Homologies and cohomologies of digraphs Yong Lin Renmin University - PowerPoint PPT Presentation

Homologies and cohomologies of digraphs Yong Lin Renmin University of China July 8, 2014

This is a joined work with: Alexander Grigor’yan, Yuri Muranov, Shing-Tung Yau 1

Definition of homologies and cohomologies of digraphs 2

A differential calculus on an associative algebra A is an algebraic analogue of the calculus of differential forms on a smooth manifold. The discrete differential calculus is based on the universal differential calculus on an associative algebra of functions on discrete set. By a natural way, we can consider this calculus as a calculus on a universal digraph(complete digraph) with the given discrete set of vertices. This approach gives an opportunity to define differential calculus for every subgraph of the universal digraph. 3

Given a finite set V , we define a p -form ω on V as K -valued function on V p + 1 . The set of all p -forms is a linear space over K that is denoted by Λ p . It has a canonical basis e i 0 ... i p . For any ω ∈ Λ p , we have � ω i 0 ... i p e i 0 ... ı p ω = i 0 ,..., i p ∈ V where ω i 0 ... i p = ω ( i 0 , . . . , i p ) . The exterior derivative d : Λ p → Λ p + 1 is defined by p + 1 � ( − 1 ) q ω i 0 ... ˆ ( d ω ) i 0 ... i p + 1 = i q ... i p + 1 q = 0 and satisfies d 2 = 0. 4

An elementary p - path is any (ordered) sequence i 0 , ..., i p of p + 1 vertices of V that will be denoted simply by i 0 ... i p or by e i 0 ... i p . We use the notation e i 0 ... i p when we consider the elementary path as an element of a linear space Λ p = Λ p ( V ) that consists of all formal linear combinations of all elementary p -paths. The elements of Λ p are called p - paths . For any v ∈ Λ p we have � v i 0 ... i p e i 0 ... i p v = i 0 ,..., i p ∈ V and a pairing with a p -path ω � ω i 0 ... i p v i 0 ... i p . ( ω, v ) = i 0 ,..., i p 5

The dual operator ∂ : Λ p + 1 → Λ p is given by p + 1 � ( − 1 ) q e i 0 ... ˆ ∂ e i 0 ... i p + 1 = i q ... i p + 1 q = 0 6

We say that a path i 0 ... i p is regular if i k � = i k + 1 for all k = 0 , ..., p − 1, and irregular otherwise. Consider the following subspace of Λ p : � e i 0 ... i p : i 0 ... i p is regular � R p = span ω ∈ Λ p : ω i 0 ... i p = 0 if i 0 ... i p is irregular � � = , where span A denotes the linear space spanned by the set A . The elements of R p are called regular p -forms. For example, ω ∈ R 1 then ω ii ≡ 0 and ω ∈ R 2 then ω iij ≡ ω jii ≡ 0 . 7

We say that an elementary p -path e i 0 ... i p is regular (or irregular) if the path i 0 ... i p is regular (resp. irregular). We would like to define the boundary operator ∂ on the subspace of Λ p spanned by regular elementary paths. Just restriction of ∂ to the subspace does not work as ∂ is not invariant on this subspace, so that we have to consider a quotient space instead. 8

Let I p be the subspace of Λ p that is spanned by irregular e i 0 ... i p . Consider the quotient spaces R p := Λ p / I p . The elements of R p are the equivalence classes v mod I p where v ∈ Λ p , and they are called regularized p -paths. 9

Lemma Let p ≥ − 1 . If v 1 , v 2 ∈ Λ p and v 1 = v 2 mod I p , then ∂ v 1 = ∂ v 2 mod I p − 1 . 10

Proof. If p ≤ 0 there is nothing to prove since I p = { 0 } . In the case p ≥ 1, it suffices to prove that if v = 0 mod I p then ∂ v = 0 mod I p − 1 . It suffices to prove that if e i 0 ... i p is non-regular then ∂ e i 0 ... i p is non-regular, too. Indeed, for a non-regular path i 0 ... i p there exists an index k such that i k = i k + 1 . Then we have ∂ e i 0 ... i p = e i 1 ... i p − e i 0 i 2 ... i p + ... + ( − 1 ) k e i 0 ... i k − 1 i k + 1 i k + 2 ... i p + ( − 1 ) k + 1 e i 0 ... i k − 1 i k i k + 2 ... i p + ... + ( − 1 ) p e i 0 ... i p − 1 . By i k = i k + 1 the two terms in the above line cancel out, whereas all other terms are non-regular, whence ∂ e i 0 ... i p ∈ I p − 1 . 11

Let i 0 . . . i p be an elementary regular p -path on V. It is called allowed if i k i k + 1 ∈ E for any k = 0 , . . . , p − 1, and non-allowed otherwise. The set of all allowed elementary p -paths will be denoted by E p ,and non-allowed by N p . For example E 0 = V and E 1 = E . Denote by A p = A p ( V , E ) the subspace of R p spanned by the allowed elementary p -paths, that is, A p = span { e i 0 ... i p : i 0 . . . i p ∈ E p } = { v ∈ R p : v i 0 ... i p = 0 ∀ i 0 . . . i p ∈ N p } The elements of A p are called allowed p -paths. 12

Similarly, denote by N p the subspace of R p , spanned by the non-allowed elementary p -forms, that is, N p = span { e i 0 ... i p : i 0 . . . i p ∈ N p } = { ω ∈ R p : ω i 0 ... i p = 0 ∀ i 0 . . . i p ∈ E p } Clearly, we have A p = ( N p ) ⊥ where ⊥ refers to the annihilator subspace with respect to the couple ( R p , R p ) of dual spaces. 13

