SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS In this section we will prove the Künneth theorem which in principle allows us to calculate the (co)homology of product spaces as soon as we know the homology of the spaces involved. The result itself is already nice as it allows for some explicit calculations. However, more importantly to us, the techniques leading to this result will play an essential role in the next section where we study product structures in cohomology. As a first step, we establish the algebraic Künneth theorem which is a statement about the homology of tensor products of chain complexes over principal ideal domains. Next, in order to be able to apply this to singular homology we establish the Eilenberg Zilber theorem according to which for two spaces X and X the chain complexes C (X) C (X ) and C (X X ) are naturally chain homotopy equivalent. The method of acyclic models used in these proofs is a very convenient tool which in many cases allows us to avoid explicit geometric or combinatorial arguments. There are further applications of this method in the following section. We begin with some basic definitions. Definition 1. Let R be a commutative ring. A (non-negatively, homologically) graded R-module M is a collection of R-modules M k, k 0. An element in M k is homogeneous of degree k. A morphism f : M N of graded R-modules is a collection of R-module maps f k : M k N k, k 0. Graded R-modules and morphisms between them assemble in a category grmod 0 (R). We know already many examples of graded abelian groups since each chain complex over R has an underlying graded R-modules, i.e., there is a forgetful functor U : Ch 0 (R) grmod 0 (R) (which we will always drop from notation). Similarly, every graded R-module gives rise to a chain complex by endowing it with the zero differentials (a construction we already used twice in this course), giving rise to a functor grmod 0 (R) Ch 0 (R). Moreover, the homology groups of a chain complex assemble to a graded R-module. In fact, there is a homology functor H : Ch 0 (R) grmod 0 (R). Our first goal is to understand the behavior of the homology functor H : Ch 0 (R) grmod 0 (R) with respect to tensor products at least for nice rings R. Definition 2. Let R be a commutative ring and let M, N grmod 0 (R). (1) The tensor product M R N grmod 0 (R) is defined by (M R N) k = M p R N q, k 0. p+q=k (2) The torsion product Tor R (M, N) grmod 0 (R) is defined by Tor R (M, N) k = Tor R (M p, N q ), k 0. p+q=k We leave it to the reader to check that we actually obtain functors of two variables R, Tor R : grmod 0 (R) grmod 0 (R) grmod 0 (R). Next, we extend the tensor product to chain complexes C, D Ch 0 (R). The underlying graded module of the tensor product C R D Ch 0 (R) is the tensor product of the underlying graded 1

2 SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS modules, i.e., we again set (C R D) k = p+q=k C p R D q, k 0. Since the differential : (C R D) k (C R D) k 1 has to be additive it is enough to define it on elements of the form c d with c C p, d D q. And in that case we set (c d) = (c) d + ( 1) p c (d). The reader checks that the fundamental relation 2 = 0: (C R D) k (C R D) k 2 is satisfied, concluding the definition of the tensor product C R D Ch 0 (R). Moreover, one easily verifies that the tensor product defines an additive functor of two variables R : Ch 0 (R) Ch 0 (R) Ch 0 (R). The sign showing up in the definition of the differential of the tensor product is a special instance of the Koszul sign convention. According to this convention, whenever the order of a homogeneous element of degree p and a homogeneous element of degree q is reversed in a formula, then a sign ( 1) pq has to show up. In our case, the differential is basically modeled by a product rule or Leibniz rule. But as the differential which is a map of degree -1 jumps over the element c of degree p this gives rise to a sign ( 1) 1 p = ( 1) p. (Although we do not want to get into this, let us mention that this sign convention is not just some heuristics but can be turned into precise mathematics.) One motivation for this definition of the tensor product comes from the first of the following examples (all of which have