As we have mentioned in the introduction, from the work of James [J2], we can see that Snhas the L00-property. However, in this subsection, we are going to use our method to show that S5 and S6 have the (L0, p)-property for all p. This method is different from James, but it gives us an idea that how to show a given group has (L0, p)-property.
Lemma 3.1.
Let V , W be simple FG-modules with dimF(V ) = 1. Then V ⊗FW is also a simple FG-module.
Lemma 3.2.
Let V , W be two K0G-modules (or KpG-modules) which afford character (or p-modular character) χV, χW respectively. Then the character (or p-modular character) of V ⊗K0W (or V ⊗KpW) is χV× χW.
Lemma 3.3.
Let µ be a partition of n. Then SµK
p is a reduction mod m of SµK
0. Proof.
Let t1, · · · ,tmbe all distinct standard µ-tableaux. Then by proposition 2.19,
SKµ
0 = K0et1⊕ · · · ⊕ K0etm.
Consider the RSn-submodule X = Ret1⊕ · · · ⊕ Retm. Then clearly X is a RSn-lattice of SµK
0,
and its reduction mod m is
X/mX ∼=KpSn Kp⊗RX ∼=KpSn Kpet1⊕ · · · ⊕ Kpetm= SµK
p.
So SµKp is a reduction mod m of SµK
0.
Proposition 3.4.
(1) If λ is a p-regular partition of n, then the character of SλK0 has the (L00, p)-property.
(2) Let µ be a partition of n. If the character of SµK
0 has the (L0, p)-property, then so does SKµ0
0. Remark.
By revising the proof of the proposition 3.4 (2), it is easy to see we can replace the term “(L0, p)-property” to “(L00, p)-property” in (2).
Proof.
(1) Let χ be the character of SλK
0, and let φµ be the p-modular character of DµKp for any p-regular partition µ of n. Then by proposition 0.10 and lemma 3.3, χ|
S(p)n is the p-modular character of SKµ
p. Now by proposition 2.24, since λ is p-regular, DλK
p occurs precisely once in the composition factors of SλKp, and the others (if exists) have the form DµK
p with µ . λ . Thus
χ |S(p)n ≥ φλ with multiplicity 1,
i.e. χ has the (L00, p)-property.
(2) Let χλ, χλ0, and χ1be characters of SλK0, SλK00, and S(1
(n))
K0 respectively. Then by proposi-tion 0.10 and lemma 3.3, χλ|
p respectively. Moreover, let φ∗ be the p-modualr character of (SλK0
p)∗. Then by lemma 2.21, since every simple KpSn-modules are self dual, all composition factors of SλK0
Kp by proposition 2.25, we obtain φ∗= χλ|
Now by the assumption, since the character of SλK
0 has the (L0, p)-property,
S(p)n are irreducible p-modular characters for all i.
To show it, observe that since S(1
(n)) Kp
∼=KpSn D(1
(n)) Kp , χ1|
S(p)n is the p-modular character
(1(n))
the p-modular character of Di⊗KpD(1
(n)) Kp is
φi× χ1|
S(p)n .
Since the dimension of D(1K(n))
p is 1, the KpSn-module Di⊗KpD(1K(n))
p is simple by lemma 3.1, and hence φi× χ1|
S(p)n is an irreducible p-modular character as we desired.
Proposition 3.5.
Let M be a simple K0G-module and let X be a RG-lattice. If the dimension of M is divisible by the largest power of p dividing the order of G, then X /mX is a simple KpG-module.
Proof.
Cf., for example, [S1] proposition 46, page 136.
Example 3.6.
We are going to show that S5has the (L0, p)-property. First observe that all partitions of S5are:
(5), (4, 1), (3, 2), (3, 1, 1),
(1, 1, 1, 1, 1), (2, 1, 1, 1), (2, 2, 1).
For the case p > 2, since (5), (4, 1), (3, 2), and (3, 1, 1) are p-regular, by proposition 3.4, we conclude S5has the (L0, p)-property for p > 2.
