From Pascal’s triangle to CR Geometry
John P. D’Angelo
University of Illinois at Urbana-Champaign
March 27, 2015
Plan of talk:
I Introduction
I Pascal’s triangle revisited
I When is the reciprocal of a sum the sum of the reciprocals?
I A refreshing pause?
I Roots of unity
I The invariant polynomial
I The other triangles
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 3 / 61
We all know the first order recurrence: n + 1 k + 1 =n k + n k + 1 .
Combinatorial proof: How many ways can we choose k + 1 people from a collection of n + 1 people?
Either I am included: nk ways, or I am excluded: n
k+1 ways. Algebraic proof: n+1 X j =0 n + 1 j xjyn+1−j = (x + y )n+1= (x + y )(x + y )n= (x + y ) n X j =0 n j xjyn−j.
We all know the first order recurrence: n + 1 k + 1 =n k + n k + 1 .
Combinatorial proof: How many ways can we choose k + 1 people from a collection of n + 1 people?
Either I am included: nk ways, or I am excluded: n
k+1 ways. Algebraic proof: n+1 X j =0 n + 1 j xjyn+1−j = (x + y )n+1= (x + y )(x + y )n= (x + y ) n X j =0 n j xjyn−j.
Now equate coefficients of xk+1yn−k.
We all know the first order recurrence: n + 1 k + 1 =n k + n k + 1 .
Combinatorial proof: How many ways can we choose k + 1 people from a collection of n + 1 people?
Either I am included: nk ways, or I am excluded: n
k+1 ways. Algebraic proof: n+1 X j =0 n + 1 j xjyn+1−j = (x + y )n+1= (x + y )(x + y )n= (x + y ) n Xn j xjyn−j.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 7 / 61
To verify this recurrence, iterate the usual one: n + 1 k =n k + n k − 1 =n k +n − 1 k − 1 +n − 1 k − 2 = k X j =0 n − j k − j .
We will describe infinitely many other triangles of integers that involve similar recurrences. These triangles will share other properties with Pascal’s triangle. There will only one other special triangle, where all the integers are positive.
A well-known fact about Pascal’s triangle:
Except for the leading and final 1, all the entries in the row 1 p p(p − 1)/2 p(p − 1)(p − 2)/6 · · · are divisible by p if and only if p is prime (or p = 1): A freshman’s dream:
(x + y )p ∼= xp+ yp mod (p) if and only if p is prime (or p = 1).
Our infinitely many other triangles satisfy this property as well.
A well-known fact about Pascal’s triangle:
Except for the leading and final 1, all the entries in the row 1 p p(p − 1)/2 p(p − 1)(p − 2)/6 · · · are divisible by p if and only if p is prime (or p = 1): A freshman’s dream:
(x + y )p ∼= xp+ yp mod (p) if and only if p is prime (or p = 1).
An alternate route to roots of unity
What about (x + y )−1= x−1+ y−1?
There are no real numbers satisfying this identity.
But, teachers sometimes see 1x +y1 = x +y1 on exam papers. Egad!
BUT, there are complex numbers that do satisfy it!
An alternate route to roots of unity
What about (x + y )−1= x−1+ y−1?
There are no real numbers satisfying this identity.
But, teachers sometimes see 1x +y1 = x +y1 on exam papers. Egad!
Let ω =−1+i
√ 3
2 be a cube root of unity. Thus ω
3 = 1.
Also ω1 = ω and hence ω is on the unit circle.
The picture, or the finite geometric series, gives 1 + ω + ω2 = 0.
1 2-1 + ⅈ3 1 2-1 - ⅈ3 -1.0 -0.5 0.5 1.0 -1.0 -0.5 0.5 1.0
Figure: Cube roots of unity
Then 11 +ω1 = 1+ω1 because
(1 + ω)(1 + 1
ω) = 1 + ω + ω + ωω = 1 + ω + ω
Can we find real linear operators (matrices) which satisfy
A−1+ B−1= (A + B)−1? (∗)
Definition
A real vector space V admits a complex structure if there is a linear map J : V → V such that J2= −I .
