Explorations with q-series

49 views Mar 19, 2023

The Gaussian q-binomial coefficients have another interpretation in terms of Young diagrams contained in a rectangle, and inversion numbers of associated paths. We give an introduction to this interesting and important chapter of combinatorics, and then connect this interpretation of Gaussian q-binomial coefficients with the one discussed in the previous video, relating to counting subspaces of a vector space over a finite field F_q of size q. It turns out that the key to connecting these two seemingly quite different interpretations is the row-echelon forms for subspaces.

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Wild Egg Maths](https://www.youtube.com/@WildEggmathematicscourses)

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hello everyone I'm Norman walburger thanks for joining me on this exploration of Q Series my last video I

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talked about gaussian Cube binomial coefficients and the connection with counting subspaces in a vector space

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over a finite field today I want to discuss a quite different interpretation of these

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interesting objects in the direction of combinatorics associated to Young diagrams and paths

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and Associated areas to such paths and we're also going to be touching on a

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notion of inversion which plays a role a big role in the theory of the symmetric group

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okay so let's start by defining a partition of a natural number n to be a

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list of strictly positive natural numbers in decreasing order which sum to

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n so for example if n is equal to 15 we could write 15 as 5 plus 5 plus 3 plus 2

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then these numbers five five three two in that order so decreasing or at least

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weekly decreasing in the sense that they could be the same is a partition of 15.

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okay now associate to such a partition there's actually a physical uh thing

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that we can create called a young diagram and it has rows of lengths the numbers

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involved reading from left to right and putting those rows from top to bottom

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and the rows are left Justified so corresponding to 5532 we have this young

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diagram five and this first row five in the second row three in the next row and

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two in the fourth row now we say that this young diagram has

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shape five five three two and its total size is well the total number of squares

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that it has which is the number n that we're talking about in 15. now occasionally there's a variant of

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this that's used called the fairer's diagram that's usually the case when instead of using squares like this we

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have little dots okay in terms of

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the convention for a notation the squares in the young diagram are read

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using standard Matrix conventions where we read the row number first and then

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the column number second so the that we're reading from top to bottom as far as rows are concerned we're reading from

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left to right as far as columns are concerned so in this example here this particular entry here would be the one

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three entry because it's in the row number one and then column number three this thing here isn't the two one entry

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because it's in row two and column one I probably should also say that there is

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actually another convention around for young diagrams this is sometimes called the English convention okay and there's

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also a French convention where we sort of Stack the the rows from the bottom to

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the top so just to alert you to there's a bit of a Divergence in the literature

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okay another interesting concept Associated to a young diagram is

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the hook length of a square okay that's the sum of the number of squares to the right below and also the square itself

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so for example in this diagram here this square right here okay has one to the

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left and including itself that's two this five here is five because it's

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Counting this Square this Square this Square this square and the square so there's a hook okay this one's six

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because there's one two three four five six so these are the various hook lengths of the squares in this young

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diagram those numbers turn out to be important in in representation Theory certain

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counting problems associated to what are called tableau so a young Tableau is a filling of the

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squares of a young diagram with strictly positive natural numbers now we're not actually going to be using Tableau here

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in this lecture but I just want to alert you to the notation because it is an

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important sort of Allied concept to that of a young diagram usually one introduces young diagrams in order to

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talk about young Tableau so I just want to briefly mention this I will talk more about these things at some other point

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probably in the context of representations of either the symmetric group or the general linear group or

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representations more generally so actually this would be a young Tableau because we filled up the squares with

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natural numbers but there's some special kinds of young Tableau that are particularly interesting the stat a

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standard young Tableau is a situation where we put in the numbers from 1 to n

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if there are altogether n boxes or n squares okay and and E so each number is

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occurring exactly once and furthermore the numbers are strictly increasing along both rows and columns so if you go

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down a row the numbers are increasing if we go across a column the numbers are increasing so listen here is an example

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this uh this young diagram has size 15 and 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

