Incidence Matrices of Finite Projective Planes and Their Eigenvalues

Incidence Matrices of Finite Projective Planes and Their Eigenvalues

191, 265]278 Ž1997. JA966919 JOURNAL OF ALGEBRA ARTICLE NO. Incidence Matrices of Finite Projective Planes and Their Eigenvalues John G. Thompson De...

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191, 265]278 Ž1997. JA966919

JOURNAL OF ALGEBRA ARTICLE NO.

Incidence Matrices of Finite Projective Planes and Their Eigenvalues John G. Thompson Department of Mathematics, Uni¨ ersity of Florida, Gaines¨ ille, Florida 32611 Communicated by Walter Feit Received October 3, 1996

1. NOTATION AND PRELIMINARY DISCUSSION If k is a commutative ring, denote by Mm, nŽ k . the k-module of all m = n matrices over k. Denote by t the transpose map M ¬t M from Mm, nŽ k . to Mn, mŽ k ., and set col nŽ k . s Mn, 1Ž k ., rownŽ k . s M1, nŽ k ., and MnŽ k . s Mn, nŽ k ., identifying k with M1Ž k .. Most of this paper concerns the case k s C. If u, ¨ g col nŽC. and ¨1

u1 . u s .. , un

0 0 .. .

¨s

,

¨n

set

Ž u, ¨ . st u ? ¨ s Ý u i¨ i . We also need the unitary inner product given by

Ž u, ¨ . 0 s Ý u i¨ i . If c1 . c s .. cn

0

and

ds

d1 .. .

0 dn

265 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.

266

JOHN G. THOMPSON

are in col nŽC., set E Ž c, d . s E Ž c1 , . . . , c n , d1 , . . . , d n . s c t d s Ž c i d j . g Mn Ž C . . Note that if a g C=, then EŽ c, d . s EŽ ac, ay1 d .. We check that if ¨1

¨s

0 .. .

g col n Ž C . ,

¨n

then E Ž c, d . ¨ s Ž d, ¨ . c.

Ž 1.1.

Also, if V s Ž ¨ i j . and U s Ž u i j . are in MnŽC., then VE Ž c1 , . . . , c n ; d1 , . . . , d n . U s E Ž a1 , . . . , a n ; b1 , . . . , bn . ,

Ž 1.2.

where ai s

bj s

Ý ¨ i k ck ,

Ý dk uk j ,

k

1 F i , j F n.

Ž 1.3.

k

If x 1 , . . . , x n g C, set d Ž x1 , . . . , x n . s

Ý x i eii g Mn Ž C. , i

the diagonal matrix whose ith diagonal entry is x i . In case x j s 1 for all j / i and x i s c, set d i Ž c . s dŽ x 1 , . . . , x n .. If C g MnŽR. and C t C st CC, C is said to be normal, and in this case there is a unitary matrix G such that Gy1 CG s d Ž Q 1 , . . . , Qn . ,

Gy1 t CG s d Ž Q1 , . . . , Qn . .

Ž 1.4.

If M g MnŽC. and x is an indeterminate, set f M Ž x . s det Ž xI y M . , let m M Ž x . be the minimal polynomial of M, and if a g C, let mŽ M, a . be the integer defined by fM Ž x . s Ž x y a .

mŽ M , a .

rŽ x. ,

r Ž x . g Cw x x ,

r Ž a . / 0. Ž 1.5.

If M g MmŽQ., let FM be the splitting field of f M Ž x . with FM : C, and set GM s Gal Ž FM rQ . .

FINITE PROJECTIVE PLANES

267

Let c M be that element of GM determined by complex conjugation, so that cM Ž a . s a

for every zero a of f M Ž x . in C.

Ž 1.6.

If s g Sn , set P Ž s . s Ý i e ii s g GLŽ n, C., so that Pn s  P Ž s .< s g Sn4 is the group of permutation matrices of GLŽ n, C.. If r, s are distinct positive integers F n, let t r s denote the transposition which interchanges r and s, and denote by Tn the set of all P Ž t r s ., 1 F r, s F n, r / s. If Ž P, L . is a finite projective plane of order n, set N s n2 q n q 1 Žs < P < s < L <., and if P s  p1 , . . . , pN 4 ,

L s  l 1, . . . , lN 4,

set ai j s

½

0,

if pi f l j ,

1,

if pi g l j ,

A s Ž ai j . .