A finite-dimensional real vector space admits a complex structure if and only if its dimension is even. (Take the determinant!)
The linear transformation J : R2 → R2 corresponding to the
complex structure is given by the matrix
J =0 −1
1 0
.
Can we find real linear operators (matrices) which satisfy
A−1+ B−1= (A + B)−1? (∗)
Definition
A real vector space V admits a complex structure if there is a linear map J : V → V such that J2= −I .
A finite-dimensional real vector space admits a complex structure if and only if its dimension is even. (Take the determinant!)
The linear transformation J : R2 → R2 corresponding to the
complex structure is given by the matrix
J =0 −1
1 0
.
Theorem
Let V be a vector space over R. Then there are invertible linear transformations A, B on V satisfying
(A + B)−1 = A−1+ B−1 (∗)
if and only if V admits a complex structure.
Proof: Invertible A, B satisfying (*) exist if and only if I = (A + B)(A−1+ B−1) = I + BA−1+ I + AB−1. Put C = BA−1. The condition (*) is therefore equivalent to finding C such that 0 = I + C + C−1, which is more easily expressed as 0 = I + C + C2. Notice that C behaves like a complex cube root of unity.
Suppose such C exists. Put J = √1 3(I + 2C ). Then we have J2 = 1 3(I + 2C ) 2 = 1 3(I + 4C + 4C 2) = 1 3(I − 4I ) = −I . Hence V admits a complex structure.
Conversely, if V admits a complex structure, then J exists with J2= −I . Put C = −I +
√ 3J
2 ; then I + C + C
2= 0.
Corollary
There are n by n matrices satisfying A−1+ B−1 = (A + B)−1 if and only if n is even!
Suppose such C exists. Put J = √1 3(I + 2C ). Then we have J2 = 1 3(I + 2C ) 2 = 1 3(I + 4C + 4C 2) = 1 3(I − 4I ) = −I . Hence V admits a complex structure.
Conversely, if V admits a complex structure, then J exists with J2= −I . Put C = −I +
√ 3J
2 ; then I + C + C
2= 0.
Corollary
There are n by n matrices satisfying A−1+ B−1 = (A + B)−1 if and only if n is even!
Second proof of the corollary:
I + C + C2 = 0 forces the only eigenvalues of C to be the cube roots of unity ω and ω. The sum of the eigenvalues is the trace of C , which is real. Thus each eigenvalue ω must have a
corresponding conjugate eigenvalue ω. Hence the dimension is even.
A pause that refreshes?
An equation with no solutions for finite-dimensional matrices. 1) There are no real numbers x and y with xy − yx = 1. 2) There are no finite-dimensional matrices A, B with [A, B] = AB − BA = I . (Take the trace!)
Finding operators with this property arises in quantum mechanics. Note: For those who like both physics and swamp rock music, the identity is called CCR.
Fact: No bounded operators satisfy [A, B] = AB − BA = I . Here is an example that does:
A = dxd and B = x on the space of differentiable functions. Then (AB − BA)(f ) = (xf )0− xf0 = f and hence [A, B] = I .
A pause that refreshes?
An equation with no solutions for finite-dimensional matrices. 1) There are no real numbers x and y with xy − yx = 1. 2) There are no finite-dimensional matrices A, B with [A, B] = AB − BA = I . (Take the trace!)
Finding operators with this property arises in quantum mechanics. Note: For those who like both physics and swamp rock music, the identity is called CCR.
Fact: No bounded operators satisfy [A, B] = AB − BA = I .
Here is an example that does:
A = dxd and B = x on the space of differentiable functions. Then (AB − BA)(f ) = (xf )0− xf0 = f and hence [A, B] = I .
Remark: AB − BA is not a Swedish pop group but ABBA is.
A pause that refreshes?
