Labelling of basis elements

I.2.6. Faithfulness of the graphical representation I.2.6.2 Definition. Let a(j) = a(j^(m))a(j^(m−1)) ⋯ a(j⁽¹⁾)a(j⁽⁰⁾) be a normal form monomial.

i) We call j<sup>(`)</sup> the`-th block ofj.

ii) A string of indices of maximal length of the formis∈j⁽⁰⁾, is+1∈j⁽¹⁾, is+2∈j⁽²⁾, . . . (moduloN) is called the s-th strand ofj.

I.2.6.3 Example. LetN =6, and consider Example I.2.5.5 once again, where j= (1 5 0 2 3 4 5 1 2 3 4 0 1 2 3 5 0 1 2 4).

The blocks arej⁽⁰⁾= (0124),j⁽¹⁾= (1235),j⁽²⁾= (2340),j⁽³⁾= (3451),j⁽⁴⁾= (502), and j⁽⁵⁾= (1). The strands are[3210],[54321],[105432]and[21054]. In particular, strands (and blocks) can have different lengths, but the longest strand has lengthm=6. ◊ Each monomial a(j) ∈ nTL̂N determines two sets

i

ⁱⁿj ,

i

^outj and an integer `_j ∈ Z_≥0 as follows:

i

ⁱⁿj = {i∈ {0,1, . . . , N−1} ∣no i−1 to the right ofiinj} (I.2.2)

i

^outj = {i∈ {0,1, . . . , N−1} ∣no i+1 to the left ofiinj}

`_j = the number of zeros inj.

These are well defined because, as in the proof of Lemma I.2.5.6, any element of nTL̂N is uniquely determined by the number of factorsai and the relative position of eachai and ai±1, for all i. The set iⁱⁿ_j equals the underlying set ofj⁽⁰⁾ in the normal form from the algorithm above. All strands ofj begin with an element in

i

ⁱⁿj and end with an element from

i

^outj .

The goal of this subsection is to show I.2.6.4 Proposition. The mapping

ψ∶ {a(j) ∈nTL̂_N in normal form} → PN ×PN ×Z_≥0 (I.2.3) a(j) ↦ (

i

ⁱⁿj ,

i

^outj , `j),

is injective, wherePN is the power set of {0,1, . . . , N−1}.

I.2.6.5 Remark. i) The map ψ is defined so that in the graphical description of the representation V of nTL̂_N, the set

i

ⁱⁿ_j equals the set of positions where a(j) expects a particle to be. The set

i

^outj equals the set of positions wherea(j)moves the particles from

i

ⁱⁿj , but each one is translated by 1, that is,

a(j) applied to a particle ati∈

i

ⁱⁿj gives a particle at j+1 for some j∈

i

^outj .

ii) The mapψ is far from being surjective. An obvious constraint is that∣

i

ⁱⁿj ∣ = ∣

i

^outj ∣, and furthermore, for some pairs (

i

ⁱⁿj ,

i

^outj ), one can only obtain sufficiently large

values `j. ◊

We start by proving injectivity of the restriction ψ0 of ψ to those monomials a(j) in normal form whose first element i₁ of j⁽⁰⁾ is 0. The proof itself will amount to counting indices.

I.2.6.6 Proposition. The map

ψ0∶ {a(j) ∈nTL̂N in normal form, withi1=0} → PN ×PN×Z_≥0, a(j) ↦ (

i

ⁱⁿj ,

i

^outj , `j)

that sends a basis element in normal form with rightmost index i₁ = 0 to the tuple (

i

ⁱⁿ_j ^,

i

^out_j ^{, `}j) defined in Equation I.2.2 is injective.

Before beginning the proof of this result, we note that for monomials a(j) withi₁ =0, the inequalityik<N−1 must hold in

i

ⁱⁿj , sincei1 =0 implies thati1−1=N−1 is not an element of

i

ⁱⁿj . Consequently, the ordering of the indices in

i

ⁱⁿj agrees with the natural ordering of Z, so we can regard (

i

ⁱⁿj ,<) as a subset of (Z,<) and replace the modular index sequence j by an integral index sequencej^Z such thatj^Z( modN) =j.

