Classification of $Q$-multiplicity-free skew Schur $Q$-functions

We classify the $Q$-multiplicity-free skew Schur $Q$-functions. Towards this result, we also provide new relations between the shifted Littlewood-Richardson coefficients.


Introduction
Schur functions form an important basis of the algebra of symmetric functions.
They appear in the study of the representations of the symmetric groups and the general linear groups. Schur P -functions and Schur Q-functions are bases of the subalgebra generated by the odd power sums. In [11], Stembridge proved a number of important properties of Schur Q-functions emphasizing that they may be viewed as shifted analogues of Schur functions. While in the classical situation Schur functions are closely related to ordinary irreducible characters of the symmetric groups, Schur Q-functions are intimately connected to irreducible spin characters of the double covers of the symmetric groups. Multiplicity-free products of Schur functions were classified by Stembridge in [10]. As a shifted analogue of Stembridge's result, Bessenrodt then classified the P -multiplicity-free products of Schur P -functions in [2]. A skew generalization of Stembridge's result was proved independently by Gutschwager in [5] and Thomas and Yong in [12]. While Gutschwager classified the multiplicity-free skew Schur functions, Thomas and Yong classified the multiplicity-free products of Schubert classes. However, what was missing was a skew analogue of Bessenrodt's result or equivalently a shifted analogue of Gutschwager's result. The main goal of this article is to provide this here, i.e., to classify the Q-multiplicity-free skew Schur Q-functions. We will heavily rely on the shifted Littlewood-Richardson rule obtained by Stembridge in [11] (another version of this rule was given by Cho [3]).
The paper is structured as follows. In the second section we provide the required definitions and some properties needed later. In the third section we prove relations between shifted Littlewood-Richardson coefficients, which will simplify proofs of the fourth section. In the fourth section we will first exclude all non-Q-multiplicity-free skew Schur Q-functions before proving the Q-multiplicity-freeness of the remaining skew Schur Q-functions to obtain our main classification result, Theorem 4.61. Note that we define some special notation for partitions with distinct parts whose shifted diagrams have at most two corners in Definition 4.19. We will use that notation in most lemmas of the fourth section.

Preliminaries
We will use the same notation as in [9]. Some of the tools introduced there will also be useful in the context here.
2.1. Partitions, diagrams and tableaux. We define a partition as a tuple λ = (λ 1 , λ 2 , . . . , λ n ) where λ j ∈ N for all 1 ≤ j ≤ n and λ i ≥ λ i+1 > 0 for all 1 ≤ i ≤ n − 1. The length of λ is (λ) := n. A partition λ is called a partition of k if |λ| := λ 1 + λ 2 + . . . + λ (λ) = k where |λ| is called the size of λ. A partition with distinct parts is a partition λ = (λ 1 , λ 2 , . . . , λ n ) where λ i > λ i+1 > 0 for all 1 ≤ i ≤ n − 1. The set of partitions of k with distinct parts is denoted by DP k . By definition, the empty partition ∅ is the only element in DP 0 and it has length 0. The set of all partitions with distinct parts is denoted by DP := k DP k . For λ ∈ DP the shifted diagram D λ is defined by Convention. In this paper we will omit the adjective shifted. This means that whenever a diagram is mentioned it is always a shifted diagram.
Each edgewise connected part of a skew diagram D is called a component. The number of components of D is denoted by comp(D). If comp(D) = 1, then D is called connected, otherwise it is called disconnected.
In the following, if components are numbered, the numbering is as follows: the first component is the leftmost component, the second component is the next component to the right of the first component etc.
A corner of a skew diagram D is a box (x, y) ∈ D such that (x+1, y), (x, y +1) / ∈ D. An unshifted diagram is a skew diagram D λ/µ with (µ) = (λ) − 1. A (skew) diagram can be depicted as an arrangement of boxes where the coordinates of the boxes are interpreted in matrix notation. .

Skew Schur Q-functions.
For λ, µ ∈ DP and a countable set of independent variables x 1 , x 2 , . . . the skew Schur Q-function is defined by where T (λ/µ) denotes the set of all tableaux of shape D λ/µ and x (c1,c2,...,c ) := x c1 1 x c2 2 · · · with c k := 0 for k > . If D µ D λ then Q λ/µ := 0. Since D λ/∅ = D λ , we denote Q λ/∅ by Q λ . Definition 2.5. Let a diagram D be such that the y th column has no box, but there are boxes to the right of the y th column and after shifting all boxes that are to the right of the y th column one box to the left we obtain a diagram D α/β for some α, β ∈ DP . Then we call the y th column empty and the diagram D α/β is obtained by removing the y th column. Similarly, let a diagram D be such that the x th row has no box, but there are boxes below the x th row and after shifting all boxes that are below the x th row one box up and then all boxes of the diagram one box to the left, we obtain a diagram D α/β for some α, β ∈ DP . Then we call the x th row empty and the diagram D α/β is obtained by removing the x th row. Definition 2.6. For λ, µ ∈ DP we call the diagram D λ/µ basic if it satisfies the following properties: This means that D λ/µ has no empty rows or columns.
For a given diagram D letD be the diagram obtained by removing all empty rows and columns of the diagram D. Since the tableau restrictions on each box in a diagram are unaffected by removing empty rows and columns, there is a contentpreserving bijection between tableaux of a given shape and tableaux of the shape obtained by removing empty rows and columns; thus we have Q D = QD. Hence, we may restrict our considerations to partitions λ and µ such that D λ/µ is basic.