Ž 1.7.

Such a matrix A is called an incidence matrix for Ž P, L ., and I denote by I Ž P, L . the set of all incidence matrices for Ž P, L .. We note that A tA st AA s nI q J ,

Ž 1.8.

where J g MN ŽC. has all entries s 1. In particular, A is normal, so there is a unitary matrix UA such that UAy1AUA s d Ž Q 1 , . . . , QN . , UAy1 tAUA s d Ž Q1 , . . . , QN . .

Ž 1.9.

Here f AŽ x . s Ł i Ž x y Qi .. From Ž8. and the fact that det A s det tA, we get det A s " Ž n q 1 . nŽ Ny1.r2 .

Ž 1.10.

Since every row sum and every column sum of A is n q 1, we assume with no loss of generality that UA has been chosen so that if u1 , . . . , u n are the columns of UA , then QN s n q 1 and uN s

1

'N

1 .. . . 1

0

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JOHN G. THOMPSON

Hence UAy1A tAUA s d Ž Q 1Q 1 , . . . , QNy1QNy1 , Ž n q 1 .

2

.

s UAy1 Ž nI q J . UA 2

s d Ž n, n, . . . , n, Ž n q 1 . . , and so Qi Qi s n,

1 F i - N.

Ž 1.11.

In particular, if Qi is real, then Qi g  'n , y'n 4 , and we conclude that det A
s Ž y1 .

m Ž A , y'n .

s Ž y1 .

m Ž A , 'n .

,

Ž 1.12.

as the nonreal zeroes of f AŽ x . occur in complex conjugate pairs, N y 1 is Ny 1 even, and as det A s Ž n q 1.Ł is1 Qi . Note that c A g Z Ž GA . . For if g g GA , and f AŽ Q . s 0 with Q / n q 1, then Q ? c AŽ Q . s n, whence g Ž Q . g Ž c AŽ Q .. s n s g Ž Q . c AŽ g Ž Q .., so g Ž c AŽ Q .. s c AŽ g Ž Q ... This holds for all zeros Q of f AŽ x . and so gc A s c A g. If P, Q g PN , then PAQ g I Ž P, L ., since A s Ý a i j e i j , and if P s P Ž s ., Q s QŽt ., s , t g SN , then Pe i j Q s e i sy1 , jt , so that PAQ s

Ý ai j ei s

y1

, jt

s

Ý bi j ei j ,

where bi j s

½

0,

if pi s f l jty1 ,

1,

if pi s g l jty1 .

Hence, PAQ is the incidence matrix which occurs if we order the points X pXi s pi s and the lines l j s l jty1 . For each A g I Ž P, L ., write f A Ž x . s Ž x y Ž n q 1. . g A Ž x . .

Ž 1.13.

Since Ž n q 1. 2 / n and mŽ A, n q 1. G 1, we conclude that mŽ A, n q 1. s 1, and so g AŽ x . g Zw x x ,

g A Ž n q 1 . / 0,

g A is monic.

Ž 1.14.

FINITE PROJECTIVE PLANES

269

2. A SPLITTING THEOREM If A g I Ž P, L ., we say that A splits if and only if c A is of order 2, and ² c A : is a direct factor of GA .

Ž 2.1.

THEOREM 1. Suppose n is not a square, A g I Ž P, L ., and A is not split. Then AT is split for all T g TN . Before giving the proof of Theorem 1, we interpolate two lemmas. LEMMA 1.

If A g I Ž P, L ., T g TN , and Q g C, then m Ž A, Q . y m Ž AT , Q . g  y1, 0, 1 4 .

Proof. Let V Ž A, Q . s  ¨ g col N Ž C . < A¨ s Q ¨ 4 , V Ž AT , Q . s  ¨ g col N Ž C . < A¨ s Q ¨ 4 . Since A and AT are semisimple, dim V Ž A, Q . s m Ž A, Q . ,

dim V Ž AT , Q . s m Ž AT , Q . .