An equation with no solutions for finite-dimensional matrices. 1) There are no real numbers x and y with xy − yx = 1. 2) There are no finite-dimensional matrices A, B with [A, B] = AB − BA = I . (Take the trace!)
Finding operators with this property arises in quantum mechanics. Note: For those who like both physics and swamp rock music, the identity is called CCR.
Fact: No bounded operators satisfy [A, B] = AB − BA = I . Here is an example that does:
A = dxd and B = x on the space of differentiable functions. Then (AB − BA)(f ) = (xf )0− xf0 = f and hence [A, B] = I .
A pause that refreshes?
An equation with no solutions for finite-dimensional matrices. 1) There are no real numbers x and y with xy − yx = 1. 2) There are no finite-dimensional matrices A, B with [A, B] = AB − BA = I . (Take the trace!)
Finding operators with this property arises in quantum mechanics. Note: For those who like both physics and swamp rock music, the identity is called CCR.
Fact: No bounded operators satisfy [A, B] = AB − BA = I . Here is an example that does:
A = dxd and B = x on the space of differentiable functions. Then (AB − BA)(f ) = (xf )0− xf0 = f and hence [A, B] = I .
Remark: AB − BA is not a Swedish pop group but ABBA is.
roots of unity
Consider a primitive m-th root of unity ω. We have
zm− 1 = (z − 1)(z − ω)(z − ω2)...(z − ωm−1). (1) Why? The roots of the equation are the powers of ω.
Replace z by z1 in (1). Multiply on the left-hand side by zm and each term on the right-hand side by z. We get
1 − zm = (1 − z)(1 − ωz)(1 − ω2z)...(1 − ωm−1z). (1.1) We will be generalizing (1.1).
The set
{1, ω, ω2, ..., ωm−1}
roots of unity
Consider a primitive m-th root of unity ω. We have
zm− 1 = (z − 1)(z − ω)(z − ω2)...(z − ωm−1). (1) Why? The roots of the equation are the powers of ω.
Replace z by 1z in (1). Multiply on the left-hand side by zm and each term on the right-hand side by z. We get
1 − zm= (1 − z)(1 − ωz)(1 − ω2z)...(1 − ωm−1z). (1.1) We will be generalizing (1.1).
The set
{1, ω, ω2, ..., ωm−1}
is a cyclic group of order m. Here ω−1 = ω = ωm−1.
roots of unity
Consider a primitive m-th root of unity ω. We have
zm− 1 = (z − 1)(z − ω)(z − ω2)...(z − ωm−1). (1) Why? The roots of the equation are the powers of ω.
Replace z by 1z in (1). Multiply on the left-hand side by zm and each term on the right-hand side by z. We get
1 − zm= (1 − z)(1 − ωz)(1 − ω2z)...(1 − ωm−1z). (1.1) We will be generalizing (1.1).
The set
Rewrite (1.1): φ(z) = 1 − m−1 Y j =0 (1 − ωjz) = 1 − (1 − zm) = zm. (2.1) The main unifying theme of this talk is the generalization of (2.1) to other groups.
Second proof of (2.1):
φ is a polynomial of degree m in z.
It is invariant under the transformation z → ωz. Hence it can contain only the monomial zm and constants. Thus
φ(z) = a + bzm.
Evaluating (2.1) at z = 0 gives 1 − 1, hence 0. Thus a = 0. Evaluating (2.1) at z = 1 gives 1 − 0, hence 1. Thus b = 1 and φ(z) = zm.
Rewrite (1.1): φ(z) = 1 − m−1 Y j =0 (1 − ωjz) = 1 − (1 − zm) = zm. (2.1) The main unifying theme of this talk is the generalization of (2.1) to other groups.
Second proof of (2.1):
φ is a polynomial of degree m in z.
It is invariant under the transformation z → ωz. Hence it can contain only the monomial zm and constants. Thus
φ(z) = a + bzm.