variants for modules over commutative rings). Example 3. (1) Recall that, in general, if X and X are CW complexes, then the product X X does not carry a natural CW structure. However, if X and X are finite CW complexes then so is X X. If we denote the set of k-cells of a CW complex X by J k (X), then for the induced CW structure on X X we have J k (X X ) = J p (X) J q (X ), k 0. p+q=k Since for sets S, S we have natural isomorphisms Z(S S ) = Z(S) Z(S ) and Z(S S ) = Z(S) Z(S ), we conclude that at least levelwise we obtain an isomorphism C cell k (X X ) = (C cell (X) C cell (X )) k. One can check that this description is also compatible with differentials, i.e., that we have an isomorphism of chain complexes C cell (X X ) = C cell (X) C cell (X ). (2) Let C Ch 0 (Z) and let A be an abelian group. If we consider A as a chain complex concentrated in degree zero (and hence necessarily with trivial differentials), then the tensor product of these chain complexes is naturally isomorphic to the levelwise tensor product of section 1. (3) Let us write I + Ch 0 (Z) for the chain complex... 0 0 Z Z Z in which the differential Z Z Z is given by z ( z, z). Then for C Ch 0 (Z) there is a natural isomorphism I + C = cyl(c). Here, cyl denotes the cylinder construction for

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS 3 chain complexes introduced on exercise sheet 2. (The motivation for the notation I + and more examples along these lines are discussed on exercise sheet 5.) Thus, if we want to study the homology of the product of two finite CW complexes then we are led to the algebraic problem of relating the homology of the tensor product of two chain complexes to the homology groups of the respective two chain complexes. At least for principal ideal domains, this relation is controlled by an algebraic Künneth theorem as we show next. The point of working with principal ideal domains here is that in that context submodules of free modules are again free (this fails for more general rings, making the situation more complicated). Interesting examples for such rings are fields and Z. The following lemma works for arbitrary commutative rings. Lemma 4. Let R be a commutative ring and let F, D Ch 0 (R). If F is levelwise free and has zero differentials then there is a natural isomorphism F R H (D) = H (F R D) of graded R-modules. Proof. Let us unravel the definition of the differential : (F R D) k (F R D) k 1. Since F has a trivial differential, one observes that on the summand F p R D q the differential is given by ( 1) p id : F p R D q F p R D q 1 followed by the inclusion F p R D q 1 (F R D) k 1. Since F is levelwise free the tensor product functors F p R ( ) are exact. This implies that at the level of cycles we have a natural isomorphism Z k (F R D) = (F R Z(D)) k. Similarly, we have natural isomorphisms B k (F R D) = (F R B(D)) k and, using the right exactness of R, at the level of homology this yields the desired natural isomorphism F R H (D) = H (F R D). There is a variant of this lemma in the case that the second factor is levelwise free and comes with trivial differentials. Theorem 5. (Algebraic Künneth theorem) Let R be a principal ideal domain and let C, D Ch 0 (R) such that C or D is levelwise free. Then there is a natural short exact sequence 0 (H(C) R H(D)) k H k (C R D) Tor R (H(C), H(D)) k 1 0. Moreover, these sequences split (but the splitting is not natural). Proof. The proof is very similar to the proofs of the algebraic universal coefficient theorems in (co)homology so that we allow ourselves to not carry out all details. We give the proof in the case that C is levelwise free and begin by considering the presentations 0 B k 1 (C) j Z k 1 (C) p H k 1 (C) 0, k 1. Since C is levelwise free and R is a principal ideal domain, this is a free presentation and it can hence be used in the calculation of Tor R (H k 1 (C), ). In the remainder of the proof short hand notations like Z k 1, Z 1 will always refer to C and not to D. As a next step, let us consider the short exact sequence, 0 Z k i C k B k 1 0, where is induced from the differential. These short exact sequences split because the B k 1 are free, and we hence obtain a levelwise split short exact sequence of chain complexes 0 Z C B 1 0,

