The Heisenberg product: from Hopf algebras and species to symmetric functions

Abstract

Many related products and coproducts (e.g. Hadamard, Cauchy, Kronecker, induction, internal, external, Solomon, composition, Malvenuto–Reutenauer, convolution, etc.) have been defined in the following objects: species, representations of the symmetric groups, symmetric functions, endomorphisms of graded connected Hopf algebras, permutations, non-commutative symmetric functions, quasi-symmetric functions, etc. With the purpose of simplifying and unifying this diversity we introduce yet, another product the Heisenberg product, that is not graded and its highest and lowest degree-terms are the classical external and internal products (and their namesakes in different contexts). In order to define it, we start from the two opposite more general extremes: species in the “commutative context”, and endomorphisms of Hopf algebras in the “non-commutative” environment. Both specialize to the space of commutative symmetric functions where the definitions coincide. We also deal with the different coproducts that these objects carry, to which we add the Heisenberg coproduct for quasi-symmetric functions, and study their Hopf algebra compatibility particularly for symmetric and non commutative symmetric functions. We obtain combinatorial formulas for the structure constants of the new product that extend, generalize and unify results due to Garsia, Remmel, Reutenauer and Solomon. In the space of quasi-symmetric functions, we describe explicitly the new operations in terms of alphabets.

Appendices

Part 4. Appendix

The proofs

In this “Appendix” we provide the postponed proofs of the technical lemmas used in the paper.

1.1 Proof of Lemma 3.2

Proof

To define the bijection $\Upsilon \rightarrow \mathcal {M}_{\alpha ,\beta }^n$, we start by splitting the interals [1, p] and [1, q] as below:

where $\alpha =(a_1,\dots ,a_k)$ and $\beta =(b_1,\dots ,b_s)$. Given an element $v=\sigma \times \tau \in S_p\times S_q$ we consider the shuffles $\zeta _\alpha (\sigma )\in {\text {Sh}}(\alpha )$ and $\zeta _\beta (\tau )\in {\text {Sh}}(\beta )$ characterized by the equations

$$\begin{aligned} \sigma = \zeta _\alpha (\sigma ) u, \qquad \tau = \zeta _\beta (\tau ) v, \end{aligned}$$

(67)

with $u\in S_\alpha$ and $v\in S_\beta$. To simplify the notation, we write $\zeta _\alpha =\zeta _\alpha (\sigma )$ and $\zeta _\beta =\zeta _\beta (\tau )$. We further split each interval $E_i$ and $F_j$ as below:

$$\begin{aligned} E_i = E'_i \sqcup E''_i, \qquad F_j = F'_j \sqcup F''_j, \end{aligned}$$

such that

for $i=1,\dots ,k$ and $j=1,\dots ,s$. Observe that with these definitions we have the decomposition of the interval [1, n] into

$$\begin{aligned}{}[1,n-q]&= \bigsqcup _{i=1}^k \zeta _\alpha (E'_i), \end{aligned}$$

(68)

$$\begin{aligned}{}[n-q+1,p]&= \bigsqcup _{i=1}^k \zeta _\alpha (E''_i) = \bigsqcup _{j=1}^s \bigl (n-q +\zeta _\beta (F'_j)\bigr ), \end{aligned}$$

(69)

$$\begin{aligned}{}[p+1,n]&= \bigsqcup _{j=1}^s \bigl (n-q +\zeta _\beta (F''_j)\bigr ). \end{aligned}$$

(70)

Define the matrix $M_{\sigma \times \tau }$ of dimension $(k+1)\times (s+1)$ whose entries are

$$\begin{aligned} \begin{array}{llll} m_{00} &{}= 0, &{}&{}\\ m_{i0} &{}= {\#}{E'_i}, &{}\qquad &{} \text {for }i=1,\dots ,k, \\ m_{0j} &{}= {\#}{F''_j}, &{}\qquad &{} \text {for }j=1,\dots ,s, \\ m_{ij} &{}= {\#}{\Bigl [ \zeta _\alpha (E''_i) \cap \bigl (n-q+\zeta _\beta (F'_j)\bigr ) \Bigr ]} &{}\qquad &{} \text {otherwise.} \end{array} \end{aligned}$$

