\section{Scratch Work --- will be removed before posting}

\subsection{An $\infty$-dimensional DD Theorem, take 1.}

Suppose $\bbA=\calU(\fraka)$ and $\bbB=\calU(\frakb)$ are Hopf algebras with their native products and with
$aS$, $a\Delta$, $bS$, $b\Delta$, etc. Suppose $\langle\cdot,\cdot\rangle\colon\frakb\otimes\fraka\to\bbQ$ is a
pairing such that:
\begin{itemize}
\item Compatibility of $[]_b$ and $a\Delta$ etc.
\item Non degeneracy.
\end{itemize}

Then
\begin{enumerate}
\item $\langle\cdot,\cdot\rangle$ extends uniquely to a non-degenerate pairing $\bbB\otimes\bbA\to\bbQ$ such that
$m$ and $\Delta$ are compatible.
\item $\bbD=\bbB\otimes\bbA$ is a Hopf algebra with the DD formulas and $\bbA\to\bbD$ and $\bbB\to\bbD$ are
Hopf morphisms.
\item If $b_i$ and $a_i$ are dual bases of $\bbB$ and $\bbA$ relative to our pairing, then $R=\sum b_\otimes
a_i$ satisfies the quasi-triangularity axioms.
\end{enumerate}

\subsection{An $\infty$-dimensional DD Theorem, take 2.}

{\bf Step 1.} Everybody knows that if $H$ is a finite dimensional Hopf algebra then $D=H^\ast\otimes$ is a
quasi-triangular Hopf algebra, with $R$, $m$, $\Delta$, $S$ given by the following formulas\ldots

{\bf Step 2.} If $A$ and $B$ are Hopf algebras over a ring $\Omega$
with a Hopf pairing $P\colon A\otimes B\to\Omega$ and $R\in B\otimes A$ contracts $P$ to the identity, the same
conclusion holds for $D=B\otimes A$.

{\bf Step 3.} Over $\Omega=\bbQ\llbracket\hbar\rrbracket$
let $\bbA=\calU(\fraka)\llbracket\hbar\rrbracket$,
$\bbA'=\langle\hbar\fraka\rangle\subset\bbA$, and
$\bbB=\calU(\frakb)$, with $P\colon\bbA'\otimes\bbB\to\Omega$ be
given by $\langle \hbar a,b\rangle=\langle\hbar x,y\rangle=1$, and let
$R=\sum_{m,n}y^nb^m\otimes(\hbar a)^m(\hbar x)^n/m![n]_q!$. Then we're
in the situation of Step 2, with $A=\bbA'$ and $B=\bbB$, and hence
$\bbD'=\bbB\otimes\bbA'$ is a quasi-triangular Hopf algebra.

{\bf Step 4.} All the formulas extend $\Omega$-linearly to
$\bbD=\bbB\otimes\bbA$ and hence all identities hold there too.

\Needspace{5.2in}
\subsection{A Naming Question} The following was posted on Facebook on May 10, 2021:

\[ \parbox{3.5in}{{\bf A naming question follows.}

Physicists (and some mathematicians) know how to integrate Gaussians
multiplied by polynomials, and they do it often, especially when they
think about ``perturbation theory''.

There are two types of Gaussians: the ``one type of variables''
kind, which looks like $\bbe^{x^TAx}$, and the ``dual variables''
kind, which looks like $\bbe^{x^TBy}$. With the first type, we study
$\int_{\bbR^n}dx\,\bbe^{x^TAx}p(x)$ where $p$ is a polynomial. With
the second type we study $\int_{\bbR^{2n}}dxdy\,\bbe^{x^TBy}f(x,y)$,
where $f$ is a polynomial. But for the second type, the answer is $0$
unless $\deg_xf=\deg_yf$, so in fact we can extend to the case where $f$
is a polynomial in (say) $x$ yet is allowed to be a power series in $y$.

{\bf Question.} What is the second type of Gaussians called? ``Polarized
Gaussians''? ``Bipartite Gaussians''? Is there a name for the fact that
perturbations in the second case vanish if not balanced? A name or a
precedent for the (trivial) fact that $f$ can be a power series in one
of its sets of variables?

Mathematicians, please don't complain about convergence. Add conditions
if you must, or think that I'm really imitating some QFT-like context
in which convergence is not an issue.
}\]

\subsection{Iterated Gaussian Integration.} We wish to compute the formal
$(2m+n)$-dimensional near-Gaussian integral $I_{\alpha\beta\xi} = \int\bbe^Lda\,db\,dx$, where
\[ L = \lambda^{ij}a_ib_j + \frac12 q^{kl}(b_j)x_kx_l + \alpha^ia_i + \beta^ib_i + \xi^kx_k, \]
and where $i,j\in\underline{m}$ and $k,l\in\underline{n}$.

