Citation: Edward F. Durner. Enhanced flowering of the F1 long-day strawberry cultivars ‘Tarpan’ and ‘Gasana’ with nitrogen and daylength management[J]. AIMS Agriculture and Food, 2017, 2(1): 1-15. doi: 10.3934/agrfood.2017.1.1
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In [22] and [23], D. Ruelle has introduced the notion of Smale space. A Smale space is a hyperbolic dynamical system with local product structure. He has constructed groupoids and its operator algebras from the Smale spaces. After the Ruelle's initial study, I. Putnam in [14] (cf. [9], [15], [16], [17], [26], etc.) constructed various groupoids from Smale spaces and studied their
In this paper, we will focus on the study of the latter class, the hyperbolic toral automorphisms from the view points of
d(x,y)=inf{‖z−w‖:q(z)=x,q(w)=y,z,w∈R2} for x,y∈T2 |
where
GaA={(x,z)∈T2×T2∣limn→∞d(Anx,Anz)=limn→∞d(A−nx,A−nz)=0} | (1) |
with its unit space
(GaA)(0)={(x,x)∈T2×T2}=T2. | (2) |
The multiplication and the inverse operation on
(x,z)(z,w)=(x,w),(x,z)−1=(z,x) for (x,z),(z,w)∈GaA. |
As in [14], the groupoid
Let
Theorem 1.1 (Theorem 2.10 and Proposition 3.1). The
U1U2=U2U1,V1V2=V2V1,V1U1=e2πiθ1U1V1,V1U2=e2πiθ2U2V1,V2U1=e2πiθ3U1V2,V2U2=e2πiθ4U2V2, |
where
θ1=12(1+a−d√Δ(A)),θ2=c√Δ(A),θ3=b√Δ(A),θ4=12(1−a−d√Δ(A)). |
The range
τ∗(K0(C∗(GaA)))=Z+Zθ1+Zθ2+Zθ3inR. | (3) |
We note that the slopes
Since the étale groupoid
As commuting matrices have common eigenvectors, we know that if two matrices
For a vector
rA:=⟨vu|vs⟩. |
Define two vectors
v1:=vu−rAvs,v2:=rAvu−vs. |
Lemma 2.1. For two vectors
(ⅰ)
(ⅱ)
(ⅲ)
Proof. For two vectors
z≡x+tvu≡x+svs(modZ2) for some t,s∈R. | (4) |
In this case, we see that
⟨tvu−svs|vu⟩=⟨(m,n)|vu⟩, | (5) |
⟨tvu−svs|vs⟩=⟨(m,n)|vs⟩ | (6) |
and we have
t=11−r2A⟨(m,n)|v1⟩,s=11−r2A⟨(m,n)|v2⟩. | (7) |
This shows the implications (ⅰ)
Assume that (ⅱ) holds. By putting
Let us define an action
αA(m,n)(x):=x+11−r2A⟨(m,n)|v1⟩vu,(m,n)∈Z2,x∈T2. |
For a fixed
Lemma 2.2. Keep the above notation.
(ⅰ) If
(ⅱ) For
Proof. (ⅰ) Suppose that
(ⅱ) Let
{x+11−r2A⟨(m,n)|v1⟩vu∣(m,n)∈Z2} |
is dense in
The action
If a discrete group
Proposition 2.3. The étale groupoid
T2×αAZ2={(x,αA(m,n)(x))∈T2×T2|(m,n)∈Z2} |
defined by the action
Proof. By the preceding discussions, a pair
Remark 2.4. Define a map
αA(m,n):=11−r2A⟨(m,n)|v1⟩vu,(m,n)∈Z2. | (8) |
It is easy to see that the étale groupoid
GaA=T2×αA(Z2) | (9) |
as a transformation groupoid.
We set
θ1:=u1s2u1s2−u2s1,θ2:=u2s2u1s2−u2s1, | (10) |
θ3:=−u1s1u1s2−u2s1,θ4:=−u2s1u1s2−u2s1. | (11) |
Lemma 2.5. The real numbers
θ2θ1=θ4θ3=u2u1,θ1θ3=θ2θ4=−s2s1, | (12) |
θ1+θ4=1. | (13) |
Conversely, if real numbers
ζ2ζ1=ζ4ζ3=u2u1,ζ1ζ3=ζ2ζ4=−s2s1, | (14) |
ζ1+ζ4=1, | (15) |
then we have
Proof. The identities (12) and (13) are immediate. Conversely, suppose that real numbers
{u2u1(−s2s1)+1}ζ4=1, |
so that
ζ4=−u2s1u1s2−u2s1 |
and hence
ζ1=u1s2u1s2−u2s1,ζ2=u2s2u1s2−u2s1,ζ3=−u1s1u1s2−u2s1. |
Proposition 2.6. For
αA(1,0)(x1,x2)=(x1+θ1,x2+θ2),αA(0,1)(x1,x2)=(x1+θ3,x2+θ4), |
and hence
αA(m,n)(x1,x2)=(x1+mθ1+nθ3,x2+mθ2+nθ4)for(m,n)∈Z2. |
Proof. We have
αA(m,n)(x1,x2)=(x1,x2)+11−r2A⟨(m,n)|vu−rAvs⟩vu=(x1,x2)+11−r2A⟨(m,n)|(u1−rAs1,u2−rAs2)⟩(u1,u2). |
In particular, for
αA(1,0)(x1,x2)=(x1+11−r2A(u1−rAs1)u1,x2+11−r2A(u1−rAs1)u2),αA(0,1)(x1,x2)=(x1+11−r2A(u2−rAs2)u1,x2+11−r2A(u2−rAs2)u2). |
We put
αA(1,0)(x1,x2)=(x1+ξ1u1,x2+ξ1u2), | (16) |
αA(0,1)(x1,x2)=(x1+ξ2u1,x2+ξ2u2). | (17) |
We then have
ξ1=11−r2A{u1−(u1s1+u2s2)s1}=11−r2A{u1(1−s21)−u2s2s1}=11−r2A(u1s2−u2s1)s2 |
and similarly
ξ2=11−r2A{u2−(u1s1+u2s2)s2}=11−r2A{u2(1−s22)−u1s1s2}=11−r2A(u2s1−u1s2)s1. |
Hence we have
ξ1u1+ξ2u2=11−r2A{(u1−rAs1)u1+(u2−rAs2)u2}=11−r2A{u21+u22−rA(u1s1+u2s2)}=11−r2A(1−r2A)=1. |
By Lemma 2.5, we have
We will next express
Lemma 2.7. The following identities hold.
(ⅰ)
aθ1+bθ2=λuθ1,aθ3+bθ4=λuθ3,cθ1+dθ2=λuθ2,cθ3+dθ4=λuθ4, |
and hence
aθ1+bθ2+cθ3+dθ4=λu. | (18) |
(ⅱ)
aθ3−bθ1=λsθ3,aθ4−bθ2=λsθ4,cθ3−dθ1=−λsθ1,cθ4−dθ2=−λsθ2, |
and hence
aθ4−bθ2−cθ3+dθ1=λs. |
Proof. By the identities
[θ1θ2]=s2u1s2−u2s1[u1u2],[θ3θ4]=−s1u1s2−u2s1[u1u2],[θ3−θ1]=−u1u1s2−u2s1[s1s2],[θ4−θ2]=−u2u1s2−u2s1[s1s2], |
with
Lemma 2.8.
(ⅰ)
(ⅱ)
Hence we have
bθ2=cθ3. |
Proof. (ⅰ) By the first and the fourth identities in Lemma 2.7 (ⅰ), we know the identity (ⅰ). The identities of (ⅱ) is similarly shown to those of (ⅰ). By (ⅰ) and (ⅱ) with the identity
Recall that
Lemma 2.9. The identities
θ1⋅θ4=θ2⋅θ3,θ1+θ4=1,(aθ1+bθ2)θ4=(cθ3+dθ4)θ1,(aθ3−bθ1)θ2=(−cθ4+dθ2)θ3 |
imply
(θ1,θ2,θ3,θ4) | (19) |
={(12(1+|a−d|√Δ(A)),|a−d|a−dc√Δ(A),|a−d|a−db√Δ(A),12(1−|a−d|√Δ(A)))or(12(1−|a−d|√Δ(A)),−|a−d|a−dc√Δ(A),−|a−d|a−db√Δ(A),12(1+|a−d|√Δ(A)))ifa≠d,(12,12√cb,12√bc,12)or(12,−12√cb,−12√bc,12)ifa=d. | (20) |
We thus have the following theorem.
Theorem 2.10. The
U1U2=U2U1,V1V2=V2V1,V1U1=e2πiθ1U1V1,V1U2=e2πiθ2U2V1,V2U1=e2πiθ3U1V2,V2U2=e2πiθ4U2V2, |
where
θ1=12(1+a−d√Δ(A)),θ2=c√Δ(A),θ3=b√Δ(A),θ4=12(1−a−d√Δ(A)). | (21) |
Hence the
Proof. As in Lemma 2.2, the action
Since the
Proposition 2.11 (Slawny [24], Putnam [14]). The
Remark 2.12. (ⅰ) We note that the simplicity of the algebra
(ⅱ) Suppose that two hyperbolic matrices
In this section, we will describe the trace values
In [20], M. A. Rieffel studied K-theory for irrational rotation
K0(C(T2)×αAZ2)≅K1(C(T2)×αAZ2)≅Z8 |
([6], cf. [24]). For
(a1,b1,a2,b2)∧(c1,d1,c2,d2)=(|a1c1b1d1|,|a1c1b2d2|,|a2c2b1d1|,|a2c2b2d2|) |
where
(Θ∧Θ)(x1∧x2)∧(x3∧x4)=12!2!∑σ∈S4sgn(σ)Θ(xσ(1)∧xσ(2))Θ(xσ(3)∧xσ(4)) |
for
exp∧(Θ)=1⊕Θ⊕12(Θ∧Θ)⊕16(Θ∧Θ∧Θ)⊕⋯:∧evenZ4⟶R |
becomes
exp∧(Θ)=1⊕Θ⊕12(Θ∧Θ). |
Let
exp∧(Θ)(∧evenZ4)=τ∗(K0(AΘ)). | (22) |
Proposition 3.1. Let
τ∗(K0(C∗(GaA)))=Z+Zθ1+Zθ2+Zθ3inR. | (23) |
Proof. Take the unitaries
ujuk=e2πiθjkukuj,j,k=1,2,3,4. |
As
θ12θ34−θ13θ24+θ14θ23=0. |
By (22) or [6] (cf. [3,2.21], [13,Theorem 3.9]), we have
τ∗(K0(C∗(GaA)))=Z+Z(θ12θ34−θ13θ24+θ14θ23)+∑1≤j<k≤4Zθjk=Z+Zθ1+Zθ2+Zθ3. |
Remark 3.2. (ⅰ) It is straightforward to see that the skew symmetric matrix
(ⅱ) Suppose that two hyperbolic toral automorphisms
τA∗(K0(C∗(GaA)))=τB∗(K0(C∗(GaB))). |
We may also find a matrix
In this section, we will present some examples.
1.
(θ1,θ2,θ3,θ4)=(12(1+1√5),1√5,1√5,12(5−1√5). | (24) |
By the formula (23), we have
τ∗(K0(C∗(GaA)))=Z+5+√510Z. |
Proposition 4.1. Let
Proof. Let
(θ1,θ2,θ3,θ4)=(θ,2θ−1,2θ−1,1−θ) |
by (24), we have
U1U2=U2U1,V1V2=V2V1,V1U1=e2πiθU1V1,V1U2=e2πi2θU2V1,V2U1=e2πi2θU1V2,V2U2=e−2πiθU2V2, |
We set
u1=U1U22,u2=U2,v1=V1V22,v2=V2. |
It is straightforward to see that the following equalities hold
u1u2=u2u1,v1v2=v2v1,v1u1=e2πi5θu1v1,v1u2=u2v1,v2u1=u1v2,v2u2=e−2πiθu2v2. |
Since the
C∗(GaA)≅C∗(u1,v1)⊗C∗(u2,v2)≅A5θ⊗Aθ. |
2.
(θ1,θ2,θ3,θ4)=(3+√36,√33,√36,3−√36) |
and
λu=aθ1+bθ2+cθ3+dθ4=2+√3,λs=aθ4−bθ2−cθ3+dθ1=2−√3. |
Since
τ∗(K0(C∗(GaA)))=Z+Zθ1+Zθ2+Zθ3=12Z+√36Z. |
Proposition 4.2. Let
Proof. Since the
τ∗(K0(C∗(GaA1)))=Z+5+√510Z,τ∗(K0(C∗(GaA1)))=12Z+√36Z, |
we see that
The author would like to deeply thank the referee for careful reading and lots of helpful advices in the presentation of the paper. This work was supported by JSPS KAKENHI Grant Numbers 15K04896, 19K03537.
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