The matrix $M_{\sigma \times \tau }$ belongs to $\mathcal {M}_{\alpha ,\beta }^n$. Assume that $i\not =0$. Since $\zeta _\alpha (E''_i)\subseteq [n-q+1,p]\subseteq \bigsqcup _{j=1}^s \bigl (n-q +\zeta _\beta (F''_j)\bigr )$, we get

$$\begin{aligned} \sum _{j=0}^s m_{ij}&= {\#}{E'_i} + \sum _{j=1}^s {\#}{\Bigl [ \zeta _\alpha (E''_i) \cap \bigl (n-q+\zeta _\beta (F'_j)\bigr ) \Bigr ]} \\&= {\#}{E'_i} + {\#}\Bigl [ \zeta _\alpha (E''_i)\cap \bigsqcup _{j=1}^s \bigl (n-q+\zeta _\beta (F'_j)\bigr ) \Bigr ] \\&= {\#}{E'_i} + {\#}\bigl (\zeta _\alpha (E''_i) \bigr ) = {\#}{E_i} = a_i. \end{aligned}$$

On the other hand, if $i=0$, then, by (70), the sum of $m_{0j}$ for $j=0,\dots ,s$, coincides with ${\#}[p+1,n]=n-p$.

Next we show that the matrix $M_{\sigma \times \tau }$ does not depend on the choice of representative of the coset $(S_p\times _nS_q)v$. Let $x\in S_{n-q}$, $y\in S_{p+q-n}$, and $z\in S_{n-p}$, so that $x\times y\times z\in S_p\times _nS_q$. Consider the representative $v'=\sigma '\times \tau '$ where

$$\begin{aligned} \sigma ' = (x\times y) \sigma \quad \text {and}\quad \tau ' = (y\times z) \tau . \end{aligned}$$

Let $\zeta '_\alpha$ and $\zeta '_\beta$ the shuffles associate to $v'$. As $\zeta _\alpha (E'_i)\subseteq [1,n-q]$, then $(x\times y) \bigl (\zeta _\alpha (E'_i)\bigr ) = x\bigl (\zeta _\alpha (E'_i)\bigr )$. But we also have $x\bigl (\zeta _\alpha (E'_i)\bigr ) = \zeta '_\alpha (E_i)=\zeta '_\alpha (\tilde{E}'_i)\sqcup \zeta '_\alpha (\tilde{E}''_i)$, where $E_i=\tilde{E}'_i\sqcup \tilde{E}''_i$ is the decomposition of $E_i$ corresponding to the shuffle $\zeta '_\alpha$, that satisfies $\zeta '_\alpha (\tilde{E}'_i) \subseteq [1,n-q]$ and $\zeta '_\alpha (\tilde{E}''_i) \subseteq [n-q+1,p]$. In summary, $x\bigl (\zeta _\alpha (E'_i)\bigr ) \subseteq \zeta '_\alpha (\tilde{E}'_i)$. Interchanging the roles of $\zeta _\alpha$ and $\zeta '_\alpha$ we obtain an equality, which implies that $m_{i0}={\#}E'_i = {\#}\tilde{E}'_i = m'_{i0}$, where $m'_{ij}$ are the entries of the matrix $M_{v'}$. This proves the equality of the first row of the matrices. The argument for the other rows is similar.

The matrix $M_v$ do not depend on the choice of representative of $v(S_\alpha \times S_\beta )$, since the shuffles satisfying (67) are the same for all the elements on this coset. In conclusion, the matrix $M_v$ depends only on the double cosets $(S_p\times _nS_q)v(S_\alpha \times S_\beta )$.

Next we show that the parabolic subgroup $S_{p(M_v)}$ is $S_\alpha \times _n^\upsilon S_\beta$. An element of $S_\alpha \times _n^\upsilon S_\beta$ can be written as $x\times y\times z$ where

$$\begin{aligned} x\times y&= \zeta _\alpha (\sigma _{a_1}\times \dots \times \sigma _{a_k})\zeta ^{-1}_\alpha , \\ y\times z&= \zeta _\beta (\tau _{b_1}\times \dots \times \tau _{b_s})\zeta ^{-1}_\beta . \end{aligned}$$

Evaluating at $\zeta _\alpha (E'_i)$ we deduce that $\zeta _\alpha \sigma _{a_i}(E'_i) = x(E'_i)$ and conclude that $\sigma _{a_i}(E'_i)=E'_i$. Proceeding in a similar manner with the other decompositions we obtain

$$\begin{aligned} \sigma _{a_i}(E'_i)= & {} E'_i, \qquad \tau _{b_j}(F'_j) = F'_j, \end{aligned}$$

(71)

$$\begin{aligned} \sigma _{a_i}(E''_i)= & {} E''_i, \qquad \tau _{b_j}(F''_j) = F''_j, \end{aligned}$$

(72)

for all $i=1,\dots ,k$ and $j=1,\dots ,s$.