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that's a young Tableau for that young diagram and these turn out to be very important

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in the representation theory of the symmetric group s n

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there's another kind of young Tableau that are very important

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they're going to call the semi-standard young Tableau in this case we're filling the uh the squares also with with

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natural numbers say natural numbers from 1 to to R Maybe here's R of five and the

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the young diagram should have no more than R rows so no more than five

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rows and the rule is that the number should be increasing down the column so

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strictly increasing okay strictly increasing down the columns but weekly increasing along the rows

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that means they can stay the same for a little while before they start increasing so one one two three four

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that's a weekly increasing sequence as is two two three five five three four

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four etcetera but down the columns they are strictly increasing and these kinds

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of young Tableau are important in the representation theory of the general linear group or the special linear group

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now I just have to very quickly mention there's a another kind of Tableau called

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quasi-standard Tableau which were introduced by DDA Arnel Nadia Bell

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Baraka and myself some years ago and uh I'm not going to tell you what these

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things are but at this point but I will tell you about them some other time they're very important variants that

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play a crucial role in understanding certain aspects of representation theory of Lee algebras

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okay so today I want to focus on young diagrams which appear in a rectangle

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okay so we're going to fix a rectangle of size say k by n and our convention is

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that that means that we have K rows and N columns so this is k equals three and

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n equals four again it's a rectangle and we're just going to look at the Young diagrams which fit into this rectangle

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which fit inside it so here in blue one two three four five

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six seventy seven squares in blue that's a young diagram that's sitting inside

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the rectangle it's sitting kind of in a Justified way so that it's bumping up

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against the left side of the rectangle and bumping up against the top of the rectangle so it's sort of in the top

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left corner if you like okay the shape of this diagram is three three one

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now Associated to such a young diagram

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inside the rectangle there's a path a path from the bottom

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left corner to the top right corner in this case this path in red okay and what

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is this path right well it's just the border between the squares which are

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involved in the young diagram and the complement in other words those squares which are not contained in

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the young diagram okay so here we have this this path this border path which

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separates the rectangle into two pieces one of which is the young diagram that

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we're considering now such a path will consist of K plus n

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segments because there's K rows and N columns so to go from here to here we're

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going to have to go up the K row so we're going to go across the N column so

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all together there are K plus n segments in this case one two three four five six

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seven and so it's natural for us to label these little segments okay by the

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numbers one through seven just in order so this is segment one segment two segment three segment four segment five

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segment six and segment seven okay and each of these segments is of

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one of two kinds it's either a right segment namely it's going horizontally to the right or it's an up

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segment going vertically up so for example in this example here here's just the path itself one three four and seven

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are the right edges or right segments and the up edges or up segments are

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numbers two five and six

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okay so now if we think about it for a second we see that such a path is actually uniquely determined by either

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one of these sets so in particular by the set of right edges if you know what

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the right edges are then you know what the up edges are because they're the other ones okay and uh these these

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numbers are telling you when you should go right so if it have the set one three four seven it means that we're going to

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go right on the first segment and also right on the third and

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the fourth and the seventh so it means on the other ones we're going to be going up so it's quite clear what the

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path is if we know the set of right edges and this set of right edges can be

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in principle any any subset of of the seven possible segments that has four

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elements because to go from here to here we are going to have to go right four times

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okay so of the seven steps that we're taking four of them have to be right steps

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and so there are seven choose four which is seven times six times five or

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three times two times one that's uh 35 such sets and therefore that's the same as

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the number of such paths that we're considering and therefore also the same as the number of young diagrams that fit

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inside this K by n rectangle so we have a

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a couple of different ways now of seeing the usual binomial coefficients okay

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like seven choose four here as counting objects which are reasonably familiar

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okay you're used to thinking of seven choose four as counting the number of subsets of four things from seven things

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that's that's very familiar but maybe a little bit less familiar is to think of

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this number is also counting the number of paths from this lower left to the upper right corner of this rectangle

