SPANNING
TREES
Richard P. Stanley
Department of Mathematics
M.I.T. 2-375
Cambridge, MA 02139
rstan@math.mit.edu
http://www-math.mit.edu/~rstan
Transparencies available at:
http://www-math.mit.edu/~rstan/trans.html
1
G
= loopless graph with verties 1; 2; : : : ; n
spanning forest:
subgraph with verties
1; : : : ; n and no yles
spanning tree:
onneted spanning forest
2
Let (G) = number of spanning trees
(or omplexity) of G.
Origin (Kirhho).
Suppose that eah
edge of G is a unit resistane. Let
u; v
Ruv (G)
= distint verties of G
= total resistane of the network
between u and v
0 = G with u and v identied:
G
Then
(G0)
:
Ruv (G) =
(G)
3
Let Kn be the omplete graph on 1; : : : ; n.
Thus (Kn) is the total number of trees
on 1; : : : ; n.
Theorem (Borhardt, Sylvester, Cay-
ley) (Kn) = nn 2.
First proof (Joyal, 1981).
The number of ways to hoose a spanning tree
T of Kn and two verties u and v is
n2(Kn).
4
u
11
13
3
10
4
v 9
6
7
1
8
14
5
2
15
12
1 3 4 7 9 10 11 13
11 13 4 10 3 7 1 9
7
1
3
4
13
9
11
10
12
6
10
1
3
5
2
15
8
4
14
9
13
11
5
7
We get the graph of a funtion
f : f1; : : : ; ng ! f1; : : : ; ng;
i.e., i!f (i). There are nn suh funtions, so
n2(Kn) = nn:
6
(Pitman, 1997). Let
Rn be the set of all rooted forests on
1; : : : ; n. Dene F to over F 0 in Rn if
F 0 an be obtained from F by removing
one edge e and rooting the \detahed"
tree at the vertex inident to e.
Seond proof
7
Rn beomes a ranked poset.
The elements of rank i are the rooted forests
with i edges.
1
2
3
2
1
3
1
3
2
2
1
2
2
3
1
3
1
1
3
3
1
1 2 3
8
2
2
3
2
1
3
2
1
3
1
2
2
3
2
3
1
3
1
2
1
3
3
1
2
Let
Mn
= #(maximal hains of Rn):
Choose a maximal element of Rn in n (Kn) ways, then remove an edge in n
1 ways, then another edge in n 2 ways,
et. Hene
Mn = (n
1)! n (Kn) = n! (Kn):
9
If F has i edges, then F
is overed by n(n i 1) elements of
Rn.
Lemma.
Proof.
Hene
Mn = n(n 1) n(n 2) n(1)
= n! nn 2:
Sine also Mn = n! (Kn), we get
(Kn) = nn 2. 2
10
The Aylotope
Let o be an orientation of G. Let
di = di(o) be the outdegree of vertex
i. Dene the outdegree sequene
d(o) = (d1; : : : ; dn) 2 R n :
Dene the aylotope AG by
AG = onvfd(o) : o is an
orientation of Gg R n :
dim AG = n
11
1
2
210
201
1
3
120
111
(0,1,2)
102
2
021
012
1
3
(1,1,1)
12
(Zaslavsky). The following are equivalent:
Theorem
(a) d(o) is a vertex of
(b) o is an
ayli
AG.
orientation.
() d(o) is uniquely realizable by o.
(Zaslavsky). AG \ Z n =
fd(o) : o is an orientation of Gg
Theorem
Let
AG; r) = # (rAG \ Z n ) 2 Q [r℄;
the Ehrhart polynomial of AG. Easy:
i( P ; r ) = V ( P ) r n + O ( r n ) :
i(
1
13
2
Let
fi (G)
= # i edge spanning forests of G:
Theorem.
i(AG; r) =
nX1
i=0
fi(G)ri.
Theorem.
V (AG) = fn 1(G) = (G)
(with respet to the integer lattie in
the aÆne span of AG).
14
210
120
201
102
111
012
021
Put r = 1 to get:
Corollary. The number of spanning forests of G equals the number
of distint outdegree sequenes of
orientations of G.
(no simple proof known)
15
The Matrix-Tree Theorem
G = (loopless) graph with verties v1; : : : ; vn
e = edge of G; xe = indeterminate
T = spanning tree with edge set E (T )
T
x =
QG(x)
Y
e2E (T )
=
X
T
xe
xT ;
summed over all spanning trees of G.
16
1
2
4
3
QG(x) = x1x2x3+x1x2x4+x1x3x4+x2x3x4
17
Dene the generi Laplaian man
of G by
trix L = Lij
1
Lij =
L0
Note.