Since T g TN , T s P Ž t r s ., where 1 F r - s F N. Let W s  w g col N Ž C . < Tw s w 4 . Since T is a reflection, dim W s N y 1, and

¡ ¢

W s~w s

w1 .. . wn

0

¦ w s w ¥. § r

s

Hence V Ž A, Q . l W is of codimension 0 or 1 in V Ž A, Q ., and V Ž AT, Q . l W is of codimension 0 or 1 in V Ž AT, Q .. If w g V Ž A, Q . l W, then Tw s w, and so ATw s Aw s Qw, whence w g V Ž AT, Q . l W. Similarly, V Ž AT, Q . l W : V Ž A, Q . l W, and so V Ž AT , Q . l W s V Ž A, Q . l W. Set d s dim V Ž AT, Q . l W. Then m Ž A, Q . s d q « ,

m Ž AT , Q . s d q h ,

where « , h g  0, 14 . The lemma follows.

270

JOHN G. THOMPSON

LEMMA 2. Suppose A g I Ž P, L ., T g TN , Q g C, and u, ¨ g col N ŽC.. Then one of the following holds: Ža. Žb. Žc. Žd.

Au / Qu. ATu s Qu. A¨ s Q ¨ . AT¨ / Q ¨ .

Proof. Suppose false, so that Ž1. Au s Qu, ATu / Qu, Ž2. A¨ / Q ¨ , AT¨ s Q ¨ . Note that u / 0, since ATu / Qu, and ¨ / 0, since A¨ / Q ¨ . Thus, f AŽ Q . s 0, f AT Ž Q . s 0. If Q s n q 1, then u s c ? t Ž1, . . . , 1. for some c g C=. But then ATu s Ž n q 1. u, and as Q s n q 1, Ž1. is violated. So < Q < s 'n . Set T s I y E, so that E s e r r q e s s y e r s y e sr . Choose a unitary U s Ž u i j . such that Uy1AU s D s dŽ Q1 , . . . , QN . is diagonal, with Q1 s ??? s Qm s Q, Qi / Q if i ) m. Set F s Uy1 EU s Ž a i bj ., where Ž1.2. and the equality Uy1 st U yield bi s a i s u r i y u si . Set u 0 s Uy1 u,

¨ 0 s Uy1 ¨ ,

S s Uy1 TU.

Then ŽUy1AU . u 0 s ŽUy1AU .Uy1 u s Uy1Au s Uy1 Qu s QUy1 u s Qu 0 , ŽUy1ATU . u 0 s ŽUy1ATU .Uy1 u s Uy1ATu / Uy1 Qu s QUy1 u s Qu 0 , so Du 0 s Qu 0 ,

DSu 0 / Qu 0 ,

Ž 2.2.

D¨ 0 / Q ¨ 0 ,

DS¨ 0 s Q ¨ 0 .

Ž 2.3.

and similarly,

Since S s I y F, Ž2.2. yields DFu 0 / 0, and in particular, Fu 0 / 0. On the other hand, Fu 0 s Ž b, u 0 . a, where a s tŽ a1 , . . . , a N . ,

u 0 s tŽ x 1 , . . . , x N . ,

Ž b, u 0 . s b1 x 1 q ??? qbN x N .

271

FINITE PROJECTIVE PLANES

In particular, Ž b, u 0 . / 0. Since Du 0 s Qu 0 , we get x i s 0 if i ) m. Hence, there is an integer i 0 with 1 F i 0 F m such that bi 0 x i 0 / 0. In particular, bi 0 / 0. Since a i s bi for all i, we get a i 0 / 0. Now we focus on Ž2.3.. Since D¨ 0 / Q ¨ 0 , we have ¨ 0 st Ž y 1 , . . . , yN . and y k / 0 for some k ) m. Since S s I y F and DS¨ 0 s Q ¨ 0 , we get D¨ 0 y DF¨ 0 s Q ¨ 0 , and since D¨ 0 / Q ¨ 0 , we get DF¨ 0 / 0. Now F¨ 0 s Ž b, ¨ 0 . a, and so Ž b, ¨ 0 . / 0, and c1 . F¨ 0 s .. , cn