For some given skew diagram D let the diagram obtained after removing empty rows and columns be D λ/µ for some λ, µ ∈ DP . Then Q D is equal to the skew Schur Q-function Q λ/µ . For a tableau T of shape D, the reading word w = w(T ) is the word obtained by reading the rows from left to right beginning with the bottom row and ending with the top row. The length (w) is the number of letters and, thus, the number n = |D| of boxes in D. Let (x(i), y(i)) denote the box of the i th letter of the reading word w(T ).
For the reading word w = w 1 w 2 . . . w n of the tableau T the statistics m i (j) are defined as follows: • m i (0) = 0 for all i.
• For 1 ≤ j ≤ n the number m i (j) is equal to the number of times i occurs in the word w n−j+1 . . . w n .
• For n + 1 ≤ j ≤ 2n we set m i (j) := m i (n) + k(i) where k(i) is the number of times i occurs in the word w 1 . . . w j−n .
Lemma 2.9. Let T be an amenable tableau. Then there are no entries greater than k in the first k rows.
Proof. Assume the opposite. Let i be the topmost row with an entry greater than i.
Let this entry be x. Then, while scanning the word w(T ) from right to left, the letter x will be scanned before any letter |x| − 1 will be scanned; a contradiction to the amenability of the tableau T .
Using Lemma 2.9, one can construct a brute force algorithm to obtain amenable tableaux of a given shape. The algorithm fills the first row with elements from where f λ µν is the number of amenable tableaux T of shape D λ/µ and content ν.
In [11,Chapter 8], Stembridge showed that the numbers f λ µν above also appear in the product of P -functions: Using this, one easily obtains the equation f λ µν = f λ νµ for all λ, µ, ν ∈ DP . A (p, q)-hook is a set of boxes More precisely, we say that the set of boxes above is a (p, q)-hook at (u, v).
Definition 2.12. Let T be a tableau of shape D λ/µ . Define T (i) by Remark 2.17. The set S λ/µ (x, y) above is the set of boxes that are simultaneously weakly above and weakly to the right of the box (x, y).
In Section 4 we will start with a specific amenable tableau for a given diagram and change some entries to obtain new tableaux. This specific tableau is obtained by an algorithm described by Salmasian in [8, Section 3.1].
Definition 2.24. Let λ ∈ DP . Then the border of λ is defined by Definition 2.25. Let λ ∈ DP . Define E λ to be the set of all partitions whose diagram is obtained after removing a corner in D λ .
Hence it will be sufficient to analyze the skew Schur Q-function of one of these (transformed) diagrams to obtain statements for all of them. This approach significantly reduces the effort in the proofs for the lemmas that lead to the classification of Q-multiplicity-free skew Schur Q-functions.    Lemma 2.32. Let λ, µ ∈ DP , ν = c(T λ/µ ) and n = (ν). Let D ot λ/µ have shape D γ/δ . Let T = T γ/δ . If U i (λ/µ) has shape D α/β then U i (γ/δ) has shape D ot α/β .
D γ/δ } and the image of this set of boxes after orthogonally transposing is given by Remark 2.33. For i = n this means that T (n) has the same shape as P ot n .

Relations between the coefficients f λ µν
In this section we will prove some inequalities satisfied by the shifted Littlewood-Richardson coefficients f λ µν that will be helpful for the proofs in Section 4.
Then, by Lemma 2.13, each box (x, y) of the ν 1 entries from {1 , 1} is such that The set of all such boxes is P 1 and we have |P 1 | = ν 1 . Hence, T (1) = P 1 . Then, by Lemma 2.13, each box (x, y) of the ν 2 entries from The set of all such boxes is P 2 and we have |P 2 | = ν 2 . Hence, T (2) = P 2 . Repeating this argument for all entries greater than 2, we see that T (j) = P j for all 1 ≤ j ≤ k − 1. Hence, after removing T (1) , T (2) , . . . , T (k−1) we obtain some tableau of U k (λ/µ) of shape D α/β .
If for each box of this tableau we replace its entry i (respectively, i ) by i − k + 1 (respectively, (i − k + 1) ) then we obtain a tableau T of D α/β of content γ. The amenability of the tableau T follows from the amenability of the tableau T . After removing the ribbon strips P j for all 1 ≤ j ≤ k − 1, the tableau T is independent of the different possible fillings of these P j . By Lemma 2.14, we have 2 comp(Pj )−1 possible fillings of P j . Thus, for each of the k−1 j=1 2 comp(Pj )−1 tableaux of content (ν 1 , . . . ν k−1 , γ 1 , . . . , γ (γ) ) with the same entries in U k (λ/µ) we obtain T . Hence, we ..,γ (γ) ) and the statement follows.
The following example illustrates how to obtain tableaux of shape D λ/µ from the tableaux of shape D α/β as explained in the proof of Lemma 3.1.  .
We obtain two amenable tableaux of D λ/µ of the same content: . Definition 3.3. Let λ, µ ∈ DP , let 2 ≤ a ≤ (µ) + 2, and let b ≥ (λ). Let Γ → a (D λ/µ ) be the diagram obtained from D λ/µ by shifting all boxes above the a th row one box to the right. Let Γ ↓ b (D λ/µ ) be the diagram obtained from D λ/µ by shifting all boxes (x, y) such that y < b one box down.
Proof. For every given amenable tableau T of shape D λ/µ one obtains an amenable the tableauT is amenable. Let T, T be two amenable tableaux of shape D λ/µ andT ,T the tableaux obtained from T, T as above.
