CHAPTER 3. GOE and GUE

Transcription

1 CHAPTER 3 GOE ad GUE We quicly recall that a GUE matrix ca be defied i the followig three equivalet ways. We leave it to the reader to mae the three aalogous statemets for GOE. I the previous chapters, GOE ad GUE matrices appeared merely as special cases of Wiger matrices for which computatios were easier. However they have a great may eat properties ot shared by other Wiger matrices. The mai fact is that the exact desity of eigevalues of GOE ad GUE ca be foud explicitly! Ad eve more surprisigly, these exact desities have a ice structure that mae them ameable to computatios. May results that are true for geeral Wiger matrices are much harder to prove i geeral but fairly easy for these two cases. Crucial to the itegrability properties of GOE ad GUE are their ivariace uder orthogoal ad uitary cojugatios respectively. Exercise 46. (a) Let X ad Y be GUE ad GOE matrices respectively. The, for ay fixed U U() ad P O(), we have U XU d = X ad P t YP d = Y. (b) If X is a radom matrix such that X i, j, i j are idepedet real valued etries ad suppose that PXP t d = X for all P O(), the show that X has the same distributio as c X where c is a costat ad X is a GOE matrix. The aalogous statemet for uitary ivariace is also true. Remar 47. This is aalogous to the followig well ow fact. Let X be a radom vector i R. The the followig are equivalet. (1) X N (0,σ 2 I) for some σ 2. (2) X are idepedet ad PX d = X for ay P O(). To see that the secod implies the first, tae for P a orthogoal matrix whose first colum is (1/ 2,1/ 2,0,...,0) to get X 1 d =(X1 +X 2 )/ 2. Further, X 1,X 2 are i.i.d - idepedece is give, ad choosig P to be a permutatio matrix we get idetical distributios. It is well ow that the oly solutios to this distributioal equatio are the N(0,σ 2 ) distributios. If ot coviced, use characteristic fuctios or otherwise show this fact. What is the use of uitary or orthogoal ivariace? Write the spectral decompositio of a GUE matrix X = V DV. For ay fixed U U(), the UXU =(UV)D(UV). By the uitary ivariace, we see that V DV has the same distributio as (UV)D(UV). This suggests that V ad D are idepedet. The oly hitch i this reasoig is that the spectral decompositio is ot exactly uique, but it ca be tae care of 1 1 The eigespace for a give eigevalue is well-defied. This is the source of o-uiqueess. The set S of Hermitia matrices havig distict eigevalues is a dese ope set i the space of all Hermitia matrices. Therefore, almost surely, a GUE matrix has o eigevalues of multiplicity more tha oe (explai why). However, eve whe the eigespace is oe dimesioal, we ca multiply the eigevector by e iθ for some θ R ad that leads to o-uiqueess. To fix this, let D() be the group of diagoal uitary matrices ad cosider the quotiet space Q = U()/D() cosistig of right cosets. The, the mappig X ([V ],D) is oe to oe ad oto o S. Now observe that for ay U, ([UV],D) d = ([V ],D) ad hece [V ] ad D are idepedet. 29

2 30 3. GOE AND GUE 1. Tridiagoalizatio Let A be a GOE. Write it as [ ] a u t A = u B so that a N(0,2), u N 1 (0,I), B GOE 1, ad all three are idepedet. Coditio o u. The pic ay orthogoal matrix P O( 1) such that Pu = u e 1. To be specific, we ca tae the trasformatio defied by Pv = v 2 v,w w,w w, with w = u e 1. For ay w 0, the trasformatio defied o the left is the reflectio across the hyperplae perpedicular to w. These are also referred to as Householder reflectios. Chec that P is ideed uitary ad that Pu = e 1. Sice P depeds o u ad B is idepedet of U, the orthogoal ivariace of GOE shows that A 1 := P t BP = d B, that is A 1 is a GOE matrix. Also A 1 is idepedet