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this three by four rectangle and correspondingly it also is really counting the number of young diagrams

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which are above and then sort of to the left of this path that we're choosing so

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all these quantities are being counted by the same binomial coefficients

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something else that we can associate to such a path okay which is also important

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which is a permutation of n in the example seven so this is obtained by

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listing in order all the right edges followed by all the up edges so in the

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example the right edges were 1 3 4 7 the up edges were two five six okay there

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they are one three four seven on the right edges two five and six are the up

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edges so we can make a permutation by listing the right edges first and then

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the up edges after that we get one three four seven two five six

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now this kind of permutation okay of course

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there are seven factorial permutations that's a lot more than 35 but the the

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permutations that we're obtaining this way have a particular property namely that there's only one inversion

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um uh spot in them there's only maybe inversion is not

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actually the right word here um there's only one decrease probably that would be a better word so um forget about this

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inversion here I should have said there's really only one decrease so we're going up one three four seven

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we're going up and then we're going down to two five and then going up five to six because the edges that are right are

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listed in order increasing order and the up edges are also listed in increasing

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order so the only possibility when we're actually going down like seven is bigger

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than two we're going down from seven to two is right at this step the the place where the the right edges are separated

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from the down edges okay so uh maybe okay the

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permutations obtained this way have inversion set I'm not again I'm not sure if I should be using inversions that probably I should be using descent set

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instead so I think about this as descent set instead contained in in four so only

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at this point here are we allowed to go down

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actually when we're not uh absolutely required to go down because if we had the sequence one two three four five six

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seven that would also be a legitimate sequence that would not have a descent at that point but if we are going to

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have a descent in this um in this permutation it has to appear right at this point from the fourth at

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the fourth element down okay so we now have another kind of

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combinatorial object that essentially being counted by these classical binomial coefficients now this uh this

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permutation uh supports a notion of inversion number okay it's a more proper

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use of the word inversion so let me Define that the inversion number of a permutation okay what is that it's the

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minimal number of swaps required to get it into the sort of standard increasing form one two three up to n

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and more precisely perhaps inversions of Pi it's the number in this set the set

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of this should be a bracket here the set of i j where I is less than J and these

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numbers are between 1 and n including one and then so I should be less than J and Pi of I

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where Pi is the permutation Pi by is bigger than Pi of J okay so for example in this example here

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the uh the inversions in in this permutation one three four seven two

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five six we look for two numbers in this range say like two three where two is less

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than three okay which are inverted in the permutation so in the permutation if

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you look for two and three three comes before two all right so two three is a kind of inversion pair for this

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permutation and other inversion pairs involve four two and four that's an

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inversion pair a two and seven they've been reversed five and seven and six and

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seven okay but for example one five is not an inversion pair because one and five are appearing in the same order as

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their natural increasing order three seven is not an inversion pair etcetera so there are five inversion pairs in

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this um in this permutation okay and so this

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number inversion in inv of Pi is five in this case

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okay so let me look at some examples using the three by four rectangle to

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illustrate these various ways of thinking about uh these young diagrams and you know

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possibilities for them just contained in this three by four rectangle here okay so here is actually

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the one that we started with so it's this path here with right edges one

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three four and seven okay here's the corresponding permutation one three four

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seven and then two five six and in versions of this this Lambda Lambda is

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the is the young diagram up here three three one okay uh the inversion number for that is five

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here's a very particularly simple one where the path just goes right for four

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steps and then up for three steps in this case the the set the subset is just the set one

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two three four the initial four right steps and the actual corresponding permutation for it is just the simplest

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one sort of the identity permutation one two three four five six seven and the inversion number of that is obviously

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zero and the corresponding young diagram has shape four four here it is

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surprising the entire rectangle here's another example where we're going up and then we're going right right up right

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right up okay so the young diagram here shape four two

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the subset that's corresponding to the right

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edges segment two segment three segment five and segment six and then the

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corresponding permutation two three five six the right edges and then one four