For every given amenable tableau T of shape D λ/µ one obtains an amenable T (x, y) = T (x, y) for all y ≥ b such that (x, y) ∈ Dα /β . By Lemma 2.19, the tableaũ T is amenable. Let T, T be two amenable tableaux of shape D λ/µ andT ,T the tableaux obtained from T, T as above. IfT =T then T (x, y) =T (x+1, y) =T (x+ all y ≥ b and, hence, T = T . Thus, the statement f λ µν ≤ fα βν follows.
Remark 3.6. The statement f λ µν ≤ f α βν in the sense of Lemma 3.5 appeared in the proof of [2, Theorem 2.2] and is, hence, due to Bessenrodt. In the same proof the statement f λ µν ≤ fα βν for b = µ 1 + 2 can be found (without explicitly stating that Lemma 3.7. Let w be an amenable word. Letw be a word such that after removing one letter 1 the word obtained is w (this means thatw can be obtained from w by adding a letter 1). Thenw is amenable.
Proof. The number of letters equal to 1 inw is greater than the number of letters equal to 1 in w. Then the wordw is not amenable only if there is some j ≥ n = (w) such that m 1 (j) = m 2 (j) and w j−n+1 is this added 1. But then for the word w we have m 1 (j − 2) < m 2 (j − 2); a contradiction to the amenability of the word w.
Proof. For this proof we will assume that for a tableau of shape D λ/µ the boxes of D µ are not removed but instead are filled with 0. Given an amenable tableau T of shape D λ/µ we obtain an amenable tableauT of shape D (λ+(1 a ))/(µ+(1 a−1 )) as follows. Insert a box with entry 0 into each of the first a − 1 rows such that the rows are weakly increasing from left to right and insert a box with entry 1 into the a th row such that this row is weakly increasing from left to right. The word w(T ) differs from w(T ) only by one added 1. By Lemma 3.7, the Gutschwager proved a similar statement for classical Littlewood-Richardson coefficients.
Proof. Again, we assume that for a tableau of shape D λ/µ the boxes of D µ are not removed but instead are filled with 0. Given an amenable tableau T of shape D λ/µ we obtain an amenable tableauT of shape D α/β as follows. Insert a box with entry 0 into each of the first b − 2 columns such that the columns are weakly increasing from top to bottom. If there is no 1 or 1 in the (b − 1) th column then insert a box with entry 1 into the (b − 1) th column such that this column is weakly increasing from top to bottom. If there is a 1 or a 1 in the (b − 1) th column then insert a box with entry 1 into the (b − 1) th column such that this column is weakly increasing from top to bottom.
LetT be the tableau defined byT (x, y) By Lemma 3.5, the tableauT is amenable. The word w = w(T ) differs from w(T ) only by an added 1 or an added 1. If a 1 is added then clearly, the tableauT is amenable. If a 1 is added then, by Lemma 3.7, the word w(T ) is amenable. Clearly, if T = T for some tableaux T, T ∈ T (λ/µ) thenT =T .

Q-multiplicity-free skew Schur Q-functions
With the tools provided in Section 3 we can finally start to prove results towards our desired classification.
Lemma 3.1 will be crucial in this chapter. It allows us to consider subdiagrams consisting of the boxes of T λ/µ with entries bigger than some given k and simplifies the proof of non-Q-multiplicity-freeness. This follows from the fact that if the skew Schur Q-function to a diagram is not Q-multiplicity-free then the same must hold for each diagram that contains this diagram as a subdiagram of T λ/µ of boxes with entries greater than k for some k.
We will analyze diagrams D λ/µ and show that the corresponding Q λ/µ are not Q-multiplicity-free by finding two different amenable tableaux of the same content, derived by changing entries in the tableau T λ/µ obtained in Definition 2.22. Using this method, we are able to find many types of diagrams D λ/µ such that the corresponding Q λ/µ are not Q-multiplicity-free; we will prove afterwards that all remaining diagrams do give Q-multiplicity-free skew Schur Q-functions, hence then our classification is complete.
Let λ, µ ∈ DP and ν = c(T λ/µ ). Lemma 2.14 and the definition of amenability, which requires P i to be fitting, state that f λ µν = 1 is only possible if each P i is connected.

Hypothesis 4.2.
From now on we will consider only diagrams that satisfy the property that each P i is connected.
The following lemmas give restrictions for the border strips P i ; they will enable us to prove Lemma 4.15 and Corollary 4.17 which lower the number of families of partitions λ we have to consider to find Q-multiplicity-free skew Schur Q-functions. Lemma 4.3. Let λ, µ ∈ DP . Let ν = c(T λ/µ ) and n = (ν). If P n is neither a hook nor a rotated hook then Q λ/µ is not Q-multiplicity-free.
Proof. By Lemma 3.1, it is enough to find two amenable tableaux of P n of the same content. Assume that the diagram P n is neither a hook nor a rotated hook.
Then we can find a subset of boxes of P n such that all but one of the boxes form a (p, q)-hook where p, q ≥ 2 and there is either a single box above the rightmost box of the hook, or a single box to the left of the lowermost box of the hook. By Lemmas 2.28, 3.5, 3.9 and 3.11, it is enough to assume that P n has shape D (4,2)/(2) .

Lemma 4.4.