0

where c i s Ž b, ¨ 0 . a i . In particular, c i 0 / 0, and so Q 1 c1 .. DF¨ 0 s , . QN c n

 0

and Qi 0 c i 0 / 0. This is false, as D¨ 0 y DF¨ 0 s Qs 0 , so that Qi 0 yi 0 y Qi 0 c i 0 s Qi 0 yi 0 Žas Q s Qi 0 .. The proof is complete. We turn to the proof of Theorem 1. h Ž x y w i .. Since A is semisimple, m AŽ x . has no Let m AŽ x . s Ł is1 repeated roots, so F s  w i <1 F i F h4 has h elements. Also GA acts as permutations of F. Let F 1 , . . . , F k be the orbits of GA on F. Also, for each i s 1, . . . , k, either all elements of Fi are real or none are. First, suppose there is i such that no roots of Fi are real, and in addition < Fi < ' 2 Žmod 4.. Since c AŽ w . s w for all w g F, it follows that c A induces an odd permutation of Fi , and so GA has a subgroup H of index 2 consisting of those elements which induce an even permutation of Fi . Since c A g ZŽ GA ., we get GA s H = ² c A : and A is split, against our assumption. It remains to treat the case where, for each i, either all elements of Fi are real or < Fi < ' 0 Žmod 4.. Since n is not a square and A g MN ŽQ., it follows that mŽ A, 'n . s mŽ A, y'n ., and so N y 1 s deg g A Ž x . ' 2 m Ž A, 'n .

Ž mod 4 . ,

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JOHN G. THOMPSON

or, equivalently, Ny1 2

' m Ž A, 'n .

Ž mod 2 . .

If T g TN , then Žy1. mŽ AT , 'n . s yŽy1. mŽ A, 'n ., and so Ny1 2

k m Ž AT , 'n .

Ž mod 2 . .

So AT is split, or every zero of g AT Ž x . is real. However, if every zero of g AT Ž x . is real, then g AT Ž x . s Ž x 2 y n.Ž Ny1.r2 , whence g AŽ x . s Ž x 2 y n.Ž Ny1.r2y1 ? hŽ x ., where hŽ x . is a quadratic polynomial whose roots are nonreal, and so < Fi < ' 2 Žmod 4., where Fi is the set of zeroes of hŽ x .. The proof is complete.

3. THE INTERLACING THEOREM THEOREM 2.

Suppose n is not a square, A g I Ž P, L ., and T g Tn . Let

g AŽ x . s DŽ x . EŽ x . ,

g AT Ž x . s D Ž x . F Ž x . ,

where DŽ x . is the greatest common di¨ isor of g AŽ x . and g AT Ž x .. Then EŽ x . and F Ž x . are square free, and on the circle of radius 'n centered at the origin, between any two successi¨ e zeroes of EŽ x . there is precisely one zero of F Ž x .. Proof. By definition of DŽ x ., it follows that EŽ x . and F Ž x . are coprime. If Q g C and < Q < s 'n , let d, e, and f be the multiplicity of Q as a zero of DŽ x ., EŽ x ., and F Ž x ., respectively. Then m Ž A, Q . s d q e,

m Ž AT , Q . s d q f ,

and so m Ž A, Q . y m Ž AT , Q . s e y f g  y1, 0, 1 4 . Since ef s 0, it follows that E and F are square free. Let T s P Ž t r, s ., and set h

EŽ x . s

Ł Ž x y wj . , js1

h

FŽ x. s

Ł Ž x y jj . . js1

FINITE PROJECTIVE PLANES

273

Then

w j s 'n e i a j ,

j j s 'n e i b j ,

1 F j F h,

and we order the zeroes so that 0 F a 1 - a 2 - ??? - a h - 2p , 0 F b 1 - b 2 - ??? - b h - 2p . For each Q g C, let V Ž A,Q . s  ¨ g col N Ž C . < A¨ s Q ¨ 4 , V Ž AT , Q . s  ¨ g col N Ž C . < AF¨ s Q ¨ 4 . I argue that V Ž A, j j . ; V Ž AT , j j . ,

1 F j F h.

Ž 3.1.