Consider a simple generalization of (2.1): (x + y )m = 1 − m−1 Y j =0 (1 − ωjx − ωjy ). (2.2) Here we have the cyclic group of order m represented as 2-by-2 matrices:
ωI =ω 0
0 ω
. The group consists of
I , ωI , ω2I , ..., ωm−1I . (x + y )m = 1 − m−1 Y j =0 (1 − ωjx − ωjy ) Thus Pascal’s triangle is related to group invariance,
Consider a simple generalization of (2.1): (x + y )m = 1 − m−1 Y j =0 (1 − ωjx − ωjy ). (2.2) Here we have the cyclic group of order m represented as 2-by-2 matrices:
ωI =ω 0
0 ω
. The group consists of
I , ωI , ω2I , ..., ωm−1I .
Invariant polynomials
The theme will be to generalize (2.2) to the abstract expression 1 −Y
γ∈Γ
(1 − hγz, zi) = ΦΓ(z, z). (3)
Here Γ will be a finite group of unitary matrices.
We cannot do much with the circle, because it is Abelian and the only finite subgroups are cyclic.
We will work with unitary matrices, usually 2-by-2. The inner product:
hz, w i =
n
X
j =1
zjwj.
U is unitary if and only if hUz, Uw i = hz, w i for all z, w .
Invariant polynomials
The theme will be to generalize (2.2) to the abstract expression 1 −Y
γ∈Γ
(1 − hγz, zi) = ΦΓ(z, z). (3)
Here Γ will be a finite group of unitary matrices.
We cannot do much with the circle, because it is Abelian and the only finite subgroups are cyclic.
We will work with unitary matrices, usually 2-by-2. The inner product:
hz, w i =
n
X
j =1
Assume that ω is a primitive p-th root of unity. For each q with 1 ≤ q ≤ p − 1 we consider a different representation of the cyclic group of order p:
γ =ω 0 0 ωq
.
Mimicking (2.2) we obtain the following function:
fp,q(x , y ) = 1 − p−1 Y j =0 (1 − ωjx − ωqjy ). 35 / 61
These group-invariant maps have remarkable number-theoretic properties. When q = 1 we get Pascal’s triangle. For q = 2, we get
1 1 1 3 1 1 5 5 1 1 7 14 7 1 1 9 27 30 9 1 1 11 44 77 55 11 1 1 13 65 156 182 91 13 1 1 15 90 275 450 378 140 15 1
These group-invariant maps have remarkable number-theoretic properties. When q = 1 we get Pascal’s triangle. For q = 2, we get
1 1 1 3 1 1 5 5 1 1 7 14 7 1 1 9 27 30 9 1 1 11 44 77 55 11 1 1 13 65 156 182 91 13 1 1 15 90 275 450 378 140 15 1 37 / 61
Hence the following polynomials equal one when we put y = 1 − x . Not obvious! x5+ 5x3y + 5xy2+ y5 x9+ 9x7y + 27x5y2+ 30x3y3+ 9xy4+ y9 x15+ 15x13y + 90x11y2+ 275x9y3+ 450x7y4+ 378x5y5+ 140x3y6+ 15xy7+ y15
The previous triangle satisfies a third order recurrence (a second order recurrence if we ignore the final 1). Can you find a way to get the next row from the two rows above it?
The polynomial fp,2 satisfies
fp,2(x , y ) ∼= xp+ yp mod (p)
if and only if p is prime.
Theorem
For each q,
fp,q(x , y ) ∼= xp+ yp mod (p)
if and only if p is prime.
The previous triangle satisfies a third order recurrence (a second order recurrence if we ignore the final 1). Can you find a way to get the next row from the two rows above it?
The polynomial fp,2 satisfies
fp,2(x , y ) ∼= xp+ yp mod (p)
if and only if p is prime.