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seven the up edges and how many inversions are there there's four plus two why is that well

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see this one here if we wanted to get it you know back to where it started we'd have to pass it across the six the five

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the three and the two so there's four inversions involved with

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this one the number of inversions involved with four well there's two of them because we have to move it past the

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six and the five okay and the Seven there's no inversions Associated to that so if there's four plus two and note

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that four plus two is also the number of um squares that are appearing below here

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in fact if we go back to this thing here we see that something similar is happening here that if we look at the

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inversions the inversions actually can be counted in the following way so these guys here are increasing order

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if we want to put the two five and six in sort of standard increasing order we have to move the two down we have to

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move the two down three steps past the seven past the four past the three so there are three inversions sort of

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coming from the number two that sort of corresponds to this row here and then

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five well five has to be moved past seven so there's one move one version Associated to five and

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six also has to be moved past the seven so another inversion for that so that sort of corresponds to

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those two things let's have a look at this other sort of extreme case where we're going up first and then right so now the right Edge set

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is four five six seven the permutation is four five six seven followed by one

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two three and now the inversions are

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what well the one has to be moved across all four of them in order to get in position then the two has to be moved

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also across all four of these and the three has to be moved also across all four of these so we get four plus four

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plus four which is twelve and that's the same as this uh

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this sort of set of squares which are to the right and and below are path

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and in this case here we're going up and then right right right right and then up

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up so the the set is two three four five okay that's the the subset of four

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things from Seven that uh that determines that this this path the

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corresponding permutation is two three four five one six seven and the inversions well we

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have to move the one over that takes four steps and that's all that's required corresponding to that four

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there in this case here the um the Young diagram has shape four four in this

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diagram here the young diagram is is sort of invisible as shape zero okay

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so these are just some of the uh I guess 35 all together that you could tabulate

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and it's useful to spend some time just tabulating more of these things so you get a sense of the the strong

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correlation between these different objects which are all being counted by

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the standard binomial coefficient in this case seven choose four okay so we

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could be talking about young diagrams we could be talking about paths we could be talking about subsets

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we could be talking about permutations that have a particular descent position

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okay and now a sort of an interesting observation that I've already alluded to

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that the inversion number of a path is the area in unit squares under the path

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of the number okay now this inversion number of four here is really counting the number of

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squares underneath it here similarly this 6 is counting the number

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of squares here and to give you an idea for that so suppose we took this one here okay

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listening here and let's just look at this particular uh part of it we're right from here to here we're going up

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and then right suppose we change that okay so we suppose we change that so that up uh is uh is that is that four

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there and that right is is at five so both we change that so to go you know

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across here and up there well in that case this this thing would look like two

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three four

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um six so there'd be a two three four six here and and uh a one five seven so

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basically the four and the five would have changed places okay so

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um what that's going to do is going to through add another inversion in into

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the into the mix okay because uh well actually in this case we're

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going to reduce the inversion because if we replace the if we interchange the four and the five then currently they're

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responsible for an inversion because 5 is bigger than four but if we interchange them then that in that

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inversion is is lost so we have one less and that corresponds exactly to this

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little square here that we are sort of evaporating if we change this little

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path to this one here okay so it's some kind of idea of why this works but it's definitely a good

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exercise to convince yourself of this and a good way to do that is just to

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work through a lot of these things here and sort of do a lot of counting and sort of see how it goes and by the time

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you've done close to 35 you will probably have a very good idea of why this works okay so now we can get to this new

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interpretation of the gaussian Q binomial coefficients okay

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so um we've seen that the classical binomial coefficients say m choose R

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corresponds to the number of subsets of size R from a set of size m

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and in terms of these rectangles it's the number of paths

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shape right and up in a rectangle of size M minus r cross R so our example

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was seven um choose four so that corresponds to a 3 times 4 rectangle

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so we could Define say p m r to be the

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set of such paths like right up pass in the rectangle of