To see this, note that m Ž A, j j . s m Ž AT , j j . y 1. Suppose there is a u g V Ž A, j j . such that Tu / u. Then Au s j j u, ATu / j j u. Since mŽ AT, j j . s 1 q mŽ A, j j ., there is a ¨ g V Ž AT, j j . such that ¨ f V Ž A, j j .. Thus, AT¨ s j j ¨ ,

A¨ / j j ¨ .

This violates Lemma 2 with j j in the role of Q, and proves Ž3.1.. By symmetry, V Ž AT , w j . ; V Ž A, w j . ,

1 F j F h.

Ž 3.2.

Now suppose DŽ Q . s 0 and EŽ Q . F Ž Q . / 0. In this case, we get V Ž A, Q . s V Ž AT , Q . , and so, equivalently, T¨ s ¨ for all ¨ g V Ž A, Q . j V Ž AT, Q .. Choose a unitary matrix U such that Uy1AU s d Ž Q 1 , . . . , QN . ,

QN s n q 1,

and let u j be the jth column of U. Then Au j s Qj u j ,

1 F j F N,

Ž 3.3.

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JOHN G. THOMPSON

as AU s UdŽ Q1 , . . . , QN .. Order the Qs so that Qj s w j , 1 F j F h. We also choose U so that Tu j s u j ,

h q 1 F j F N,

Ž 3.4.

1 F j F h.

Ž 3.5.

and are then guaranteed that Tu j / u j ,

Let T s I y E, where E s e r r q e s s y e r s y e sr . Set F s Uy1 EU, S s U TU s I y F, so that y1

F s Ž a i bj . , where, by Ž2.2., together with the equality Uy1 st U, bj s a j s u r j y u s j , and u1 j .. uj s . , uN j

0

1 F j F N.

By our choice of U, we get a j s bj s 0,

h q 1 F j F N.

Thus, d Ž Q 1 , . . . , QN . s D 1 [ D 2 , where D 1 s dŽ w 1 , . . . , w h ., D 2 s dŽ Q hq1 , . . . , QN ., and Fs

ž

F1 0

0 , 0

/

Ss

ž

S1 0

0 , I

/

where F1 and S1 are h = h matrices with F1 s Ž a i bj ., 1 F i, j F h, and S1 s I y F1. This implies that D 1 S1 is unitarily equivalent to dŽ j 1 , . . . , j h .. Note that a j s bj s u r j y u s j / 0,

1 F j F h.

Ž 3.6.

For each j s 1, . . . , h, choose wj g col hŽC., wj / 0, with D 1 S1wj s j j wj .

Ž 3.7.

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FINITE PROJECTIVE PLANES

Set w1 j .. wj s . . wh j

b1 . b s .. , bh

a1 . a s .. , ah

0

0 0

Since S1 s I y F1 , we get D 1wj y D 1 F1wj s j j wj ,

Ž 3.8.

F1wj s Ž b, wj . a,

Ž 3.9.

and by Ž1.1., which yields

w i wi j y w i Ž b, wj . a i s j j wi j ,

1 F i F h, 1 F j F h.

Ž 3.10.

Thus, wi j s

w i Ž b, wj . a i wi y j j

.

Ž 3.11.

Since wj / 0, we have Ž b, wj . / 0, 1 F j F h. Set c j s Ž b, wj ., so that wi j s

wi c j ai wi y j j

.

Ž 3.12.

Since Ž b, wj . s b1w 1 j q ??? qbh w h j , Ž12. yields h

cj s

Ý

bi a i w i c j

is1

wi y j j

,

Ž 3.13.

and since c j / 0, Ž13. yields h

1s

Ý is1

bi a i w i

wi y j j

1 F j F h.

,

Ž 3.14.

We now introduce a rational function of x, setting h

RŽ x . s 1 q

Ý is1

ai ai wi x y wi

.

Ž 3.15.

Since bi s a i / 0, Ž3.14. and Ž3.15. show that RŽ x . has a simple pole at w i and a zero at x s j j , 1 F i, j F h. Hence RŽ x . s

FŽ x. EŽ x .

