There is a CW complex 𝒯(X), which gives a
rational homotopical classification of almost free toral actions on spaces in the
rational homotopy type of X associated with rational toral ranks and also
presents certain relations in them. We call it the rational toral rank complex
of X. It represents a variety of toral actions. In this note, we will give effective
2-dimensional examples of it when X is a finite product of odd spheres. This
is a combinatorial approach in rational homotopy theory.

1. Introduction

Let X be a simply connected CW complex with dimH*(X;ℚ)<∞ and r0(X) be the rational toral rank of X, which is the largest integer r such that an r-torus Tr=S1×···×S1 (r-factors) can act continuously on a CW-complex Y in the rational homotopy type of X with all its isotropy subgroups finite (such an action is called almost free) [1]. It is a very interesting rational invariant. For example, the inequalityr0(X)=r0(X)+r0(S2n)<r0(X×S2n)
can hold for a formal space X and an integer n>1 [2]. It must appear as one phenomenon in a variety of almost free toral actions. The example (*) is given due to Halperin by using Sullivan minimal model [3].

Put the Sullivan minimal model M(X)=(ΛV,d) of X. If an r-torus Tr acts on X by μ:Tr×X→X, there is a minimal KS extension with |ti|=2 for i=1,…,r(Q[t1,…,tr],0)⟶(Q[t1,…,tr]⊗∧V,D)⟶(∧V,d)
with Dti=0 and Dv≡dv modulo the ideal (t1,…,tr) for v∈V which is induced from the Borel fibration [4]X⟶ETr×TrμX⟶BTr.
According to [1, Proposition 4.2], r0(X)≥r if and only if there is a KS extension of above satisfying dimH*(ℚ[t1,…,tr]⊗∧V,D)<∞. Moreover, then Tr acts freely on a finite complex that has the same rational homotopy type as X. So we will discuss this note by Sullivan models.

We want to give a classification of rationally almost free toral actions on X associated with rational toral ranks and also present certain relations in them. Recall a finite-based CW complex 𝒯(X) in [5, Section 5]. Put 𝒳r={(ℚ[t1,…,tr]⊗∧V,D)} the set of isomorphism classes of KS extensions of M(X)=(ΛV,d) such that dimH*(ℚ[t1,…,tr]⊗∧V,D)<∞. First, the set of 0-cells 𝒯0(X) is the finite sets {(s,r)∈ℤ≥0×ℤ≥0} where the point Ps,r of the coordinate (s,r) exists if there is a model (ΛW,dW)∈𝒳r and r0(ΛW,dW)=r0(X)-s-r. Of course, the model may not be uniquely determined. Note that the base point P0,0=(0,0) always exists by X itself.

Next, 1-skeltons (vertexes) of the 1-skelton 𝒯1(X) are represented by a KS-extension (ℚ[t],0)→(ℚ[t]⊗ΛW,D)→(ΛW,dW) with dimH*(ℚ[t]⊗∧W,D)<∞ for (ΛW,dW)∈𝒳r, where W=ℚ(t1,…,tr)⊕V and dW|V=d. It is given as

where P exists by (ΛW,dW), and Q exists by (ℚ[t]⊗ΛW,D). The 2 cell is given if there is a (homotopy) commutative diagram of restrictions

which represents (a horizontal deformation of)

Here Pa exists by (ΛW,dW), Pb(or Pd) by (ℚ[tr+1]⊗ΛW,Dr+1), Pc by (ℚ[tr+1,tr+2]⊗ΛW,D), and Pd(or Pb) by (ℚ[tr+2]⊗ΛW,Dr+2). Then we say that a 2 cell attaches to (the tetragon) PaPbPcPd. Thus, we can construct the 2-skelton 𝒯2(X).

Generally, an n-cell is given by an n-cube where a vertex of (ℚ[tr+1,…,tr+n]⊗ΛW,D) of height r+n, n-vertexes {(ℚ[tr+1,…,tr+i∨,…,tr+n]⊗ΛW,D(i))}1≤i≤n of height r+n-1, …, a vertex (ΛW,dW) of height r. Here ∨ is the symbol which removes the below element, and the differential D(i) is the restriction of D.

We will call this connected regular complex 𝒯(X)=∪n≥0𝒯n(X) the rational toral rank complex (r.t.r.c.) of X. Since r0(X)<∞ in our case, it is a finite complex. For example, when X=S3×S3 and Y=S5, we haveT(X)∨T(Y)=T1(X)∨T1(Y)=T1(X×Y)=T(X×Y),
which is an unusual case. Then, of course, r0(X)+r0(Y)=r0(X×Y). Recall that r0(S3×S3)+r0(S7)=r0(S3×S3×S7) but 𝒯1(S3×S3)∨𝒯1(S7)⊊𝒯1(S3×S3×S7) [5, Example 3.5]. In Section 2, we see that r.t.r.c. is not complicated as a CW complex but delicate. We see in Theorems 2.2 and 2.3 that the differences between X=Z×S7 and Y=Z×S9 for some products Z of odd spheres make certain different homotopy types of r.t.r.c., respectively. Remark that the above inequality (*) is a property on 𝒯0(X) or 𝒯1(X) as the example of Theorem 2.4(1). We see in Theorem 2.4(2) an example that 𝒯1(X)=𝒯1(X×ℂPn) but 𝒯2(X)⊊𝒯2(X×ℂPn), which is a higher-dimensional phenomenon of (*).

2. Examples

In this section, the symbol PiPjPkPl means the tetragon, which is the cycle with vertexes Pi, Pj, Pk, Pl, and edges PiPj, PjPk, PkPl, PlPi.

In general, it is difficult to show that a point of 𝒯0(X) does not exist on a certain coordinate. So the following lemma is useful for our purpose.

Lemma 2.1.

If X has the rational homotopy type of the product of finite odd spheres and finite complex projective spaces, then (1,r)∉𝒯0(X) for any r.

Proof.

Suppose that X has the rational homotopy type of the product of n odd spheres and m complex projective spaces. Put a minimal model A=(ℚ[t1,…,tn-1,x1,…,xm]⊗Λ(v1,…,vn,y1,…,ym),D) with |t1|=⋯=|tn-1|=|x1|=⋯=|xm|=2 and |vi|,|yi| odd. If dimH*(A)<∞, then A is pure; that is, Dvi,Dyi∈ℚ[t1,…,tn-1,x1,…,xm] for all i. Therefore, from [2, Lemma 2.12], r0(A)=1. Thus, we have (1,r0(X)-1)=(1,n-1)∉𝒯0(X).

Theorem 2.2.

Put X=S3×S3×S3×S7×S7 and Y=S3×S3×S3×S7×S9. Then 𝒯1(X)=𝒯1(Y). But 𝒯(X) is contractible and 𝒯(Y)≃S2.

Proof.

Let M(X)=(ΛV,0)=(Λ(v1,v2,v3,v4,v5),0) with |v1|=|v2|=|v3|=3 and |v4|=|v5|=7. Then
T0(X)={P0,0,P0,1,P0,2,P0,3,P0,4,P0,5,P2,1,P2,2,P2,3,P3,1,P3,2}.
For example, they are given as follows.

(0) P0,0 is given by (ΛV,0).

𝒯1(X) is given as

For example, the edges (1 simplexes)
(2.3){P0,0P0,1,P0,1P0,2,P0,2P0,3,P0,3P0,4,…,P0,0P3,1,P3,1P3,2}
are given as follows.(1)

P0,1P3,2 is given by the projection (ℚ[t1,t2]⊗ΛV,D)→(ℚ[t1]⊗ΛV,D1) where Dv1=Dv2=Dv3=0, Dv4=v1v2t2+t14, Dv5=v1v3t2+t24, and D1v1=D1v2=D1v3=D1v5=0 and D1v4=t14.

(2)

P2,1P3,2 is given by Dv1=Dv2=Dv3=0, Dv4=v1v2t1+t14, and Dv5=v1v3t2+t24.

(3)

P3,1P3,2 is given by Dv1=Dv2=Dv3=0, Dv4=v1v2t1+t14, and Dv5=v1v3t1+t24.

𝒯2(X) is given as follows.

(1)

P0,0P2,1P3,2P3,1 is attached by a 2 cell from Dv1=Dv2=Dv3=0, Dv4=v1v2(t1+t2)+t14 and Dv5=v1v3t2+t24. (Then P2,1 is given by D1v4=v1v2t1+t14, D1v5=0, and P3,1 is given by D2v4=v1v2t2, D2v5=v1v3t2+t24.)

(2)

P0,0P0,1P3,2P3,1 is attached by a 2 cell from Dv1=Dv2=Dv3=0, Dv4=v1v2t2+t14, and Dv5=v1v3t2+t24.