of u ad a. Thus, [ ][ ][ ] [ 1 0 t a u t 1 0 t a r C := 0 P t = 1 e t ] 1 u B 0 P r 1 e 1 A 1 where A 1 GOE 1, a N(0,1) ad r 1 = u are all idepedet. Sice C is a orthogoal cojugatio of A, the eigevalues of A ad C are exactly the same. Observe that C j,1 = C 1, j = 0 for 2 j. Note that r1 2 = u 2 has χ 2 1 distributio. Now A 1 is a GOE matrix of oe less order. We ca play the same game with A 1 ad get a matrix D which is cojugate to A 1 but has D 1, j = D j,1 = 0 for 2 j 1. Combiig with the previous oe, we get C 2 := a r 1 0 t r 1 a r 2 e t 1 0 r 2 e 1 D with the followig properties. C 2 is cojugate to A ad hece has the same eigevalues. D GOE 2, a,a N(0,2), r1 2 χ2 1, r2 2 χ2 2, ad all these are idepedet. It is clear that this procedure ca be cotiued ad we ed up with a tridiagoal matrix that is orthogoally cojugate to A ad such that a 1 b b 1 a 2 b b 2 a 3 b (20) T = b 2 a 1 b b 1 a where a N(0,2), b 2 χ2, ad all these are idepedet. Exercise 48. If A is a GUE matrix, show that A is cojugate to a tridiagoal matrix T as i (20) where a,b are all idepedet, a N(0,1) ad b 2 Gamma(,1). Recall that χ 2 p is the same as Gamma( p 2, 1 2 ) or equivaletly, the distributio of 2Y where Y Gamma( p 2 ). Thus, we arrive at the followig theorem2. 2 The idea of tridiagoalizig the GOE ad GUE matrices was origially due to Hale Trotter?. Part of his origial motivatio was to give a simple proof of the semicircle law for GOE ad GUE matrices.

3 2. TRIDIAGONAL MATRICES AND PROBABILITY MEASURES ON THE LINE 31 Theorem 49. Let T be a tridiagoal matrix as i (20) where a,b are all idepedet. (1) If a N(0,2) ad b 2 χ2, the the vector of eigevalues of T has the same distributio as the vector of eigevalues of a GOE matrix. (2) If a N(0,2) ad b 2 χ2 2( ), the the vector of eigevalues of T has the same distributio as the eigevalues of a GUE matrix scaled by a factor of Tridiagoal matrices ad probability measures o the lie Our objective is to fid eigevalue desity for certai radom matrices, ad hece we must fid 1 auxiliary parameters i additio to the eigevalues (sice there are 2 1 parameters i the tridiagoal matrix) to carry out the Jacobia computatio. The short aswer is that if UDU is the spectral decompositio of the tridiagoal matrix, the p j = U 1, j 2,1 j 1 are the right parameters to choose. However, there are may coceptual reasos behid this choice ad we shall sped the rest of this sectio o these cocepts. Fix 1 ad write T = T (a,b) for the real symmetric tridiagoal matrix with diagoal etries T, = a for 1 ad T,+1 = T +1, = b for 1 1. Let T be the space of all real symmetric tridiagoal matrices ad let T 0 be those T (a,b) i T with distict eigevalues. Let P be the space of all probability measures o R whose support cosists of at most distict poits ad let P 0 be those elemets of P whose support has exactly distict poits. Tridiagoal matrix to probability measure: Recall that the spectral measure of a Hermitia operator T at a vector v is the uique measure ν o R such that T p v,v = R x p ν(dx) for all p 0. For example, if T is a real symmetric matrix, write its spectral decompositio as T = λ u u. The {u } is a ONB of R ad λ are real. I this case, the spectral decompositio of T at ay v R is just ν = v,u 2 δ λ. Thus ν P (observe that the support may have less tha poits as eigevalues may coicide). I particular, the spectral measure of T at e 1 is p j δ λ j where p j = U 1, j 2 (here U 1, j is the first co-ordiate of u j ). Give a real symmetric tridiagoal matrix T, let ν T be the spectral measure of T at the stadard uit vector e 1. This gives a mappig from T ito P which maps T 0 ito P 0. Probability measure