? R0 Ž x . ,

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JOHN G. THOMPSON

where R 0 Ž x . g Cw x x. Since EŽ x . and F Ž x . are monic of the same degree, Ž3.15. implies that R 0 Ž x . ' 1. On the other hand, the partial fraction decomposition of F Ž x .rEŽ x . is FŽ x. EŽ x .

s R1 Ž x . q

Ý i

ei

e i g C, Ž R1 Ž x . g C w x x . , Ž 3.16.

,

x y wi

and since F Ž x . and EŽ x . are monic of the same degree we get R1Ž x . s 1, while e i s lim

F Ž x . Ž x y wi . EŽ x .

xª w i

s

F Ž wi . E9 Ž w i .

,

and so, by the uniqueness of the partial fraction decomposition, ai ai wi s

F Ž wi . E9 Ž w i .

.

Ž 3.17.

Hence F Ž wi . E9 Ž w i . ? w i

is a positive real number, 1 F i F h.

Ž 3.18.

Replacing the pair Ž A, T . by the pair Ž AT, T . if necessary, and using our assumption that n is not a square, we assume with no loss of generality that EŽ x . and x 2 y n are coprime. Hence 0 - a 1 - ??? - a h - 2p and

aj / p ,

all j s 1, . . . , h. Ž 3.19.

Since Žy1. mŽ A, 'n . s yŽy1. mŽ AT , 'n ., we conclude that x 2 y n divides F Ž x ., so

b 1 s 0,

bŽ hq2.r2 s p ,

Ž 3.20.

the second equality holding since F Ž x . has Ž h y 2.r2 zeroes with positive imaginary part. h Ž w j y j k ., Now F Ž w j . s Ł ks1 h

E9 Ž w j . w j s w j Ł Ž w j y w k . , ks1 k/j

277

FINITE PROJECTIVE PLANES

and so F Ž wj . E9 Ž w j . w j

h Ł ks1 Ž 1 y exp i Ž bk y a j .

s

h Ł ks1, k/ j

Ž 1 y exp

.

iŽ ak y a j .

.

) 0.

We use the fact that if a g R, then 1 y e i a s y2 i sin

a 2

e i a r2 ,

and we get F Ž wj . E9 Ž w j . w j

s

h y2 i Ł ks1 sin h Ł ks1, k/ j

h

ž

? exp i s

Ý

sin Ž a k y a j . r2

bk y a j 2

ks1

h y2 i Ł ks 1 sin h Ł ks 1, k / j

Ž b k y a j . r2 h

/ ž

? exp yi

Ý

ak y a j 2

ks1

Ž b k y a j . r2

sin Ž a k y a j . r2

h

ž

? exp i

Ý

/

bk y a j

ks1

2

/

.

Now h

Ý a k s 2p ? ks1

h 2

,

h

Ý b k s p q 2p

Ž h y 2. 2

ks1

,

so Ý hks 1 Ž a k y b k . s p and eyi p r2 s yi, whence F Ž wj . E9 Ž w j . w j

s

h y2 Ł ks1 sin h Ł ks1, k/ j

Ž b k y a j . r2

sin Ž a k y a j . r2

) 0.

Now Ž b k y a j .r2 g Žyp , 0. j Ž0, p ., sin x ) 0 if x g Ž0, p ., - 0 if x g Žyp , 0., and so h Ł ks 1 sin h Ł ks 1

sin

Ž b k y a j . r2 Ž b k y a j . r2

where A j s < k < b k y a j - 04<.

s Ž y1 .

Aj

,

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JOHN G. THOMPSON

Since 0 - a 1 - a 2 - ??? - a h - 2p , h Ł ks 1, k / j sin Ž a k y a j . r2 h Ł ks 1, k / j sin Ž a k y a j . r2

s Ž y1 .

jy1

,

Hence Aj ' j

Ž mod 2 . .

In particular, A jq1 ' 1 q A j

Ž mod 2 . .

Since A jq1 y A j s  k < a j - b k - a jq1 4 ,

1 F j - h,

we conclude that there are an odd number of b s between successive a s. Since there are h a s and h b s, the theorem follows. Remarks. With mild modifications, the results of this paper can be adapted to the case where n is a square. When n is not a square, one may interpret Theorem 2 as providing constraints on the unitary matrices which diagonalize A. In order to study these constraints systematically, it is relevant to study the set of all AT, where T ranges over TN .