For the case p = 2, observe that all 2-regular partitions are
(5), (4, 1), (3, 2),
and their conjugate partitions are
(1, 1, 1, 1, 1), (2, 1, 1, 1), (2, 2, 1)
respectively. So again by proposition 3.4, all characters of their corresponding Specht modules have the (L0, 2)-property. So it remains to show the character of S(3,1,1)K
0 has the (L0, 2)-property.
To show this, we want to find dimensions of S(3,1,1)K
0 , D(5)K
2, D(4,1)K
2 , and D(3,2)K
2 . Observe that the number of all standard (3, 1, 1)-tableaux is 4!
2! 2! = 6. So by proposition 2.19,
dimK0(S(3,1,1)K
0 ) = 6.
On the other hand, since D(5)K
2 is the trivial K2S5-module, its dimension is 1, and by using proposition 2.22, we can calculate the dimensions of D(4,1)K
2 and D(3,2)K
Now we show the character of SK(3,1,1)
0 has the (L0, 2)-property. Let χ be the
we have
6 = χ|
S(2)5 (e) = aφ(5)(e) + bφ(4,1)(e) + cφ(3,2)(e) = a + 4b + 4c.
Moreover, since a, b, c ∈ N ∪ {0}, the equation 6 = a + 4b + 4c implies that
a= 2 or 6.
Therefore, to show the character χ of S(3,1,1)K
0 has the (L0, 2)-property, it suffices to show a6= 6.
Assume a = 6. Denote g = (123)(4)(5) ∈ S5, then
χ (g) = χ |
S(2)5 (g) = 6φ(5)(g) = 6.
Let r1, · · · , r6∈ K0be all eigenvalues of the matrix representation ρg of g with respect to any fixed basis of S(3,1,1)K0 . Then since ρg3is the identity matrix, every riare roots of unity in K0. Since χ(g) = 6, we have
r1+ r2+ · · · + r6= 6,
and hence ri= 1 for all i. This means ρg is the identity matrix, i.e. g · et = et for all (3, 1, 1)-tableaux t. But this is impossible clearly. Therefore a 6= 6, and hence χ has the (L0, 2)-property.
We therefore conclude that S5 has the (L0, 2)-property, and hence S5 has the (L0, p)-property for all prime p.
Remark.
In fact, this method can show that S5 has the (L00, p)-property. Indeed, we can use proposition 3.4 to show that all irreducible characters of S5 have (L00, p)-property for all p, except the case that the irreducible character χ of S(3,1,1)K
0 has (L00, 2)-property. However, if we use same notations and the same method as in the above example, we will obtain 6 = χ|
S(2)5 (e) = a + 4b + 4c, and we have shown that a = 2. This means b = 1 or c = 1, and hence χ has (L00, 2)-property.
Example 3.7.
We are going to show that S6has the (L0, p)-property. Observe that all partitions of S6 are
(6), (5, 1), (4, 2), (4, 1, 1), (3, 3), (3, 2, 1),
(1, 1, 1, 1, 1, 1), (2, 1, 1, 1, 1), (2, 2, 1, 1), (3, 1, 1, 1), (2, 2, 2).
For the case p > 2, since (6), (5, 1), (4, 2), (4, 1, 1), (3, 3), (3, 2, 1) are p-regular, we conclude that S6has (L0, p)-property for p > 2 by proposition 3.4. It is the same method as in example 3.6.
For the case p = 2, observe that all 2-regular partitions are
(6), (5, 1), (4, 2), (3, 2, 1),
and their conjugate partitions are
(1, 1, 1, 1, 1, 1), (2, 1, 1, 1, 1), (2, 2, 1, 1), (3, 2, 1)
respectively. Again by proposition 3.4, all characters of their corresponding Specht mod-ules have the (L0, 2)-property. To show the other characters have the (L0, 2)-property, by corollary 3.4 (2), it suffices to show characters of S(4,1,1)K
0 and S(3,3)K
0 have the (L0,
2)-Again, we calculate dimensions of S(4,1,1)K
0 , SK(3,3)
0 , and all simple KpS6-modules. Ob-serve that the number of all standard (4, 1, 1)-tableaux is 5!