to Tridiagoal matrix: Now suppose a measure µ P 0 is give. Write µ = p 1 δ λ p δ λ where λ j are distict real umbers ad p j > 0. Its momets are give by α = p j λ j. Let h (x) =x, so that {h 0,h 1,...,h 1 } is a basis for L 2 (µ) (how do you express h as a liear combiatio of h 0,...,h 1?). Apply Gram-Schmidt to the sequece h 0,h 1,... by settig ϕ 0 = ψ 0 = h 0, ad for 1 iductively by 1 ψ = h j=0 ψ h,ϕ j ϕ j, ϕ =. ψ L 2 (µ) This process is stopped whe ψ = 0. Here are some elemetary observatios. (a) Sice {h 0,...,h 1 } is a liear basis for L 2 (µ), it follows that {ϕ 0,...,ϕ 1 } are welldefied ad form a ONB for L 2 (µ). (b) For 0 1, ϕ is a polyomial of degree ad is orthogoal to all polyomials of degree less tha. (c) As h is a liear combiatio of h 0,...,h 1 (i L 2 (µ)), we see that ψ is well-defied but ψ = 0 ad hece ϕ is ot defied. Note that ψ = 0 meas that ψ (λ )=0

4 32 3. GOE AND GUE for all, ot that ψ is the zero polyomial. I fact, ψ is moic, has degree ad vaishes at λ,, which implies that ψ (λ)= j=1 (λ λ j). Fix 0 1 ad expad xϕ (x) as xϕ (x) L2 (µ) = c, j ϕ j (x), c, j = xϕ (x)ϕ j (x)dµ(x). j=0 Now, xϕ j (x) has degree less tha if j < ad xϕ (x) has degree less tha j if < j. Hece, c, j = 0 if j 2 or if j + 2. Further, c,+1 = c +1, as both are equal to R xϕ (x)ϕ +1 (x)dµ(x). Thus, we get the three term recurreces (21) L 2 (µ) xϕ (x) = b 1 ϕ 1 (x)+a ϕ (x)+b ϕ +1 (x), 0 where a = xϕ (x) 2 dµ(x), b = xϕ (x)ϕ +1 (x)dµ(x). We adopt the covetio that ϕ 1, ϕ, b 1 ad b 1 are all zero, so that these recurreces also hold for = 0 ad =. Sice ϕ all have positive leadig co-efficiets, it is ot hard to see that b is oegative. From µ P 0 we have thus costructed a tridiagoal matrix T µ := T (a,b) T (cautio: here we have idexed a,b startig from = 0). If µ P 0 m for some m <, the T µ costructed as before will have size m m. Exted this by paddig m colums ad rows of zeros to get a real symmetric tridiagoal matrix (we abuse otatio ad deote it as T µ agai) i T. Thus we get a mappig µ T µ from P ito T. The followig lemma shows that T ν T is a bijectio, ad relates objects defied o oe side (matrix etries, characteristic polyomials, eigevalues) to objects defied o the other side (the support {λ j }, the weights p j, associated orthogoal polomials). Lemma 50. Fix 1. (a) The mappig T ν T is a bijectio from T 0 ito P 0 whose iverse is µ T µ. (b) Let T = T (a,b) ad let µ = ν T. For 0 1 P be the characteristic polyomial of the top pricipal submatrix of T ad let ψ,ϕ be as costructed earlier. The ψ = P for ad hece there exist costats d such that ϕ = d P (for 1). (c) The zeros of ϕ are precisely the eigevalues of T. (d) If T = T (a,b) ad ν T = p jδ λ j, the (22) I particular, T 0 b 2( +1) = p λ i λ j 2. i< j gets mapped ito P 0 (but ot oto). PROOF. (a) Let µ = j=1 p jδ λ j P ad let T = T µ. For 0 1, let u =( p 1 ϕ 0 (λ ),..., p ϕ 1 (λ )) t. The three-term recurreces ca be writte i terms of T as T u = λ u. Thus, u is a eigevector of T with eigevalue λ. If U is the matrix with colums u, the the rows of U are orthoormal because ϕ are orthogoal polyomials of µ. Thus UU = I ad hece also U U = I, that is {u } is a ONB of R. Cosequetly, T = λ u u is the spectral decompositio of T. I particular, T p e 1 =,1 u 2 λ p = p λ p

