David M. BressoudMacalester College, St. Paul, MNTalk given MAA KY sectionMarch 28, 2008
We mathematicians often deludeourselves into thinking that we createproofs in order to establish truth.
In fact, that which is "proven" is oftennot true, and mathematical results areoften known with certainty to be truelong before a proof is found.
Imre Lakatos, 1922–1974
Born Imre Lipschitz, Jewish, Hungarian
Changed name to Imre Molnar during war soas not to be identified as Jewish
Mother and grandmother died in Auschwitz
1944, graduated college with degrees inmathematics, physics, and philosophy
After war changed his name to Imre Lakatos(Locksmith) because he had shirtsmonogrammed IL.
Imre Lakatos, 1922–1974
Studied in Budapest and at Moscow StateUniversity
1947: position in Ministry of Education.
1950: arrested, three years in prison
1956: Hungarian uprising, flees to Vienna,then on to England
1961: PhD in Philosophy at Cambridge
Sir Karl Popper, The Logic of Scientific Discovery
Science advances by
1. Observing nature
2. Creating a theory to explain what is happening
3. Looking for consequences of this theory
4. Testing the predicted consequences
5. Adjusting the theory when predictions do not pan out
In science, nothing can be proven to be true. Realprogress in science comes from establishing thatsomething is false.
Lakatos: Is this relevant to Mathematics?
Lakatos: Is this relevant to Mathematics?
Icosahedron
20 faces
30 edges
12 vertices
20 – 30 + 12 = 2
Theorem (Euler): For allpolyhedra, V – E + F = 2.
Lakatos: Is this relevant to Mathematics?
Icosahedron
20 faces
30 edges
12 vertices
20 – 30 + 12 = 2
Theorem (Euler): For allpolyhedra, V – E + F = 2.
Definition: A polyhedron is a solid whosesurface consists of polygonal faces.
V = 16
E = 24
F = 12
16 – 24 + 12 = 4
V = 7
E = 12
F = 8
7 – 12 + 8 = 3
Small StellatedDodecahedron
12 faces (pentagrams)
30 edges
12 vertices
12 – 30 + 12 = –6
For any polygon,V = E.
Can even be aself-intersectingpolygon:
Great Stellated Dodecahedron
Great Stellated Dodecahedron
Great Stellated Dodecahedron
V = 20
E = 3×20÷2 = 30
F = 2×30÷5 = 12
20 – 30 + 12 = 2
Great Stellated Dodecahedron
V = 20
E = 30
F = 12
20 – 30 + 12 = 2
What’s different?
Great Stellated Dodecahedron
V = 20
E = 30
F = 12
20 – 30 + 12 = 2
What’s different?
Every closed curve made up ofedges is the boundary of achain of contiguous faces(faces that share an edge).
Here we have aclosed circuit of facesthat is not theboundary of the solid.
Here we have a closedcircuit of edges that isnot the boundary of achain of contiguousfaces.
The appendix from Lakatos’sProof and Refutations wouldbe the inspiration for my ownA Radical Approach to RealAnalysis
Cauchy, Cours d’analyse, 1821
“…explanations drawn from algebraictechnique … cannot be considered, in myopinion, except as heuristics that willsometimes suggest the truth, but whichaccord little with the accuracy that is sopraised in the mathematical sciences.”
1789–1857
Niels Henrik Abel (1826):
“Cauchy is crazy, and there is no way ofgetting along with him, even though rightnow he is the only one who knows howmathematics should be done. What he isdoing is excellent, but very confusing.”
1802–1829
Cauchy, Cours d’analyse, 1821, p. 120
Theorem 1. When the terms of a series are functions ofa single variable x and are continuous with respect tothis variable in the neighborhood of a particular valuewhere the series converges, the sum S(x) of the series isalso, in the neighborhood of this particular value, acontinuous function of x.
S x( ) = fk x( )k=1
!
" , fk continuous # S continuous
Sn x( ) = fk x( )k=1
n
! , Rn x( ) = S x( ) " Sn x( )
Convergence # can make Rn x( ) as small as we wish
by taking n sufficiently large. Sn is continuous for n < $.