Let λ, µ ∈ DP . Let ν = c(T λ/µ ) and n = (ν) > 1. Let P n be a (p, q)-hook or a rotated (p, q)-hook where p, q ≥ 3. Suppose the last box of P n−1 is not in the row directly above the row of the last box of P n . Then Q λ/µ is not Proof. We may assume that P n is a (p, q)-hook where p, q ≥ 3. Otherwise, P n is a rotated (p, q)-hook where p, q ≥ 3 and we may consider D ot λ/µ since if D ot λ/µ has shape D α/β then, by Lemma 2.32, the set of boxes T By Lemma 3.1, we may assume that n = 2. Let (x, y) be the last box of P 2 . By Lemmas 3.5, 3.9 and 3.11, we may assume that (x, y − 1) is the last box of P 1 . We but there is no 2 in the (y − 1) th column. However, there are at least two 2s with no 3 below them in the first two boxes of P 2 . Hence, by Lemma 2.19, this tableau is amenable.
By Corollary 2.20, this tableau is m-amenable for m = 2, 3. Since there is a 1 but no 2 in the y th column, 2-amenability follows. We have T (x, y) = 3 but there is no 2 in the y th column. Also, we have T (x − 1, y) = 3 and T (x − 2, y − 1) = 2 .
However, in the first two boxes of P n are 2s with no 3 below. Additionally, there is another 2 with no 3 below in the (y − 1) th column. Thus, by Lemma 2.19, 3-amenability follows.
Proof. Let k, i be maximal with respect to these conditions and let j = min{k, i}.
By Lemma 3.1, we may assume that j = 1. First, we assume that i ≤ k. Then letk be minimal such that the last box of Pk is in a row strictly lower than the last box of P n . Let (u, v) be the lowermost box in the rightmost column with a box of Pk in a row strictly lower than the last box of Pk +1 . Let x = u −k + i and y) is the lowermost box of P i in the y th column. We get a new tableau T if after the (i − 1) th step of the algorithm of Definition 2.22 we use Clearly, by Corollary 2.20, this tableau is m-amenable for m = i + 1. We possibly have T (x, y) = (i + 1) and But there is an i with no i + 1 below in the column of the first box of P i . Thus, by Lemma 2.19, (i + 1)-amenability follows.
Let (c, d) be the last box of Pk +1 . We get another tableau T with the same is an unshifted diagram and we showed that two amenable tableaux of U k (λ/µ) t with the same content exist. By Lemma 2.28, the statement follows. .
Let there be some k < n such that there is a corner, (x, y) say, in P k above the boxes of P n and let there be some i ≤ k such that the first box of P i is above the (x − k + i) th row.
Proof. Let k be minimal and i be maximal with respect to these conditions. Then for all i + 1 ≤ a ≤ k the first box of P a has no box of P a below. Let (x − k + a, y) be the first box of P a for i + 1 ≤ a ≤ k − 1 and let (x − k + i, y) be the rightmost box of However, we have T (x − 1, y) = k and there is no k + 1 in the y th column. Hence, by Lemma 2.19, T is m-amenable for all m.
Let there be some k > 1 such that the first box of P k−1 is to the right of the column of first box of P k , and P k−1 is not a hook. Then Q λ/µ is not Q-multiplicity-free.
Proof. Let k be maximal with respect to this property. By Lemma 3.1, we may assume that k = 2. If the first box of P 1 is not a corner then, by Lemma 4.8, Q λ/µ is not Q-multiplicity-free. Thus, consider that the first box of P 1 is a corner. If the first box of P 1 is not in the row above the first box of P 2 then an orthogonally transposed version of Lemma 4.8 states that Q λ/µ is not Q-multiplicity-free. Since and the first box of P 1 is not in the w th column. Let v be maximal with respect to this property.
We get a new tableau T if we use However, in the column containing the first box of P 1 there is a 1 and no 2. Thus, by Lemma 2.19, this tableau is amenable.
Proof. Let k be maximal with respect to this property. Let (u, v) be the lowermost box of P k in the i th column and let (a r , b r ) be the first box of P r for all r. We get Corollary 2.20, T is amenable.
Corollary 4.14. Let λ, µ ∈ DP . Let ν = c(T λ/µ ) and n = (ν) > 1. Let P n be a (p, q)-hook where p, q ≥ 2 and let (x, y) be the first box of P n . Let there be some k < n and some i ≥ x such that there are at least two boxes of P k in the i th row. Then Q λ/µ is not Q-multiplicity-free.
Now we are able to show an intermediate result that bounds the number of corners of D λ/µ in the case of Q-multiplicity-freeness and, hence, of D λ for µ = ∅, (1). The number of corners of D µ is then also bounded for most D λ/µ because of orthogonal transposition. This restricts the number of cases we have to analyze.
If D λ has more than two corners then Q λ/µ is not Q-multiplicity-free.
Proof. Assume D λ has more than two corners, µ = ∅, (1), and Q λ/µ is Q-multiplicityfree. We will construct two amenable tableaux of shape D λ/µ and of the same content to arrive at a contradiction. Let ν = c(T λ/µ ) and n = (ν). Let k be maximal such that U k (λ/µ) has at least three corners. Thus, at least one corner is in P k .
By Lemma 4.3 P n must be a hook or a rotated hook, so P n can have at most two corners and, hence, k < n. By Lemma 4.6 either the uppermost or the lowermost corner must be in P n , so we only consider diagrams such that the uppermost or the lowermost corner is in P n . Without loss of generality we may assume that the lowermost corner of U k (λ/µ) is in P n , otherwise U k (λ/µ) is an unshifted diagram and we may transpose U k (λ/µ). Thus, the uppermost corner is in P k . By Lemma 4.8, which forbids to have boxes of P k to the left and above a corner in P k at the same time, the uppermost corner is the first box of P k and it is the only corner of the diagram U k (λ/µ) that is in P k .
Case 1: two corners are in P n .
