Decomposing inversion sets of permutations and applications to faces of the Littlewood–Richardson cone

Abstract

If $\alpha \in S_n$ is a permutation of $\{1, 2,\ldots ,n\}$, the inversion set of $\alpha$ is $\Phi (\alpha ) = \{ (i, j) \,| \, 1 \leqslant i < j \leqslant n, \alpha (i) > \alpha (j)\}$. We describe all r-tuples $\alpha _1, \alpha _2, \ldots , \alpha _r \in S_n$ such that $\Delta _n^+ = \{ (i, j) \, | \, 1 \leqslant i < j \leqslant n\}$ is the disjoint union of $\Phi (\alpha _1), \Phi (\alpha _2), \ldots , \Phi (\alpha _r)$. Using this description, we prove that certain faces of the Littlewood–Richardson cone are simplicial and provide an algorithm for writing down their sets of generating rays. We also discuss analogous problems for the Weyl groups of root systems of types B, C and D providing solutions for types B and C. Finally, we provide some enumerative results and introduce a useful tool for visualizing inversion sets.

Appendix: Sign diagrams

This appendix is devoted to sign diagrams, a method of displaying type A inversion sets which in some sense extends to complete flag varieties the Young diagrams used when describing Schubert cycles on Grassmannians. Although the use of sign diagrams is not necessary for the proofs of the theorems, many of our arguments have been guided by diagrammatic thinking and their point of view makes several statements in the paper transparent.

1.1 Basic definition

In order to display the inversion set of an element $\alpha \in S_n$, we start by listing the numbers 1,..., n across the page, and draw a triangular grid of squares below them, as illustrated at right in the case $n=6$. Every square in the grid corresponds to exactly one (i, j) with $1\leqslant i< j\leqslant n$; the square corresponding to (i, j) is the unique square which is directly southeast of i and directly southwest of j. In the picture we have labeled the sample squares (a) (1,6); (b) (2,4); and (c) (4,5).

Given $\alpha$ we then mark all the squares corresponding to $(i,j)\in {\Phi ({\alpha })}$ with a shaded “−” (to indicate that the positive root (i, j) is sent to a negative root by $\alpha$), and mark those $(i,j)\not \in {\Phi ({\alpha })}$ with an unshaded “$+$” (to indicate that (i, j) is sent to a positive root by $\alpha$). In order to reduce clutter in the diagram, we sometimes simply omit the $+/-$ signs or the numbers 1,..., n at the top, since these may be deduced from the size and shading of the diagram. Here is the sign diagram for the inversion set of $\alpha =(1, 6, 3, 5, 2, 4)\in S_6$ displayed using the two different conventions.

The main problem motivating the paper is describing decompositions of $\Delta ^{+}_{n}$. Here are the sign diagrams for such a decomposition with $n=21$, reduced in scale to fit the page.

For large n the inversion sets can become quite intricate, revealing patterns reminiscent of cellular automata.

1.2 Connection with Young diagrams

Let G(r, n) denote the Grassmannian of r-planes through the origin in $\mathbb {C}^n$ (with $1\leqslant r\leqslant n$). The cohomology ring of G(r, n) has a $\mathbb {Z}$-basis consisting of Schubert cycles: cohomology classes Poincaré dual to particular Zariski-closed subsets of G(r, n). Fixing a complete flag in $\mathbb {C}^n$ (equivalently a Borel subgroup B of ${\text {GL}}_n(\mathbb {C})$), the subsets are the closures of the points in G(r, n) parameterizing those r-planes intersecting the elements of the flag in fixed dimensions (equivalently the closures of the B-orbits). The combinatorial object parameterizing the data of how the r-planes meet the fixed flag, and therefore parameterizing the cohomology classes, are the Young diagrams which fit into an $r\times (n-r)$ box.

A similar construction works for the variety $X={\text {GL}}_n(\mathbb {C})/B$ parameterizing complete flags in $\mathbb {C}^n$. Here the subsets are the Zariski closures of the set of points in X where the elements of the flag meet elements of the fixed flag in prescribed dimensions, or equivalently, the B-orbits on X. The combinatorial objects parameterizing the B-orbits in this case are the elements of $S_{n}$, the Weyl group of ${\text {GL}}_n(\mathbb {C})$.