3! 2! = 10, and all standard
On the other hand, since the dimension of D(6)K
2 is 1 since it is the trivial K2S5-module, and by using proposition 2.22, we can calculate dimensios of D(5,1)K
2 and D(4,2)K
To see the dimension of D(3,2,1)K
2 , of course we can use proposition 2.22 to calculate. How-ever, it is a little complicate. So we want to use another way to find its dimension. Observe that the number of all standard (3, 2, 1)-tableaux is 16. So the dimension of S(3,2,1)K0 is 16.
Since |S6| = 16 × 45, by proposition 3.5, any reduction mod m of S(3,2,1)K
0 is a simple K2S6 -module. Moreover, by lemma 3.3, we know S(3,2,1)K
2 is a reduction mod m of S(3,2,1)K
Now we show that characters of S(4,1,1)K
0 and S(3,3)K
0 have the (L0, 2)-property. Let χ(4,1,1), χ(3,3) be characters of S(4,1,1)K
0 , S(3,3)K
0 respectively, and let φ(6), φ(5,1), φ(4,2), φ(3,2,1) be 2-modular characters of D(6)K
2, D(5,1)K
2 , D(4,2)K
2 , D(3,2,1)K
2 respectively. Then we
can write
χ(4,1,1)|
S(2)6 = a1φ(6)+ a2φ(5,1)+ a3φ(4,2)+ a4φ(3,2,1) χ(3,3)|
S(2)6 = b1φ(6)+ b2φ(5,1)+ b3φ(4,2)+ b4φ(3,2,1)
for some ai, bi∈ N ∪ {0}. Let e be the identity of S5. Then since
dimK0(S(4,1,1)K
0 ) = 10, dimK0(S(3,3)K
0 ) = 5, dimK2(D(6)K
2) = 1, dimK2(D(5,1)K
2 ) = 4, dimK2(D(4,2)K
2 ) = 4, dimK2(D(3,2,1)K
2 ) = 16,
we have
10 = χ(4,1,1)|
S(2)6 (e) = a1φ(6)(e) + a2φ(5,1)(e) + a3φ(4,2)(e) + a4φ(3,2,1)(e)
= a1+ 4a2+ 4a3+ 16a4,
and
5 = χ(3,3)|
S(2)6 (e) = b1φ(6)(e) + b2φ(5,1)(e) + b3φ(4,2)(e) + b4φ(3,2,1)(e)
= b1+ 4b2+ 4b3+ 16b4.
Since ai, bi∈ N ∪ {0}, the above equations imply that
a1= 2, 6, or 10, and b1= 1 or 5
Therefore, to show χ(4,1,1)and χ(3,3)have the (L0, 2)-property, it suffices to show a16= 10 and b16= 5.
in example 3.6, χ(4,1,1)(g) = 10 will imply that g · et = et for all (4, 1, 1)-tableaux t. But clearly this is impossible. So a16= 10. Using the same method, we can also conclude that b16= 5. Hence characters of S(4,1,1)K
0 and S(3,3)K
0 have the (L0, 2)-property.
We therefore concldue that S6 has the (L0, 2)-property, and hence S6 has the (L0, p)-property for all prime p.
Remark.
Let λ be a partition of n such that it is not 2-regular or their conjugate partitions. Let χ be an irreducible character of SλK0, and φ be the trivial 2-modular character of Sn. In the above two examples and in the case p = 2, we calculate the dimension of SλK
0 and dimension of all distinct simple K2Sn-modules, and using them to show χ|
Sn(2)≥ φ . Then by a little work, we conclude that
χ |S(2)n 6= χ(e)φ ,
where e is the identity of Sn, and this implies that χ has the (L0, 2)-property.