4 SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS in which the graded R-modules Z and B 1 are considered as chain complexes with trivial differentials. An application of the additive functor ( ) R D yields a further levelwise split short exact sequence of chain complexes 0 Z R D i id C R D id B 1 R D 0. A typical part of the associated long exact homology sequence looks like... H k+1 (B 1 R D ) δ k+1 H k (Z R D ) H k (C R D ) H k (B 1 R D ) δ k... In order to simplify this a bit, note that the chain complexes B 1 and Z are levelwise free and come with trivial differentials. Thus, Lemma 3 implies that this sequence can be rewritten as δ k+1 δ... (B R H (D)) k (Z R H (D)) k H k (C R D ) (B R H (D)) k k 1... and hence gives rise to a short exact sequence 0 cok(δ k+1 ) H k (C R D ) ker(δ k ) 0. The explicit description of the connecting homomorphism δ k in this long exact sequence allows us to identify δ k+1 as δ k+1 = (j id) : (B R H (D)) k (Z R H (D)) k. Note that j shows up in the free presentation of H (C) and observe that this allows us to make the identifications cok(δ k+1 ) = (H(C) R H(D)) k and ker(δ k ) = Tor R (H(C), H(D)) k 1. Thus, we obtain the desired short exact sequence and we let the reader verify the statement concerning the splitting. In the special case that R = Z and that the chain complex D is simply an abelian group A considered as a chain complex concentrated in degree zero, we have just reproduced the algebraic universal coefficient theorem for homology from section 2. Theorem 6. (Geometric Künneth theorem for finite CW complexes) Let X, X be finite CW complexes and let R be a principal ideal domain. Then there is a natural short exact sequence 0 (H cell (X; R) R H cell (X ; R)) k H cell k (X X ; R) Tor R (H cell (X; R), H cell (X ; R)) k 1 0. Moreover, this short exact sequence splits (but the splitting is not natural in the spaces). Proof. We apply the algebraic Künneth theorem to the levelwise free chain complexes C cell (X; R) and C cell (X ; R). That way we obtain a similar short exact sequence with the group in the middle replaced by H k (C cell (X; R) R C cell (X ; R)). But using elementary properties of the tensor product and Example 3.(1) we obtain isomorphisms of chain complexes C cell (X; R) R C cell (X ; R) = (C cell (X) Z R) R (C cell (X ) Z R) = C cell (X) Z C cell (X ) Z R = C cell (X X ) Z R = C cell (X X ; R). Thus, by passing to homology we obtain the desired short exact sequence.

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS 5 Using the isomorphism between cellular and singular homology, this already allows us to calculate the homology of certain product spaces (see the end of this section and the exercises). Instead, we would like to get rid of the restrictive assumption on the spaces and obtain a similar result for arbitrary spaces. More importantly, the techniques which allow us to do so will play a key role in the next section in which we introduce product structures in singular cohomology. Thus, let us consider spaces X, X and for simplicity let us work with integral coefficients. Then the algebraic Künneth theorem applied to the levelwise free chain complexes C (X), C (X ) in principle allows us to calculate the homology of C (X) C (X ). But instead of this we would prefer to learn something about the homology of the product space, i.e., about the homology of C (X X ). Contrary to the cellular context, the relation between the chain complexes C (X) C (X ) and C (X X ) is more subtle, and our next goal is to investigate this relation. Building on the algebraic Künneth theorem we are able to relate these two chain complexes by first gaining an additional insight about contractible spaces. If X and X are contractible spaces, then the algebraic Künneth theorem implies that the homology of C (X) C (X ) vanishes in positive dimensions. Just to be clear, so far we only knew this for the homology of C (X X ) since a product of contractible spaces is again contractible. But in what follows it is essential that this is also the case for the homology of C (X) C (X ). With this observation we will now show that for arbitrary spaces X, X the chain complexes C (X) C (X ) and C (X