Then P n is a (p, q)-hook where p ≥ 2 and q ≥ 2. By Lemma 4.12 and Corollary 4.14, which in this case for all k ≤ i ≤ n − 1 forbid to have more than one box of P i in the column of the first box of P n and in the row of the last box of P n , all P i are hooks. Also by Corollary 2.20, the tableau T 1 is also (k + 1)-amenable because in the column of the first box of P k is a k and no k + 1.
We get another tableau T 1 if we set T 1 (u n − 1, v n ) = n and T 1 (r, s) = T 1 (r, s) for every other box (r, s) ∈ D λ/µ . By Corollary 2.20, T 1 is m-amenable for m = n + 1.
We have T 1 (u n , v n ) = n + 1 and T 1 (u n − 1, v n ) < n, however, there is an n with no n + 1 below in the first box of P n , and we have T 1 (u n−1 , v n−1 ) = n. Thus, by Lemma 2.19, (n + 1)-amenability follows. We have c(T 1 ) = c(T 1 ).
Case 1.2: the last box of P n−1 is in the row above the row of the last box of P n .
We get another tableau T 2 if we set T 2 (u n − 2, v n ) = n and T 2 (r, s) = T 2 (r, s) for every other box (r, s) ∈ D λ/µ . By Lemma 2.19, it is clear that T 2 is amenable if T 2 is amenable. We have c(T 2 ) = c(T 2 ).
Case 2: only one corner is in P n .
Let the second uppermost corner be in P i . Then by Lemma 4.8, the second uppermost corner is the first box of P i and the uppermost corner is the first box of P k . If P i has all boxes in a row then µ = ∅; a contradiction. Thus, the diagram P i has at least two corners. By Lemma 4.10, P i is a hook. Then for all i ≤ j < n each P j is a (p, q)-hook for some p, q ≥ 2. We get a new tableau T 4 if for all k ≤ a ≤ i − 1 we set T 4 (c a , d a ) = a + 1,  .
The case that the diagram D λ or the diagram D µ has at most two corners also occurs in the classical setting of Schur functions s λ/µ . Gutschwager proved [5, We will introduce some new notation for partitions in DP with at most two corners and then obtain restrictions until we can exclude all non-Q-multiplicityfree skew Schur Q-functions in Proposition 4.34.  In particular, we see that for λ = (8, 7, 6, 5) we have π λ = [4,5], and for λ = (8, 7, 6, 3, 2) we have π λ = [3, 2, 2, 2] Remark 4.21. For a given λ ∈ DP ≤2 the cardinality of the border B λ can be derived from the shape path.
Notation. From now on we will identify a partition λ ∈ DP ≤2 with at most two corners with its shape path π λ . Lemma 4.23. Let µ ∈ DP , λ ∈ DP ≤2 , and suppose λ = [a, b] where b ≤ 2. If D µ has more than two corners then Q λ/µ is not Q-multiplicity-free.
Proof. For each corner (x, y) of D µ except for the lowermost, there is a box (x + ∈ D λ/µ and there is no box above because (1, w) is in the first row. After transposing this diagram orthogonally, the image of these boxes are corners of D ot λ/µ . The diagram D ot λ/µ has shape D α/β where β = ∅, (1) and D α has more than two corners. By Corollary 4.17, Q λ/µ is not Q-multiplicity-free. Proof. The leftmost box of the first row of D λ/µ , which is (1, w + x + y + z), has no box to the left or above. Also, the leftmost box of the (w + 1) th row of D λ/µ , which is (w + 1, w + y + z), has no box to the left or above. In addition, the leftmost box of the (w +y +1) th row of D λ/µ , which is (w +y +1, w +y +1), has no box to the left or above. After transposing this diagram orthogonally, the images of these boxes are corners of D ot λ/µ . Then the diagram D ot λ/µ has shape D α/β where β = ∅, (1) and D α has more than two corners. By Corollary 4.17, Q λ/µ is not Q-multiplicity-free.  (p, q). Then (p, q) ∈ P k and also (p − 1, q), (p, q − 1) ∈ P k . Let n = (c(T λ/µ )).
If P n is not a rotated hook, then by Lemma 4.3, Q λ/µ is not Q-multiplicity-free.
If P n is a rotated (l, m)-hook where l, m ≥ 2 then, since λ = [a, b, c, d], there is some j < n such that either the first box of P j is in a column to the right of the boxes of P n or the last box of P j is in a row below the boxes of P n . Let j be maximal with respect to this condition.
We may assume that the first box of P j is in a column to the right of the boxes of P n , otherwise U j (λ/µ) is unshifted and we may consider U j (λ/µ) t . By Lemma 2.32, if D ot λ/µ has shape D α/β then T (n) α/β is an (m, l)-hook where l, m ≥ 2 and the diagram U j (α/β) satisfies the conditions of Lemma 4.12. By Lemma 3.1, it follows that Q λ/µ is not Q-multiplicity-free.
If U k+1 (λ/µ) has at least two components then the last box of the second component can be filled with (k + 1) or k + 1 and, by Lemma 2.19, Q λ/µ is not Then either U k (λ/µ) or U k (λ/µ) t satisfies the conditions of Lemma 4.12 and Q λ/µ is not Q-multiplicity-free.
Case 2.2: only one corner is in P n .
Let (f, g) be this corner. Then there is some e such that there are two boxes of P e either in a row weakly below the f th row or in a column weakly to the right of g th column. There is also some h such that either the first box of P h is to the right of the g th column or the last box of P h is below the f th row. Let e, h be maximal with respect to these conditions. By orthogonal transposition, transposition or rotation of U min{e,h} (λ/µ), we may assume that h ≤ e and that the first box of P h is to the right of the g th column. By Lemma 4.8, if h = e then Q λ/µ is not Q-multiplicity-free. Hence, we assume h < e. There is a box (r, u) Then we obtain T =  Proof. We will show that for case a = 3 and for case w = a + c − 2 the statement holds. Afterwards we will explain case a > 3 and w < a + c − 2 by these two cases.