However, this method does not always work. For example, consider n = 7. The dimension of S(4,1,1,1)K
0 is 20, and dimensions of all distinct 2-modular characters of S7are 1, 6, 8, 14, 20. The equation
20 = a1+ 6a2+ 8a3+ 14a4+ 20a5, ai∈ N ∪ {0}
can not guarantee that there exists i such that ai> 0.
Even though the method does not always work, it induce we think about a question, that is, for any character χ of G, can we find a p-modular character φ of G such that either χ|G(p)= φ if χ(e) = φ (e), or χ|G(p) ≥ φ with multiplicity smaller than χ(e)/φ (e) if χ (e) 6= φ (e)? In theorem 3.17, we will see that for any prime p and for any character χ of
Sn, there always exists a p-modular character φ of G such that χ|
S(p)n ≥ φ with multiplicity 1.
3.2 Diagrams
In this subsection, we will introduce some tools which will be used to show theorem 3.17, that is, for any prime p and for any irreducible character χ of Sn, there always exists an irreducible p-modular character φ of G such that χ|
S(p)n ≥ φ with multiplicity 1.
Definition 3.8.
(1) Consider a fixed origin, and a first axis pointing south and a second axis pointing east.
These axes construct a coordinate system, and we define vertices to be elements of {(i, j) | i, j ∈ N}.
We say a vertex (i, j) is higher than (k, l) if i < k. Similary define “lower than”, “to the right of”, and “to the left of”.
(2) We say D is a diagram (for Sn) if D is a set which collects n vertices such that if (i, j) ∈ D, then (i, j − 1), (i − 1, j) ∈ D. The vertices which belong to D is called the nodes of D.
Let µ = (µ1, · · · , µm) be a partition of n and denote
Dµ = {(i, j) : vertex | 1 ≤ i ≤ m, 1 ≤ j ≤ µi}.
Then it is easy to see Dµ is a diagram, and for any diagram D, there is an unique partition µ such that D = Dµ. In this case, we say the diagram D corresponds to µ, or µ corresponds to D.
Let a be a fixed number such that the set {(a, j) | if (a, j) ∈ D} is not empty. Then the set is called the a-th row of D, and its cardinality is called the length of the a-th row.
Let a be a fixed number such that the set {( j, a) | if ( j, a) ∈ D} is not empty. Then the set is called the a-th column of D, and its cardinality is called the length of the a-th column.
We say a diagram is p-regular if no p rows of it have the same length; otherwise the diagram is called p-singular.
For two diagrams D1, D2, we say D1D D2if for every j,
i≤ j∑
(length of the i-th row of D1) ≥ ∑
i≤ j
(length of the i-th row of D2)
We say D1. D2if D1D D2and D16= D2.
(3) If D is a diagram for Snwhich corresponds to the partition µ of n, then we denote χD as the character of SµK0.
If D is a p-regular diagram for Snwhich corresponding to the p-regular partition µ of n, then we denote φDas the p-modular character of DµKp.
Definition 3.9.
(1) A ladder is a straight line joining the vertex (i, 1) to the point (1, i− 1
p− 1+ 1). The vertices which a ladder passes through will be called the rungs of the ladder.
Note that the rungs of a ladder are (i, 1), (i − (p − 1), 2), (i − 2(p − 1), 3), and so on.
For example, let p = 3 and consider the ladder passing through (6, 1). Then the rungs of the ladder are (6, 1), (4, 2), (2, 3).
(2) A subset of the rungs of a ladder is called a complete k subset if it consists of the top k rungs of the ladder. We say a vertex x is the 1-st rung of a ladder l if x belongs to the complete 1 subset, and x is the k-th (k > 1) rung of l if x belongs to the complete ksubset of l but not in the complete k − 1 subset of l.
For example, let p = 2 and consider the ladder passing through (3, 1), (2, 2), (1, 3).
Then
{(1, 3)} is the complete 1 subset, {(2, 2), (1, 3)} is the complete 2 subset, {(3, 1), (2, 2), (1, 3)} is the complete 3 subset, and
{(3, 1), (2, 2)} is not a complete k subset for any k ∈ N.
