MT581 Advaced Combiatorics 6 Gaussia coefficiets The -aaolgue of a combiatorial formula is, loosely speaig, a formula ivolvig a parameter, which teds to the origial formula as 1 However, this is much too vague to be a defiitio, ad it turs out that there are some very specific -aalogues which crop up i several differet fields Most importat of these are the Gaussia, or -biomial, coefficiets, which we discuss i this chapter Amog other places, they come up i the followig areas: Combiatorics of vector spaces over fiite fields It turs out that there are close aalogies betwee sets ad subsets, o oe had, ad vector spaces ad subspaces, o the other The coutig formulae replace the biomial coefficiets by their -aalogues Lattice paths We ow that the umber of lattice ( paths ) from (0,0) to (m,) m + (usig oly orthward ad eastward steps) is To cout these paths by the area uder them, we itroduce a ew variable to give a geeratig fuctio, ad the formula becomes the -aalogue of the biomial coefficiet No-commutative geometry I will ot give a detailed accout of this, but ote a couple of occurreces The simplest to describe is the biomial theorem for idetermiates x, y which satisfy yx = xy; the biomial coefficiets i the expasio are replaced by their -aalogues There are several further applicatios of the -biomial coefficiets, amog them uatum calculus ad braided categories I will ot discuss these, but there is a very accessible boo o uatum calculus if you are iterested: V Kac ad P Cheug, Quatum Calculus, Spriger, New Yor, 00 1
61 The defiitio Let ad be o-egative itegers The Gaussia or -biomial coefficiet is defied to be = ( 1)( 1 1) ( +1 1) ( 1)( 1 1) ( 1) I other words, we tae the formula for the biomial coefficiet, ad replace each factor m i either umerator or deomiator by m 1 For example, [ ] 4 = (4 1)( 3 1) ( 1)( 1) = ( + 1)( + + 1), a polyomial of degree 4 i Puttig = 1, we obtai 3 = 6, which is eual 4 to Propositio 61 lim 1 = Proof By L Hôpital s Rule, we have a 1 lim 1 b 1 = lim a a 1 1 b b 1 = a b Now brea the fractioal expressio givig the Gaussia coefficiet ito the product of fractios, of which the ith teds to ( i + 1)/( i + 1) as 1 The result follows 6 Vector spaces over fiite fields The umber of elemets i a fiite field is ecessarily a prime power, ad there is a uiue field (up to isomorphism) of ay give prime power order The field with elemets is deoted by GF(), or sometimes F
From elemetary liear algebra, we ow that a -dimesioal vector space V over GF() has a basis v 1,,v such that every vector v has a uiue expressio as a liear combiatio of basis vectors: v = c 1 v 1 + c v + + c v, c i GF() So, as usual, V ca be idetified with the space GF() of all -tuples of elemets of GF(), with coordiatewise operatios I particular, we see: Propositio 6 If V is a -dimesioal vector space over GF(), the V = Now the coectio with Gaussia coefficiets is the followig: Theorem 63 The umber [ ] of -dimesioal subspaces of a -dimesioal vector space over GF() is Proof We specify a -dimesioal subspace by givig a basis, a liearly idepedet -tuple (v 1,,v ) of vectors i V The umber of choices for v 1 is 1 (ay vector except 0); for v is (ay vector except oe of the multiples of v 1 ); for v 3 is (ay vector except oe of the liear combiatios of v 1 ad v ); ad so o; the total umber of choices is ( 1)( ) ( 1 ) However, we have over-couted, sice a give -dimesioal subspace has may bases How may? The umber is foud from the same formula usig istead of, sice we are worig withi a -dimesioal space So the umber of -dimesioal subspaces is ( 1)( ) ( 1 ) ( 1)( ) ( 1 ) Cacellig powers of, this reduces to the formula for the Gaussia coefficiet This result has aother coutig iterpretatio Recall that a matrix A over a field F is i reduced echelo form if 3
if a row is ot idetically zero, the first o-zero elemet i it is 1; the leadig oes i the o-zero rows occur further to the right as we go dow the matrix; all the other elemets i the colum of a leadig oe are zero Now stadard liear algebra shows that ay matrix ca be put ito reduced echelo form by elemetary row operatios, which do ot chage the row space of the matrix So ay -dimesioal subspace of F has a basis whose vectors are the rows of a matrix i reduced echelo form Moreover, it is easy to see that the reduced echelo basis of a give subspace is uiue So: Propositio 64 The umber of matrices [ ] over GF() which have o zero rows ad are i reduced echelo form is The defiitio of reduced echelo does ot deped o properties of fields; ay alphabet cotaiig distiguished elemets called 0 ad 1 will do So we have a coutig iterpretatio of the Gaussia coefficiets for arbitrary positive itegers > 1 Example For = 4 ad =, the matrices i reduced echelo form are show ca be ay elemet of the field Beside each matrix is the umber of matrices of this form over GF() 1 0 We fid, as before, that 4 0 1 1 0 0 0 1 3 1 0 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 1 0 0 1 0 1 0 0 0 1 [ ] 4 = ( + 1)( + + 1) 4