I.2.6.7 Definition. Assume j = j^(m)⋅. . .⋅j⁽¹⁾⋅j⁽⁰⁾ is a normal form sequence with j⁽⁰⁾= {0=i₁ <. . . <i_k <N−1} and j⁽ⁿ⁾ = (i_h1+n, . . . , i_hk(n)+n) ⊆ (i₁+n, . . . , i_k+n), where indices in j⁽ⁿ⁾ are modulo N and 1 ≤ k(n) ≤ k for all 1 ≤ n≤ m. The integral normal form sequence forj is

j^Z= (j^(m))^Z⋅. . .⋅ (j⁽¹⁾)^Z⋅j⁽⁰⁾ where (j⁽ⁿ⁾)^Z ∶= (i_h1+n, . . . , i_hk(n)+n) ∈Z^k(n) forn=1, . . . , m.

I.2.6.8 Example. We illustrate our notation with our running Example I.2.5.5 forN = 6.

If j= (1 5 0 2 3 4 5 1 2 3 4 0 1 2 3 5 0 1 2 4),

then j^Z = (7 5 6 8 3 4 5 7 2 3 4 6 1 2 3 5 0 1 2 4). ◊ Our proof of Proposition I.2.6.6 will hinge upon the following technical lemma.

I.2.6.9 Lemma. Let j^Z be the integral normal form sequence forj and let[is, . . . , is+ ns]fors=1, . . . , k be the strands ofj^Z in the sense of Definition I.2.6.2. Assume i1 =0.

Then

I.2.6. Faithfulness of the graphical representation i) n₁ =i₁+n₁ < i₂+n₂ < . . . < i_k+n_k,

ii) i_k+n_k<i₁+n₁+N =n₁+N.

Assume this lemma for the moment. We postpone the proof of this result and proceed directly to proving the proposition.

Proof (Proposition I.2.6.6). Since we will fix the sequence j throughout the proof, we will drop the subscript j on

i

ⁱⁿj ,

i

^outj ,`_j. To show the injectivity ofψ₀, we consider the factorizationψ₀=γ○β○α given by

ψ₀∶ a(j) z→^α a(j^Z)

β

z→ ((

i

^in)Z,(

i

^{out)Z) z→γ (iin,iout, `),}

where(

i

^{in)Z=iin and(}

i

^{out)Z = {i∈jZ ∣no i+}1 to the left ofi}similar to the definition of i^out. The map α replaces indices in Z/NZ by indices in Z as in Definition I.2.6.7 above. The mapβ is given by reading off(

i

^{out)Z and(}

i

^{in)Z from jZ. The map γ sends}

the pair ((

i

^in)Z,(

i

^out)Z) to a triple consisting of the respective images iⁱⁿ,i^out modulo N of the pair and the integer ` = 1+ ∑`_r where `_r = ⌊^j_N^r⌋ for each j_r ∈ (

i

^{out)Z. The}

summand 1 corresponds to 0=i1; all other occurrences of 0 are counted by∑`r. Now we check injectivity of all factors ofψ0 in the factorisationψ0=γ○β○α.

The mapα is clearly injective sincej^Z ↦ j^Z( modN) is a left inverse map.

To see that β is injective, we need to know that j^Z can be uniquely reconstructed from ((

i

^in)Z,(

i

^out)Z). Observe that j^Z is determined by knowing all the “strands”

is, is+1, is+2, . . . , is+ns for 1≤s≤k, hence by assigning an elementis+ns∈ (

i

^{out)Z to}

each i_s∈ (

i

^in)Z. But it follows from Lemma (I.2.6.9.i) that i₁+n₁ must be the smallest element of(

i

^out)Z,i2+n₂ the second smallest, etc., so that the elementi_s+n_sis assigned to thes-th element in iⁱⁿ, that is, toi_s.