The vertex (1, 3) is the 1-st rung of l, (2, 2) is the 2-nd, and (3, 1) is the 3-rd.
(3) Suppose we have p colors, which we shall call 0, 1, · · · , p − 1, Color the vertices by letting (i, j) have the color which is the smallest non-negative residue of j − i mod p.
We say two diagrams D1, D2 belong to the same block if and only if they have the same color content, that is, for every i, the number of nodes of D1colored i is equal to the number of nodes of D2colored i.
Proposition 3.10.
(1) All the rungs of a ladder have the same color.
(2) A diagram D is p-regular if and only if the following happens for each ladder l:
If the k-th rung of l belongs to D, then so does the k − 1-th rung of l.
(3) Let D1 and D2 be two diagrams correspending to partitions µ1and µ2 respectively.
Then D1and D2belong to the same block if and only if χµ1 and χµ2 lies in the same p-block, that is, there exists an irreducible p-modular character φ such that
χµ1|
S(p)n ≥ φ and χµ2|
S(p)n ≥ φ . Proof.
(1) let l be a ladder and let (a, b) be a rung of l. Then other rungs of l can be written as the form (a − c(p − 1), b + c) for some integer c. Clearly, (a − c(p − 1), b + c) and (a, b) have the same color.
(2) Assume D is p-singular. Then there are p rows of D, say a+1-th, a+2-th,· · · ,a+ p-th rows, such that they have the same length, say m. So
(a + p, m) ∈ D and (a + 1, m + 1) 6∈ D.
If we say (a + p, m) is the k-th rung of a ladder l, then the k − 1-th rung of l is (a + p − (p − 1), m + 1) = (a + 1, m + 1). So the k-th rung of l belong to D, but the k− 1-th does not.
Assume D is p-regular. Let l be a ladder and its k-th rung (k > 1) is (i, j). Then the k− 1-th rung of l is (i − (p − 1), j + 1). Let m be the length of the i-th row. Then since D is p-regular, the number of rows with length m is < p. So the length of the i− (p − 1)-th row must be ≥ m + 1. Hence (i − (p − 1), j + 1) ∈ D, i.e. the k − 1-th rung of l belongs to D.
(3) To show this, we have to introduce “p-hook” and “p-core”. A p-hook of a diagram D is a connected part with p-nodes of the rim of D beginning with the last node of
any row and ending with the last node of an earlier column, and these nodes can be removed to leave a diagram. For example, consider following four diagrams. In (a) and (b), the nodes which are connected by lines can be 5-hooks of D(4,4,3). In (c) and (d), the nodes which are connected by lines can not be 5-hooks of D(4,4,3)
(a) (b) (c) (d)
Moreover, we say a diagram is a p-core of D if we remove p-hooks in succession from D as many as we can. For example,
The last diagram is the 5-core of D(4,4,3). Note that a p-core of a diagram is unique (cf., for example, [R] 4.53, page 85).
Now to see (3), we have to introduce two theorems:
Theorem. (Nakayama’s Conjecture)
Let D1, D2 be two diagrams for Sn. Then χD1, χD2 are in the same p-block if and only if D1and D2have the same p-core.
Proof. Cf., for example, [R], 5.36, page 98.
Theorem.
The p-cores of two diagrams are equal if and only if they are in the same block.
Proof. Cf., for example, [R], 5.42, page 99.
So we can see that (3) holds by this two theorem.
Definition 3.11.
Let D be a diragram. Construct a new set Dr from D as follows. For each ladder l, if lhits D in k nodes, replace these nodes by the complete k subset of l.
For example, let p = 3.
If D =
· · ·
· ·
· ·
· ·
, then Dr=
· · ·
· · ·
· ·
· .
Proposition 3.12.
(1) Dr is a p-regular diagram.
(2) D and Dr belong to the same block.
Proof.
