On a representation of the automorphism group of a graph in a unimodular group
István Estelyi, Ján Karabáš, Alexander Mednykh, Roman Nedela
aa r X i v : . [ m a t h . C O ] F e b ON REPRESENTATION OF THE AUTOMORPHISM GROUPOF A GRAPH IN A UNIMODULAR GROUP
ISTV ´AN EST´ELYI, J ´AN KARAB ´AˇS, ROMAN NEDELA, AND ALEXANDER MEDNYKH
Abstract.
We investigate a representation of the automorphism group of a connectedgraph X in the group of unimodular matrices U β , of rank β , where β is the Betti numberof X . We classify the graphs for which the automorphism group does not embed into U β .It follows that if X has no pendant vertices, then the representation is faithful, except X is a simple cycle. The latter statement can be viewed as a 1-dimensional analogue of aclassical theorem on Riemann surfaces of genera > Introduction
Throughout we assume that X is a simple connected graph. By a dart of X we meanan edge endowed with one of the two possible orientations. Thus every edge uv gives riseto two darts ( u, v ) and ( v, u ) distinguished just by the orientation. If x is dart, then theoppositely oriented dart underlying the same edge will be denoted by x − . For a dart x = ( u, v ) denote by I ( x ) = u the origin of x . Hence I : D ( X ) → V ( X ) is a function fromthe set of darts onto the set of vertices. An oriented simple cycle in X is a cyclic sequenceof darts C = ( x , x , . . . , x k − ), k ≥
2, such that • I ( x i +1 ) = I ( x − i ) for every i ∈ Z k , • I ( x i ) = I ( x j ) for every i = j , i, j ∈ Z k .By Z k we mean the additive cyclic group of order k . If C = ( x , x , . . . , x k − ) is an orientedsimple cycle, then C − = ( x − k − , x − k − , . . . , x − , x − ) is the associated inverse cycle. Boththe cycle and its inverse give rise to the same (unoriented) simple cycle. An (unoriented) simple cycle is a connected subgraph of X , where every vertex has degree 2. In what follows,we freely use both oriented and unoriented simple cycles. To stress that we consider theunderlying (unoriented) subgraph of the oriented simple cycle C , we write the cycle symbolin bold, i.e. C . Vice-versa, if C is an unoriented simple cycle, then C denotes one of thetwo oriented cycles arising from C . In particular, we use the above mentioned conventionfor fundamental cycles defined with respect to a spanning tree of X .An automorphism of a graph is a permutation of the vertices taking adjacent verticesonto adjacent vertices. In this note we consider as a rule the induced action of graphautomorphisms on the set of darts. Hence automorphisms of a graph are considered aspermutations of the set of darts. Mathematics Subject Classification.
Key words and phrases. graph, automorphism, unimodular matrix.
Let T be a spanning tree of X . For every automorphism f ∈ Aut X we can constructa matrix M = M T ( f ) with entries in { , , − } as follows. Fix a linear order of the co-tree edges e , e , . . . , e β , where β = e ( X ) − v ( X ) + 1, the Betti number of X . For everyco-tree edge e i choose one of the two underlying darts, and denote it by x i . Denote by D T = { x , x , . . . , x β } the set of co-tree darts. Clearly, each x i determines a unique simpleoriented cycle C i = ( x i , y , y , . . . , y k ), where y y . . . y k is the unique path in T joiningthe terminal vertex of x i to the initial vertex of x i . Set Row( i ) = D T ∩ D ( f ( C i )). Set M T ( f ) = ( m i,j ), to be a β × β matrix, defined by setting m i,j = 1 if x i ∈ Row( i ), m i,j = − x − i ∈ Row( i ), and m i,j = 0 otherwise.Recall that a square matrix is unimodular if its determinant is ±
1. The unimodularmatrices of rank β (with standard multiplication) form a group, here denoted U β . Byconvention, if β = 0, the group U β is trivial. The following result is well-known. Theorem 1 ([8]) . The assignment Θ T : f M T ( f ) defines a homomorphism of Aut X into the group of unimodular matrices. Observe that if X is a tree with a non-trivial automorphism group, or if X is a simplecycle, then Θ = Θ T is not injective for any spanning tree T of X . In what follows, weconsider the following problem: Problem.