Now to see that γ is injective, we need to recover ((

i

^in)Z,(

i

^out)Z) in a unique way from (iⁱⁿ,i^out, `). Write iⁱⁿ = {0 = i1 < . . . < i_k < N −1}, and set (

i

^{in)Z ∶= iin. By}

Lemma (I.2.6.9.i), we know that(

i

^out)Z is of the form(i₁+n₁<. . .<i_k+n_k), and since the elements ofi^out have to be equal to the elements of(

i

^{out)Z moduloN}, we can write i_r+n_r=N `<sub>r</sub>+d<sub>r</sub> for `_r= ⌊^ir+_N^nr⌋ and some d_r∈i^out. Comparing`<sub>r</sub> and `_s forr <s, we have

N `r ≤ N `r+dr = ir+nr < is+ns = N `s+ds ≤ N(`s+1).

So`<sub>r</sub><`_s+1, i.e. `<sub>r</sub>≤`_s. Similarly, we obtain from Lemma (I.2.6.9.ii) that`<sub>k</sub>≤`₁+1.

As a result,

N `<sub>k</sub> ≤ N `_k+d_k = i_k+n_k < i₁+n₁+N = N(`<sub>1</sub>+1) +d<sub>1</sub> ≤ N(`₁+2),

i.e. `<sub>k</sub> <`₁+2. Together we have `<sub>1</sub> =. . . = `_s <`<sub>s+1</sub> = . . . =`₁+1 for some 1< s≤k (where we treat the case s=kby `<sub>1</sub>=. . .=`_k). Set̃`∶=`₁. Then

i_r+n_r = Ñ`+d<sub>r</sub> for 1≤r≤s, ir+nr = N(̃`+1) +ds for s+1≤r≤k.

As a first consequence,

` = 1+ ∑

r

`r = 1+k̃`+ (k−s), which determines̃`= ⌊<sup>`−1</sup>

k ⌋, and hence all`r, as well as the indexs. Using Lemma I.2.6.9, we determine that

is+1+ns+1 < . . . < ik+nk < i1+n1+N < . . . < is+ns+N, and so

N(̃`+1) +ds+1 < . . . < N(̃`+1) +d_k < N(̃`+1) +d1 < . . . < N(̃`+1) +ds. Therefore, ds+1 < . . . < dk < d1 < . . . < ds, which fixes the choice of dr for all r. We conclude that given(iⁱⁿ,i^out, `), we can reconstruct(

i

^{out)Z by settingi}r+nr∶=N `r+dr.

This completes the proof of Proposition I.2.6.6. ◻

Proof (Lemma I.2.6.9). i) Let j^Z be a nonempty integral normal form sequence with 0 = i1 < . . . < i_k ≤ N −1 and strands [ir, . . . , ir+nr] for 1 ≤ r ≤ k (recall Definitions I.2.6.2 and I.2.6.7). Assume that there is some index 1≤t≤k−1 such that i_t+n_t ≥ i_t+1+n_t+1. Since i_t<i_t+1, we haven_t>n_t+1. So

j^Z = . . .(. . . it+nt . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

thentth bracket

. . . (. . . it+nt+1 it+1+nt+1 . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

then_t+1th bracket

. . . .

Fromi_t+n_t+1 < i_t+1+n_t+1 ≤ i_t+n_tit follows that there is some integern_t+1<p≤n_t such that i_t+1+n_t+1=i_t+p appears in the strand [i_t, . . . , i_t+n_t], i.e.

j^Z = . . .(. . . it+nt . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

thentth bracket

. . .(. . . it+p . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

thepth bracket

. . . (. . . it+nt+1 it+1+nt+1 . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

then_t+1th bracket

. . .

withi_t+p=i_t+1+n_t+1. But by the definition of the strands, there is noi_t+1+n_t+1+1 appearing to the left of it+1 +nt+1. Due to Lemma I.2.5.2, we know that (even modulo N) there is no repetition of it+1+nt+1 to the left. Thus it+p=it+1+nt+1

is not possible, and we obtain i₁+n₁ < i₂+n₂ < . . . < i_k+n_k.