5 3. TRIDIAGONAL MATRIX GENERALITIES 33 because u,1 = p ϕ 0 (λ )= p (as h 0 = 1 is already of uit orm i L 2 (µ) ad hece after Gram-Schmidt ϕ 0 = h 0 ). Thus, T p e 1,e 1 = R x p µ(dx) which shows that ν T = µ. This proves the first part of the lemma. (b) We saw earlier that ϕ is zero i L 2 (µ). Hece ϕ (λ j )=0 for 1 j. Thus, ϕ ad P are o-zero polyomials of degree both of which vaish at the same poits. Hece, ϕ = d P for some costat P. If S is the top pricipal submatrix of T, the it is easy to see that the first orthogoal polyomials of ν S are the same as ϕ 0,...,ϕ (which were obtaied as orthogoal polyomials of ν T ). This is easy to see from the three-term recurreces. Thus the above fact shows ϕ = d P for some costat d. (c) By the first part, ν T = p j δ λ j where λ j are the eigevalues of T ad p j > 0. The footote o the previous page also shows that ϕ vaishes at λ j, j. Sice it has degree, ϕ has oly these zeros. (d) This proof is tae from Forrester s boo. Let T deote the bottom ( ) ( ) pricipal submatrix of T. Let Q be its characteristic polyomial ad let λ () j,1 j be its eigevalues. I particular, T 0 = T. If T = λ u u is the spectral decompositio of T ad λ is ot a eigevalue of T, the (λi T ) 1 = (λ λ ) 1 u u. Hece, (λi T )1,1 = (λi T ) 1 e 1,e 1 = j p j (λ λ j ) 1 for λ {λ j }. But we also ow that (λi T ) 1,1 is equal to det(λi T 1 )/det(λi T )=Q 1 (λ)/q 0 (λ). Let λ approach λ to see that (23) p = lim λ λ (λ λ )(λi T ) 1,1 = lim λ λ (λ λ ) Q 1(λ) Q 0 (λ) = Q 1 (λ ). (λ λ j ) j=1 j Tae product over to get (the left side is positive, hece absolute values o the right) p i< j (λ i λ j ) 2 = Q 1 (λ (0) ). Let A be ay matrix with characteristic polyomial χ A ad eigevalues λ i. Let B be a m m matrix with characteristic polyomial χ B ad eigevalues µ j. The we have the obvious idetity i=1 Apply to T 0 ad T 1 to get χ B (λ i ) = m i=1 j=1 Q 1 (λ (0) µ j λ i = ) = 1 by the first row, we also have the idetity m j=1 Q 0 (λ) = (λ a 1 )Q 1 (λ) b 2 1Q 2 (λ). Therefore Q 0 (λ (1) )=b 2 1 Q 2(λ (1) ) for 1. Thus χ A (µ j ). Q 0 (λ (1) ). But by expadig det(λi T ) Q 1 (λ (0) ) = b Q 2 (λ (1) ). The right side is of a similar form to the left side, with matrix size reduced by oe. Thus, iductively we get Q 1(λ (0) ) = 1 b2 2. Pluggig ito (23) we get the statemet of the lemma. 3. Tridiagoal matrix geeralities Fix 1 ad write T = T (a,b) for the real symmetric tridiagoal matrix with diagoal etries T, = a for 1 ad T,+1 = T +1, = b for 1 1. Let T

6 34 3. GOE AND GUE be the space of all real symmetric tridiagoal matrices ad let T 0 be those T (a,b) i T with b strictly positive. Let P be the space of all probability measures o R whose support cosists of at most distict poits ad let P 0 be those elemets of P whose support has exactly distict poits. Give a real symmetric tridiagoal matrix T, let ν T be the spectral measure of T at the stadard uit vector e 3 1. This gives a mappig from T ito P. For future purpose, we also give the followig idea to fid eigevalues of T. Fix some λ R ad suppose we wat to fid a vector v such that T v = λv. This meas b 1 v 1 + a v + b v +1 = λv = v +1 = λv b 1 v 1 a v b. where we adopt the covetio that b 0 = 0. We have also assumed that b 0 for all (if b = 0, the matrix splits ito a direct sum of two matrices). Thus, we set v 1 = x to be arbitrary (o-zero)ad solve for v 1,v 2,... successively. Deote these as v 1 (x),v 2 (x),... Therefore, Now suppose a measure µ P 0 is give. We ca costruct a tridiagoal matrix T as follows. Write µ = p 1 δ λ p δ λ where λ j are distict real umbers ad p j > 0. The momets are give by α = p j λ j. Let h (x)=x, so that {h 0,h 1,...,h 1 } is a basis for L 2 (µ). [Q: How do you express h as a liear combiatio of h 0,...,h 1?]