(1) We are going to show Dr is a diagram. Assume x = (i, j) ∈ Dr. Let l be the ladder in which x lies, and say x is the k-th rung of l. First we claim that (i − 1, j) (if i > 1) belong to Dr. Let l1be the ladder in which (i − 1, j) lie. Observe that the 1-st rung of l is
(i − (k − 1)(p − 1), j + (k − 1)).
Assume
i− (k − 1)(p − 1) = 1.
Then the 1-st rung of l1is
(i − (k − 2)(p − 1) − 1, j + (k − 2)).
So (i − 1, j) is the k − 1-th rung of l1. Therefore, to show (i − 1, j) ∈ Dr in this case, we have to find k − 1 nodes of D lie in l1.
Since x is the k-th rung of l, we can find k rungs of l belong to D, say
(a, b), (a − c2(p − 1), b + c2), · · · , (a − ck(p − 1), b + ck)
with ci∈ N and c2< · · · < ck. Note that for convinence, we denote c1= 0. So for any z= 1, · · · , k − 1,
a− cz(p − 1) > a − ck(p − 1) ≥ i − (k − 1)(p − 1) = 1.
Now since D is a diagram, by the above inequality, we can guarantee that
(a − 1, b), (a − c2(p − 1) − 1, b + c2), · · · , (a − ck−1(p − 1) − 1, b + ck−1) ∈ D,
and since these nodes lie in l1, we conclude that l1hits D in at least k − 1 nodes, and hence (i − 1, j) ∈ Dr.
On the other hand, assume
i− (k − 1)(p − 1) > 1.
Then the 1-st rung of l1is
(i − (k − 1)(p − 1) − 1, j + (k − 1)).
in l1. Observe that for any z = 1, · · · , k,
a− cz(p − 1) ≥ i − (k − 1)(p − 1) > 1.
So we can guarantee that
(a − 1, b), (a − c2(p − 1) − 1, b + c2), · · · , (a − ck(p − 1) − 1, b + ck) ∈ D,
and hence l1hits D in at least k nodes, i.e. (i − 1, j) ∈ Dr.
Next we claim that (i, j − 1) (if j > 1) belong to Dr. Let l2 be the ladder in which (i, j − 1) lies. Since
(i − (k − 1)(p − 1), j + (k − 1))
is the 1-st rung of l, the 1-st rung of l2is
(i − (k − 1)(p − 1), j + (k − 1) − 1).
So (i, j − 1) is the k-th rung of l2. Hence to show (i, j − 1) ∈ Dr, we have to find k nodes of D lies in l2.
Assume b > 1. Then since
(a, b − 1), (a − c2(p − 1), b + c2− 1), · · · , (a − ck(p − 1), b + ck− 1) ∈ D
and they are all lie in l2, we can see that l2hits D in at least k nodes, i.e. (i, j − 1) ∈ Dr. Assume b = 1. Observe that since 1 + cz≤ ckfor all z = 1, · · · , k − 1, we can guarantee
that the following k − 1 rungs of l2exist and belong to D:
(a − (p − 1), b), (a − (1 + c2)(p − 1), b + c2), · · · , (a − (1 + ck−1)(p − 1), b + ck−1).
If there is an i between 2 and k such that 1 + ci−1< ci, then the rung of l2
(a − ci(p − 1), (b + ci) − 1) ∈ D
is distinct to the above k − 1 rungs. So l2hits D in at least k nodes, i.e. (i, j − 1) ∈ Dr. If we can not find such i, then it means
1 + ci−1= cifor all i = 2, · · · , k.
Note that if (a − ck(p − 1), b + ck) is not the 1-st rung of l, then the rung of l2
(a − (1 + ck)(p − 1), b + ck)
exists and belongs to D. Hence l2hits D in at least k nodes, i.e. (i, j − 1) ∈ Dr. So it remains to consider the case
1 + ci−1= cifor all i = 2, · · · , k and (a − ck(p − 1), b + ck) is the 1-st rung of l.