This decomposition can be further refined. Evaluating as above at the subsets $X_{ij}=\zeta _\alpha (E''_i)\cap \zeta _\beta (F'_j)$, we obtain the equality

$$\begin{aligned} \zeta _\alpha \sigma _{a_i} \bigl ( \zeta _\alpha ^{-1}(X_{ij}) \bigr ) = y(X_{ij}) = \zeta _\beta \tau _{b_j} \bigl ( \zeta _\beta ^{-1}(X_{ij}) \bigr ). \end{aligned}$$

Now, $\zeta _\alpha \sigma _{a_i} \bigl ( \zeta _\alpha ^{-1}(X_{ij}) \bigr ) \subseteq \zeta _\alpha (E''_i)$ and also $\zeta _\beta \tau _{b_j} \bigl ( \zeta _\beta ^{-1}(X_{ij}) \bigr ) \subseteq \zeta _\beta (F'_j)$. From the above equality we conclude that $\zeta _\alpha \sigma _{a_i} \bigl ( \zeta _\alpha ^{-1}(X_{ij}) \bigr ) \subseteq \zeta _\alpha (E''_i)\cap \zeta _\beta (F'_j)$, and then $\sigma _{a_i} \bigl ( \zeta _\alpha ^{-1}(X_{ij}) \bigr ) \subseteq \zeta _\alpha ^{-1}(X_{ij})$. This inclusion is actually an equality, since both sets have the same cardinality. Therefore, we get the following refinment of (71)

$$\begin{aligned} \begin{aligned} \sigma _{a_i}(E'_i)&= E'_i,&\qquad \sigma _{a_i}\bigl (\zeta ^{-1}_\alpha (X_{ij})\bigr )&= \zeta ^{-1}_\alpha (X_{ij}), \\ \tau _{b_j}(F''_j)&= F''_j,&\qquad \tau _{b_j}\bigl (\zeta ^{-1}_\beta (X_{ij})\bigr )&= \zeta ^{-1}_\beta (X_{ij}). \end{aligned} \end{aligned}$$

Note that ${\#}X_{ij}=m_{ij}$, and thus the previous decomposition shows that $x\times y\times z$ belongs to $S_{p(M)}$.

The map $\upsilon \mapsto M_\upsilon$ is invertible, since from the entries of the matrix $M_\upsilon$ we can recover the shuffles $\zeta _\alpha$ and $\zeta _\beta$, which are in the same double coset as $\upsilon$. $\square$

1.2 Proof of Lemma 7.5

Proof

As $\eta$ and $\tau$ are fixed throughout this lemma, we write $\varphi =\varphi _{\eta ,\tau }$. Let $x,y\in F_i\cap \varphi ^{-1} E_j$ with $x<y$. Consider z such that $x<z<y$. Therefore, $x,y\in F_i$ and, since $F_i$ is an interval, we conclude that $z\in F_i$.

On the other hand $\varphi (x), \varphi (y)\in E_j$. Since $\tau \in \mathcal {B}_\beta$, then $\mathrm {Id}\times \tau$ is increasing in $F_i$:

$$\begin{aligned} (\mathrm {Id}\times \tau )(x)< (\mathrm {Id}\times \tau )(z) < (\mathrm {Id}\times \tau )(y). \end{aligned}$$

(73)

In order to prove that $\varphi (z)$ also belongs to $E_j$, we consider the following cases:

• (1) Assume that $j=0$. Then, $\varphi (x),\varphi (y)\in E_0=[p+1,n]$. Since $(\eta \times \mathrm {Id})$ is the identity on that interval, this implies that $\beta _0(\mathrm {Id}\times \tau )(x)$ and $\beta _0(\mathrm {Id}\times \tau )(y)$ are in $[p+1,n]$. But $\beta _0^{-1}[p+1,n] = [n-q+1, 2n-p-q]$ and $\beta _0$ is increasing in that set. Therefore, the three terms in (73) belong to $[n-q+1,2n-p-q]$ and, applying $(\eta \times \mathrm {Id})\beta _0$, which is increasing on this set, we obtain that $\varphi (x)<\varphi (z)<\varphi (y)$.

