $V$ as having a basis consisting of all subsets of $[2n]=\{1,2,\dots,2n\}$. Thus $V$ has a grading $V_0\oplus V_1\oplus\cdots \oplus V_{2n}$, where $V_i$ has a basis consisting of all $i$-element subsets of $[2n]$. The symmetric group $\mathfrak{S}_{2n}$ acts on $V$ and preserves the grading. Write $M_\lambda$ for the irreducible $\mathfrak{S}_{2n}$-module indexed by the partition $\lambda$ of $2n$. Then (as is well-known) $V_i$ decomposes as $M_{(2n)}\oplus M_{(2n-1,1)} \oplus \cdots\oplus M_{(2n-i,i)}$ for $0\leq i\leq n$, and $V_i\cong V_{2n-i}$ (as $\mathfrak{S}_{2n}$-modules).

Assuming that $\dim\ker A_{2n}=C_n$, it is natural to conjecture that as an $\mathfrak{S}_{2n}$-module, $\ker A_{2n}$ is isomorphic to the irreducible module $M_{(n,n)}$. The operator $\frac{1}{(2n)!}\sum_{w\in\mathfrak{S}_{2n}}\chi^{(n,n)}(w)w$ will then project $V_n$ to $\ker A_{2n}$. Here $\chi^{(n,n)}$ is the irreducible character of $\mathfrak{S}_{2n}$ indexed by the partition $(n,n)$. There is something similar for $M_{2n+1}$.

Answer 2

Only a suggestion: This smells of the Temperley-Lieb algebra. Eigenspaces of $M_n$ are invariant under the obvious action of the symmetric group $S_n$. The $2-$eigenspace of $M_n$ is thus a decomposition of irreps of $S_n$. The sum of squared dimensions for irreps with Young diagrams having at most $2$ rows of $S_n$ is the $n-$th Catalan number.

The relation is probably not completely straightforward: Since Catalan numbers grow like $4^nn^{-3/2}$ (up to a constant), your matrix $M_n$ should roughly involve the Temperley Lie algebra of $S_{\lfloor n/2\rfloor}$.

In any case, the determination of the involved irreps of $S_n$ gives probably an indication of what is going on.