8
>
>
>
>
>
<
>
>
>
>
>
:
X
xe; i 6= j
inident
to vi ;vj
e
X
e
xe; i = j
inident
to vi
= L with last row and
last olumn removed
det L = 0
Theorem
(Kirhho, : : :).
det L0 = QG(x):
18
An equivalent form:
Let 1; : : : ; n be the
eigenvalues of L with n = 0. Then
1
QG(x) = (1 n 1) :
n
Theorem.
Speial ase (all xe = 1):
Let (i; j ) be the number of edges between vi and vj . Dene
Corollary.
(
L(1)ij =
(i; j ); i 6= j
deg(vi); i = j:
Then (G) = det L(1)0.
19
Example.
ube.
Cn = graph of the n-
C3
Symmetry of Cn ) eigenvalues
of L(1)
are 2k with multipliity nk , where 0 k n.
20
Corollary.
(Cn) = 2
n
2
n
1
Is there a ombinatorial proof?
21
n
Y
k ( ):
n
k
k=1
A Conjeture of Kontsevih
q
G(q)
= prime power
= # solutions to QG(x) 6= 0 over F q
QG(x) = x1 + x2 + x3 + x4
G(q) = q3(q 1) 2 Z [q℄
22
G (q ) = q (q
1)(q 2
23
2) 2 Z [q ℄
(M. Kontsevih, 8 De.
1997). For any G,
Conjeture
G(q) 2 Z [q℄:
(still open)
Note.
G (q ) = q jE j + O q jE j
G(q) 2 Q [q℄ ) G(q) 2
1
Z [q ℄
(by rationality of the zeta funtion
of a variety=F q )
24
Let L be the generi Laplaian
of G. Matrix-Tree Theorem )
Note.
G(q) = number of solutions to
det L0 6= 0 over F q
25
Suppose vn is an apex, i.e., is adjaent to v1; : : : ; vn 1.
4
5
3
4
1
1
2
3
2
3
2
x1 + x3
x1
0
7
L0 = 6
x2 5
x1 x1 + x2 + x4
4
0
x2
x2 + x5
Can hange signs of o-diagonal entries
and kill non-red diagonal entries without aeting set of L0's over F q .
26
Let G be simple (no
multiple edges), and let vn be an apex
of G. Then
Corollary.
1) (n 1) nonsingular
symmetri matries M over F q
suh that
G(q) = # (n
Mij = 0 if i 6= j; ij 62 E:
Example. G = Kn (omplete graph)
)
1) (n 1) nonsingular
symmetri matries=F q :
K (q) = # (n
n
27
This number was omputed by Carlitz
(1954) and MaWilliams (1969).
Theorem.
qm(m
1)
K (q ) =
n
(q 1)(q 3 1) (q 2m
qm(m+1)(q 1)(q3 1) (q2m
1
1
1); n = 2m
1);
n = 2m + 1;
so K (q ) 2 Z [q ℄.
n
Stembridge showed Kontsevih's onjeture holds for all graphs with
12 edges.
Note.
28
Related Problems
Let A be an n n matrix over a eld
K . Dene
supp(A) = f(i; j ) : Aij =
6 0g:
Let S f1; : : : ; ng f1; : : : ; ng
suh that (i; j ) 2 S , (j; i) 2 S . Let
1.
eS (q) = # nn nonsingular symmetri
matries M over F q ; supp(M ) S:
Is eS (q ) 2 Q [q ℄? (Kontsevih onjeture ) yes if eah (i; i) 2 S .)
No: If n is odd with Aii = 0, then
det A = 0 when q = 2j . There are
also examples for whih eS (q ) is a polynomial for odd q but not all q , and
eS (2j ) 6= 0.
29
Is eS (q ) a polynomial for
(open)
2.
Let S
Dene
3.
odd
q?
f1; : : : ; ng f1; : : : ; ng.
dS (q) = # nn nonsingular matries M
over F q ; supp(M ) S:
Is dS (q ) 2 Q [q ℄?
Kontsevih laimed 9 ounterexample.
One was found by Stembridge, with n =
7 and #S = 21.
4. Is dS (q ) a polynomial for odd q ?
(Open | andidate for ounterexample
with n = 13, #S = 52.)
30
Are (q ), dS (q ), eS (q ) quasipolynomials, i.e., for some N are they polynomials on residue lasses modulo N ?
(open)
6. Is Kontsevih's onjeture true for
bases of matroids? (False, even for regular matroids.)
5.
31