Then replace the entry 3 in the last box of P 3 with 3 and set T 1 (w + 1, w + 1) = 3.
Afterwards fill the remaining boxes using the algorithm of Definition 2.22 starting with k = 4. By Corollary 2.20, it is clear that T 1 is m-amenable for m = 3, 4.
There is a 3 but no 2 in the (w + 1) th column. However, there is a 2 and no 3 in the column of the last box of P 3 and there is a 2 and no 3 in the column to the left of it. Thus, by Lemma 2.19, this tableau is 3-amenable. In the (w + 2) th column and possibly in the (w + 3) th column, there are 4s and no 3s. However, there are 3s and no 4s in the columns of the first two boxes of P 3 . We have T 1 (w + 1, w + 2) = 4 and T 1 (w, w + 1) = 3 . However, if (y, z) is the third box of P 3 then we either have T 1 (y, z) = 3 and there is no 4 in the z th column or if w = a + c − 2 we have T 1 (y, z) = 3 and T 1 (y+1, z +1) = 4 . If w < a+c−2 then we have T 1 (w, w+3) = 4 and T 1 (w −1, w +2) = 3 . However, we have T 1 (w −1, w +3) = 3 . Thus, by Lemma 2.19, this tableau is 4-amenable.
There is a 3 with no 2 above in the (w + 2) th column. However, there is a 2 with no 3 below in the column of the last box of P 3 . Thus, by Lemma 2.19, this tableau is 3-amenable. By Lemma 2.19, it is clear that T 1 is 4-amenable if T 1 is.
The 5-amenability is clear for P 5 = ∅. If P 5 = ∅ then there is a 4 and no 5 in the (z + 2) th column. Thus, by Lemma 2.19, this tableau is 5-amenable.
We get another tableau T 2 of the same content if we set T 2 (w + 1, w + 1) = 2, T 2 (w, w + 2) = 3 and T 2 (r, s) = T 2 (r, s) for every other box (r, s) ∈ D λ/µ . By Corollary 2.20, T 2 is m-amenable for m = 2, 3, 4. There is a 1 and no 2 in the (w + 2) th column. Thus, by Lemma 2.19, 2-amenability follows. There is a 3 and no 2 in the (w + 2) th column. However, there is a 2 with no 3 below in the z th column. Thus, by Lemma 2.19, this tableau is 3-amenable. By Lemma 2.19, it is Case 3: a > 3 and w < a + c − 2.
Either we have a = a − 1 = 3 or w = a + c − 2 or else there is some j such that U j (λ/µ) has shape D λ /µ where λ = [a , b, c, 1] where a = a − j such that either a = 3 or w = a + c − 2. Then, by Case 1 and Case 2, we find two different amenable tableaux of the same content and, by Lemma 3.1, Q λ/µ is not Q-multiplicity-free.

Lemma 4.31. Let
Proof. Let n = (c(T λ/µ )). First, we assume w = a + c − 1 and prove the result when a = 2 or d = 2. Then, we show that the case where a, d ≥ 3 can be explained by the cases where a = 2 or d = 2. Afterwards, we tackle the case w < a + c − 1 using the case w = a + c − 1 where we first prove the subcase d = 2 and then show how to add boxes with entries to obtain diagrams such that d > 2.
We may assume a = 2, otherwise we transpose the diagram. If d = 2 then P n is a (b + 1, c + 1)-hook and, by Lemma 4.4, which in this case forbids to have a box directly to the left of the last box of P n , Q λ/µ is not Q-multiplicity-free. Thus, we may now assume d ≥ 3.
The box (w + 1, w + 1) is the last box of P 1 . We get a new tableau T 1 if we set We get another tableau T 1 if we set T 1 (w + 1, w + 1) = 2, T 1 (w, w + 2) = 3 and T 1 (r, s) = T 1 (r, s) for every other box (r, s) ∈ D λ/µ . By Corollary 2.20, T 1 is m-amenable for m = 2, 3. In the (w + 2) th column is a 1 with no 2 below. Thus, by Corollary 2.20, 2-amenability follows. We have T 1 (w, w + 2) = 3 and T 1 (w − 1, w + 1) = 2 and there is a 3 and no 2 in the (w + 2) th column. However, there are two 2s and no 3s in the columns of the first two boxes of P 2 . Thus, by Lemma 2.19, 3-amenability follows. It is clear that T 1 has the same content as T 1 . Hence, Q λ/µ is not Q-multiplicity-free. The box (w + 1, w + 1) is the last box of P 1 . We get a new tableau T 2 as follows: In the algorithm of Definition 2.22 use P 1 := P 1 \ {(w + 1, w + 1)} and P 2 := P 2 \ {(w + 1, w + 2), (w + 2, w + 2)} instead of P 1 and P 2 , respectively. By Corollary 2.20, T 2 is m-amenable for m = 3. There is a 3 and no 2 in the (w + 1) th column. However, there are 2s and no 3s in the columns of the first two boxes of P 2 . Thus, by Lemma 2.19, 3-amenability follows. We get another tableau T 2 as follows: • Set T 2 (r, s) = T 2 (r, s) for every (r, s) ∈ P 1 ∪ (P 2 \ {(w, w + 2)}) where P 1 and P 2 as above.