. Apply Gram-Schmidt to the sequece h 0,h 1,... to get a orthoormal basis {ϕ :0 1} of L 2 (µ). It is easy to see that ϕ is a polyomial of degree exactly, ad is orthogoal to all polyomials of degree less tha. Fix ay ad write xϕ (x) L2 (µ) = j=0 c, j ϕ j (x), c, j = xϕ (x)ϕ j (x)dµ(x). Now, xϕ j (x) has degree less tha if j < ad xϕ (x) has degree less tha j if < j. Hece, c, j = 0 if j 2 or if j + 2. Further, c,+1 = c +1, as both are equal to R xϕ (x)ϕ +1 (x)dµ(x). Thus, we get the three term recurreces (24) L 2 (µ) xϕ (x) = b 1 ϕ 1 (x)+a ϕ (x)+b ϕ +1 (x), 0 where a = xϕ (x) 2 dµ(x), b = xϕ (x)ϕ +1 (x)dµ(x). We adopt the covetio that ϕ 1, ϕ, b 1 ad b 1 are all zero, so that these recurreces also hold for = 0 ad =. From µ P 0 we have thus costructed a tridiagoal matrix T µ := T (a,b) T (cautio: here we have idexed a,b startig from = 0). If µ P 0 m for some m <, the T µ costructed as before will have size m m. Exted this by paddig m colums ad rows of zeros to get a real symmetric tridiagoal matrix (we abuse otatio ad deote it as T µ agai) i T. Thus we get a mappig µ T µ from P ito T. Lemma 51. Fix 1. (a) The mappig T ν T is a bijectio from T ito P whose iverse is µ T µ. 3 The spectral measure of a Hermitia operator T at a vector v is the uique measure ν o R such that T p v,v = R x p ν(dx) for all p 0. For example, if T is a real symmetric matrix, write its spectral decompositio as T = λ u u. The {u } is a ONB of R ad λ are real. I this case, the spectral decompositio of T at ay v R is just ν = v,u 2 δ λ. Thus ν P (observe that the support may have less tha poits as eigevalues may coicide). I particular, if T = UDU the spectral measure of T at e 1 is ν T = p i δ λi, where p i = U 1,i 2.

7 4. MORE ON TRIDIAGONAL OPERATORS* 35 (b) Let µ = ν T. Write P for the characteristic polyomial of the top submatrix of T for ad let ϕ be the orthogoal polyomials for µ as defied earlier. The ϕ = d P for for costats d. I particular, zeros of ϕ are precisely the eigevalues of T. (c) If T = T (a,b) ad µ = p jδ λ j correspod to each other i this bijectio, the (25) I particular, T 0 b 2( +1) = p λ i λ j 2. i< j gets mapped ito P 0 (but ot oto). PROOF. (a) Let µ = j=1 p jδ λ j P ad let T = T µ. For 0 1, let u =( p 1 ϕ 0 (λ ),..., p ϕ 1 (λ )) t. The three-term recurreces ca be writte i terms of T as T u = λ u. Thus, u is a eigevector of T with eigevalue λ. If U is the matrix with colums u, the the rows of U are orthoormal because ϕ are orthogoal polyomials of µ. Thus UU = I ad hece also U U = I, that is {u } is a ONB of R. Cosequetly, T = λ u u is the spectral decompositio of T. I particular, T p e 1 =,1 u 2 λ p = p λ p because u,1 = p ϕ 0 (λ )= p (h 0 = 1 is already of uit orm i L 2 (µ) ad hece after Gram-Schmidt ϕ 0 = h 0 ). Thus, T p e 1,e 1 = R x p µ(dx) which shows that ν T = µ. This proves the first part of the lemma. (b) By part (a), the coefficiets i the three term recurrece (24) are precisely the etries of T. Note that the equality i (24) is i L 2 (µ), which meas the same as sayig that equality holds for x = λ,1. Here is a way to fid T (c) Let A be ay matrix with characteristic polyomial χ A ad eigevalues λ i. Let B be a m m matrix with characteristic polyomial χ B ad eigevalues µ j. The we have the obvious idetity i=1 χ B (λ i )= m i=1 j=1 (µ j λ i ) = ( 1) m m χ A (µ j ) If b are all positive, the right had side of (??) is o-zero ad hece λ must be distict. This shows that T 0 gets mapped ito P 0. It is obviously ot oto (why?). Lemma 52. For T = T (a,b) havig the spectral measure +1 j=1 p iδ λ j at e 0, we have the idetity 1 =0 b 2(+1 ) +1 = i=1 p i j=1 (λ i λ j ) 2. i< j More o tridiagoal operators* This sectio may be omitted as we shall ot use the cotets i this course. However, as we are this close to a very rich part of classical aalysis, we state a few iterestig facts. The followig four objects are show to be itimately coected. (1) Positive measures µ P (R) havig all momets.

