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An Induction without Hypothesis

I think Nate and Eli will get this without a lot of background set up.

Let $$\omega(x) = \sum_{\mathfrak t : \underline L \nearrow \underline{NL}} \mathrm{Wr}(\mathbf p_{\mathfrak t}; x) \mathbf e_{\mathfrak t} \in \Lambda^L(V),$$ where $V$ is an $LN$-dimensional vector space with basis $\mathbf e_1, \ldots, \mathbf e_{LN}$ and $p_{\mathfrak t}$ is a subvector of a complete family of monic polynomials indexed by $\mathfrak t$. I’m using $\mathbf e_{\mathfrak t}$ instead of $\epsilon_{\mathfrak t}$ for basis vectors in the exterior algebra because I think it is more consistent notation.

Anyway, given the measure $\mu$, write $\mu[ \omega ] := \int \omega d\mu$ for the $L$-form formed by integrating the coefficients of $\omega(x)$ with respect to $\mu$. (The notation $\mu[ \omega ]$ here is thanks to Çinlar).

We also have the inner product on $L^2(\mu)$ which we will denote by by $\langle f | g \rangle = \mu[fg] = \int f g \, d\mu$. It’s conceivable that this should be Hermitian, but I am thinking more algebraically at the moment so we keep it simple.

We use the Laplace expansion on the Wronskian (along the row/column with the highest normalized derivative—in our case $D^{L-1}$ to see $$\omega(x) = \sum_{\mathfrak t : \underline L \nearrow \underline{NL}} \sum_{\ell=1}^L (-1)^{\ell} D^{L-1} p_{\mathfrak t(\ell)}(x) \mathrm{Wr} (\mathbf p_{\mathfrak t \setminus \{\mathfrak t(\ell)\} }; x) \mathbf e_{\mathfrak t},$$ And, $$\mu[\omega] = \sum_{\mathfrak t : \underline L \nearrow \underline{NL}} \sum_{\ell=1}^L (-1)^{\ell} \left\langle D^{L-1} p_{\mathfrak t(\ell)} \big| \mathrm{Wr} (\mathbf p_{\mathfrak t \setminus \{\mathfrak t(\ell)\} }) \right\rangle \mathbf e_{\mathfrak t}, $$

Let’s change notation a bit. Given a set $X$ and a positive integer, let ${X \choose L}$ represent the set of subsets of $X$ with exactly $L$ elements. So that $\mathfrak t \in {\underline{NL} \choose L}$ can be identified by with $\mathfrak t : \underline{L} \nearrow \underline{NL}$. It’s easy to see that $\mathfrak t’$ is the complement of $\mathfrak t$ in $\underline{NL}$. Using this notation, and the Laplace expansion of the Wronskian with the $(L-1)$st and $(L-2)$nd derivatives (the last two rows or columns, depending on how you orient your Wronskian) to find $$\mu[\omega] = \sum_{\mathfrak w \in {\underline{NL} \choose 2}}\sum_{\mathfrak u \in {\mathfrak w’ \choose {L – 2}}} \left\langle D^{L-1} p_{\mathfrak{t}(1)} D^{L-2} p_{\mathfrak{t}(2)} – D^{L-1} p_{\mathfrak{t}(2)} D^{L-2} p_{\mathfrak{t}(1)} \big| \mathrm{Wr}(\mathbf p_{\mathfrak u}) \right\rangle \mathbf e_{\mathfrak t} \wedge \mathbf e_{\mathfrak u}.$$

In particular, for $L = 4$, we see $$\mu[\omega] = \sum_{\mathfrak w \in {\underline{NL} \choose 2}}\sum_{\mathfrak u \in {\mathfrak w’ \choose {2}}} \left\langle D^3 p_{\mathfrak{t}(1)} D^2 p_{\mathfrak{t}(2)} – D^3 p_{\mathfrak{t}(2)} D^2 p_{\mathfrak{t}(1)} \big| D p_{\mathfrak{u}(1)} \cdot p_{\mathfrak{u}(2)} – D p_{\mathfrak{u}(2)} \cdot p_{\mathfrak{u}(1)} ) \right\rangle \mathbf e_{\mathfrak t} \wedge \mathbf e_{\mathfrak u}.$$ Both of the arguments in the inner product are the integrand in the skew-inner products in the $L=2$ ensemble.

So what’s the inductive hypothesis on the Wronskian, or the skew-orthogonal polynomials, or something that allows us to complete the induction or simplify the Gram-form?

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