However, in this case, we have c2= 1, c3= 2, · · · , ck= k − 1. So the 1-st rung of l is
(a − ck(p − 1), b + ck) = (a − (k − 1)(p − 1), 1 + (k − 1)) = (a − (k − 1)(p − 1), k),
and hence the k-th rung of l is (a, 1), which means j = 1, which is impossible. Now
we complete the proof of Dr is a diagram.
Finally, to see Dris p-regular, it is clearly by proposition 3.10 (2).
(2) It is clearly by proposition 3.10 (1).
Definition 3.13.
(1) Let D be a diagram for Sn and x ∈ D. If D \ {x} is a diagram for Sn−1, then x is called a removable node of D, and write D − x for D \ {x}. The node x is called regular-removable if D − x is a p-regular diagram for Sn−1.
(2) The node x of D is called a shadow node if x is regular-removable and no node higher than or equal to x can be raised, retaining its color (the term “raised” means if y is removable in D and there is a vertex z which is higher than y and it can be added to D− y to give a diagram, then we say y is raised to z).
We say D − x is a shadow of D if x is a shadow node of D.
Proposition 3.14.
(1) A diagram D has a shadow node if and only if D is p-regular.
(2) Let D1 and D2are p-regular diagrams in the same block and x, y are shadow nodes of D1, D2respectively, If D1, D2are distinct, then D1− x 6= D2− y.
Proof.
(1) Assume D is p-regular. Consider the longest ladder l which hits D. Then since no nodes of D are to the right of this ladder or lower than this ladder, all rungs of l in D are removable. Let x be the lowest rung of l in D. Denote m as the length of the row
in which x lies. Since x is the lowest rung of l, the number of rows of D with length
= m − 1 is < p − 1. So the number of rows of D − x with length = m − 1 is < p.
Therefore D − x is p-regular since D is p-regular, and hence x is regular-removable.
Now we claim that x is a shadow node of D.
Observe that all removable nodes higher than or equal to x are rungs of l higher than or equal to x, and so have the same color as x. Let y be one rung of them and we want to raise y. Then positions which y can be raised to are only such vertices which higher than y and below a rung of l, or to the right of the most right node in the first row of D. If we raise y to z which below a rung t of l, then z can not have the same color as ysince the color of t is the same as y. In the other case, we raise y to z which is to the right of the most right node in the first row of D. Denote (i, j) as the highest rung of l. Then no any node in D can righter than (i, j), and since D is p-regular, we have i< p. Since all rungs of l have the same color, the color of y is j − i (mod p), which is not equal to j (mod p) because i < p. Now since no any node in D can righter than (i, j), the coordinate of z is (1, j + 1), and hence its color is j (mod p), which is different than y. Hence y can not be raised, rataining its color, i.e. x is a shadow node of D.
Assume D is p-singular, and x is a regular-removable node of D. Let (i, j) be the coordinate of x. Then it means x lies in the i-th row of D whose length is j. Since D is p-singular and x is regular-removable, the number of rows of D with length j is p.
So i ≥ p and lengths of i-th, i − 1-th,· · · , i − (p − 1)-th rows are all j. If i = p, then xcan be raised to (1, j + 1) = (i − (p − 1), j + 1). If i > p, then since the number of rows of D with length j is p, the length of the i − p-th row is > j, and hence x can be raised to (i − (p − 1), j + 1). In both cases, we all conclude that x can be raised
to (i − (p − 1), j + 1), whose color is j − i (mod p). So x can be raised, retaining its color, i.e. x is not a shadow node of D.
(2) If D1− x = D2− y, then x, y have the same color since D1, D2are in the same block.
Since D16= D2and D1−x = D2−y, x can not as high as y. Without loss of generality, we may assume y is highter than x. Then it means x ∈ D1can be raised to y retaining its color, which is impossible since x is a shadow node. Therefore D1− x 6= D2− y.
Definition 3.15.
Let D be a p-regular diagram, and let l be the longest ladder which hits D. Let x be the lowest rung of l in D. In the proof of proposition 3.14 (1), we see that x is a shadow node of D. The shadow node x is called the first shadow node of D, and D − x is called the first shadow of D.