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size M minus r cross R so we can think about a combinatorial object which is

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the set of all such paths okay well then

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what M choose R is okay one way of saying what it is is

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we're counting the number of such things and you know from our point

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of view it's sort of like a in in sort of uh box arithmetic language

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if you allow me to just momentarily do this basically what we're doing is we're replacing each of these paths with zero

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thinking of having an M set this is a set but we can think of as an amp set and we're just replacing each element by

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zero and once you've done that then basically you've counted the number of of such paths so for example when we

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have um m equals four at R equals two then we're talking about a two by two square

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and here are the six possible paths from the lower left to the upper right

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this one here this one here this one here this one here this one here and

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this one here here are the associated um

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subsets okay corresponding to the right edges here the right edges are numbers three and four here they are two and

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four here they are two and three here they are one and four here they are one and three and here they are one and two

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okay and when we also record the corresponding

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permutations okay that's the permutation you get by listing the right edges first

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three four and then the up Edge is one two and here two four and then one three and then here two three and one four

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here one four and two three here one three and two four here one two followed by three four okay so this set the four

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choose two what that is counting is just counting the number of elements in this set just replacing each of them by zero

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in a box arithmetic point of view I've been following the math foundations series on unboxed arithmetic but in any

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case that's sort of obvious what we're doing we're just counting these things okay now

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the proposition now is that these gaussian binomial coefficients are also

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counting these these objects in in this set but

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counting in a somewhat more sophisticated way what we're doing is we're looking at the inversions

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of of uh of these paths so let me explain

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that so the we have this inversion operator okay it takes a path and

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assigns a number to it so we could if we're sort of happy with working with multi-sets think about applying

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something like inversion to the entire set so this is a multi-set of paths or

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permutations if you like and we can apply the inversion function to this

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this set or this mset by applying it to each individual element

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okay so what are you we're going to get well we have to count the number of

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inversions of each of these things so let's just count how many inversions does this thing have well one way of

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seeing that is just to count the number of squares to the right and under this thing so this one has four this thing

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here has an inversion number of three this one two this one two this one one

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and this one zero okay so the inversion number of this permutation is four this

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one is three two two one zero so what we get when we apply the inversion to this

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set is we get this m set 4 3 2 2 1 0

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we're just recording the numbers which correspond to the inversions of each of the individual elements of this

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set now from a box arithmetic point of view and I do

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hope that you're watching those box arithmetic videos this is

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an M set okay which is an mset of natural numbers and so it corresponds to

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a polynomial or poly number and if you haven't watched those what

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we're just doing here is we're just interpreting each of these numbers here as a power of alpha so there's an alpha

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to the four represented by this thing an alpha cubed an alpha squared an alpha squared an alpha and an alpha to the

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zero or one so we add them all up we get 1 plus Alpha plus two alpha squared plus Alpha cubed plus Alpha to the fourth

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so we can do something a little bit more sophisticated if we do the simplest

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thing we're just counting the number in in this set of paths a little bit more sophisticated is we count or compute the

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inversion numbers of each of them all right now if we take this polynomial

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and we uh evaluated at one we're going to get one

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plus one plus two plus one plus one we're going to get six that's basically reverting to the counting

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if we evaluate a q which just means replacing the alpha with Q then we get what we usually call the gaussian Q

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binomial coefficient 4 choose two Q which is the polynomial 1 plus Q Plus 2q

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squared plus Q cubed plus Q to the fourth

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Okay so this is a way of thinking about what the

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the step from ordinary binomial coefficients to gaussian binomial coefficients involves instead of just

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sort of counting the number of possible paths okay we're counting the we're

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calculating the inversion numbers of each of these paths or if you prefer geometrically we're counting areas that

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are obtained by all these paths okay we're looking at how much what's the area of this path what's the area

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this path what's the area of this path and recording those various areas in this polynomial or polynomial

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so now we have two quite different interpretations of these gaussian Q binomial coefficients