The Grassmannian G(r, n) may be realized as ${\text {GL}}_n(\mathbb {C})/P$, where P is a maximal parabolic subgroup containing B (which maximal subgroup depends on the value of r). We therefore have a quotient map $\pi :X\longrightarrow G(r,n)$, and this gives rise to the following procedure. Start with a Young diagram $\lambda$ fitting in an $r\times (n-r)$ box, take the corresponding Schubert class $[\Sigma _{\lambda }]$ on G(r, n), pull this back via $\pi$ to a cohomology class $[\Sigma _{\alpha }]$ on X (with $\alpha \in S_{n}$), and finally take the inversion set of $\alpha$, as represented by a sign diagram. Skipping the cohomology classes and showing only the combinatorial objects (Young diagram, element of $S_{n}$, and inversion set) here is an example from the cohomology of G(3, 7):

The conclusion suggested by this example holds in general: the inversion set associated with a Young diagram $\lambda$ by this procedure is that same Young diagram, rotated $45^{\circ }$. For a class on G(r, n) the top corner of the Young diagram appears between the labels r and $r+1$.

1.3 Inflation

The graphical description of inflation follows easily from the “shuffling cards” model. It is again easiest to explain with an example.

In this example the fact that $\beta _1$,..., $\beta _4$ are elements of $S_3$, $S_4$, $S_5$, and $S_3$, respectively, tells us that the resulting inflation is an element of $S_n$ with $n=3+4+5+3=15$, and that we should divide $\{1,\ldots , 15\}$ into the consecutive subsets $U_1=\{1,2,3\}$, $U_2=\{4,5,6,7\}$, $U_3=\{8,9,10,11,12\}$, and $U_4=\{13,14,15\}$ of lengths 3, 4, 5, and 3, respectively.

The large blocks of $+$ and − signs (indicated by the large blocks with a single $+$ or −) result from permuting the subsets $U_1$,..., $U_4$ as prescribed by $\sigma \in S_4$. Explicitly, setting $\alpha =\sigma [\beta _1,\beta _2,\beta _3,\beta _4]$, for every $(i,j)\in {\Phi ({\sigma })}$, we have $(a,b)\in {\Phi ({\alpha })}$ for all $a\in U_i$, $b\in U_j$, and similarly for $(i,j)\notin {\Phi ({\sigma })}$. Each element (i, j) of ${\Phi ({\sigma })}$ therefore inflates to give an $|U_i|\times |U_j|$ block in ${\Phi ({\alpha })}$ (length $|U_i|$ in the northeast-southwest direction, $|U_j|$ in the northwest-southeast direction). For each $(a,b)\in \Delta ^{+}_n$ with a and b in different intervals, we thus know whether (a, b) is in ${\Phi ({\alpha })}$ or not. However, as part of inflation we also permute each $U_i$ using $\beta _i$, and this tells us how to decide on the status of those (a, b) with a, b in the same interval. Visually this amounts to simply inserting the sign diagram for ${\Phi ({\beta _i})}$ in the appropriate empty space left by the inflation process. This procedure is the graphical translation of Lemma 3.2.

After inflating, we may leave the large blocks in the diagram to remind us of the inflation, or subdivide them into the usual smaller squares, depending on the situation. Thus the inflation above may be represented (again reduced in scale to fit the page) by

1.4 Relation with ideas from the text

In this subsection we use sign diagrams to illustrate some of the ideas from the main article.

If $\sigma$’s and $\beta$’s give a decomposition, so do the inflations As in Sect. 1.4, suppose that we divide $\{1,\ldots , n\}$ into m consecutive intervals $U_1$,..., $U_m$, choose $\sigma _i\in S_m$, $i=1$,..., r such that $\Delta ^{+}_m = \sqcup _i {\Phi ({\sigma _i})}$, and furthermore choose $\beta _{ij}\in S_{|U_j|}$ for $i=1,\ldots , r$, $j=1,\ldots , m$ such that $\Delta ^{+}_{|U_j|}=\sqcup _i {\Phi ({\beta _{ij}})}$ for each j. Then it should be clear from the visual description of the inflation procedure that this implies the decomposition $\Delta ^{+}_n = \sqcup _i \sigma _i[\beta _{i1},\ldots , \beta _{im}]$.

As an exercise the decomposition from Example 1.12, which is constructed in such a manner, is pictured below. The reader is invited to identify the diagrams inflated and inserted in each of the three pieces of the decomposition and check that they satisfy the hypotheses above.

Theorem 1.11, the central result of the paper, shows conversely that every decomposition of $\Delta ^{+}_n$ admits a recursive description by inflations satisfying the above conditions. The result of the theorem is more precise, identifying a canonical such description satisfying additional properties well suited to recursive analysis.