\noindent{\bf Method 1.} First compute the $ab$-integral.
\[ \int\bbe^L da\,db
  = \bbe^{\xi^kx_k}\exp\left(\frac12q^{kl}(\partial_{\beta^j})x_kx_l\right)
    \int\exp\left(\lambda^{ij}a_ib_j + \alpha^ia_i + \beta^ib_i\right)da\,db
\]
\[ = \bbe^{\xi^kx_k}\exp\left(\frac12q^{kl}(\partial_{\beta^j})x_kx_l\right)\exp\left(-\lambda_{ij}\alpha^i\beta^j\right)
    \int\exp\left(
      \lambda^{ij}(a_i+\lambda_{ii'}\alpha^{i'})(b_j+\lambda_{j'j}\beta^{j'})
    \right)da\,db
\]
\[ = \det(\lambda)^{-1}\bbe^{\xi^kx_k}
    \exp\left(\frac12q^{kl}(\partial_{\beta^j})x_kx_l\right)\exp\left(-\lambda_{ij}\alpha^i\beta^j\right)
\]
\[ = \det(\lambda)^{-1}\bbe^{\xi^kx_k}
    \exp\left(\frac12q^{kl}(-\lambda_{ij}\alpha^i)x_kx_l\right)\exp\left(-\lambda_{ij}\alpha^i\beta^j\right)
\]
\[ = \det(\lambda)^{-1}\exp\left(-\lambda_{ij}\alpha^i\beta^j\right)
    \exp\left(\frac12q^{kl}(-\lambda_{ij}\alpha^i)x_kx_l + \xi^kx_k\right)
\]

\begin{minipage}{4.2in}{\bf Seeking a Precedent --- Two-Stage Gaussian Integration?} (Posted at
\url{https://mathoverflow.net/questions/395934}).

Sometimes, by iteration, linear algebra can be used to solve non-linear equations. For example, consider the system
\[ Ax=a \qquad B(x)y=b(x), \]
where $a$ is a vector with scalar entries, $A$ is a matrix with scalar entries, $b(x)$ is a vector whose entries are
functions of $x$, and $B(x)$ is a matrix whose entries are functions of $x$. This system can be solved by first solving
$Ax=b$, then substituting the solution into the second equation $By=b$, and then solving the second equation. The system
can also be solved by first solving $By=b$ over the ring of functions of $x$, and then solving the first equation.

Similarly, formal$^*$ Gaussian integration techniques can
sometimes be used iteratively to compute the {\em exact} integrals of
non-Gaussian integrands. Here's a 3D example in the variables $a,b,x$;
it is easy to raise this example to higher dimensions by replacing
scalars with vectors and matrices.

Let $L=\lambda ab+\frac12 q(b)x^2+\alpha a+\beta b+\xi x$, where all
the letters represent scalars except for $q(b)$ which is a function of
$b$. We wish to compute $I \coloneqq \int \bbe^Lda\,db\,dx$. This is
not a Gaussian integral because the $q(b)x^2$ term is not quadratic in
the integration variables.

Yet first computing the $ab$ integral we get
\[ I(x) \coloneqq \int \bbe^Lda\,db =
  \bbe^{\xi x}\bbe^{q(\partial_\beta)x^2/2}\int\bbe^{\lambda ab+\alpha a+\beta b}da\,db
\]
\[ = \frac{2\pi}{\lambda}\bbe^{\xi x}\bbe^{q(\partial_\beta)x^2/2}\bbe^{-\alpha\beta/\lambda}
  = \frac{2\pi}{\lambda}\bbe^{-\alpha\beta/\lambda + \xi x + q(-\alpha/\lambda)x^2/2}.
\]

Thus $I(x)$ is a Gaussian with respect to $x$, so we can (formally) compute
\[ I = \int I(x)dx = \frac{(2\pi)^{3/2}}{\lambda\sqrt{q(-\alpha/\lambda)}}
  \bbe^{-\alpha\beta/\lambda - q(-\alpha/\lambda)^{-1}\xi^2/2}.
\]

We could have arrived at the same result by first computing the $x$ integral as a formal Gaussian over the ring of
functions of $b$ and then computing the $ab$ integral.

{\bf Question.} Is there a precedent for this procedure? A name? Is there a place where people routinely iterate
Gaussian integration to integrate non-Gaussians?

\rule{0.25\linewidth}{0.5pt}

$^*${\footnotesize Meaning, applying standard formulas without worrying about convergence. Add conditions
if you must, or think that I'm really imitating some QFT-like context
in which convergence is not an issue.}
\end{minipage}