• (2) Assume that $j>0$. Consider the cases:

  • (a) Assume $i=0$. In this case we have $x,z,y\in F_0=[1,n-q]$. Then, applying $\mathrm {Id}\times \tau |_{F_0}=\mathrm {Id}$ we continue in the same set. The permutation $\beta _0$ sends increasingly $[1,n-q]$ into $[p+q-n+1,p]$. In this last interval, $\eta$ is also increasing. Thus, the inequality (73) implies that $\varphi (x)<\varphi (z)< \varphi (y)$.

  • (b) Assume $i>0$. We have that $x,y,z\in F_j\subset [n-q+1,n]$. Applying $\mathrm {Id}\times \tau$ we have that the terms of (73) are also in $[n-q+1,n]$. If $(\mathrm {Id}\times \tau )(x) \in [n-q+1,2n-p-q]$, then $\beta _0(\mathrm {Id}\times \tau )(x)\in [p+1,n]$ and $\varphi (x)\in [p+1,n]=E_0$, which contradicts the assumption $j>0$. Therefore, the terms in (73) belong to $[2n-p-q+1, n]$. The permutation $\beta _0$ maps increasingly this interval into $[1,p+q-n]$, and $\eta$ is also increasing in that image. Thus, we conclude that $\varphi (x)<\varphi (z)<\varphi (y)$.

In all the cases we obtain that $\varphi (x)<\varphi (z)<\varphi (y)$, and since $\varphi (x)$ and $\varphi (y)$ belong to the interval $E_j$, we deduce that $\varphi (z)\in E_j$. This proves that $F_i \cap \varphi ^{-1}E_j$ is an interval.

Notice that along the way we also proved that $\varphi$ is increasing in the intervals $F_i \cap \varphi ^{-1}E_j$ as well as the assertions concerning the images.

The fact that the intervals $F_i \cap \varphi ^{-1}E_{j}$ are disjoint follows immediately from the fact that the sets $E_{j}$, for $j=0,\dots ,r$, and the sets $F_{i}$, for $i=0,\dots ,s$, are disjoint. This finishes the proof. $\square$

1.3 Proof of Lemma 7.6

Proof

For the matrix $M=\{m_{ij}\}$, denote by $s_{ij}$ the sum of the entries $m_{k\ell }$ of M for $(k,\ell )\le (i,j)$ with respect to the lexicographical order of pairs. We define $R_{00} = [1,s_{00}]$ and $R_{ij} = [s_{k\ell }, s_{ij}]$ where $s_{ij}$ covers $s_{k\ell }$. Observe that some of the intervals $R_{ij}$ may be empty. Also note that ${\#}R_{ij}=m_{ij}$.

The sequence $(R_{00},R_{01},\ldots , R_{sr})$ is a pseudo-partition of the interval [n] and $\gamma \in \mathcal {B}_{c(M)}$ if and only if $\gamma$ is increasing in $R_{ij}$ for all $i\in \{0,\ldots ,s\}$ and $j\in \{0,\dots ,r\}$.

Since $M\in \mathcal {M}_{\alpha ,\beta }$ and therefore, $\sum _{j}{\#}(R_{ij})=\sum _j m_{ij} = {\#}F_i$, it follows that

$$\begin{aligned} F_i = \bigcup _j R_{ij}. \end{aligned}$$

(74)

Moreover, if $\eta \in {\text {Sh}}(p+q-n,n-q)$ and $\tau \in \mathcal {B}_\beta ^\eta$, then $F_i\cap \varphi _{\eta ,\tau }^{-1}E_j = R_{ij}$. This can be seen from the fact both sets are intervals with the same cardinal and from the following relation:

$$\begin{aligned} \bigcup _j (F_i\cap \varphi ^{-1}_{\eta ,\tau }E_j) = F_i = \bigcup _jR_{ij}. \end{aligned}$$

In particular, we deduce that $\varphi _{\eta ,\tau }$ is increasing in $R_{ij}$.

Given $(\xi ,\eta ,\sigma ,\tau )\in S_{\alpha ,\beta }(M)$, we will show that $g_{\xi ,\eta }(\sigma ,\tau )\in \mathcal {B}_{c(M)}$. To prove this, since $\varphi _{\eta ,\tau }|_{R_{ij}}$ is increasing and $\varphi _{\eta ,\tau }R_{ij} \subseteq E_j$, we observe that

$$\begin{aligned} (\sigma \times \mathrm {Id})(\eta \times \mathrm {Id})\beta _0(\mathrm {Id}\times \tau )_{|_{R_{ij}}} \end{aligned}$$

is also increasing. According to Lemma 7.5, the images of $R_{ij}$ under the previous permutation are in [1, p] or $[p+1,n]$, where $\xi$ is increasing. Therefore, left multiplying by $\xi$ we deduce that $g_{\xi ,\eta }(\sigma ,\tau )$ is increasing in $R_{ij}$, which proves that it belongs to $\mathcal {B}_{c(M)}$.