63 Relatios betwee Gaussia coefficiets The -biomial coefficiets satisfy a aalogue of the recurrece relatio for biomial coefficiets [ ] [ ] [ ] [ ] [ ] 1 1 Propositio 65 = = 1, = + for 0 < < 0 1 Proof This comes straight from the defiitio Suppose that 0 < < The [ ] [ ] ( 1 )[ ] 1 1 = 1 1 1 1 ( = )[ ] 1 1 1 1 = 1 The array of Gaussia coefficiets has the same symmetry as that of biomial coefficiets Propositio 66 [ ] [ ] = The proof is a exercise from the formula Note that, i the vector space iterpretatio, we have a differet way to see this Give a -dimesioal vector space V over GF(), it has a dual space V, the space of liear maps from V to GF() Now ay -dimesioal subspace of V has a ( )-dimesioal aihilator i V, ad the correspodece betwee -dimesioal subspaces of V ad ( )-dimesioal subspaces of V is bijective From this we ca deduce aother recurrece relatio [ ] [ ] [ ] [ ] [ ] 1 1 Propositio 67 = = 1, = 0 1 + for 0 < < 5
Proof = = = [ ] [ ] 1 1 + 1 [ ] [ ] 1 1 + 1 We come ow to the -aalogue of the biomial theorem, which states the followig Theorem 68 For a positive iteger, a real umber 1, ad a idetermiate x, we have (1 + i 1 x) = i=1 =0 ( 1)/ x Proof The proof is by iductio o ; startig the iductio at = 1 is trivial Suppose that the result is true for 1 For the iductive step, we must compute ( 1 [ ] ) 1 (1 ( 1)/ x + 1 x ) =0 The coefficiet of x i this expressio is [ ] [ ] 1 1 ( 1)/ + ( 1)( )/+ 1 1 ( [ ] 1 = ( 1)/ ] [ 1 + 1 ] ) = ( 1)/ [ by Propositio 67 6
64 Lattice paths We cosider lattice paths from the origi to the poit (m,), where m ad are o-egative ( itegers, ) ad the allowable steps are east ad orth The umber of m + paths is, sice we have to tae m + steps altogether, of which m must m be easterly ad ortherly For each such path p, there is a certai area A(p) eclosed betwee the path, the X-axis, ad the lie x = m Figure 1 shows the six paths for m = = ad the area eclosed i each case Path Area 0 1 3 4 Figure 1: Lattice paths If we tae a geeratig fuctio i the variable for these areas, that is, a path with area A cotributes A to the sum, we obtai [ ] 4 1 + + + 3 + 4 = 7
This is uite geeral: Theorem 69 Let P be the set of lattice paths from (0,0) to (m,) usig ortherly ad easterly steps oly For p P, let A(p) be the area eclosed by p, the X-axis ad the lie x = m The [ ] m + A(p) = p P Proof Call the left-had side F(m,) It is clear that F(0,) = F(m,0) = 0 Now cosider F(m,) There are two cases: If the last step o the path p is ortherly, the it is a path from (0,0) to (m, 1) followed by a ortherly step, ad the last step does t chage the area If the last step is easterly, the p is a path from (0,0) to (m 1,) followed by a easterly step, which adds to the area m So F(m,) = F(m, 1) + F(m 1,) Now a simple iductio usig Propositio 65 gives the result Exercises 61 Show that the Gaussia coefficiet ( ) is a polyomial i with degree 6 Use the iterpretatio i terms of matrices i reduced echelo form to show the recurrece relatio of Propositio 65 63 Show that the polyomial i the first problem has coefficiets which are symmetric, that is, if ( ) = a i i, the a i = a ( ) i i=0 8
64 A matrix is said to be i echelo form if it satisfies the first two of the three coditios for reduced echelo form Show that, if is a iteger greater tha, the right-had side of the - biomial theorem with x = 1 couts the umber of matrices i echelo form 65 Let x ad y be elemets of a algebra, which satisfy yx = xy Prove that (x + y) = x y =0 66 Let X be the set of 1-dimesioal subspaces of a -dimesioal vector space V over GF(), where 3 For ay -dimesioal subspace W of V, let B(W) be the set of 1-dimesioal subspaces cotaied i W, ad let B be the collectio of all blocs or subsets of X arisig i this way Prove that (X,B) is a (, + 1,( 1)/( 1)) Steier system (This system is ow as the ( 1)- dimesioal projective space over GF()) 9