Der topologische Zyklenraum

Der topologische Zyklenraum Zur Verallgemeinerung auf unendliche G von (z.b.) Satz 4.5.1. (MacLane 1937) Ein Graph ist genau dann plättbar, wenn sein Zyklenraum eine schlichte Basis besitzt. Interesting thm: captures the geometric sparseness of planar graphs algebracically, and so well that it becomes a defining/equivalent criterion for planarity.

Der topologische Zyklenraum Zur Verallgemeinerung auf unendliche G von (z.b.) Satz 4.5.1. (MacLane 1937) Ein Graph ist genau dann plättbar, wenn sein Zyklenraum eine schlichte Basis besitzt.... brauchen wir unendliche Kreise, und unendliche Summen: e We need 1 cycles and for the trivial forward implication, to generate C from face bdries: e lies on a finite cycle (which has to be generated) but on no finite face bdry, and once we use the infinite face bdry we need an infinite sum to get rid of the unwanted fat edges. Also, there s no other thin generating set for the finitary cycle space than the finite face bdries: since G is 3-connected, any thin generating set would be the face bdries in any drawing. (See remark in book after MacLane s thm.) I don t know whether a graph whose finitary cycle space has a sparse basis of finite cycles must be planar. I could imagine that the topologcial ML proof yields a surface that s just a punctured sphere.

Der topologische Zyklenraum C = C(G) E(G): Durch dünne Summen aus topologischen Kreisen erzeugbare Kantenmengen We d like to take the just circuits as generators and generally allow infinite thin sums. However, in order to get a vector space (or a group) we have to build 1 sums in to single elements, so that for those it s enough to take finite sums (to generate all edge sets we want to generate).

Der topologische Zyklenraum C = C(G) E(G): Durch dünne Summen aus topologischen Kreisen erzeugbare Kantenmengen + - abgeschlossen: vereinige zwei dünne Summen zu einer so the (edges sets built from) thin sums do indeed form a vector space Note that we can t prove 1 closure (under infinite thin sums) in the same way: combining those thin families may give a non-thin family. Example: let every summand in a thin family contain some fixed edge e twice

Der topologische Zyklenraum C = C(G) E(G): Durch dünne Summen aus topologischen Kreisen erzeugbare Kantenmengen + - abgeschlossen: vereinige zwei dünne Summen zu einer Beispiele: Quadrate der Leiter endl. Gebietsränder beim wilden Kreis. These are examples for sparse generating sets, as for MacLane.

Der topologische Zyklenraum C = C(G) E(G): Durch dünne Summen aus topologischen Kreisen erzeugbare Kantenmengen + - abgeschlossen: vereinige zwei dünne Summen zu einer C fin (G): the finite sets in C (clearly a subspace of C(G)) C fin is the set of finite edge sets D that are sums, maybe infinite sums, of circuits (maybe infinite circuits). However, we ll show later that D is also a disjoint union of circuits, which must be finite if D if finite. Hence C fin is also the finitary cycle space of G, the space of finite sums of finite circuits.

Der topologische Zyklenraum C = C(G) E(G): Durch dünne Summen aus topologischen Kreisen erzeugbare Kantenmengen + - abgeschlossen: vereinige zwei dünne Summen zu einer C fin (G): the finite sets in C (clearly a subspace of C(G)) The cut space B(G) E(G): B(G): subspace consisting of all cuts (finite or infinite) B fin (G): the finite sets in B (clearly a subspace of B(G)) Closed under finite sums (proof as for finite G: diagonal pairs of corners) We know neither of C nor of B (yet) that they re closed under 1 thin sums.

Theorem 2.1. [Proof in survey Locally finite graphs with ends ] (i) Given an ordinary spanning tree of G, its fundamental cuts generate B(G). Its fundamental circuits generate C fin (G) but not necessarily C(G). Move on without proof: both statements are analogous/dual to (ii) proved below. generate C fin : because, as we ll show later, C fin = finitary cycle space (which we know is generated by the fund.cycles of ordinary sp.trees). C e! finite cycles C! C fin So far, we don t know yet that C fin is generated by the finite circuits (which are in turn generated by the fundamental circuits of any ordinary sp.tree, with the same proof as for finite graphs). We ll be able to mimic the proof that the finite cycles generate every D 2 C fin once we know that these D meet every finite cut evenly: then they decompose greedily into (finite) by applying this to the cuts E(v). not nec : if T contains an infinite circle, then this cannot be a sum of C e s (since those e s would still be in the sum).

Theorem 2.1. (ii) Given a TST of G, say T, its fundamental circuits C e generate C(G). Its fundamental cuts D f generate B fin (G) but not necessarily B(G). Proof of (ii), first statement:

Theorem 2.1. (ii) Given a TST of G, say T, its fundamental circuits C e generate C(G). Its fundamental cuts D f generate B fin (G) but not necessarily B(G). Proof of (ii), first statement: We show that the following are equivalent for edge sets D: (1) D 2 C(G); (the equivalence (1)$(2) will be (2) D meets every finite cut F of G evenly; needed all the time) (3) D is a thin sum of fundamental circuits. (1)!(2): jumping arc ) F meets each C i in D = P i C i evenly ) F finite F meets D evenly. F meets only finitely many of the C i in D = P i C i, since this is a thin sum. Since F meets each C i evenly, it also meets P i C i evenly, because addition and multiplication in the ring E(G)! F 2 are distributive: if F \ C i and F \ C j are even then so is their symmetric di erence (F \ C i ) (F \ C j ). But this equals F \ (C i C j ), which we wanted to be even. (Ring E(G)! F 2 : addition = of sets; multiplication = \ of sets.)

Theorem 2.1. (ii) Given a TST of G, say T, its fundamental circuits C e generate C(G). Its fundamental cuts D f generate B fin (G) but not necessarily B(G). Proof of (ii), first statement: We show that the following are equivalent for edge sets D: (1) D 2 C(G); (the equivalence (1)$(2) will be (2) D meets every finite cut F of G evenly; needed all the time) (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) This choice of which C e to use is not a clever idea but the only chance, for any set D: any sum P C e = D must include all C e with e 2 D (can t get those e 2 D from any other C e 0), and must not include any C e with e /2 D (we couldn t get rid of those e again by adding other C e 0).

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin, i.e. well defined:

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin, i.e. well defined: Edges e 0 /2 T lie in C e only for e 0 = e. Edges f 2 T lie in C e only for e 2 D f, and D f is finite (TST). So every edge of G lies in only finitely many C e, so P C e is well-defined.

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin. We want to prove that D + P e2dre(t ) C e = ; :

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin. We want to prove that D + P e2dre(t ) C e = ; : As seen earlier, D + P C e E(T ). Edges e 0 /2 T only lie in C e for e 0 = e, but then they re also in D and hence not in the sum D + P C e (= D + P e2dre(t ) C e). For finite G 9 trick: E(T ) implies D + P C e = ;: by (2), D + P C e induces even degrees and hence lies in C, but every nontrivial subgraph of a finite tree has a vertex of degree 1, so D + P C e 2 C = ;. We still know that that D + P C e induces even degrees D does by (2), and the C i do but an infinite TST can contain edge sets with all degrees even. Moreover, inducing even degrees doesn t mean 2 C now (double rays needn t lie in C). For infinite G, it s conceivable that T, which we know contains no circuit, might however contain a thin sum of circuits.

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin. We want to prove that D + P e2dre(t ) C e = ; : As seen earlier, D + P C e E(T ). So for every f 2 E(T ), D + P C e meets the finite cut D f in an even subset of D f \ E(T ) = {f}, D + P e2dre(t ) C e meets every finite cut evenly, because D and all the C e do.

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin. We want to prove that D + P e2dre(t ) C e = ; : As seen earlier, D + P C e E(T ). So for every f 2 E(T ), D + P C e meets the finite cut D f in an even subset of D f \ E(T ) = {f}, i.e. in ;. Hence D + P C e contains no f 2 T, and hence no edge at all, sinc it s E(T ).

(2) D meets every finite cut F of G evenly; (3) D is a thin sum of fundamental circuits. (2)!(3): show that D = X C e. e2dre(t ) P C e is thin. We want to prove that D + P e2dre(t ) C e = ; : As seen earlier, D + P C e E(T ). So for every f 2 E(T ), D + P C e meets the finite cut D f in an even subset of D f \ E(T ) = {f}, i.e. in ;. We have thus shown the first statement of Theorem 2.1 (ii):

Theorem 2.1. [Proof in survey Locally finite graphs with ends ] (i) Given an ordinary spanning tree of G, its fundamental cuts generate B(G). Its fundamental circuits generate C fin (G) but not necessarily C(G). (ii) Given a TST of G, say T, its fundamental circuits C e generate C(G). Its fundamental cuts D f generate B fin (G) but not necessarily B(G). Proof of (ii), second statement: The D f generate B fin (G), since every finite cut F is F = P f 2F \E(T ) D f. As before, this is the only chance to generate F from D f s. It happens to work when F is finite: a given edge e /2 T lies in D f for precisely the f 2 C e, and although C e can now be infinite, it meets the (finite!) cut F evenly by jumping arc; so the number of f 2 F for which e 2 D f is even if e /2 F, and odd if e 2 F, as needed. But when F is infinite and has no edge in T it does not work, see!

Theorem 2.1. [Proof in survey Locally finite graphs with ends ] (i) Given an ordinary spanning tree of G, its fundamental cuts generate B(G). Its fundamental circuits generate C fin (G) but not necessarily C(G). (ii) Given a TST of G, say T, its fundamental circuits C e generate C(G). Its fundamental cuts D f generate B fin (G) but not necessarily B(G). Proof of (i): almost analogous/dual to (ii). First assertion of (i): one now has to show that the D f do not generate more than B, ie that a thin sum of cuts is another cut. (In (ii) this was automatic, by def of C. But B is not defined as the thin sums of cuts but simply as all cuts.) We ll prove this below. Second assertion of (i): Every D 2 C, as we saw, meets every finite cut evenly. Every such D is a disjoint union of circuits: this was an exercise, and will be proved again soon. For such D 2 C fin we can even decompose D greedily: can t get stuck because it meets every E(v) evenly. And finite circuits are known to be generated by C e s of ordinary sp.trees. not necesarily C(G) : An infinite circuit contained in T will never be a sum of fundamental circuits C e, whose e /2 T we wouldn t be able to shed.