X ) are naturally chain homotopy equivalent. Before reading further in this section, we suggest that the reader again has a look at the construction of the chain level cross product in section 5 of the previous course. The methods used in that context are very similar to the ones in some of the following proofs, the difference being that we have now more advanced algebraic tools. (This similarity can be made precise by means of the so-called acyclic models theorem, but we will not get into that.) Thus, we want to find a relation between the functors Top Top Ch 0 (Z) given by (X, X ) C (X X ) and (X, X ) C (X) C (X ). In dimension zero the respective functors are given by C 0 (X X ) and C 0 (X) C 0 (X ). Since the free abelian group functor sends products to tensor products, these two functors are naturally isomorphic. Denoting by κ x : 0 X the map picking out a specified point x X, such a natural isomorphism C 0 (X X ) = C 0 (X) C 0 (X ) is given by the linear extension of the correspondence C 0 (X X ) κ (x,x ) κ x κ x C 0 (X) C 0 (X ). Let us refer to these isomorphisms as the canonical isomorphisms. Although the situation is more complicated in larger dimensions there is the following result. Proposition 7. There is a natural transformation φ: C (X X ) C (X) C (X ), X, X Top, which in degree zero is the canonical isomorphism C 0 (X X ) = C 0 (X) C 0 (X ). Moreover, any two such natural transformations φ, φ are naturally chain homotopic. Proof. We want to construct φ inductively over the dimension of the chains. In dimension zero there is no choice, since the transformation φ 0 : C 0 (X X ) C 0 (X) C 0 (X ) has to be the canonical one. We will continue inductively and, given k 1, we assume that natural transformations φ l : C l (X X ) (C (X) C (X )) l, l k 1, are already constructed such that they assemble to partial chain maps. In dimension k, we begin by considering the special case of

6 SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS X = X = k (the so-called models) and we thus want to construct a map φ k : C k ( k k ) (C ( k ) C ( k )) k. In k k there is the particular k-chain given by the diagonal map d k : k k k and first construct φ k (d k ). Whatever this element is, since we want to construct a partial chain map, it has to satisfy the relation φ k (d k ) = φ k 1 (d k ) in (C ( k ) C ( k )) k 1. There are now two cases. (1) Case k = 1: In this case we are looking for an element φ 1 (d 1 ) (C ( 1 ) C ( 1 )) 1 with prescribed boundary equal to φ 0 ( d 1 ) = κ e1 κ e1 κ e0 κ e0. If ι 1 : 1 1 denotes the identity map, then such an element is given by κ e0 ι 1 + ι 1 κ e1. In fact, this follows from the calculation (κ e0 ι 1 + ι 1 κ e1 ) = κ e0 (ι 1 ) + (ι 1 ) κ e1 = κ e0 (κ e1 κ e0 ) + (κ e1 κ e0 ) κ e1 = κ e1 κ e1 κ e0 κ e0. (2) Case k 2: The boundary of the element φ k (d k ) we want to construct has to be φ k 1 ( d k ). By inductive assumption we already have a partial chain map up to dimension k 1, thus we calculate (φ k 1 ( d k )) = φ k 2 ( ( (d k ))) = 0. But this is to say that φ k 1 ( d k ) (C ( k ) C ( k )) k 1 is a positive dimensional cycle of C ( k ) C ( k ). Since k is contractible, the algebraic Künneth theorem implies that this tensor product has trivial homology in positive dimensions. Thus, the cycle φ k 1 ( d k ) has to be a boundary, and any chosen preimage of φ k 1 ( d k ) under will be used to define φ k (d k ). Now, if we are given an arbitrary k-simplex (σ, τ): k X X for arbitrary spaces X, X, then we observe that this simplex factors as (σ τ) d k : k k k X X. But this means that (σ, τ) = (σ τ) (d k ) and the desired naturality of φ k hence forces us to set φ k (σ, τ) = (σ τ )(φ k (d k )). We conclude the definition of φ k on chains by a linear extension and the definition of φ by induction. By construction, φ is natural, consists of group homomorphisms, and it agrees with the canonical isomorphism in dimension zero. It remains to check that φ is compatible with the boundary operators. For this purpose let us again consider an arbitrary k-simplex (σ, τ): k X X and calculate (φ(σ, τ)) = ( (σ τ )(φ(d k )) ) = (σ τ )( φ(d k )) = (σ τ )(φ( d k )) = φ ( (σ τ) (d k ) ) = φ ( (σ τ) (d k ) ) = φ( (σ, τ)). Thus we established the existence of such a natural transformation φ. Let us now show that two such transformations φ, φ are naturally chain homotopic, i.e., we want to construct natural maps s k : C k (X X ) (C (X) C (X )) k+1, k 0,