• Fill the remaining boxes using the algorithm of Definition 2.22 starting with k = 3.
For d > 3, since the (y + 1) th column has the same entries in both tableaux, the algorithm fills the other d − 3 columns in the same amenable way. Clearly, the contents ofT 2 andT 2 are equal. .
Then we add two columns using the algorithm of Lemma 4.31: Proof. If λ, µ satisfy these properties then the diagram D ot λ/µ is equal to D α/β where α = [a , b , c , d ] and β = [w , 1] The number a is the number of boxes of the first row of D λ/µ and can be calculated by a = λ 1 λ/µ is not Q-multiplicity-free and, thus, Q λ/µ is not Q-multiplicity-free.
As we will see soon we have already determined all non-Q-multiplicity-free skew Schur Q-functions. The following proposition gives a list of all skew Schur Qfunctions that are possibly Q-multiplicity-free. This is half of the classification of Q-multiplicity-free skew Schur Q-functions. Proposition 4.34. Let λ, µ ∈ DP such that D λ/µ is basic. Let a, b, c, d, w, x, y ∈ N. If Q λ/µ is Q-multiplicity-free then λ and µ satisfy one of the following conditions: (i) λ is arbitrary and µ ∈ {∅, (1)}, Some of these cases overlap.
The cases (iii) -(vi) are depicted as diagrams in the remark after the proof of this proposition.

We want to note that Case (i) is the orthogonal transposition of Case (ii). Also, Case (iii) is the orthogonal transposition of Case (iv). Case (v) is the orthogonal transposition of Case (vi) for x > 1. The orthogonal transposition of Case (vi) for
x = 1 is also covered in Case (vi).
Proof. If µ = ∅, (1) we have no restrictions for λ. We also have no restrictions for  a, b, c, d, w, x, y ∈ N. Note that in the last case if w ≥ a+c then (µ) ≥ (λ) and the diagram D λ/µ is either not defined or is not basic since it has an empty column. Hence, we will only consider w ≤ a + c − 1.
To show that the list in Proposition 4.34 gives the classification of Q-multiplicityfree skew Schur Q-functions we have to prove the Q-multiplicity-freeness in each of these cases. We will do this in the following until we are able to state the classification result as Theorem 4.61.
The next lemma shows the Q-multiplicity-freeness of 4.34 (i).
If λ is arbitrary and µ = (1) then where E λ is the set from Definition 2.25. In particular, Q λ/µ is Q-multiplicity-free.
The case µ = (1) is a case of Proposition 2.26.
The Q-multiplicity-freeness as well as the decomposition of Q λ/µ is is proved in [9,Lemma 4.19].
Remark 4.41. In [8] there is already a proof for the statement that Q [a,1]/µ = Q ν for some ν. Using Lemma 2.31, this statement follows immediately and Lemma 4.40 helps to obtain this ν.
We postpone to prove the Q-multiplicity-freeness of 4.34 (iii). We will first show the Q-multiplicity-freeness of 4.34 (iv) and then prove that 4.34 (iii) is the orthogonally transposed version of 4.34 (iv). Proof. Let the diagram be shifted such that the uppermost leftmost box is (1, 1), the uppermost rightmost box is (1, a + b − w − 1) and the lowermost rightmost box is the box (a, a + b − w − 1). Let T be an amenable tableau of D.
Then the uppermost leftmost box is (1, 1), the uppermost rightmost box is (1, b), the lowermost leftmost box is (a, 1) and the lowermost rightmost box is (a, b). Let T (j) be the subtableau of T consisting of the boxes with their entries of the first j rows. We need to show that T (j) ∩ T (i) is a (j + 1 − i, b + 1 − i)-hook at (i, i) for 1 ≤ i ≤ min{b, j} where T (i) is as in Definition 2.12.
Then we have T (j, 1) > 1. Let t = T (j, 1). For t ∈ {j , j}, by Lemma 2.9, all boxes in the j th row are then filled with entries from {j , j}. The remark after Definition 2.7 implies that c (u) (T ) j = b ≥ c (u) (T ) j−1 ; a contradiction to Lemma 2.8.Thus, we have 1 < t < j . Then the last box of T (j−1) ∩ T (t) contains a |t| , for otherwise, by the remark after Definition 2.7, we have at least as many |t|s as (|t| − 1)s, which contradicts Lemma 2.8. We have |T (j, 2)| > |t|. Otherwise, we would have at least as many |t|s as (|t| − 1)s, which contradicts Lemma 2.8.
Repeating this argument, we get |T (j, s)| > |T (j, s − 1)| for 2 ≤ s ≤ r where r is such that T (j, r + 1) is the leftmost box with an entry that does not appear in the first (j − 1) th rows.
Let j be minimal with respect to this property. By Case 1.1, we may assume that v > 1. Let v be minimal with respect to this property. Then we may take T (j) , remove P 1 , P 2 , . . . , P v−1 , and replace each entry In this way, we get a tableau U of shape Using the same argument, P 2 is a (w + 1, a + b − w − 2)-hook at (2,2).
Repeating this argument, we find that all non-empty P i s are hooks at (i, i) and, therefore, the filling of the boxes of the first k rows of D in any amenable tableau T is unique up to marks. Then the difference between these two tableaux are marks since the content of the (a+ 1) th row and, therefore, the filling of this row up to marks is determined. Thus, there is a minimal k such that an entry k is in the lowermost row and there is a box (a, k) with entry k in T 1 , say, and with entry k in T 2 . Since the k in the (a + 1) th row must be in a column to the left of the k th column, we have k > 1. In which is a contradiction to Lemma 2.8. Thus, there is a k − 1 in the (a + 1) th row in a box to the left of the (k − 1) th column. If there is no k − 2 in the (a + 1) th row then  Q (9,8,6,4) +Q (9,8,6,3,1) +Q (9,8,5,4,1) +Q (9,8,5,3,2) +Q (9,7,6,4,1) +Q (9,7,6,3,2) +Q (9,7,5,4,2) .