Sn x( ) = fk x( )k=1
n
! , Rn x( ) = S x( ) " Sn x( )
Convergence # can make Rn x( ) as small as we wish
by taking n sufficiently large. Sn is continuous for n < $.
S continuous at a if can force S(x) - S(a)
as small as we wish by restricting x ! a .
Sn x( ) = fk x( )k=1
n
! , Rn x( ) = S x( ) " Sn x( )
Convergence # can make Rn x( ) as small as we wish
by taking n sufficiently large. Sn is continuous for n < $.
S continuous at a if can force S(x) - S(a)
as small as we wish by restricting x ! a .
S x( ) ! S a( ) = Snx( ) + R
nx( ) ! S
na( ) ! R
na( )
" Snx( ) ! S
na( ) + R
nx( ) + R
na( )
Abel, 1826:
“It appears to me that thistheorem suffers exceptions.”
sin x !1
2sin2x +
1
3sin 3x !
1
4sin 4x +!
Sn x( ) = fk x( )k=1
n
! , Rn x( ) = S x( ) " Sn x( )
Convergence # can make Rn x( ) as small as we wish
by taking n sufficiently large. Sn is continuous for n < $.
S continuous at a if can force S(x) - S(a)
as small as we wish by restricting x ! a .
S x( ) ! S a( ) = Snx( ) + R
nx( ) ! S
na( ) ! R
na( )
" Snx( ) ! S
na( ) + R
nx( ) + R
na( )
x depends on n n depends on x
Sn x( ) = fk x( )k=1
n
! , Rn x( ) = S x( ) " Sn x( )
Convergence # can make Rn x( ) as small as we wish
by taking n sufficiently large. Sn is continuous for n < $.
S x( ) ! S a( ) = Snx( ) + R
nx( ) ! S
na( ) ! R
na( )
" Snx( ) ! S
na( ) + R
nx( ) + R
na( )
x depends on n n depends on x
Uniform convergence eliminates thepossibility that n depends on x.
The existence of a proof does not(necessarily) imply the truth of amathematical statement. But also,the absence of a proof does not(necessarily) mean that we are in anydoubt about the truth of amathematical statement.
Proofs and Confirmations: The Storyof the Alternating Sign MatrixConjecture
Alternating sign matrix:
•Square matrix of 1’s, –1’s,and 0’s
•Each row and columnadds to 1
•Nonzero entries in anyrow or column alternate insign
n
1
2
3
4
5
6
7
8
9
An
1
2
7
42
429
7436
218348
10850216
911835460
= 2 × 3 × 7
= 3 × 11 × 13
= 22 × 11 × 132
= 22 × 132 × 17 × 19
= 23 × 13 × 172 × 192
= 22 × 5 × 172 × 193 × 23
How many n × n alternating signmatrices?
n
1
2
3
4
5
6
7
8
9
An
1
2
7
42
429
7436
218348
10850216
911835460
= 2 × 3 × 7
= 3 × 11 × 13
= 22 × 11 × 132
= 22 × 132 × 17 × 19
= 23 × 13 × 172 × 192
= 22 × 5 × 172 × 193 × 23
n
1
2
3
4
5
6
7
8
9
An
1
2
7
42
429
7436
218348
10850216
911835460
There is exactly one 1in the first row
n
1
2
3
4
5
6
7
8
9
An
1
1+1
2+3+2
7+14+14+7
42+105+…
There is exactly one 1in the first row
1
1 1
2 3 2
7 14 14 7
42 105 135 105 42
429 1287 2002 2002 1287 429
1
1 1
2 3 2
7 14 14 7
42 105 135 105 42
429 1287 2002 2002 1287 429
+ + +
1
1 1
2 3 2
7 14 14 7
42 105 135 105 42
429 1287 2002 2002 1287 429
+ + +
1 0 0 0 0
0 ? ? ? ?
0 ? ? ? ?
0 ? ? ? ?