I.2.6. Faithfulness of the graphical representation ii) For the second statement of Lemma I.2.6.9, assumei_k+n_k≥i₁+n₁+N. It is true generally thatN >i_k, so we geti_k+n_k≥i₁+n₁+N >i_k+n₁. Hencei₁+n₁+N =i_k+b for somen1<b≤nk, i.e. i1+n1+N appears in the strand [ik, . . . , ik+nk]and we have

j^Z = . . .(. . . ik+nk)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

thenkth bracket

. . . (. . . ik+b . . .)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

thebth bracket

. . . (i1+n1 . . . ik+n1)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

then1th bracket

. . . .

Here it may be that the n_k-th bracket and the b-th bracket coincide, but in any case, we find thati_k+b = i₁+n₁+N = i₁+n₁ modN, and soi_k+b appears to the left ofi₁+n₁. By the definition of the strands, there is no i₁+n₁+1 to the left ofi1+n1, and from Lemma I.2.5.2 we deduce that in j=j^Z mod N there is no i1+n1 mod N to the left ofi1+n1 allowed, which leads to a contradiction. Hence

i_k+n_k<i1+n1+N must hold. ◻

Having established that ψis injective when restricted to sequences with i₁ =0, we now show the injectivity ofψ in general.

Proof (Proposition I.2.6.4). We have the following disjoint decompositions accord-ing to the smallest valuei₁ in j⁽⁰⁾ forj:

{a(j) in normal form} = ∐

i

{a(j) in normal form, i₁=i} {(

i

ⁱⁿj ,

i

^outj , `j)} = ∐

i

{(

i

ⁱⁿj ,

i

^outj , `j) ∣i1=i∈

i

ⁱⁿj }

ψ = ∐

i

ψ_i

whereψ_i∶ {a(j) in normal form, i₁ =i} → {(

i

ⁱⁿj ,

i

^outj , `_j) ∣i₁=i∈

i

ⁱⁿj }. By Proposition I.2.6.6, the mapψ0 ∶ a(j) ↦ (

i

ⁱⁿj ,

i

^outj , `j) restricted to those a(j) with i1 =0 is injective. We argue next that by an index shift this result is true for all other ψ_i. It follows from Proposition I.2.6.6 that the map ψ̂₀ defined by

ψ̂0∶ {a(j) ∈nTL̂N in normal form, withi1=0} → {(

i

ⁱⁿj ,

i

^outj ,`̂j) ∣i1 =0∈iin} is injective, where`̂_j= ̂`_j(i)counts the occurences of N−iin j. Recall that

`_j = ∑

r

`<sub>r</sub>+1 where `_r is the number of 0 in the rth strand [i_r, . . . , i_r+n_r]ofjmod N.

Now observe that we can obtain`<sub>j</sub> from`̂_j as

`j = ̂`j− ∣{dr∈

i

^outj ∣dr≥N −i}∣ + ∣{ir∈

i

ⁱⁿj ∣ir>N−i}∣ +1,

which follows from a computation using`̂<sub>j</sub> = ∑<sub>r</sub>`̂_r and

`̂_r = the number of N−i in therth strand [i_r, . . . , i_r+n_r] mod N

=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⌊^ir+nr+i

N ⌋ ifir≤N−i

⌊^ir+nr+i

N ⌋ −1 ifir>N−i

=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⌊^{N `r+dr+i}

N ⌋ ifir≤N−i

⌊^{N `r+dr+i}

N ⌋ −1 ifir>N−i

=

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪

⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

`r+1 ifir≤N −iand dr+i≥N

`r ifir≤N −iand dr+i<N

`r ifir>N −iand dr+i≥N

`r−1 ifir>N −iand dr+i<N.

We obtain ψ_i by first shifting the indices of j by subtracting i from each index, j− (i, . . . , i), then applying ψ̂₀, and finally shifting the indices from

i

ⁱⁿ_j ^and

i

^out_j ^{by adding}

i to each. Hence, ψ_i is injective for each i, and ψ is injective because the unions are

disjoint. ◻

I.2.6.2. Description and linear independence of the matrices

Im Dokument Affine nilTemperley-Lieb Algebras and Generalized Weyl Algebras: (Seite 68-74)