4. FUNCIONES DE LIBRERÍA
4.5. Funciones que actúan sobre matrices
Given any two Young diagrams, and , either 1. and _ are inseparable.
2. _ and are inseparable.
3. =.
It therefore follows that if and _ are separable and
6
=, then _ and are
inseparable.
Proof.
Given any two Young diagrams, either case 3 holds or > or< . By Lemma 4.5.4, if > , then case 1 holds and if < , then case 2 holds.
The following lemma is at the crux of proving that the quasi-idempotents are orthogonal. Roughly speaking, the lemma says that if two strings leave a box labelled ai and arrive at a box labelledbj then, whatever happens in between,
then the braid combination is equal to 0 as an element of the Hecke algebra.
4.5.6 Lemma.
Given any two Young diagrams, 6=with jj=jj, if and are inseparable
then
E(a)HnE(b) = 0
and
E(b)HnE(a) = 0:
Proof.
We will show that E(a)HnE(b) = 0. The proof of the secondstatement is similar.
Since Hn is spanned by positive permutation braids, it is enough to show
that E(a)!E(b) = 0 for an arbitrary positive permutation braid, !.
Since and are inseparable, there are two cells in some row of , the lth say, which are sent to two cells in the same row of , thepth say, by !.
Suppose that the two cells in the lth row of are not adjacent. We can nd a transposition, , for which !! sends two adjacent cells of the lth row of
to the pth row of. Now preserves the rows of i.e. 2R(). In fact only
permutes strings in the lth row of . Hence al! =(!)al. Therefore, at the
expense of a scalar, we can assume that the two cells are an adjacent pair,i and
i+ 1 say.
Similarly, we can suppose the two cells inare an adjacent pair j and j+ 1, again at the expense of some scalar. Note that
i! =!j: (4.1)
By Proposition 4.4.5 we know that
al =a(i) l(1 a 1 i) = a 1a (i) l(i a); and bp = (1 b 1 j)b(j) p = b 1( j b)b(j) p :
Hence E(a)!E(b) has the term (i a)!(j b) in its expression. This is
demonstrated pictorially in Figure 4.3. By equation 4.1 (i a)! =!(j a),
therefore, we can rewrite this subexpression as !(j a)(j b).
Now (j a)(j b) = 02Hn, therefore, E(a)HnE(b) = 0.
4.5.7 Lemma.
Let be a permutation which separates and _. We can nd permutations
1 2R() and 2 2R( _) for which 12 =.
Proof.
Recall that j is the number of rows in _ with at least j cells.Since separates and _, exactly one cell from the jth row of must be sent
to each of the rst j rows of_.
µ b( j )p ( ) aλil π ω i b a - - j σ σ = ( ) aλl i a - σj σj-b µ b( j )p ωπ Figure 4.3: A subexpression of E(a)!E(b).
(This can be seen by considering each row of in turn. There are exactly 1
rows in _. Exactly one cell from the rst row of must be sent to each row
of _ by , since separates. Only the rows of _ with more than one cell will
have spaces left after we have dealt with the image of the rst row of under . The number of such rows is exactly 2 and so forth.)
We can nd a permutation 1;j
2 R() which reorders the cells in the jth
row of, so that , for 1ij, the ith cell of the row is sent to the ith row of
_ by 1;j.
Note that, since 1;j only permutes cells in the jth row of , all the cycles
in 1;j are disjoint from those in 1;j 0 if j
6
= j0 and so they commute. Set
1 = Q
kj=11;j. Then 1
2R().
Similarly, for the jth row of _, we can nd a permutation 2;j
2 R( _),
which reorders the cells in the jth row of _, so that the ith cell is the image of
a cell in theith row of, under 12;j. Let 2 = Q 1 j=1 2;j 2R( _). Then 12
sends the ith cell of thejth row of to the jth cell of the ith row of _, i.e. it
interchanges rows and columns. Hence =12 as required.
4.5.8 Example.
In this example, we show how to calculate1 and 2 for the Young diagram .