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS 7 such that s k + s k 1 = φ k φ k : C k (X X ) (C (X) C (X )) k, k 0, where we set s 1 = 0. In the case of k = 0, we can simply set s 0 = 0 since the maps φ 0 and φ 0 agree. Let now k 1 and let us assume inductively that we constructed natural partial chain homotopies up to dimension k 1. In order to define s k, we again stick to the special case of X = X = k and to the particular k-simplex d k : k k k. Whatever the element s k (d k ) is, its boundary would have to agree with (φ k φ k s k 1 )(d k ). But the boundary of this expression can be calculated as (φ φ s k 1 )(d k ) = ( (φ φ) s k 1 ) ) (d k ) = ( (φ φ) + s k 2 (φ φ) ) (d k ) = 0. Thus, (φ k φ k s k 1 )(d k ) (C ( k ) C ( k )) k is a positive dimensional cycle. Since all positive dimensional homology groups of this chain complex vanish, this cycle has to be a boundary and we choose any preimage of it under as the definition of s k (d k ) (C ( k ) C ( k )) k+1. Given an arbitrary k-simplex (σ, τ): k X X, then the desired naturality forces us to set s k (σ, τ) = (σ τ )(s k (d k )). We conclude the definition of s k by linear extension and the definition of s by induction. By construction s consists of natural homomorphisms of abelian groups and it remains to check that we indeed get chain homotopies this way. But considering (σ, τ): k X X this follows from the following calculation which finally concludes the proof. ( s + s )(σ, τ) = (σ τ )s(d k ) + s(σ τ) (d k ) = (σ τ ) s(d k ) + (σ τ )s (d k ) = (σ τ )(φ φ)(d k ) = (φ φ)(σ τ) d k = (φ φ)(σ, τ), We refer to the strategy of this proof as the method of acyclic models. Although we will not make this method precise the reader will develop a good feeling for what is meant. As a further illustration we will give most details in one more such proof. Later, we will allow us to be more and more sketchy. For now, let us recall from section 5 of the previous course that we also have a map in the opposite direction given by the chain level cross product. Proposition 8. There is a natural transformation : C (X) C (X ) C (X X ), X, X Top, such that for x X, x X, σ : p X, and τ : q X we have κ x τ : q = 0 q κ x τ X X and σ κ x : p = p 0 σ κ x X X. Moreover, any two such natural transformations are naturally chain homotopic. The existence of such a natural transformation was already established in the previous course. Here, the result is reformulated using tensor products of chain complexes (which takes care of the biadditivity and the derivation property in the older formulation). Note that the cross product coincides in degree zero with the canonical isomorphism C 0 (X) C 0 (X ) = C 0 (X X ). As for the uniqueness up to homotopy, this is a consequence of the following stronger result.

8 SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS Proposition 9. Let F, G: Top 2 Ch 0 (Z) be any of the two functors (X, X ) C (X) C (X ) and (X, X ) C (X X ). If α, β : F G are natural transformations (1) such that α 0 = β 0 is the canonical isomorphism if F and G are different or (2) such that α 0 = β 0 = id if F and G agree, then α and β are naturally chain homotopic. Proof. We only show that any natural transformations ψ, ψ : C (X) C (X ) C (X X ) which both coincide with the canonical isomorphism in degree zero are naturally chain homotopic. The case of the other direction is already taken care of by Proposition 7 and the remaining two cases are left as exercises. Thus, we want to construct maps s k : (C (X) C (X )) k C k+1 (X X ), k 0, such that s k + s k 1 = ψ ψ : (C (X) C (X )) k C k (X X ), where we set s 1 = 0. In the case of k = 0, we can simply set s 0 = 0 since ψ and ψ coincide in that dimension. Let now k 1 and let us assume inductively that we constructed natural partial chain homotopies up to dimension k 1. A construction of s k is equivalent to the construction of natural maps s k : C p (X) C q (X ) C k+1 (X X ), p + q = k. We first consider the special case of X = p and X = q. In that case there is the universal