0 ? ? ? ?
!
"
######
$
%
&&&&&&
1
1 2/2 1
2 2/3 3 3/2 2
7 2/4 14 14 4/2 7
42 2/5 105 135 105 5/2 42
429 2/6 1287 2002 2002 1287 6/2 429
1
1 2/2 1
2 2/3 3 3/2 2
7 2/4 14 5/5 14 4/2 7
42 2/5 105 7/9 135 9/7 105 5/2 42
429 2/6 1287 9/14 2002 16/16 2002 14/9 1287 6/2 429
2/2
2/3 3/2
2/4 5/5 4/2
2/5 7/9 9/7 5/2
2/6 9/14 16/16 14/9 6/2
2
2 3
2 5 4
2 7 9 5
2 9 16 14 6
1+1
1+1 1+2
1+1 2+3 1+3
1+1 3+4 3+6 1+4
1+1 4+5 6+10 4+10 1+5
Numerators:
1+1
1+1 1+2
1+1 2+3 1+3
1+1 3+4 3+6 1+4
1+1 4+5 6+10 4+10 1+5
Conjecture 1:
Numerators:
An,k
An,k+1
=
n ! 2k !1
"#$
%&'+
n !1k !1
"#$
%&'
n ! 2n ! k !1
"#$
%&'+
n !1n ! k !1
"#$
%&'
Conjecture 1:
Conjecture 2 (corollary of Conjecture 1):
An,k
An,k+1
=
n ! 2k !1
"#$
%&'+
n !1k !1
"#$
%&'
n ! 2n ! k !1
"#$
%&'+
n !1n ! k !1
"#$
%&'
An =3 j +1( )!
n + j( )!j=0
n!1
" =1!#4!#7!! 3n ! 2( )!
n!# n +1( )!! 2n !1( )!
Conjecture 2 (corollary of Conjecture 1):
An =3 j +1( )!
n + j( )!j=0
n!1
" =1!#4!#7!! 3n ! 2( )!
n!# n +1( )!! 2n !1( )!
Exactly the formula foundby George Andrews forcounting descending planepartitions.
George AndrewsPenn State
Exactly the formula foundby George Andrews forcounting descending planepartitions.
George AndrewsPenn StateThe attempt to find the relationship between
descending plane partitions and alternating signmatrices would lead to proofs of important openproblems in the study of plane partitions, withconnections to representation theory.
December, 1992
Zeilberger announces aproof that # of ASM’sequals
3 j +1( )!
n + j( )!j=0
n!1
"
1995 all gaps removed, published as “Proof ofthe Alternating Sign Matrix Conjecture,” Elect.J. of Combinatorics, 1996.
Zeilberger’s proof is an 84-pagetour de force, but it still left openthe original conjecture:
An,k
An,k+1
=
n ! 2k !1
"#$
%&'+
n !1k !1
"#$
%&'
n ! 2n ! k !1
"#$
%&'+
n !1n ! k !1
"#$
%&'
1996 Kuperbergannounces a simple proof
“Another proof of the alternatingsign matrix conjecture,”International MathematicsResearch Notices
Greg Kuperberg
UC Davis
1996 Kuperbergannounces a simple proof
“Another proof of the alternatingsign matrix conjecture,”International MathematicsResearch Notices
Greg Kuperberg
UC Davis
Physicists have been studying ASM’s fordecades, only they call them square ice(aka the six-vertex model ).
Anatoli Izergin Vladimir Korepin, SUNY Stony Brook
Rodney J. Baxter
AustralianNationalUniversity
1996
Doron Zeilberger usesthe connection tostatistical mechanicsto prove the originalconjecture
“Proof of the refined alternating sign matrixconjecture,” New York Journal of Mathematics
Soichi Okada,Nagoya University
2004, Okada shows that the formulas forcounting ASM’s, including those subjectto symmetry conditions, are simply thedimensions of certain irreduciblerepresentations, i.e. specializations ofWeyl Character formulas.
This PowerPoint will be available atwww.macalester.edu/~bressoud/talks