For which connected graphs X does the homomorphism Θ T determine anembedding of Aut X into the group U β ?The problem is motivated by the classical Hurwitz theorem which states that the au-tomorphism group of a closed Riemann surface S of genus g > H ( S , Z ), see Farkas and Kra [4, Theorem V.3.1, p. 270]. A.M.MacBeath in [5] proved that the action remains to be faithful on the homology group H ( S , Z p ) , p > , but fails for H ( S , Z ), see also [4, p. 276]. Observe that U β acts asa group of automorphisms on the homology group of a graph X . Another motivation toinvestigate the problem comes from the applications of the matrix representation of theautomorphism group of a graph in the lifting automorphism problem [6, 7, 8]. Recentlywe have found an application of the result in investigation of structure of Jacobian of agraph, see [1] for definitions and further details.We will see that the answer to the posed problem depends on the structure of blocksof X . To this end, we recall a few related basic results on the 2-connectivity of graphs,that can be found e.g. in monographs [2, Chapter 5] and [3, Chapter 3]. A graph with atleast two vertices is , if it cannot be disconnected by the removal of a singlevertex, or in other words, if it has no cutvertices . By a block of a graph X we mean amaximal 2-connected induced subgraph of X . A block with at least three vertices is called nontrivial . Blocks of an arbitrary simple graph determine a decomposition of the set ofedges. More precisely, the following properties hold true:(P1) any two blocks of X have at most one vertex in common,(P2) every edge of X belongs to exactly one block of X ,(P3) each cycle of X is contained in exactly one block of X . N REPRESENTATION OF THE AUTOMORPHISM GROUP. . . 3
The block tree B = B ( X ) of a connected graph X is a bipartite graph with the vertex set V ( B ) formed by the blocks of X and by the cutvertices of X . A cutvertex v ∈ V ( B ) isadjacent to a block B ∈ V ( B ) if and only if v ∈ B . It can be easily seen that B is a tree [3,Proposition 3.1.2]. It is well-known [3, Chapter 3] that the centre of B is a single vertex,that can either be a block or a cut-vertex of X . In case it is a block, it is called the centralblock of X .All leafs of B are blocks and the centre of B is always formed by one vertex that can beeither a block, or a cutvertex.If X is a graph and Y is its subgraph, then an ear of Y in X is a path P whoseendvertices lie in Y but its inner vertices do not. An ear-decomposition of a graph X is anested sequence ( X , X , . . . , X k ) of graphs, i.e, X i − ⊂ X i , 1 ≤ i ≤ k , such that • X = C is a simple cycle, • X i = X i − ∪ P i − , where P i − is an ear of X i − in X , 1 ≤ i ≤ k , • X k = X ,The following characterisation is a reformulation of Theorem 5.8 of [2], see Proposition3.1.3 in [3] as well. Proposition 2.
A simple graph X on at least vertices is -connected if and only if ithas an ear decomposition. Moreover, an ear-decomposition can be started with any cycle X of X . Lemma 3.
Let T be an arbitrary spanning tree of X and B be a block of X . Then B contains a fundamental cycle C with respect to T and the subgraph T [ B ] ≤ T induced bythe vertices of B is a spanning tree of B . In particular, C is a fundamental cycle withrespect to T [ B ] .Proof. Clearly, T [ B ] being a subgraph of T is a spanning forest of B . In order to derive acontradiction, assume that T [ B ] is disconnected. Then there exist vertices u and v in B belonging to different components of T [ B ]. Choose u and v such that the distance d T ( u, v )in T is minimal. Then the unique path P T in T joining u to v in T is internally disjointfrom B . Since B is connected (even 2-connected), there is a u − v path P B in B as well.Now the (unoriented) cycle C = P B ∪ P T has non-empty intersection with B , but is notcontained in it, violating property (P3) of block-decompositions. Thus T [ B ] is a spanningtree of B . Let x i be a co-tree dart in B determining the fundamental cycle C i . Since twoblocks can intersect in at most 1 vertex, C i is a subgraph of B . (cid:3) Main result
In what follows, by a cycle in a graph X we mean a subgraph which vertices have evendegrees. Cycles of X , considered as vectors of edges, form a vector space C ( X ) of rank β = e ( X ) − v ( X ) + 1 over Z with the operation of symmetric difference as addition. It iswell-known [9] that any base of the cycle space C ( X ) is formed by the fundamental cycleswith respect to a spanning tree of X , see also [3, Theorem 1.9.6]. In particular, everysimple cycle in X can be expressed as a linear combination of fundamental cycles. I. EST´ELYI, J. KARAB ´AˇS, R. NEDELA, AND A. MEDNYKH
Lemma 4.
Let T be an arbitrary spanning tree of X and let f ∈ ker Θ T . Then f ( C ) = C or f ( C ) = C − for any oriented simple cycle C . In particular, f ( B ) = B , for eachnontrivial block B in X .Proof. The kernel of Θ T consists of exactly those graph automorphisms f that are fixing thedarts of each fundamental cycle C , C , . . . , C β , in particular f ( C i ) = C i , for i = 1 , , . . . , β ,with respect to the spanning tree T . Taking the fundamental cycles C i and C j as sets ofedges, it follows, that both C i ∪ C j , C i ∩ C j , and consequently, the symmetric difference C i △ C j are preserved by f ∈ ker Θ T . Since every (unoriented) simple cycle in X canbe obtained from the unoriented fundamental cycles by using the symmetric differenceoperation, it follows that for every oriented simple cycle C , either f ( C ) = C , or f ( C ) = C − holds. (cid:3) Proposition 5.