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on the one hand M choose r q counts the number of r-dimensional subspaces of an

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M dimensional Vector space over the finite field fq and on the other hand

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it's a generating function for inversions or perhaps areas of paths in

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an M minus r times r rectangle

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so our natural question is can we somehow connect these two things directly so in come to Torx when we have

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two sets that are sort of maybe the same size or counted by the same number

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it's natural to try to ask whether we can establish a natural bijection between the the sets that sort of

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explains you know why these two things are going to be counted by the same numbers

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so I think this is not well understood and so I'm going to give you my take on

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on what the connection is so I do believe that there's a connection and the key is to consider these row Echelon

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forms as intermediaries between these two things okay so let me remind you

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that uh let's say in the the case when we have four dimensions and we're looking

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at two dimensional subspaces that um we have row Echelon forms okay

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um so let's just look at this thing here okay so we may remember that if we have

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um four-dimensional space okay over some finite field and we saw that two-dimensional subspaces of that could

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be put into row Echelon form so again using gaussian elimination and that

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these raw Echelon forms had these six different types or shapes okay and each

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one of these shapes sort of corresponds to a power of Q because if we look at rho Echelon matrices of this kind

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there's sort of a four-dimensional family of them because these Stars can be substituted with any one of the Q

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numbers in the field so there's Q to the fourth different possibilities for a row

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Echelon form of this kind well over here there's Q cubed possibilities for row

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Echelon form of this kind and so on okay so what I want to do is I want to

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connect this subspaces of four-dimensional space with this story

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of paths in a two by two rectangle and what we're going to do is we're simply going to

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um let me perhaps explain just about here so we're going to remove these leading columns the leading columns are

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the columns in this raw Echelon form that pass through the leading entries of

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the rows namely the ones that start the rows okay and um

35:17

so when we're going to chop off those columns wherever they are and then we're going to separate the remaining entries

35:24

into the zeros and the stars and we're going to see that we're going to get paths so let's see how that works in

35:31

this example here so here if we chopped off these two columns here then we would

35:36

just get this two by two Matrix of All Stars okay

35:42

maybe it's not so clear let's go to this next one momentarily so here we remove this First Column and this third column

35:48

we get this remainder star star zero star now if we want to separate the

35:54

stars from the zeros okay then this path now going from the top

36:00

uh left to the bottom right in red here okay that's separating the stars from

36:05

the zero over here there weren't any zeros so we could put that um that

36:10

corresponding path like this over here after we've removed these two columns we get star star 0 0 for the Matrix and

36:18

then the path that separates the zeros from the stars of this one over here we remove these two internal columns we get

36:24

zero star and zero star and here's the corresponding separating path for this one we remove these two columns

36:31

we have zero star zero zero they're separating path is this one and here uh

36:37

we remove these ones here we just get a Matrix of zeros and the the path is this one here

36:43

so in each case what we've done we've obtained a path in a two by two Matrix

36:51

that separates one side from the other end the path goes from the top left to the bottom

36:57

right and this way we associate paths to subspaces

37:04

with areas to the right corresponding to powers of Q so see the the power of q that um that we

37:13

were that we're getting is essentially counting uh the the area to one on one

37:18

side of of these paths okay so we could go from this picture to the other one if

37:24

we sort of rotated everything by a quarter turn okay if we rotated

37:29

everything by a quarter turn then we would end up with a path from the uh bottom left to the top right and then

37:37

these uh these star areas here would just be the um the stuff below or to the right of

37:43

the past that we've already been considering so the fact that um the the areas of those past or these

37:51

inversion numbers are exactly corresponding to in the sub base case the degrees of freedom of the

37:58

possibilities of our plugging in values of Q okay so uh

38:05

in sort of an exact correspondence why these two things are going to be the

38:11

same so uh that's a pretty interesting kind of

38:17

enlargement of the world from ordinary binomial coefficients to these Q binomial coefficients and and it's it's

38:24

quite a natural and pleasant kind of enlargement