Since f α βν = f α νβ , we need to look at amenable tableaux of shape D α/ν and content (c + 2, c + 1). The boxes with an entry from {2 , 2} form a border strip (in fact a rotated hook) where marks are determined. In every column with a box of this border strip there is a box filled with 2. To obtain an amenable tableau in each of these columns there must be a box filled with a 1. Above the uppermost box filled with a 1 there cannot be a box filled with a 1 . Otherwise, if w is the reading word of this tableau and the uppermost box filled with 1 is (x(j), y(j)) then c + 1 = m 2 ( (w) + j − 1) ≥ m 1 ( (w) + j − 1) and w j = 1; a contradiction to the amenability of the tableau.
Suppose we have two amenable tableaux T and T of shape D λ/ν . If there are boxes (x, y) such that T (x, y) ∈ {2 , 2} and T (x, y) ∈ {1 , 1} then one of these boxes is either the first or the last box of T (2) . But then there is a box (r, s) such that T (r, s) ∈ {1 , 1} and T (r, s) ∈ {2 , 2} is the last box or the first box of T (2) , respectively. Without loss of generality we may assume that (x, y) is the first box of T (2) . Then T (x − 1, y) = 1 and (x − 2, y) is not part of the diagram. Since T (x, y) ∈ {1 , 1}, we have T (x − 1, y) = 1 ; a contradiction to the fact that there cannot be a box filled with a 1 above the uppermost box filled with a 1.
Then, by Lemma 2.13, we have |T (a + 2, a + 2)| > k. Since (a, a + 1) ∈ D λ/µ we have k > 1. If k or k occur in the first a rows, it follows that m k (2n) ≥ m k−1 (2n); a contradiction to the amenability of T . Thus, k = j + 1, where j = min{a, b + 3}. This is only possible if there are at least three unmarked js, otherwise there is no amenable tableau with these properties. Then T (a+2, a+2) = k +1 = j +2 follows and T (a + 1, a + 1), T (a + 1, a + 2) and T (a + 2, a + 2) are unmarked. Additionally, each of the entries in the a th row is unmarked and, therefore, there is no other amenable tableau of the same content.
Since (a, a + 1) ∈ D λ/µ we have k > 1. If k or k occur in the first a rows it follows that T (a + 1, a + 1) = k − 1, otherwise m k (2n) ≥ m k−1 (2n); a contradiction to the amenability of T . Assume there are two different amenable tableaux T and T of D λ/µ of the same content such that |T (a+1, a+1)| = |T (a+1, a+1)| = k −1, |T (a + 1, a + 2)| = |T (a + 1, a + 2)| = k and |T (a + 2, a + 2)| = |T (a + 2, a + 2)| = k. It follows that these tableaux differ only by markings. Then there is some i such that T (y, z) = i , say, and T (y, z) = i. It follows that y = a since the entries in the other rows are determined. It also follows that there is an i in a box which is lower and to the left of (a, z). Thus, we have i ∈ {k − 1, k} and, therefore, k > 2. Then we have u < v < t. Assume there are two different amenable tableaux T and T of D λ/µ of the same content in which the boxes (a + 1, a + 1), (a + 1, a + 2) and (a + 2, a + 2) are filled as above. It follows that these tableaux differ only by markings. Then there is some i such that T (y, z) = i , say, and T (y, z) = i. It follows that y = a since the entries in the other rows are determined. It also follows that there is an i in a box which is lower and to the left of the box (a, z). The only possible case is that i ∈ {u, v, t}. Arguing as in the cases above, we see that for T we either have m t−1 (n) = m t (n) or m v−1 (n) = m v (n) or m u−1 (n) = m u (n). This contradicts Lemma 2.8.
Since f λ µν = f λ νµ , we just need to look at tableaux of shape D λ/ν and content µ = (3, 2, 1). By Lemma 2.8, all entries must be unmarked. Assume there are two different amenable tableaux T 1 , T 2 of D λ/ν with content µ for some ν ∈ DP . Thus, we get one tableau from the other by interchanging some entries in certain boxes.
Suppose the 3 is in one of these boxes. Let (a, x) be the upper corner (where x = a + b + c) and let (e, e) be the lower corner (where e = a + c). Since the 3 is the greatest entry it must be either in (a, x) or in (e, e). Thus, we have T 1 (a, x) = 3, say, and T 2 (e, e) = 3. Then, by Lemma 2.9 and since T 1 is amenable, we have a ≥ 3, and v < f . The remaining 1 must be in a box to the right and above (v, g). If T 1 (a − 1, x − 1) = T 2 (a − 1, x − 1) = 1 then T 1 (a, x − 1) = T 2 (a, x − 1) = 2 and both tableaux are the same; a contradiction. Thus, we have T 1 (a, x−1) = T 2 (a, x−1) = 1.
The remaining entries must be in two corners below (a, x − 1). However, there is only one corner (namely (e, e)), thus, there are no two different amenable tableaux such that T 1 (a, x) = T 2 (a, x) = 3. Therefore, we have T 1 (e, e) = T 2 (e, e) = 3.
With the same argument as above we see that T 1 (a, x − 1) = T 2 (a, x − 1) = 1.