Theorem
For each q,
fp,q(x , y ) ∼= xp+ yp mod (p)
What happens for q = 3? 1 1 1 2 1 1 3 − 3 1 1 4 − 2 1 1 5 5 1 1 6 3 2 − 3 1 1 7 7 − 7 1 1 8 12 − 2 8 1 1 9 18 3 9 − 3 1 1 10 25 10 2 − 15 1 1 11 33 22 − 11 11 1
We get some negative coefficients. Grundmeier proved: asympotically 14 of the coefficients are negative.
The primality property holds and there is a 7-th order recurrence.
What happens for q = 3? 1 1 1 2 1 1 3 − 3 1 1 4 − 2 1 1 5 5 1 1 6 3 2 − 3 1 1 7 7 − 7 1 1 8 12 − 2 8 1 1 9 18 3 9 − 3 1 1 10 25 10 2 − 15 1 1 11 33 22 − 11 11 1
What happens for q = 3? 1 1 1 2 1 1 3 − 3 1 1 4 − 2 1 1 5 5 1 1 6 3 2 − 3 1 1 7 7 − 7 1 1 8 12 − 2 8 1 1 9 18 3 9 − 3 1 1 10 25 10 2 − 15 1 1 11 33 22 − 11 11 1
We get some negative coefficients. Grundmeier proved: asympotically 14 of the coefficients are negative.
The primality property holds and there is a 7-th order recurrence.
Why do the polynomials for different groups have anything to do with each other?
Think of f (wj) = (1 − xwj − ywqj) as a polynomial in w , evaluated at wj. We have 1 − p−1 Y 0 f (wj) = 1 − p−1 Y 0 q Y k=1 (1 − ck(x , y )wj).
Interchange order of product.
= 1 − q Y 1 p−1 Y 0 (1 − ck(x , y )wj).
But we know how to find the inner (no pun intended) product.
p−1
Y
j =0
Why do the polynomials for different groups have anything to do with each other?
Think of f (wj) = (1 − xwj − ywqj) as a polynomial in w , evaluated at wj. We have 1 − p−1 Y 0 f (wj) = 1 − p−1 Y 0 q Y k=1 (1 − ck(x , y )wj).
Interchange order of product.
= 1 − q Y 1 p−1 Y 0 (1 − ck(x , y )wj).
But we know how to find the inner (no pun intended) product.
p−1
Y
j =0
(1 − zwj) = 1 − zp (1.1)
Why do the polynomials for different groups have anything to do with each other?
Think of f (wj) = (1 − xwj − ywqj) as a polynomial in w , evaluated at wj. We have 1 − p−1 Y 0 f (wj) = 1 − p−1 Y 0 q Y k=1 (1 − ck(x , y )wj).
Interchange order of product.
= 1 − q Y 1 p−1 Y 0 (1 − ck(x , y )wj).
But we know how to find the inner (no pun intended) product.
Putting it all together we get fp,q(x , y ) = 1 − q Y 1 (1 − ck(x , y )p) =Xckp−X(ckcl)p+ X (ckclcm)p− · · ·
Everything comes down to finding the (reciprocals of the) roots of (1 − xw − ywq).
No closed formula in general, of course.
Conclusion: Everything works for any polynomial whose
coefficients are regarded as variables. We can express everything in terms of the roots, but there is no closed formula for the roots as functions of the coefficients.
Pascal’s triangle used (1 − xw − yw ). The next triangle used (1 − xw − yw2). The third triangle used (1 − xw − yw3).
In principle one can do something similar for any finite subgroup of unitary matrices.
Conclusion: Everything works for any polynomial whose
coefficients are regarded as variables. We can express everything in terms of the roots, but there is no closed formula for the roots as functions of the coefficients.
Pascal’s triangle used (1 − xw − yw ). The next triangle used (1 − xw − yw2). The third triangle used (1 − xw − yw3).
In principle one can do something similar for any finite subgroup of unitary matrices.
Conclusion: Everything works for any polynomial whose
coefficients are regarded as variables. We can express everything in terms of the roots, but there is no closed formula for the roots as functions of the coefficients.