Rules for indecomposibility If $\alpha =\sigma [\beta _1,\ldots , \beta _m]$ with each $\beta _i\in S_{z_i}$, then it is clear from the graphical procedure for inflation that we may use this description to decompose ${\Phi ({\alpha })}$, as shown in the following example.

In formulas this kind of decomposition is written

For the element $\alpha$ to be irreducible, it follows from the inflation decomposition that at most one of $\sigma$, $\beta _1$,..., $\beta _m$ can be different from the identity, and that this non-trivial element must itself be irreducible. This is the content of Corollary 3.5.

Rules for uniqueness in inflations The sign diagram for consists entirely of minus signs. If $\alpha$ is of the form then of course these minus signs are inflated when making the sign diagram of $\alpha$, and surround the sign diagrams of $\beta _1$,..., $\beta _m$. If some $\beta _j$ also has this form (i.e., for some $m'$), then some of the minus signs from ${\Phi ({\beta _j})}$ may be merged with the minus signs from the inflation, as in the following example.

There is an identical problem (with the roles of the + and - signs reversed) for permutations of the form $\alpha ={ I}_m[\beta _1,\ldots , \beta _m]$, where some $\beta _j$ is also of the form $\beta _j={ I}_{m'}[\tau _1,\ldots , \tau _{m'}]$. In such cases we obtain uniqueness of the representation as an inflation by requiring that the diagram of or ${ I}_m$ which is inflated account for as many of the − or $+$ signs in ${\Phi ({\alpha })}$ as possible (i.e., that m be as large as possible). In the example considered, the diagram on the right is the one corresponding to the maximal , with $m=5$.

Somewhat the opposite problem occurs for representations of the form $\alpha =\sigma [\beta _1,\ldots , \beta _m]$ with . In this case it may be that $\sigma$ can itself be represented in a non-trivial way as an inflation, and this description can then be propagated upwards to give a different representation of $\alpha$ as an inflation (i.e., if $\sigma =\gamma [\delta _1,\ldots , \delta _r]$ then we may write $\alpha =\gamma [\tau _1,\ldots , \tau _r]$ for some $\tau _i$). Here is an example where this occurs.

In these cases we obtain uniqueness of the representation by requiring that $\sigma$ be simple. This amounts to looking for $\sigma \in S_m$ with m as small as possible. In the example considered the diagram on the right is the one with smallest m, with $m=4$. The two goals (m as large as possible and m as small as possible) seem to be in opposition, however, as Theorem 1.8 guarantees, every $\alpha$ has a unique representation as in inflation $\alpha =\sigma [\beta _1,\ldots , \beta _m]$ with either $\sigma$ simple with $m \geqslant 4$ or $\sigma$ one of ${ I}_m$ or and m as large as possible.

Recursion for type A maximal decompositions In Sect. 6 we considered the problem of enumerating the decompositions of $\Delta ^{+}_n$ of maximal length, i.e., into a decomposition of $n-1$ nonempty inversion sets. Here is a picture of such a decomposition with $n=8$.

The example is relatively small, but is enough to infer the general structure of the problem.

The key is the diagram containing the highest root (1, n) (i.e, the bottom vertex of the triangle), which in the example is the diagram at lower left. Because the inversion set also contains exactly one simple root it follows that it must consist of the entire rectangle with corners (1, n) and that simple root. To see why we look at the example. In the diagram at lower left, the only simple root inverted is (5, 6). This means that the numbers $\{1,2,3,4,5\}$ all retain their relative order when $\alpha$ is applied, and that the same holds for $\{6,7,8,9\}$. Combined with the fact that the inversion set contains (1, 9), so that $\alpha (9)<\alpha (1)$, we deduce that $\alpha$ swaps the two intervals, i.e., that $\alpha =(6,7,8,9,1,2,3,4,5)$, and therefore that ${\Phi ({\alpha })}$ is the rectangle with corners (5, 6) and (1, 9).

Returning to the general case, removing the rectangle containing the highest root disconnects the diagram into two smaller diagrams, each of which must be filled in by the other parts of the decomposition. The number of maximal decompositions of each of these smaller rectangles may be computed inductively. Thus if we organize the counting of the number of maximal decompositions of $\Delta ^{+}_{n+1}$ by the rectangle containing the highest root, we immediately arrive at the recursive relation ${\text {Cat}_\text {A}(n)} = \sum _{k=1}^{n} {\text {Cat}_\text {A}(k-1)}{\text {Cat}_\text {A}(n-k)}$. This leads quickly to the result that the enumerative problem is solved by the Catalan numbers.