Let X be a simple bridgeless graph with at least vertices, other than asimple cycle. Then the homomorphism Θ T : Aut X → U β is injective for every spanningtree T of X .Proof. First assume that X is 2-connected. By the assumptions, X is neither K nor asimple cycle. Consider an ear-decomposition ( X , X , . . . , X k ) of X , starting with X = C i ,an arbitrary fundamental cycle of X . This is possible, as X can be any cycle of X , seeProposition 2. By induction with respect to k , we will prove that f (cid:12)(cid:12) X k = id, for f ∈ ker Θ T .Since X is neither K nor a cycle, k > k = 1. Suppose X = X ∪ P , where P is an ear of X in X . Denote by C thesimple cycle formed by P and a u − v path P in X . Since f ∈ ker Θ T , by Lemma 4 theautomorphism f fixes the dart set of X pointwise. Further, by Lemma 4, f ( C ) is either C or C − . It follows that f either fixes or inverts P as P = X ∩ C . Since f keeps theorientation of X = C i , f must fix P pointwise. Thus f fixes both C and X pointwise, as P is contained in both of them. Hence f (cid:12)(cid:12) X = id. Assume k >
1. By induction hypothesis, f (cid:12)(cid:12) X k − = id. Now, the induction step is done using the same argument as in case k = 1.Now suppose that X has cutvertices. By the assumptions, no block of X is K . ByLemma 3, every block B of X contains a fundamental cycle with respect to T . FromLemma 4 it follows that f ( B ) = B . It follows that the restriction f (cid:12)(cid:12) B is an automorphismin ker Θ T (cid:12)(cid:12) B .If B is not a simple cycle, by the previous part of the proof, f (cid:12)(cid:12) B fixes B pointwise.Suppose B = C is a simple cycle. By Lemma 4, f fixes C setwise. If B is not a centralblock, then B contains a cutvertex fixed by f . Since C is a fundamental cycle, B = C isfixed by f pointwise. Assume B = C is the central block. Since X is not a cycle, thereexists a non-central block B sharing a vertex v in common with B . By the precedingargument, B is fixed pointwise. Thus f ( B ) = B and f ( v ) = v . Hence either C is fixedpointwise, or f takes C to the inverse C − . Because B = C is a fundamental cycle, thelatter does not happen.Since B was an arbitrary block, f ∈ ker Θ T is the identity. (cid:3) N REPRESENTATION OF THE AUTOMORPHISM GROUP. . . 5
Let w ∈ V ( X ) be a cutvertex of X such that there exist a bridge B and a nontrivialblock B ≇ K , both containing w . By a pendant tree of X rooted at w we mean a subgraphof X induced by { w } ∪ V ( F w ), where F w is the maximal acyclic subgraph of X − w . Apendant tree will be called rigid , if its automorphism group is trivial. A unicyclic graphwill be called periodic if it admits a non-trivial automorphism f rotating the unique cycle. Example 6.
All periodic unicyclic graphs can be constructed as follows. Choose twopositive integers n, k such that n > k | n , and k < n . Further, choose a sequenceof rooted trees S , S , . . . , S k − with the corresponding roots w , w , . . . , w k − . Let C =( v , v , . . . , v n − ) be a simple cycle with vertices v i , i ∈ { , , . . . , n − } . Form a graph X by taking the cycle C and attach to every vertex v j of C a copy of the tree S i , where j ≡ i (mod k ), by identifying v j with the root w i of S i , see e.g. Figure 1. S S X Figure 1.
An example of a periodic unicyclic graph X with n = 4 and k = 2.It is obvious that X admits an automorphism ̺ of order nk > C with period k . Hence, ̺ is a nontrivial element in the kernel of Θ T , for any spanning tree T of X .Therefore, Θ T is not injective.Now we are ready to prove the main result. Theorem 7.