Recall that = . The permutation =
1234567 7635142 separates and _. 2 3 4 5 6 7 1 ! 5 7 3 6 4 2 1
Note that sends the third cell of the rst row of to the rst row of _
whereas sends the rst cell of the rst row to the rst row of _. Therefore
1;1 must send 1 to 3. By similar arguments we nd that 1;1 =
1234567 3421567
and
1;2 =1;3 = 1. Hence 1 =1;1 and 1 is as follows,
2 3 4 5 6 7 1 1 ! 4 3 1 2 5 6 7 ! 5 7 1 6 2 3 4 :
(Note that the permutations refer to the position of the cell in the diagram rather than the labels the cells carry.)
We then nd that 2;1 = 1234567 2314567 and 2;2 = 1234567 1235467 . Noting that 2;3 =2;4 = 1, we have that 2 =
1234567 2315467 . 5 7 1 6 2 3 4 2 ! 1 5 7 2 6 3 4 Therefore, 12 = (1324)(172645)(123)(45) = (247365) = :
4.5.9 Corollary.
With, , 1 and 2 as dened in Lemma 4.5.7, we have that
! =! 1 1 !!
1 2 :
Proof.
From Lemma 4.5.7, we have that = 1 1 1 2 . We must prove that ! 1 1 !! 12 is a positive permutation braid. We know that ! 1 1 2 R(), ! 1 2 2 R(
_) and they are both positive permutation braids. Note that !
doesn't cross strings belonging to the same row of or the same row of _.
Suppose ! 1 1
crosses the ith and the jth string. Since 1 1
2R(), the ith and
jth string must belong to the same row of . The braid! does not cross these
two strings since it doesn't cross any two strings in the same row of . Since
! separates and _, the ith and jth strings will end up in dierent rows of
_. Since !
1 2
2 R(
_) it will not cross over two strings in dierent rows of
_. Similar arguments show that if one of the three braids, !
1 1 , !
1
2 and !
crosses a pair of strings, neither of the other two braids will. Therefore, since 76
the three braids are positive permutation braids, ! 1 1 !!
1
2 is a braid which
crosses no two strings more than once and all the crossings have positive sign. Hence ! and ! 1
1 !! 1
2 are two positive permutation braids which represent
the same permutation and must, therefore, be the same braid.
4.5.10 Proof of Theorem 4.4.11.
We will rst show that e is quasi-idempotent.
We know that e2 =E(a)hE_(b)! 1 _ where h=!E_(b)! 1 _E(a)!:
We can express h as a linear combination of positive permutation braids,
h= X 2Sn !: Then e2 = X 2Sn E(a)!E_(b)! 1 :
A similar argument to Lemma 4.5.6 shows that if doesn't separate and _,
then E(a)!E_(b) = 0. Hence, we need only consider those which separate
and _. In this case we can write !
= ! 1 1 !! 1 2 for some 1 2 R() and 2 2R( _), by Lemma 4.5.7. Now, by Theorem 4.4.7 E(a)! 1 1 =b(! 1 1 )E(a) and ! 1 2 E(b) = a(! 1 2 )E(b): Therefore, if we set () = b(! 1 1 )a(! 1 2 ) and = X 2Sn () we have thate2 =e as required.
Suppose now that 6=. We can see readily thatee= 0. For
ee =E(a)!E_(b)! 1 _E(a)!E _(b)! 1 _: 77
Since 6= , either and
_ are inseparable or _ and are inseparable by
Corollary 4.5.5. If _ and are inseparable, then by Lemma 4.5.6,
E_(b)(! 1
_)E(a) = 0:
If and _ are inseparable, then by Lemma 4.5.6,
E(a) !E_(b)! 1 _E(a)! E_(b) = 0:
Hence, if 6=, then e and e are orthogonal as required.
4.5.11 Comment.
The orthogonality of the quasi-idempotents is proved in [Y, Proposition 2.9], also using the fact that a symmetriser and and anti-symmetriser must be joined by two strings.