We would like to reduce the space R p of regular p -forms so that the non-allowed forms can be treated as zeros. Consider the following subspaces of spaces R p J p ≡ J p ( V , E ) := N p + d N p − 1 that are d -invariant, and define the space Ω p of p -forms on the digraph ( V , E ) by Ω p ≡ Ω p ( V , E ) := R p / J p Then d is well-defined on Ω p and we obtain a cochain complex d d d d 0 → Ω 0 → Ω p → Ω p + 1 − → . . . − − − → . . . Shortly we write Ω • = R • / J • where Ω is the complex and R and J refer to the corresponding cochain complexes. If the digraph ( V , E ) is complete, that is, E = V × V \ diag then the spaces N p and J p are trivial, and Ω p = R p . 14

Consider the following subspaces of A p Ω p ≡ Ω p ( V , E ) = { v ∈ A p : ∂ v ∈ A p − 1 } that are ∂ -invariant. Indeed, v ∈ Ω p ⇒ ∂ v ∈ A p − 1 ⊂ Ω p − 1 . The elements of Ω p are called ∂ -invariant p -paths. We obtain a chain complex Ω • ∂ ∂ ∂ ∂ ∂ 0 ← − Ω 0 ← − Ω 1 ← − . . . ← − Ω p − 1 ← − Ω p ← − . . . that,in fact, is dual to Ω • . 15

By construction we have Ω 0 = A 0 and Ω 1 = A 1 so that dim Ω 0 = | V | dim Ω 1 = | E | , and while in general Ω p ⊂ A p . Let us define the cohomologies and homologies of the digraph ( V , E ) by H p ( V , E ) := H p (Ω) = ker d | Ω p / Im d | Ω p − 1 . and � H p ( V , E ) := H p (Ω) = ker ∂ | Ω p Im ∂ | Ω p + 1 . Recall that H p ( V , E ) and H p ( V , E ) are dual and hence their dimensions are the same. The values of dimH p ( V , E ) can be regarded as invariants of the digraph ( V , E ) . Note that for any p ≥ 0 dimH p (Ω) = dim Ω p − dim ∂ Ω p − dim ∂ Ω p + 1 16

Sometimes it is useful to be able to determine the homology groups H n directly via the spaces A n , without Ω n , as in the next statement. Proposition We have H n = ker ∂ | A n / ( A n ∩ ∂ A n + 1 ) (1) and dim H n = dim A n − dim ∂ A n − dim ( A n ∩ ∂ A n + 1 ) . (2) 17

Let us define the Euler characteristic of the digraph ( V , E ) by n � ( − 1 ) p dimH p (Ω) χ ( V , E ) = p = 0 provided n is so big that dimH p (Ω) = 0 for all p > n . 18

Example Consider the graph of 6 vertices V = { 0 , 1 , 2 , 3 , 5 } with 8 edges E = { 01 , 02 , 13 , 14 , 23 , 24 , 53 , 54 } . 19

Let us compute the spaces Ω p and the homologies H p (Ω) . We have Ω 0 = A 0 = span { e 0 , e 1 , e 2 , e 3 , e 4 , e 5 } , dim Ω 0 = 6 Ω 1 = A 1 = span { e 01 , e 02 , e 13 , e 14 , e 23 , e 24 , e 53 , e 54 } , dim Ω 1 = 8 A 2 = span { e 013 , e 014 , e 023 , e 024 } , dim A 2 = 4 The set of semi-edges is S = { e 03 , e 04 } so that dim Ω 2 = dim A 2 − | S | = 2. The basis in Ω 2 can be easily spotted as each of two squares 0 , 1 , 2 , 3 and 0 , 1 , 2 , 4 determine a ∂ -invariant 2-path, whence Ω 2 = span { e 013 − e 023 , e 014 − e 024 } Since there are no allowed 3-paths, we see that A 3 = Ω 3 = { 0 } . It follows that χ = dim Ω 0 − dim Ω 1 + dim Ω 2 = 6 − 8 + 2 = 0 20

Let us compute dim H 2 : dimH 2 = dim Ω 2 − dim ∂ Ω 2 − dim ∂ Ω 3 = 2 − dim ∂ Ω 2 . The image ∂ Ω 2 is spanned by two 1-paths ∂ ( e 013 − e 023 ) = e 13 − e 03 + e 01 − ( e 23 − e 03 + e 02 ) = e 13 + e 01 − e 23 − e 02 ∂ ( e 014 − e 024 ) = e 14 − e 04 + e 01 − ( e 24 − e 04 + e 02 ) = e 14 + e 01 − e 24 − e 02 that are clearly linearly independent. Hence, dim ∂ Ω 2 = 2 whence dimH 2 = 0. 21

The dimension of H 1 can be computed similarly, but we can do easier using the Euler characteristic: dimH 0 − dimH 1 + dimH 2 = χ = 0 whence dim H 1 = 1. In fact, a non-trivial element of H 1 is determined by 1-path v = e 13 − e 14 − e 53 + e 54 Indeed, by a direct computation ∂ v = 0, so that v ∈ ker ∂ | Ω 1 while for v to be in Im ∂ | Ω 2 , it should be a linear combination of ∂ ( e 013 − e 023 ) and ∂ ( e 014 − e 024 ) , which is not possible since they do not have the term e 54 . 22

Example 4 3 1 2 5 Figure: planar graph with non-trivial homology group H 2 23

A direct computation shows that H 1 ( G , K ) = { 0 } and H 2 ( G , K ) ∼ = K , where H 2 ( G ) is generated by e 124 + e 234 + e 314 − ( e 125 + e 235 + e 315 ) . It is easy to see that G is a planar graph but nevertheless its second homology group is non-zero. This shows that the digraph homologies “see” some non-trivial intrinsic dimensions of digraphs that are not necessarily related to embedding properties. 24