Let X be a connected simple graph and let T be a spanning tree of X . Then Θ T is not injective if and only if at least one of the following statements holds.(i) X is a tree and Aut X = 1 ,(ii) X contains a pendant tree S such that Aut S = 1 ,(iii) X is periodic unicyclic.Proof. The direction ( ⇐ ) is straightforward. If X is a tree, then β ( X ) = 0 and U β istrivial, hence Θ T is not injective whenever Aut X = 1. If (ii) holds, then there existsan automorphism τ ∈ ker Θ T fixing the complement of the pendant tree S pointwise andacting nontrivially on S . The case (iii) was treated in Example 6.( ⇒ ) In order to derive a contradiction we assume that there exists a nontrivial automor-phism f ∈ ker Θ T and none of (i), (ii), or (iii) holds true. In particular, we assume that X is not a tree.If X contains some pendant trees S , S ,. . . , S k − , we reduce X to a graph X ′ byremoving the trees S ,. . . , S k − . Since each of S i is rigid, we have ker Θ T (cid:12)(cid:12) X ′ ≥ ker Θ T .More precisely, we have f (cid:12)(cid:12) X ′ ∈ ker Θ T (cid:12)(cid:12) X ′ and f (cid:12)(cid:12) X ′ = id X ′ . Consider the block tree I. EST´ELYI, J. KARAB ´AˇS, R. NEDELA, AND A. MEDNYKH B ( X ′ ) = B of X ′ . Either B has a central block B , or a central cutvertex v . By definition,every pendant block is not isomorphic to K . By Lemma 4, f ( B ) = B for every pendantblock B of X ′ . It follows that the action of f on the block tree B is trivial. In particular, f ( B ) = B for every block of X ′ .We distinguish the following three categories of blocks B of X ′ :(a) B is isomorphic neither to a simple cycle C nor K ,(b) B is isomorphic to C or to K , and B is not central,(c) B = B is central and B is isomorphic to C or to K . Case (a).
By Proposition 5, the automorphism f fixes B pointwise, hence f (cid:12)(cid:12) B = id B . Case (b).
The automorphism f fixes the unique cutvertex v B ∈ B on the shortest pathconnecting B to the central block B (or to the central cutvertex v ). If B ∼ = K , then B is fixed pointwise. If B is a simple cycle, then by Lemma 3, the block B is a fundamentalcycle. Since f ( v B ) = v B , we have f (cid:12)(cid:12) B = id B . Case (c). If B is nontrivial, then B contains a cutvertex u belonging to a non-centralblock, as well. By the above argument f ( u ) = u . It follows that f fixes B pointwise.We are left with the case that the block tree B is trivial. That means X ′ is either K ,or a simple cycle. In the first case X is a tree, a contradiction. In the second case X isunicyclic, and the unique cycle is the central block. Since f is non-trivial, X is periodicunicyclic, a contradiction.We have proved that f (cid:12)(cid:12) B = id B for each block B of X ′ . Since every vertex belongs to ablock, we have f (cid:12)(cid:12) X ′ = id X ′ , a final contradiction. (cid:3) Corollary 8.
Let X be a simple connected graph of minimum degree at least and T beits spanning tree. If X is not a simple cycle, then the homomorphism Θ T is injective.Proof. Suppose Θ T is not injective. By Theorem 7, one of (i), (ii), or (iii) holds. By theassumptions, the cases (i) and (ii) are excluded. If (iii) holds, then X is periodic unicyclic.Since X does not admit vertices of degree one, X is a simple cycle, a contradiction. (cid:3) Acknowledgements
The first three authors were supported by the grant GACR 20-15576S. The first authoracknowledges the financial support of Sz´echenyi 2020 under the EFOP-3.6.1-16-2016-00015grant. The second and third author were supported by the grant No. APVV-19-0308 ofSlovak Research and Development Agency. The fourth author was supported by Mathemat-ical Center in Akademgorodok under agreement No. 075-15-2019-1613 with the Ministryof Science and Higher Education of the Russian Federation.
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NTIS, University of West Bohemia, Technick´a 8, 301 00 Plzeˇn 3,Czech republic (I. Est´elyi)
Faculty of Information Technology, University of Pannonia, Egyetem u. 10.,8200 Veszpr´em, Hungary
Email address , I. Est´elyi: [email protected] (J. Karab´aˇs)
Department of Computer Science, Faculty of Natural Sciences, Matej BelUniversity, Tajovsk´eho 40, 97401 Bansk´a Bystrica, Slovakia
Email address , J. Karab´aˇs: [email protected] (R. Nedela)
Department of Mathematics, Faculty of Applied Sciences, University of WestBohemia, Technick´a 8, 301 00 Plzeˇn 3, Czech republic (R.Nedela)
Mathematical Institute, Slovak Academy of Sciences, ˇDumbierska 1, 97411 Bansk´aBystrica, Slovakia
Email address , R. Nedela: [email protected] (A. Mednykh)
Sobolev Institute of Mathematics, Pr. Koptyuga 4, Novosibirsk, 630090,Russia (A. Mednykh)
Novosibirsk State University Pirogova 2, Novosibirsk, 630090, Russia
Email address , A. Mednykh:, A. Mednykh: