— Divisors equivalent to 0 also form a group, Prin G . (This was S 0 in our previous example.) Div 0 G — The Jacobian of a graph is Jac G Prin G . (This was 3 in our previous example.) The Jacobian provides structure to the divisors, which helps us study them. the jacobian group From an algebraic perspective: — Divisors of degree zero form a group, Div 0 ( G ) . — Toppling is an equivalence relation between divisors. 17
Div 0 G — The Jacobian of a graph is Jac G Prin G . (This was 3 in our previous example.) The Jacobian provides structure to the divisors, which helps us study them. the jacobian group From an algebraic perspective: — Divisors of degree zero form a group, Div 0 ( G ) . — Toppling is an equivalence relation between divisors. — Divisors equivalent to 0 also form a group, Prin ( G ) . (This was S 0 in our previous example.) 17
The Jacobian provides structure to the divisors, which helps us study them. the jacobian group From an algebraic perspective: — Divisors of degree zero form a group, Div 0 ( G ) . — Toppling is an equivalence relation between divisors. — Divisors equivalent to 0 also form a group, Prin ( G ) . (This was S 0 in our previous example.) — The Jacobian of a graph is Jac ( G ) = Div 0 ( G ) / Prin ( G ) . (This was Z 3 in our previous example.) 17
the jacobian group From an algebraic perspective: — Divisors of degree zero form a group, Div 0 ( G ) . — Toppling is an equivalence relation between divisors. — Divisors equivalent to 0 also form a group, Prin ( G ) . (This was S 0 in our previous example.) — The Jacobian of a graph is Jac ( G ) = Div 0 ( G ) / Prin ( G ) . (This was Z 3 in our previous example.) The Jacobian provides structure to the divisors, which helps us study them. 17
Surprisingly, the number of elements in Jac G is the number of spanning trees! Our definition didn’t relate to spanning trees at all, but suddenly they’re involved. Why? spanning trees A subgraph with n − 1 edges that connects every vertex is called a spanning tree. 18
spanning trees A subgraph with n − 1 edges that connects every vertex is called a spanning tree. Surprisingly, the number of elements in Jac ( G ) is the number of spanning trees! Our definition didn’t relate to spanning trees at all, but suddenly they’re involved. Why? 18
biconnected graphs We take the wedge sum of two graphs by taking two vertices and making them the same: ∧ = It is known that Jac ( G 1 ∧ G 2 ) = Jac ( G 1 ) ⊕ Jac ( G 2 ) . 19
These parts can’t be split into any more parts, so they’re called biconnected. Therefore, to study the Jacobian, we only need to study the Jacobians of biconnected graphs. biconnected graphs We can split a graph into parts that we can wedge sum together: 20
Therefore, to study the Jacobian, we only need to study the Jacobians of biconnected graphs. biconnected graphs We can split a graph into parts that we can wedge sum together: These parts can’t be split into any more parts, so they’re called biconnected. 20
biconnected graphs We can split a graph into parts that we can wedge sum together: These parts can’t be split into any more parts, so they’re called biconnected. Therefore, to study the Jacobian, we only need to study the Jacobians of biconnected graphs. 20
For example, the exponent of 5 is 5, because for any a 5 , we have 5 a 0 Similarly, the exponent in 2 is 2: 2 0 1 0 1 0 0 1 0 1 0 0 0 1 1 1 1 0 0 Since Jac G is a finite abelian group, it has an exponent. a a a a a the exponent The exponent of a group H , is the smallest natural number k such that for all a in H , ka = 0. 21
Similarly, the exponent in 2 is 2: 2 0 1 0 1 0 0 1 0 1 0 0 0 1 1 1 1 0 0 Since Jac G is a finite abelian group, it has an exponent. the exponent The exponent of a group H , is the smallest natural number k such that for all a in H , ka = 0. For example, the exponent of Z 5 is 5, because for any a ∈ Z 5 , we have a + a + a + a + a = 5 a = 0 . 21
Since Jac G is a finite abelian group, it has an exponent. the exponent The exponent of a group H , is the smallest natural number k such that for all a in H , ka = 0. For example, the exponent of Z 5 is 5, because for any a ∈ Z 5 , we have a + a + a + a + a = 5 a = 0 . Similarly, the exponent in Z 2 ⊕ Z 2 is 2: ( 0 , 1 ) + ( 0 , 1 ) = ( 0 , 0 ) ( 1 , 0 ) + ( 1 , 0 ) = ( 0 , 0 ) ( 1 , 1 ) + ( 1 , 1 ) = ( 0 , 0 ) . 21
the exponent The exponent of a group H , is the smallest natural number k such that for all a in H , ka = 0. For example, the exponent of Z 5 is 5, because for any a ∈ Z 5 , we have a + a + a + a + a = 5 a = 0 . Similarly, the exponent in Z 2 ⊕ Z 2 is 2: ( 0 , 1 ) + ( 0 , 1 ) = ( 0 , 0 ) ( 1 , 0 ) + ( 1 , 0 ) = ( 0 , 0 ) ( 1 , 1 ) + ( 1 , 1 ) = ( 0 , 0 ) . Since Jac ( G ) is a finite abelian group, it has an exponent. 21
Note that this can’t be true when the graph isn’t biconnected. If Jac G 1 2 and Jac G 2 2 , then Jac G 1 2 , which G 2 2 has exponent 2 as well. So we can make infinitely graphs that have exponent 2, but when it’s biconnected, there may be finitely many. We now show one way of proving this conjecture for k 2 and k 3. the main conjecture Conjecture For every positive integer k , there are finitely many biconnected graphs G such that the exponent of Jac ( G ) is k . 22
So we can make infinitely graphs that have exponent 2, but when it’s biconnected, there may be finitely many. We now show one way of proving this conjecture for k 2 and k 3. the main conjecture Conjecture For every positive integer k , there are finitely many biconnected graphs G such that the exponent of Jac ( G ) is k . Note that this can’t be true when the graph isn’t biconnected. If Jac ( G 1 ) = Z 2 and Jac ( G 2 ) = Z 2 , then Jac ( G 1 ∧ G 2 ) = Z 2 ⊕ Z 2 , which has exponent 2 as well. 22
We now show one way of proving this conjecture for k 2 and k 3. the main conjecture Conjecture For every positive integer k , there are finitely many biconnected graphs G such that the exponent of Jac ( G ) is k . Note that this can’t be true when the graph isn’t biconnected. If Jac ( G 1 ) = Z 2 and Jac ( G 2 ) = Z 2 , then Jac ( G 1 ∧ G 2 ) = Z 2 ⊕ Z 2 , which has exponent 2 as well. So we can make infinitely graphs that have exponent 2, but when it’s biconnected, there may be finitely many. 22
the main conjecture Conjecture For every positive integer k , there are finitely many biconnected graphs G such that the exponent of Jac ( G ) is k . Note that this can’t be true when the graph isn’t biconnected. If Jac ( G 1 ) = Z 2 and Jac ( G 2 ) = Z 2 , then Jac ( G 1 ∧ G 2 ) = Z 2 ⊕ Z 2 , which has exponent 2 as well. So we can make infinitely graphs that have exponent 2, but when it’s biconnected, there may be finitely many. We now show one way of proving this conjecture for k = 2 and k = 3. 22
Consider a divisor with degree zero. Let’s name one vertex q : Because it has degree zero, we know what q is. Think of the label of each vertex as the number of firefighters on that vertex. dhar’s burning algorithm We introduce Dhar’s burning algorithm with the following example. 23
Think of the label of each vertex as the number of firefighters on that vertex. dhar’s burning algorithm We introduce Dhar’s burning algorithm with the following example. Consider a divisor with degree zero. Let’s name one vertex q : 1 2 0 2 0 q Because it has degree zero, we know what q is. 23
dhar’s burning algorithm We introduce Dhar’s burning algorithm with the following example. Consider a divisor with degree zero. Let’s name one vertex q : 1 2 0 2 0 q Because it has degree zero, we know what q is. Think of the label of each vertex as the number of firefighters on that vertex. 23
The top four vertices are protected, but the lower-right vertex is not protected. It will burn the next turn. dhar’s burning algorithm We start a fire at vertex q . The fire spreads through the edges. 1 2 0 2 q 0 A vertex is safe as long as the number of firefighters is at least the number of burning edges next to it. 24
dhar’s burning algorithm We start a fire at vertex q . The fire spreads through the edges. 1 2 0 2 q 0 A vertex is safe as long as the number of firefighters is at least the number of burning edges next to it. The top four vertices are protected, but the lower-right vertex is not protected. It will burn the next turn. 24
dhar’s burning algorithm In this graph, everything eventually burns: 1 1 1 2 0 2 0 2 0 2 2 2 q 0 q 0 q 0 1 1 1 2 0 2 0 2 0 2 2 2 q 0 q 0 q 0 25
— The number assigned to all vertices other than q is non-negative. — Applying Dhar’s burning algorithm on vertex q burns everything. For a graph, it is known that a bijection exists between q -reduced divisors and the elements of Jac G . So every q -reduced divisor corresponds to a different element in Jac G . q -reduced divisor Let q be a vertex. We say a divisor with degree zero is q -reduced if: 26
— Applying Dhar’s burning algorithm on vertex q burns everything. For a graph, it is known that a bijection exists between q -reduced divisors and the elements of Jac G . So every q -reduced divisor corresponds to a different element in Jac G . q -reduced divisor Let q be a vertex. We say a divisor with degree zero is q -reduced if: — The number assigned to all vertices other than q is non-negative. 26
For a graph, it is known that a bijection exists between q -reduced divisors and the elements of Jac G . So every q -reduced divisor corresponds to a different element in Jac G . q -reduced divisor Let q be a vertex. We say a divisor with degree zero is q -reduced if: — The number assigned to all vertices other than q is non-negative. — Applying Dhar’s burning algorithm on vertex q burns everything. 26
So every q -reduced divisor corresponds to a different element in Jac G . q -reduced divisor Let q be a vertex. We say a divisor with degree zero is q -reduced if: — The number assigned to all vertices other than q is non-negative. — Applying Dhar’s burning algorithm on vertex q burns everything. For a graph, it is known that a bijection exists between q -reduced divisors and the elements of Jac ( G ) . 26
q -reduced divisor Let q be a vertex. We say a divisor with degree zero is q -reduced if: — The number assigned to all vertices other than q is non-negative. — Applying Dhar’s burning algorithm on vertex q burns everything. For a graph, it is known that a bijection exists between q -reduced divisors and the elements of Jac ( G ) . So every q -reduced divisor corresponds to a different element in Jac ( G ) . 26
If D is the top center divisor, then 2 D , 3 D , 4 D are not zero. exponent from maximum degree Let ∆ be the maximum degree. Here’s ∆ q -reduced divisors: 0 0 0 0 0 0 0 0 0 0 1 2 q 0 q 0 q 0 0 0 0 0 0 0 3 4 q 0 q 0 27
exponent from maximum degree Let ∆ be the maximum degree. Here’s ∆ q -reduced divisors: 0 0 0 0 0 0 0 0 0 0 1 2 q 0 q 0 q 0 0 0 0 0 0 0 3 4 q 0 q 0 If D is the top center divisor, then 2 D , 3 D , 4 D are not zero. 27
exponent from maximum degree Let ∆ be the maximum degree. Here’s ∆ q -reduced divisors: 0 0 0 0 0 0 0 0 0 0 1 2 q 0 q 0 q 0 0 0 0 0 0 0 3 4 q 0 q 0 So D has order ≥ ∆ , so the exponent of Jac ( G ) ≥ ∆ . 27
Suppose the exponent is 3, then 3 . It’s known if equality holds, then the graph has 2 vertices and edges. Otherwise, 2, so it must be C 3 . So the graphs with exponent 3 are: We now explain a different way to prove the conjecture for k 2. k = 2 and k = 3 Suppose the exponent of Jac ( G ) is 2. As the exponent is at least ∆ , 2 ≥ ∆ . The graph is biconnected, so it must be a cycle. It’s known Jac ( C n ) = Z n . So the only graph with exponent 2 is: 28
We now explain a different way to prove the conjecture for k 2. k = 2 and k = 3 Suppose the exponent of Jac ( G ) is 2. As the exponent is at least ∆ , 2 ≥ ∆ . The graph is biconnected, so it must be a cycle. It’s known Jac ( C n ) = Z n . So the only graph with exponent 2 is: Suppose the exponent is 3, then 3 ≥ ∆ . It’s known if equality holds, then the graph has 2 vertices and ∆ edges. Otherwise, ∆ = 2, so it must be C 3 . So the graphs with exponent 3 are: 28
k = 2 and k = 3 Suppose the exponent of Jac ( G ) is 2. As the exponent is at least ∆ , 2 ≥ ∆ . The graph is biconnected, so it must be a cycle. It’s known Jac ( C n ) = Z n . So the only graph with exponent 2 is: Suppose the exponent is 3, then 3 ≥ ∆ . It’s known if equality holds, then the graph has 2 vertices and ∆ edges. Otherwise, ∆ = 2, so it must be C 3 . So the graphs with exponent 3 are: We now explain a different way to prove the conjecture for k = 2. 28
Cycle and cut spaces
Here’s an example of an incidence matrix. e 1 e 2 e 3 e 4 e 5 1 0 1 0 0 v 1 1 1 0 0 1 v 2 0 1 1 1 0 v 3 0 0 0 1 1 v 4 D incidence matrix The incidence matrix describes which vertices are connected by which edges. v 2 e 1 e 5 v 1 e 2 v 4 e 3 e 4 v 3 29
incidence matrix The incidence matrix describes which vertices are connected by which edges. Here’s an example of an incidence matrix. v 2 e 1 e 2 e 3 e 4 e 5 e 1 e 5 − 1 0 1 0 0 v 1 v 1 e 2 v 4 1 − 1 0 0 1 v 2 D = e 3 e 4 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 3 v 4 29
Here are two vectors in C 1 : The inner product of these two is 3 1 3 1 2 2 1 1 1 1 0 So these two vectors are orthogonal as their inner product is 0. edge space The edge space C 1 is the vector space of functions f : E → R . An inner product is given by multiplying corresponding edges then adding them. We can think of this as assigning a number to each edge, but we have an inner product. 30
So these two vectors are orthogonal as their inner product is 0. edge space The edge space C 1 is the vector space of functions f : E → R . An inner product is given by multiplying corresponding edges then adding them. We can think of this as assigning a number to each edge, but we have an inner product. Here are two vectors in C 1 : − 1 3 1 1 2 − 2 3 − 1 1 1 The inner product of these two is 3 · 1 + 3 · 1 + 2 · ( − 2 ) + ( − 1 ) · 1 + ( − 1 ) · 1 = 0 . 30
edge space The edge space C 1 is the vector space of functions f : E → R . An inner product is given by multiplying corresponding edges then adding them. We can think of this as assigning a number to each edge, but we have an inner product. Here are two vectors in C 1 : − 1 3 1 1 2 − 2 3 − 1 1 1 The inner product of these two is 3 · 1 + 3 · 1 + 2 · ( − 2 ) + ( − 1 ) · 1 + ( − 1 ) · 1 = 0 . So these two vectors are orthogonal as their inner product is 0. 30
The cycle space Z is formed by linear combinations of z Q . For example, the third vector is the first cycle plus the second cycle, so it’s in Z . cycles − 1 1 Let Q be a cycle. The vector z Q ∈ C 1 is: 0 1 − 1 0 if e is not in Q , 1 if e is aligned with Q , z Q ( e ) = 1 0 − 1 if e goes opposite as Q . 1 1 0 − 1 2 1 2 − 1 31
cycles − 1 1 Let Q be a cycle. The vector z Q ∈ C 1 is: 0 1 − 1 0 if e is not in Q , 1 if e is aligned with Q , z Q ( e ) = 1 0 − 1 if e goes opposite as Q . 1 The cycle space Z is formed by linear 1 0 combinations of z Q . For example, the third vector is the first cycle plus the second − 1 2 cycle, so it’s in Z . 1 2 − 1 31
For the first z Q , 0. D e 1 D e 2 D e 3 So for every z Z , D z 0. It is known that every vector that satisfies 0 is in Z . So ker D Z . D v D z Q cycles D can be seen as a function R | E | → R | V | . e 1 e 5 e 1 e 2 e 3 e 4 e 5 − 1 0 1 0 0 e 2 v 1 1 − 1 0 0 1 v 2 e 3 e 4 D = 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 4 1 0 1 1 0 − 1 1 0 1 − 1 32
So for every z Z , D z 0. It is known that every vector that satisfies 0 is in Z . So ker D Z . D v cycles D can be seen as a function R | E | → R | V | . e 1 e 5 e 1 e 2 e 3 e 4 e 5 − 1 0 1 0 0 e 2 v 1 1 − 1 0 0 1 v 2 e 3 e 4 D = 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 4 1 0 1 For the first z Q , 1 0 D ( z Q ) = D ( e 1 ) + D ( e 2 ) + D ( e 3 ) = 0. − 1 1 0 1 − 1 32
So for every z Z , D z 0. It is known that every vector that satisfies 0 is in Z . So ker D Z . D v cycles D can be seen as a function R | E | → R | V | . e 1 e 5 e 1 e 2 e 3 e 4 e 5 − 1 0 1 0 0 e 2 v 1 1 − 1 0 0 1 v 2 e 3 e 4 D = 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 4 1 0 1 For the second z Q , 1 0 D ( z Q ) = D ( e 1 ) − D ( e 5 ) − D ( e 4 ) + D ( e 3 ) = 0. − 1 1 0 1 − 1 32
It is known that every vector that satisfies 0 is in Z . So ker D Z . D v cycles D can be seen as a function R | E | → R | V | . e 1 e 5 e 1 e 2 e 3 e 4 e 5 − 1 0 1 0 0 e 2 v 1 1 − 1 0 0 1 v 2 e 3 e 4 D = 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 4 1 0 1 It can similarly be shown for every z Q that 1 0 D ( z Q ) = 0. − 1 1 So for every z ∈ Z , D ( z ) = 0. 0 1 − 1 32
cycles D can be seen as a function R | E | → R | V | . e 1 e 5 e 1 e 2 e 3 e 4 e 5 − 1 0 1 0 0 e 2 v 1 1 − 1 0 0 1 v 2 e 3 e 4 D = 0 1 − 1 − 1 0 v 3 0 0 0 1 − 1 v 4 1 0 1 It can similarly be shown for every z Q that 1 0 D ( z Q ) = 0. − 1 1 So for every z ∈ Z , D ( z ) = 0. 0 It is known that every vector that satisfies 1 − 1 D ( v ) = 0 is in Z . So ker D = Z . 32
The cut space B is formed by linear combinations of b U . For example, the third vector is the sum of the first cut and the second cut, so it’s in B . cuts 1 0 Let U be a subset of vertices. Then the vector b U ∈ C 1 is: − 1 0 1 1 if e ’s head is in U , − 1 if e ’s tail is in U , b U ( e ) = 0 0 0 otherwise. − 1 1 1 1 0 − 2 1 2 33
cuts 1 0 Let U be a subset of vertices. Then the vector b U ∈ C 1 is: − 1 0 1 1 if e ’s head is in U , − 1 if e ’s tail is in U , b U ( e ) = 0 0 0 otherwise. − 1 1 1 The cut space B is formed by linear combinations of b U . For example, the third vector is the sum of the first cut and the 1 0 second cut, so it’s in B . − 2 1 2 33
Let z Z and b B . Say z 2 z Q z Q and b b U . Then 2 z Q 2 z Q b U z Q b U z Q b U 2 z Q b U z Q b U 2 0 0 By writing out elements as linear combinations of cycles and cuts, we can prove Z and B are orthogonal spaces. cycles and cuts It is known that any cycle z Q and any cut b U are orthogonal: − 1 1 0 0 − 1 0 1 − 1 1 1 34
Then 2 z Q 2 z Q b U z Q b U z Q b U 2 z Q b U z Q b U 2 0 0 By writing out elements as linear combinations of cycles and cuts, we can prove Z and B are orthogonal spaces. cycles and cuts It is known that any cycle z Q and any cut b U are orthogonal: − 1 1 0 0 − 1 0 1 − 1 1 1 Let z ∈ Z and b ∈ B . Say z = 2 z Q − z Q ′ and b = b U . 34
By writing out elements as linear combinations of cycles and cuts, we can prove Z and B are orthogonal spaces. cycles and cuts It is known that any cycle z Q and any cut b U are orthogonal: − 1 1 0 0 − 1 0 1 − 1 1 1 Let z ∈ Z and b ∈ B . Say z = 2 z Q − z Q ′ and b = b U . Then ⟨ 2 z Q − z Q ′ , b U ⟩ = ⟨ 2 z Q , b U ⟩ + ⟨− z Q ′ , b U ⟩ = 2 ⟨ z Q , b U ⟩ − ⟨ z Q ′ , b U ⟩ = 2 · 0 − 0 . 34
cycles and cuts It is known that any cycle z Q and any cut b U are orthogonal: − 1 1 0 0 − 1 0 1 − 1 1 1 Let z ∈ Z and b ∈ B . Say z = 2 z Q − z Q ′ and b = b U . Then ⟨ 2 z Q − z Q ′ , b U ⟩ = ⟨ 2 z Q , b U ⟩ + ⟨− z Q ′ , b U ⟩ = 2 ⟨ z Q , b U ⟩ − ⟨ z Q ′ , b U ⟩ = 2 · 0 − 0 . By writing out elements as linear combinations of cycles and cuts, we can prove Z and B are orthogonal spaces. 34
So anything in C 1 is a sum of two orthogonal vectors: one in Z and one in B . orthogonal decomposition Not only that, but it is known that C 1 = Z ⊕ B . 35
orthogonal decomposition Not only that, but it is known that C 1 = Z ⊕ B . So anything in C 1 is a sum of two orthogonal vectors: one in Z and one in B . 35
orthogonal decomposition Not only that, but it is known that C 1 = Z ⊕ B . So anything in C 1 is a sum of two orthogonal vectors: one in Z and one in B . 3 − 1 1 1 2 0 2 2 2 2 − 1 0 = 1 + 2 0 3 − 1 1 1 2 2 2 2 = = 3 1 2 2 3 − 1 2 2 3 1 2 2 + + − 1 1 2 2 − 1 − 1 2 2 − 1 1 2 2 35
— C 1 is the edge space, like “write a number on each edge”. — Z is the cycle space, formed by sums of z Q s. It’s also ker D , where D is the incidence matrix. — B is the cut space, formed by sums of b U s. Z and B are orthogonal spaces, and C 1 B . Z orthogonal decomposition Let’s recap: 36
— Z is the cycle space, formed by sums of z Q s. It’s also ker D , where D is the incidence matrix. — B is the cut space, formed by sums of b U s. Z and B are orthogonal spaces, and C 1 B . Z orthogonal decomposition Let’s recap: — C 1 is the edge space, like “write a number on each edge”. 36
— B is the cut space, formed by sums of b U s. Z and B are orthogonal spaces, and C 1 B . Z orthogonal decomposition Let’s recap: — C 1 is the edge space, like “write a number on each edge”. — Z is the cycle space, formed by sums of z Q s. It’s also ker D , where D is the incidence matrix. 36
Z and B are orthogonal spaces, and C 1 B . Z orthogonal decomposition Let’s recap: — C 1 is the edge space, like “write a number on each edge”. — Z is the cycle space, formed by sums of z Q s. It’s also ker D , where D is the incidence matrix. — B is the cut space, formed by sums of b U s. 36
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