8 36 3. GOE AND GUE (2) Positive defiite sequeces α =(α ) 0 such that (α i+ j ) i, j 0 is o-egative defiite (that is every pricipal submatrix has o-egative determiat). (3) Orthogoal polyomials. Give a ier product o the vector space of all polyomials, oe ca obtai a orthoormal basis {ϕ } by applyig Gram-Schmidt process to the basis {h } 0 where h (x)=x. The sequece {ϕ } (which may be fiite) is called a orthogoal polyomial sequece. (4) Real symmetric tridiagoal matrices. We ow cosider semi-ifiite matrices, that is T, = a, T,+1 = T +1, = b, for 0. Fiite matrices are a subset of these, by paddig them with zeros at the ed. Measure to Positive defiite sequeces: If µ is a measure that has all momets, defie the momet sequece α = R x dµ(x). The for ay for ay m 1 ad ay u R m+1, we have u t (α i+ j ) 0 i, j m u = α i+ j u i u j = u i x i 2 µ(dx) 0. i, j m i=0 Hece α is a positive defiite sequece. It is easy to see that µ is fiitely supported if ad oly L 2 (µ) is fiite dimesioal if ad oly if (α i+ j ) i, j 0 has fiite ra. Positive defiite sequece to orthogoal polyomials: Let α be a positive defiite sequece For simplicity we assume that (α i+ j ) i, j 0 is strictly positive defiite. The the formulas h i,h j = α i+ j defie a valid ier product o the vector space P of all polyomials. Complete P uder this ier product to get a Hilbert space H. I H, h are liearly idepedet ad their spa (which is P ) is dese. Hece, applyig Gram-Schmidt procedure to the sequece h 0,h 1,... give a sequece of polyomials ϕ 0,ϕ 2,... which form a orthoormal basis for H. Clearly ϕ has degree. Orthogoal polyomials to tridiagoal matrices: Let ϕ be a ifiite sequece of polyomials such that ϕ has degree exactly. The it is clear that ϕ are liearly idepedet, that h is a liear combiatio of ϕ 0,...,ϕ. Cosider a ier product o P such that ϕ,ϕ l = δ,l. The same reasoig as before gives the three term recurreces (24) for ϕ. Thus we get a R ad b > 0, 0. Form the ifiite real symmetric tridiagoal matrix T = T (a,b). Symmetric tridiagoal operators to measures: Let T be a semi-ifiite real symmetric tridiagoal matrix. Let e be the co-ordiate vectors i l 2 (N). Let D = {x e : x 0 fiitely ofte}. This is a dese subspace of l 2 (N). T is clearly well-defied ad liear o D. It is symmetric i the sese that Tu,v = u,tv for all u,v D ad the ier product is i l 2 (N). Suppose T is a self-adjoig extesio of T. That is, there is a subspace D cotaiig D ad a liear operator T : D R such that T D = T ad such that T is self-adjoit (we are talig about ubouded operators o Hilbert spaces, hece self-adjoitess ad symmetry are two distict thigs, ad this is ot the place to go ito the defiitios. Cosult for example, chapter 13 of Rudi s Fuctioal Aalysis). The it is a fact that T has a spectral decompositio. The spectral measure of T at e 0 is a measure. I geeral there ca be more tha oe extesio. If the extesio is uique, the µ is uiquely defied. This cycle of coectios is quite deep. For example, if we start with ay positive defiite sequece α ad go through this cycle, we get a OP sequece ad a tridiagoal symmetric operator. The spectral measure of ay self-adjoit extesio of this operator has the momet sequece α. Further, there is a uique measure with momets α if ad oly if T has a uique self-adjoit extesio!

9 4. MORE ON TRIDIAGONAL OPERATORS* 37 Remar 53. I the above discussio we assumed that α is a strictly positive defiite sequece, which is the same as sayig that the measure does ot have fiite support or that the orthogoal polyomial sequece is fiite or that the tridigoal matrix is essetially fiite. If we start with a fiitely supported measure, we ca still go through this cycle, except that the Gram-Schmidt process stops at some fiite etc.