By induction one also deduces that every diagram in a maximal decomposition is a rectangle. In the example all but two of these rectangles are reduced to lines or single squares, but this is simply because the example is small.

1.5 Diagrams for types B and C

For us the sign diagrams have been an extremely useful method of visualizing or discovering arguments in the type A case, and so it is natural to try and extend them to other types. Our method of displaying the type A inversion sets arose from picturing what Weyl group elements w do to an upper triangular Borel subgroup (this perspective has not been explained in the appendix), and one could try and repeat this idea in the other cases. However, in types B / C it turns out to be easier to use the group homomorphisms $\iota : {\mathcal W}(B_n)\hookrightarrow S_{2n+1}$ and $\iota : {\mathcal W}(C_n)\hookrightarrow S_{2n}$ from Sect. 5 and, rather than try and picture the inversion set of an element $\alpha \in {\mathcal W}(B_n)\cong {\mathcal W}(C_n)$ directly, to instead study the inversion set of $\iota (\alpha )$, the image of $\alpha$ under one of the homomorphisms. We first briefly recall the groups and the homomorphisms.

The Weyl groups ${\mathcal W}(B_n)$ and ${\mathcal W}(C_n)$ can be identified with the signed permutations of $\varepsilon _1$,..., $\varepsilon _n$, i.e., we are allowed not only to permute the elements, but also multiply them by ±1. Here is a sample element $\alpha$ of ${\mathcal W}(B_3)\cong {\mathcal W}(C_3)$:

$$\begin{aligned} \alpha :\left\{ \begin{array}{c} \varepsilon _1 \longrightarrow -\varepsilon _2 \\ \varepsilon _2 \longrightarrow \phantom {-}\varepsilon _3 \\ \varepsilon _3 \longrightarrow \phantom {-}\varepsilon _1 \\ \end{array} \right. . \end{aligned}$$

We can promote $\alpha \in {\mathcal W}(C_n)$ to an element $\iota (\alpha ) \in S_{2n}$ by considering $\varepsilon _1$, $\varepsilon _2$, ..., $\varepsilon _n$, $-\varepsilon _n$, $-\varepsilon _{n-1}$,..., $-\varepsilon _1$ to be distinct symbols, and using the rule given by $\alpha$ (and “linearity”) to deduce a permutation of these 2n elements. For example, the element $\alpha \in {\mathcal W}(C_3)$ shown above corresponds to

$$\begin{aligned} \iota (\alpha ):\left\{ \begin{array}{c} \phantom {-}\varepsilon _1 \longrightarrow -\varepsilon _2 \\ \phantom {-}\varepsilon _2 \longrightarrow \phantom {-}\varepsilon _3 \\ \phantom {-}\varepsilon _3 \longrightarrow \phantom {-}\varepsilon _1 \\ -\varepsilon _3 \longrightarrow -\varepsilon _1 \\ -\varepsilon _2 \longrightarrow -\varepsilon _3 \\ -\varepsilon _1 \longrightarrow \phantom {-}\varepsilon _2 \\ \end{array} \right. . \end{aligned}$$

Using the order $\varepsilon _1,\varepsilon _2, \varepsilon _3, -\varepsilon _3,-\varepsilon _2, -\varepsilon _1$, this is the element $\iota (\alpha )=(5,3,1,6,4,2)\in S_{6}$. We can similarly obtain an element in $S_{2n+1}$ by adding the element 0 (in the order $\varepsilon _1$, ..., $\varepsilon _n$, 0, $-\varepsilon _n$, ..., $-\varepsilon _1$) and simply fixing 0. In the example considered, this gives the element $(6,3,1,4,7,5,2)\in S_{7}$. (The way we have presented this rule ‘adding 0’ seems somewhat arbitrary, but it does make sense from the natural description of the complete flag variety of $B_n$ type as a subvariety of the complete flag variety of $A_{2n}$ type.) We will use the symbol $\iota (\alpha )$ for the image of $\alpha \in {\mathcal W}(C_n)\cong {\mathcal W}(B_n)$ in $S_{2n}$ or $S_{2n+1}$ under either of these homomorphisms and trust that the resulting permutation (of either an even or odd number of elements) will reveal which homomorphism was intended.

Here are the sign diagrams for $\iota (\alpha )$ (in $S_6$ and $S_7$) for the sample element of ${\mathcal W}(C_3)$ considered above.

The images of the injective homomorphisms ${\mathcal W}(C_n) \longrightarrow S_{2n}$ and ${\mathcal W}(B_n)\longrightarrow S_{2n+1}$ turn out to be precisely those elements whose sign diagram is symmetric about the vertical centerline, and the basic idea is to simply study such ‘symmetric’ sign diagrams and the corresponding elements of $S_{2n}$ and $S_{2n+1}$.

One point is worth stating explicitly: given $\alpha \in {\mathcal W}(C_n)$ or ${\mathcal W}(B_n)$, the inversion set ${\Phi ({\alpha })}$ is a subset of $\Delta ^{+}_{C_n}$ (or $\Delta ^{+}_{B_n}$), while the inversion set ${\Phi ({\iota (\alpha )})}$ is a subset of $\Delta ^{+}_{2n}$ or $\Delta ^{+}_{2n+1}$, and these sets can be quite different. For instance these sets almost never have the same number of elements. (This is evident in the example above, where inversion sets for $\iota (\alpha )$ in $S_6$ and $S_7$ do not have the same number of elements as each other and so could not both agree with the number of elements in ${\Phi ({\alpha })}$.) More importantly, the ideas of “indecomposable”, “decomposition”, “disjoint”, or “simple” could potentially be quite different in $\Delta ^{+}_{C_n}$ and $\Delta ^{+}_{2n}$ (or $\Delta ^{+}_{B_n}$ and $\Delta _{2n+1}^+$), and one of the main things we need to check is that in fact they are not.

In order to even define “simple” we need to have a notion of inflation, and to do this we simply use the inflation procedure in $S_{2n}$ or $S_{2n+1}$ but require that all the data describing the inflation be ‘symmetric’. Below is an example.

For the data describing an inflation $\iota (\alpha )=\sigma [\beta _1,\ldots , \beta _m]$ to be symmetric means that: (i) $\sigma$ is symmetric; (ii) the collection of intervals $U_1$,..., $U_m$ are symmetric (i.e, interchanged by the operation of reversing 1, ..., 2n or 1, ..., $2n+1$); (iii) if there is an interval $U_j$ which is itself symmetric (i.e., straddles the centerline), then the corresponding $\beta _j$ must be symmetric; and (iv) for all other intervals $U_j$ the sign diagram for $\beta _j$ must be the mirror image of the sign diagram for $\beta _{m+1-j}$. The example presented above has all these features. Conditions (ii), (iii) and (iv) may be summarized by the condition that for $j=1$,..., m.

Propositions 4.1 and 5.1 contain the useful result that if a symmetric $\iota (\alpha )$ can be represented nontrivially as an inflation, it can be represented nontrivially as an inflation with symmetric data. With the idea of inflation in place, we now define an element $\alpha \in {\mathcal W}(B_n)$ or ${\mathcal W}(C_n)$ to be simple if the corresponding $\iota (\alpha )$ is simple in $S_{2n}$ or $S_{2n+1}$.

In Sect. 5 the following results are established showing that the intrinsic notions for an inversion set ${\Phi ({\alpha })}$ with $\alpha \in {\mathcal W}(C_n)\cong {\mathcal W}(B_n)$ in type B / C agree with the type A notions of the corresponding element $\iota (\alpha )$ in $S_{2n}$ or $S_{2n+1}$.

• (i) Corollary 5.2: $\alpha$ is irreducible if and only if $\iota (\alpha )$ is irreducible; ${\Phi ({\alpha _1})}$ and ${\Phi ({\alpha _2})}$ are disjoint if and only if ${\Phi ({\iota (\alpha _1)})}$ and ${\Phi ({\iota (\alpha _2)})}$ are disjoint; $\Delta ^{+}_{B_n} = \sqcup _i {\Phi ({\alpha _i})}$ if and only if $\Delta _{2n+1}^{+}=\sqcup _i {\Phi ({\iota (\alpha _i)})}$ (respectively $\Delta ^{+}_{C_n} = \sqcup _i {\Phi ({\alpha _i})}$ if and only if $\Delta _{2n}^{+}=\sqcup _i {\Phi ({\iota (\alpha _i)})}$).

• (ii) Proposition 5.3: $\alpha$ is simple if and only if $\iota (\alpha )$ is simple.

With these results, one deduces Theorem 5.4 which is the type B / C version of Theorem 1.11. The arguments and pictures used in appendix section “Relation with ideas from the text” also extend in an appropriate way to the B / C case. For instance, one may also deduce a uniqueness statement for representation as a symmetric inflation, paralleling that of Theorem 1.8, or recursion relations for the type B / C Catalan numbers (Proposition 6.2).