(3)

P0,0P0,1P2,2P2,1 is attached by a 2 cell from Dv1=Dv2=Dv3=0, Dv4=v1v2t2+t24, and Dv5=t14.

(4)

P0,1P0,2P2,3P2,2 is attached by a 2 cell from Dv1=Dv2=0, Dv3=t32, Dv4=v1v2t2+t24, and Dv5=t14.

(5)

P0,0P0,1P3,2P2,1 is not attached by a 2 cell. Indeed, assume that a 2 cell attaches on it. Notice that P3,2 is given by (ℚ[t1,t2]⊗ΛV,D) with Dv1=Dv2=Dv3=0 and
(2.4)Dv4=α(v1,v2,v3)+f,Dv5=β(v1,v2,v3)+g,
where α,β∈(v1,v2,v3) and {f,g} is a regular sequence in ℚ[t1,t2]. Since P0,1P3,2∈𝒯1(X), both α and β must be contained in the ideal (ti) for some i. Also they are not in (t1t2) by degree reason. Furthermore, since P2,1P3,2∈𝒯1(X), we can put that both α and β are contained in the monogenetic ideal (vivj) for some 1≤i<j≤3 without losing generality. Then, dimH*(ℚ[t1,t2,t3]⊗ΛV,D̃)<∞ for a KS extension
(2.5)(Q[t3],0)⟶(Q[t1,t2,t3]⊗ΛV,D̃)⟶(Q[t1,t2]⊗ΛV,D),
by putting D̃vk=t32 for k∈{1,2,3} with k≠i,j and D̃vn=Dvn for n≠k. Thus, we have r0(ℚ[t1,t2]⊗ΛV,D)>0. It contradicts to the definition of P3,2.

Notice there is no 3 cell since it must attach to a 3 cube (in graphs) in general. Thus, we see that 𝒯(X)=𝒯2(X) is contractible.

On the other hand, let M(Y)=(ΛW,0)=(Λ(w1,w2,w3,w4,w5),0) with |w1|=|w2|=|w3|=3, |w4|=7 and |w5|=9. Then we see that 𝒯1(X)=𝒯1(Y) from same arguments. But, in 𝒯2(Y), P0,0P0,1P3,2P2,1 is attached by a 2 cell since we can put Dw1=Dw2=Dw3=0 and
(2.6)Dw4=w1w2t2+t24,Dw5=w1w3t1t2+t15,
by degree reason. Here P0,1 is given by D1w4=0, D1w5=t15, and P2,1 is given by D2w4=w1w2t2+t24, D2w5=0. Others are same as 𝒯2(X). Then three 2 cells on P0,0P0,1P3,2P2,1, P0,0P2,1P3,2P3,1, and P0,0P0,1P3,2P3,1 in 𝒯2(Y) make the following:to be homeomorphic to S2. Thus 𝒯(Y)=𝒯2(Y)≃S2.

Theorem 2.3.

Put X=S3×S3×S3×S3×S7×S7 and Y=S3×S3×S3×S3×S7×S9. Then 𝒯1(X)=𝒯1(Y). But 𝒯(X)≃S2 and 𝒯(Y)≃∨i=16Si2.

Proof.

We see as the proof of Theorem 2.2 that
(2.7)T0(X)={P0,0,P0,1,P0,2,P0,3,P0,4,P0,5,P0,6,P2,1,P2,2,P2,3,P2,4,P3,1,P3,2,P3,3,P4,1,P4,2}
and both 𝒯1(X) and 𝒯1(Y) are given as

For all tetragons in 𝒯1(X) except the following 4 tetragons: (1)P0,0P0,1P3,2P2,1, (2) P0,1P0,2P3,3P2,2, (3)P0,0P0,1P4,2P2,1, and (4) P0,0P0,1P4,2P3,1, 2 cells attach in 𝒯2(X). The proof is similar to it of Theorem 2.2. Thus we see that 𝒯2(X) is homotopy equivalent to which is homeomorphic to S2. For example, when M(X)=(ΛV,0)=(Λ(v1,v2,v3,v4,v5,v6),0) with |v1|=|v2|=|v3|=|v4|=3 and |v5|=|v6|=7, 2 cells attach P0,0P2,1P4,2P3,1, P0,0P3,1P4,2P4,1 and P0,0P2,1P4,2P4,1 from Dv1=⋯=Dv4=0,
(2.8)Dv5=v1v2t1+t14,Dv6=v1v3t1+v2v4t2+t24,Dv5=v1v2t1+t14,Dv6=v1v3(t1+t2)+v2v4t2+t24,Dv5=v1v2t1+t14,Dv6=v1v3t2+v2v4t2+t24,
respectively.