element ι p ι q C p (X) C q (X ) given by the tensor product of the two respective identities. Whatever the element s k (ι p ι q ) is, its boundary has to be (ψ ψ s k 1 )(ι p ι q ). But, using the inductive assumption, the boundary of this expression can be calculated as (ψ ψ s k 1 )(ι p ι q ) = ( (ψ ψ) s k 1 ) ) (ι p ι q ) = ( (ψ ψ) + s k 2 (ψ ψ) ) (ι p ι q ) = 0. Thus, (ψ ψ s k 1 )(ι p ι q ) C k ( p q ) is a positive dimensional cycle. Since p q is contractible, all its positive dimensional homology groups vanish and this cycle is hence a boundary. Any preimage of (ψ ψ s k 1 )(ι p ι q ) under is used as the definition of s k (ι p ι q ). Given arbitrary simplices σ : p X, τ : q X with p + q = k, the naturality of the chain homotopy forces us to set s k (σ τ) = (σ τ) (s k (ι p ι q )). A linear extension and induction concludes the definition of the homomorphisms s k, k N. By construction these define natural homomorphisms of abelian groups and we only have to check that we indeed obtain chain homotopies s: ψ ψ this way. So, for σ : p X and τ : q X let us calculate concluding the proof. ( s + s )(σ τ) = (σ τ) s(ι p ι q ) + s(σ τ ) (ι p ι q ) = (σ τ) s(ι p ι q ) + (σ τ) s (ι p ι q ) = (σ τ) (ψ ψ)(ι p ι q ) = (ψ ψ)(σ τ )(ι p ι q ) = (ψ ψ)(σ τ), The previous three propositions taken together yield the following.

SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS 9 Theorem 10. (Eilenberg Zilber theorem) Let X and X be topological spaces. The chain level cross product and the morphisms φ of Proposition 7 are inverse natural chain homotopy equivalences : C (X) C (X ) C (X X ) and φ: C (X X ) C (X) C (X ). Proof. By Proposition 7 and Propositon 8 both natural transformations coincide with the canonical isomorphisms in degree zero. Thus, the composition of them and the identity transformation C (X) C (X ) C (X X ) φ C (X) C (X ) and C (X) C (X ) id C (X) C (X ) coincide with the identities in degree zero. Proposition 9 implies that there is a natural chain homotopy s: φ id. Similarly, we also obtain a natural chain homotopy s : φ id, and hence conclude that these are inverse natural chain homotopy equivalences. Thus, although the chain complexes C (X) C (X ) and C (X X ) are not isomorphic they are chain homotopy equivalent. The algebraic Künneth theorem then has the following additional geometric variant. Theorem 11. (Geometric Künneth theorem for singular homology) Let X and X be topological spaces and let R be a principal ideal domain. Then there is a natural short exact sequence 0 (H (X; R) R H (X ; R)) k H k (X X ; R) Tor R (H (X; R), H (X ; R)) k 1 0. Moreover, this short exact sequence splits (but the splitting is not natural in the spaces). Proof. By the algebraic Künneth theorem applied to the levelwise free chain complexes C (X; R) and C (X ; R) we know that there is such a short exact sequence with the middle term replaced by H k (C (X; R) R C (X ; R)). But by the Eilenberg Zilber theorem we have a natural chain homotopy equivalence C (X; R) R C (X ; R) = (C (X) Z R) R (C (X ) Z R) = C (X) Z C (X ) Z R C (X X ) Z R = C (X X ; R), specializing to a natural isomorphism H k (C (X; R) R C (X ; R)) = H k (X X ; R). Thus, in particular, if the homology of one of the spaces involved is free then the torsion term vanishes and the homology of the product is just the tensor product of the homologies of the two spaces. Over a field, the torsion group is always trivial so that we obtain the following consequence. Corollary 12. Let X, X be topological spaces and let F be a field. Then the homology cross product induces a natural isomorphism of graded F-vector spaces : H (X; F) F H (X ; F) = H (X X ; F). Among other examples, the reader is asked to establish the following one in the exercises. Example 13. Let T n = S 1... S 1 be the n-dimensional torus, i.e., the n-fold product of S 1 with itself. For an arbitrary abelian group A there is a natural isomorphism H k (T n ; A) = A (n k), 0 k n.

10 SECTION 5: EILENBERG ZILBER EQUIVALENCES AND THE KÜNNETH THEOREMS This isomorphism is not only an abstract isomorphism but it also tells us what the generators look like. They are given by cross products of the corresponding generators of the homology groups of the circles.