We prove now that $\psi$ is bijective. Given $\gamma \in \mathcal {B}_{w(M)}$, we show that there exists a unique quadruple $(\xi ,\eta ,\sigma ,\tau )\in S_{\alpha ,\beta }(M)$ such that $\psi (\xi ,\eta ,\sigma ,\tau )=\gamma$.

Assume there exists such a quadruple. Using the fact that $E_j=\bigcup _i\varphi _{\eta ,\tau }R_{ij}$, we deduce that

$$\begin{aligned} \xi (\sigma \times \mathrm {Id})E_j = \gamma \bigl (\bigcup _iR_{ij}\bigr ). \end{aligned}$$

(75)

This proves the uniqueness of the permutation $\xi (\sigma \times \mathrm {Id})$, in other words, it is the only permutation which maps $E_j$ increasingly into the set on the right side; and this implies the uniqueness of $\xi$ and $\sigma$. Therefore, we have that $(\eta \times \mathrm {Id})\beta _0(\mathrm {Id}\times \tau ) = (\sigma \times \mathrm {Id})^{-1}\xi ^{-1}\gamma$.

Thus, $\eta$ is characterized by the image of $[1,n-q]$ under the permutation on the right, which is $\eta [p+q-n+1,p]$. The uniqueness of $\tau$ follows immediately.

Given $\gamma \in \mathcal {B}_{c(M)}$, to construct $(\xi ,\eta ,\sigma ,\tau )$ we note that

$$\begin{aligned} {\#}(E_j) = \sum _i m_{ij} = {\#}\Bigl (\bigcup _i R_{ij}\Bigr ) = {\#}\biggl ( \gamma \Bigl (\bigcup _i R_{ij}\Bigr ) \biggr ), \end{aligned}$$

(76)

and, thus, we can construct a permutation $\mu$ such that (75) is verified, increasingly mapping $E_j$ into $\gamma \bigl (\bigcup _i R_{ij}\bigr )$. This permutation can be written as $\mu =\xi (\sigma \times \mu ')$ with $\xi \in {\text {Sh}}(p,n-p)$, $\sigma \in S_p$ and $\mu '\in S_{n-p}$. Since $\mu$ is increasing on $E_0=[p+1,n]$ we conclude that $\mu '=\mathrm {Id}_{n-p}$, and from the monotony on $E_j$ with $j>0$ we deduce that $\sigma \in \mathcal {B}_\alpha$. In the same way as before, we construct $\eta$ by mapping the interval $[1,n-q]$ and for this, we will show that

$$\begin{aligned} (\sigma \times \mathrm {Id})^{-1}\xi ^{-1}\gamma \text { is increasing in }F_i\text { for all }i. \end{aligned}$$

(77)

In particular, for $i=0$, we obtain the desired property to define $\eta$. We then consider $\beta _0^{-1} (\eta \times \mathrm {Id})^{-1} (\sigma \times \mathrm {Id})^{-1}\gamma$, which equals $\mathrm {Id}\times \tau$ for some $\tau \in S_p$. Using (77) for $i>0$ we conclude that $\tau \in \mathcal {B}_\beta$; and it follows from  (76) that the constructed $\tau$ belongs to $\mathcal {B}_\beta ^\eta (M)$.

It remains to prove (77). Take $x_1,x_2\in F_i$ with $x_1<x_2$. Then, $x_1\in R_{ij_1}$ and $x_2\in R_{ij_2}$ for some $j_1\le j_2$. Assume that $j_1=j_2$, then $\gamma (x_1)<\gamma (x_2)$. In this case, we have $\gamma (x_1)=\xi (\sigma \times \mathrm {Id})(e_1)$ and $\gamma (x_2)=\xi (\sigma \times \mathrm {Id})(e_2)$ with $e_1,e_2\in E_j$. Since $\sigma$ is increasing in $E_j$ we obtain that $e_1<e_2$ as desired.

On the other hand, if $j_1<j_2$, then $e_1\in E_{j_1}$ and $e_2\in E_{j_2}$ and the conclusion follows easily as all the elements of $E_{j_1}$ are smaller than those of $E_{j_2}$. $\square$