In 𝒯2(Y), 2 cells attach all tetragons in 𝒯1(Y) by degree reason. For example, when M(Y)=(ΛW,0)=(Λ(w1,w2,w3,w4,w5,w6),0) with |w1|=|w2|=|w3|=|w4|=3, |w5|=7 and |w6|=9, put Dw1=Dw2=Dw3=0 and(1)

Dw4=0,Dw5=w1w3t2+t24,Dw6=w2w3t1t2+t15,

(2)

Dw4=t32,Dw5=w1w3t2+t24,Dw6=w2w3t1t2+t15,

(3)

Dw4=0,Dw5=w1w2t2+t24,Dw6=w3w4t1t2+t15,

(4)

Dw4=0,Dw5=w1w3t2+t24,Dw6=w1w4t22+w2w3t1t2+t15,

for (1)~(4) of above. Then we can check that 𝒯(Y)≃∨i=16Si2 (𝒯(Y) cannot be embedded in ℝ3).Theorem 2.4.

Even when r0(X)=r0(X×ℂPn) for the n-dimensional complex projective space ℂPn, it does not fold that 𝒯(X)=𝒯(X×ℂPn) in general. For example,(1)

When X=S3×S3×S3×S3×S7 and n=4, then 𝒯1(X)⊊𝒯1(X×ℂP4).

(2)

When X=S3×S3×S3×S7×S7 and n=4, then 𝒯1(X)=𝒯1(X×ℂP4) but 𝒯2(X)⊊𝒯2(X×ℂP4).

Proof.

Put M(ℂPn)=(Λ(x,y),d) with dx=0 and dy=xn+1 for |x|=2 and |y|=2n+1. Put (ℚ[t1,…,tr]⊗ΛV⊗Λ(x,y),D) the model of a Borel space ETr×Tr(X×ℂPn) of X×ℂPn.

(1) 𝒯1(X) and 𝒯1(X×ℂP4) are given as respectively. For M(X)=(ΛV,0)=(Λ(v1,v2,v3,v4,v5),0) with |v1|=|v2|=|v3|=|v4|=3 and |v5|=7. Here P4,1 is given by Dvi=0 for i=1,2,3,4 and Dv5=v1v2t1+v3v4t1+t14. It is contained in both 𝒯0(X) and 𝒯0(X×ℂP4). On the other hand, P3,2 is given by Dvi=0 for i=1,2,3, Dv4=t22, Dv5=v1v2t1+t14, Dx=0, and Dy=x5+v1v3t12. Then P3,1 is given by Dvi=0 for i=1,2,3,4, Dv5=v1v2t1+t14, Dx=0, and Dy=x5+v1v3t12. They are contained only in 𝒯0(X×ℂP4).

(2) Both 𝒯1(X) and 𝒯1(X×ℂP4) are same as one in Theorem 2.2. Notice that P0,0P0,1P3,2P2,1 is attached by a 2 cell in 𝒯2(X×ℂP4) from Dvi=0 for i=1,2,3, Dv4=v1v2t1+t14, Dv5=t24, Dx=0, and Dy=x5+v1v3t1t2. So 𝒯(X×ℂP4)=𝒯(Y) for Y=S3×S3×S3×S7×S9.

Remark 2.5.

The author must mention about the spaces X1 and X2 in [5, Examples 3.8 and 3.9] such that 𝒯1(X1)=𝒯1(X2). We can check that 2 cells attach on both P0P5P9P8 of them (compare [5, page 506]).

Remark 2.6.

In [5, Question 1.6], a rigidity problem is proposed. It says that does 𝒯0(X) with coordinates determine 𝒯1(X)? For 𝒯(X), it is false as we see in above examples. But it seems that there are certain restrictions. For example, is 𝒯2(X) simply connected?

Acknowledgment

This paper is dedicated to Yves Félix on his 60th birthday.

1HalperinS.Rational homotopy and torus actions2JessupB.LuptonG.Free torus actions and two-stage spaces3FélixY.HalperinS.ThomasJ.-C.4FélixY.OpreaJ.TanréD.5YamaguchiT.A Hasse diagram for rational toral ranks