Pascal’s triangle used (1 − xw − yw ). The next triangle used (1 − xw − yw2). The third triangle used (1 − xw − yw3).
In principle one can do something similar for any finite subgroup of unitary matrices.
How do these ideas lead to several complex variables and CR Geometry?
Unitary maps preserve the unit sphere. Let Γ ⊆ U(n) be a finite subgroup. It makes sense to consider invariant maps on the unit sphere.
For what finite subgroups Γ of U(n) is there a non-constant invariant rational map taking the unit sphere to some unit sphere?
Theorem (Lichtblau)
(1992) Assume Γ ⊂ U(n) is a finite subgroup. If there is a
non-constant Γ-invariant rational holomorphic map from S2n−1 to some S2N−1, then Γ is cyclic.
Theorem
If there is a non-constant complex analytic Γ-invariant map from S2n−1 to a sphere, then Γ must be represented in a special way. Let η be a primitive p-th root of unity. The only possibilities are the groups generated by:
ηI (9.1) η 0 0 η2 p odd (9.2) η 0 0 0 η2 0 0 0 η4 p = 7 (9.3)
When n = 2, (9.1) leads to fp,1(x , y ) = (x + y )p = (|z|2+ |w |2)p= 1. (9.2) leads to fp,2(x , y ) = ( x +px2+ 4y 2 ) p+ (x − p x2+ 4y 2 ) p+ yp.
These are the polynomials we saw before such as x5+ 5x3y + 5xy2+ y5.
(9.3) is a kind of conglomeration of (9.2). Works only if η7 = 1.
CR Geometry
CR Geometry studies real objects in complex spaces. Basic example: odd dimensional unit sphere S2n−1.
A CR mapping is an analogue of a complex analytic mapping. It depends on z but not on z.
In a good complex variables course, one can cop z’s but not z-bars....
The interaction between the degrees of rational maps from S2n−1 to S2N−1in terms of n, N leads to a new field of mathematics, CR
complexity theory. Furthermore, the group invariant maps fp,2 are
extremal. Proof uses directed graphs, sources, and sinks. They are of degree p = 2r + 1 and have r + 2 terms.
To obtain group-invariant maps, one must replace the target sphere with a hyperquadric.
Q(A, B) = {z : A X j =1 |zj|2− A+B X j =A+1 |zj|2= 1}. Theorem
Given any finite subgroup Γ of U(n) there is a Γ-invariant polynomial (complex analytic) map from S2n−1 to some hyperquadric.
We need enough eigenvalues of both signs. My former student Grundmeier has done lots of work on this problem.
For example: the target hyperquadric for the binary icosahedral group has 40 positive, 22 negative eigenvalues.
These invariant polynomials have taken us from Pascal’s triangle to CR Geometry!
Return to our first triangle
Putting x = |z1|2 and y = |z2|2 yields a polynomial
I f (x , y ) = 1 on x + y = 1. (sphere)
I f (ωx , ω2y ) = f (x , y ). (invariance)
I f has degree p.
I f has r + 2 positive terms and no negative terms if p = 2r + 1 is odd.
I f has r + 1 positive and 1 negative term (namely −yp) if p = 2r is even.
I f (x , y ) has integer coefficients.
I f (x , y ) is congruent to xp+ yp modulo p if and only if p is prime (or p = 1).
In fact, f has the explicit formula (x + p x2+ 4y 2 ) p+ (x − p x2+ 4y 2 ) p+ (−1)p+1yp.
Same ideas works for all representations of cyclic subgroups of U(2), but the formulas are not explicit.
In fact, f has the explicit formula (x + p x2+ 4y 2 ) p+ (x − p x2+ 4y 2 ) p+ (−1)p+1yp.
Same ideas works for all representations of cyclic subgroups of U(2), but the formulas are not explicit.
Let’s all enjoy the rest of the meeting!
