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Let us call a pair of two real polynomials $(P,Q)$ interlacing if $\deg(P)=\deg(Q)+1$, both polynomials have strictly positive leading coefficients and $P,Q$ have only real roots which interlace weakly, ie. $$\rho_1\leq \rho'_1\leq \rho_2\leq \dots\leq \rho'_{n-1}\leq \rho_n$$ where $\rho_1,\dots,\rho_n$ are all roots (including multiple roots which are repeated) of $P$ and $\rho'_1,\dots,\rho'_{n-1}$ are all roots of $Q$. We admit the pair $(\lambda,0)$ as interlacing for $\lambda>0$.

It is easy to show (use for example the fact that $(P,Q)$ is interlacing if and only if $Q/P$ has $n$ real poles and is decreasing on intervals without poles) that the set of all interlacing polynomials is a monoid for the product given by $$(P_1,Q_1)\cdot (P_2,Q_2)=(P_1P_2,P_1Q_2+P_2Q_1)\ .$$ The identity is of course the degenerate pair $(1,0)$.

I could find no mention of this structure in the literature on interlacing polynomials. Did it appear somewhere?

Remarks: The usual definition of interlacing pairs is slightly different: positivity of leading coefficients is not required but roots have to interlace strictly. The above definition is of course taylored for the monoid structure (which creates multiple roots by taking powers of pairs).

It is possible to work with equivalence classes of pairs by considering $(P,Q)\sim (\lambda P,\lambda Q)$ for $\lambda >0$.

The mononoid structure does not extend to "triplets" of interlacing polynomials.

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  • $\begingroup$ As a monoid, it is an extension over polynomials with multipilcation, and fibre polynomials with addition. Off the top of my head I don't see an isomorphism with the product extension, but perhaps it is the product? I'm ignoring your interlacing condition for now, apologies. $\endgroup$ Dec 3, 2014 at 17:39
  • $\begingroup$ This is quite interesting! Wonder how this plays together with the differential operator. $\endgroup$ Dec 3, 2014 at 19:31
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    $\begingroup$ I can find counter-examples which show that the obvious thing for triples: $(P_1,Q_1,R_1)\cdot(P_2,Q_2,R_2)=(P_1P_2,P_1Q_2+P_2Q_1,P_1R_2+2Q_1Q_2+P_2R_1)$ can fail. Do you have a proof that nothing can work? I suppose having $(1,0,0)$ be the identity precludes something silly like changing the third component to $P_1Q_2'+P_1'Q_2+Q_1P_2'+P_2Q_1'.$ $\endgroup$ Dec 3, 2014 at 21:43
  • $\begingroup$ Indeed, the natural candidate associates to a sequence $P_0,P_1,\dots$ of interlacing polynomials the series $\sum_k\geq 0}P_k t^k/k!$ (where $t$ is an additional variable). This candidate does not work. I thought a little bit about different possible twists and convinced myself that it is hopeless. $\endgroup$ Dec 4, 2014 at 9:03

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