Theorem 2.1. (iii) Given a normal spanning tree of G, its fundamental circuits generate C(G), and its fundamental cuts generate B(G). Corollary. [Thm 2.2 in survey Locally finite graphs with ends ] (i) C(G) is generated by finite circuits and is closed under infinite thin sums. (ii) B(G) is generated by finite bonds and is closed under infinite thin sums. Proof. The generation part of both (i),(ii) follows from Thm 2.1 (iii). The closure part follows from these characterizations: Lemma 2.2. (i) C(G) = edge sets meeting every finite bond evenly (see earlier) (ii) B(G) = edge sets meeting every finite circuit evenly (dual of ") (These latter types of edge sets are clearly closed under thin sums.) Finitary characterizations of C and B implying at once they re closed under thin sums! It s clear that bonds meet finite circuits evenly. But the converse, that every edge set meeting all finite circuits evenly is a cut needs proof. The proof is analogous to, or the dual of, our earlier proof of the analogous characterization of C(G). There, we showed that if D meets every finite cut evenly (2) then it s a thin sum of fundamental circuits (3) which by definition implied that D 2 C. The analogous reasoning now is that if F meets every finite circuit evenly (2 0 ) then it s a thin sum of fundamental cuts (3 0 ), but we now have to show that that s again in B. Let s prove the non-trivial -part of Lemma 2.2 (ii):!

Lemma 2.2 (ii). If F E(G) meets every finite circuit in G evenly, then F is a cut. Proof. ASK: how would you prove this? Ideas? If true, we ll have F splitting G into two halves each containing lots of stu we don t want to know about ) contract that stu!

Lemma 2.2 (ii). If F E(G) meets every finite circuit in G evenly, then F is a cut. Proof. Contract the edges not in F, keeping loops and multiple edges: H := G (E r F ). (Branch sets: the components of the graph (V, E r F ).) Branch sets are the maximal edge sets spanning connected subgraphs (connected as subgraphs, not just defining connected standard subspaces). Although we keep loops, there aren t any: they d make a finite cycle in G with 1 edge in F.

Lemma 2.2 (ii). If F E(G) meets every finite circuit in G evenly, then F is a cut. Proof. Contract the edges not in F, keeping loops and multiple edges: H := G (E r F ). (Branch sets: the components of the graph (V, E r F ).) Every finite circuit C in H is even, since it extends to a finite circuit in G meeting F precisely in C (so C \ F is even by assumption). Hence H is bipartite. Its bipartition defines a bipartition of G with cut F. The edges of H are precisely those in F

Lemma 2.2 (ii). If F E(G) meets every finite circuit in G evenly, then F is a cut. Proof. Contract the edges not in F, keeping loops and multiple edges: H := G (E r F ). (Branch sets: the components of the graph (V, E r F ).) Every finite circuit C in H is even, since it extends to a finite circuit in G meeting F precisely in C (so C \ F is even by assumption). Hence H is bipartite. Its bipartition defines a bipartition of G with cut F. Corollary. B(G) is closed under thin sums. We still needed to show this, for our proof that the fundamental cuts of an ordinary sp.tree generate B (not more than B).

Atomic bond of G: one contained in E(v) for some v 2 V (G) If v is a cutvertex, then E(v) partitions into more than one atomic bond.

Atomic bond of G: one contained in E(v) for some v 2 V (G) Theorem 2.3. [From survey Locally finite graphs with ends ] (iii) Every element of B(G) is a disjoint union of bonds. (iv) The atomic bonds of G generate B(G). Proof. (iii), (iv) as for finite graphs. (iii): Each component C of the LHS of a cut F against each of the components of G can meet both sides of F ). C (which (iv) Sum (clearly thin) of all atomic bonds contained in cuts E(v) with v on the LHS of a given cut F.

Atomic bond of G: one contained in E(v) for some v 2 V (G) Peripheral circuit: Edge set of chordless circle whose vertices do not separate G Theorem 2.3. [From survey Locally finite graphs with ends ] (i) Every element of C(G) is a disjoint union of circuits. (ii) If G is 3-connected, its peripheral circuits generate C(G). (iii) Every element of B(G) is a disjoint union of bonds. (iv) The atomic bonds of G generate B(G). (ii) is dual to (iv): For plane duals G, G, the peripheral circuits of G are the face boundaries, which are the atomic bonds of G. Open problem: Can we dualise the very simple proofs of (iii),(iv) to get simple proofs of (i),(ii)? The main task seems to be to translate the use of vertices. See Problem 128, and the Hint of the exercise setting this open problem. Assign seminar presentation of Henning s proof of (ii) if this is to be covered!