Concrete Nonsense » Combinatorics http://concretenonsense.wordpress.com A group blog about mathematics Mon, 30 Nov 2009 21:48:46 +0000 http://wordpress.com/ en hourly 1 http://www.gravatar.com/blavatar/2b77f97368a020180b21c93e224b3ae1?s=96&d=http://s.wordpress.com/i/buttonw-com.png Concrete Nonsense » Combinatorics http://concretenonsense.wordpress.com Finite field counts and the Grothendieck ring of varieties http://concretenonsense.wordpress.com/2009/11/23/finite-field-counts-and-the-grothendieck-ring-of-varieties/ http://concretenonsense.wordpress.com/2009/11/23/finite-field-counts-and-the-grothendieck-ring-of-varieties/#comments Mon, 23 Nov 2009 15:59:40 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=830 ]]>

Lately some of us at MIT have been thinking about counting \mathbf{F}_q-rational points on some classes of varieties related to linear algebra that provide natural q-analogues for various classes of permutations. One thing we came across was some classes that have the same counts over every finite field. Yan wanted me to post about the following, so I’ll delay my post on the K-theory of the Grassmannian until next time.

We’ll consider varieties defined over a fixed field K. Form the free Abelian group on the isomorphism classes of such varieties. If Z is a closed subvariety of X, then we impose the relation

[X] = [Z] + [X \setminus Z].

We can put a product structure on this group via

[X] \cdot [Y] = [X \times Y]

though it will not be relevant for this post. Related to this product structure is a paper by Bjorn Poonen which shows that if the characteristic is 0, then this ring is not an integral domain. And presumably the result is true over positive characteristic also, but the paper uses the existence of resolution of singularities. This is the Grothendieck ring of varieties. This is at least one way to make sense of statements of the form: \mathbf{P}^2 = \mathbf{A}^2 + \mathbf{P}^1 = \mathbf{A}^2 + \mathbf{A}^1 + \mathbf{A}^0.

In particular this works over a finite field, and then the equations can be turned into equations of numbers by replacing [X] with \#X(\mathbf{F}_q).

Here’s one such example. Let G be a semisimple group defined over an algebraically closed field K of characteristic p > 0, let B be a Borel subgroup, and let X = G/B be its full flag variety. Then B acts on X by left multiplication, and the orbits are indexed by elements of the Weyl group W. In particular, each orbit is an affine space, and its dimension is the length of the corresponding element. So we get a decomposition

\displaystyle [X] = \sum_{w \in W} [\mathbf{A}^{\ell(w)}].

All of this stuff is defined over the integers, so we can actually work in a finite field. Then we get the equation

\displaystyle \#X(\mathbf{F}_q) = \sum_{w \in W} q^{\ell(w)}.

The odd spin group \mathbf{Spin}_{2n+1}(K) (the double cover of the special orthogonal group \mathbf{SO}_{2n+1}(K)) and the symplectic groups \mathbf{Sp}_{2n}(K) have the same Weyl groups, so in particular, their flag varieties are equal in the Grothendieck ring of varieties, even though the varieties themselves are not isomorphic for n \ge 3. One way of seeing this is via the Borel–Weil construction: the global sections of a line bundle on G/B is either 0 or an indecomposable module called a Weyl module. The dimension of the Weyl module (after we pick a way to index the line bundles) is independent of characteristic since it has a \mathbf{Z}-form which is a free Abelian group. In general the sections contain a unique irreducible representation as a submodule, and all irreducible representations arise in this way.

In characteristic 0, indecomposable is the same thing as irreducible, so we can calculate the dimensions using the Weyl character formula. Since these multisets of dimensions are different, we can’t have an isomorphism.

Does anyone know of a better reason why they are not isomorphic varieties?

-Steven

]]> http://concretenonsense.wordpress.com/2009/11/23/finite-field-counts-and-the-grothendieck-ring-of-varieties/feed/ 0 masnevets GLFq III: characteristic map http://concretenonsense.wordpress.com/2009/10/12/glfq-iii-characteristic-map/ http://concretenonsense.wordpress.com/2009/10/12/glfq-iii-characteristic-map/#comments Mon, 12 Oct 2009 15:47:33 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=710 ]]>

In the last post of this series, I gave some definitions and facts regarding the Hall–Littlewood functions. I also sketched the relationship between symmetric functions and representations of the symmetric group. Now we’ll see how this works for the finite general linear groups.

We want to imitate the Frobenius character that is used to relate the characters of the symmetric group to the ring of symmetric functions. But since the description of the conjugacy classes of the finite general linear group (and hence the parametrization of its irreducible characters) are more complicated than the description for the symmetric group, we’ll need a bigger ring to work with.

We continue the notation from the first post. For each irreducible polynomial f \in \Phi and each positive integer i>0, we have a variable X_{i,f}, whose degree we set to be deg(f). For any symmetric function u, we set u(X_f) = u(X_{1,f}, X_{2,f}, \dots). The graded ring we work in is B = {\bf C}[e_n(X_f) \mid n \ge 1,\ f \in \Phi], where e_n denotes the elementary symmetric function. In other words, elements of B are functions which are symmetric in each family of variables X_f.

Recall from last time that for a partition \lambda, P_\lambda(x;t) and Q_\lambda(x;t) are the Hall–Littlewood and augmented Hall–Littlewood functions. We use these to define elements in B:

\tilde{P}_\lambda(X_f) = q^{-\deg(f) n(\lambda)} P_\lambda(X_f; q^{-\deg(f)})

\tilde{Q}_\lambda(X_f) = q^{\deg(f)(|\lambda| + n(\lambda))} Q_\lambda(X_f; q^{-\deg(f)}),

where n(\lambda) = \sum_i (i-1)\lambda_i. For a partition-valued function \boldsymbol{\mu}, we set

\displaystyle \tilde{P}_{\boldsymbol{\mu}} = \prod_{f \in \Phi} \tilde{P}_{\boldsymbol{\mu}(f)}(X_f),

\displaystyle \tilde{Q}_{\boldsymbol{\mu}} = \prod_{f \in \Phi} \tilde{Q}_{\boldsymbol{\mu}(f)}(X_f).

We use these two bases to define a (complex) inner product on B:

\langle \tilde{P}_{\boldsymbol{\lambda}}, \tilde{Q}_{\boldsymbol{\mu}} \rangle = \delta_{\boldsymbol{\lambda}, \boldsymbol{\mu}}.

Now we need to construct the representation ring of the finite general linear groups G_n over {\bf F}_q, where now q will remain fixed. This will be very similar to what happens for the symmetric groups. Given characters u and v for G_n and G_m, respectively, let P be the parabolic subgroup of G_{n+m} consisting of matrices of the form

g(A,B,C) = \begin{bmatrix} A & B \\ 0 & C \end{bmatrix}

where A is an n \times n matrix, B is an n \times m matrix, and C is an m \times m matrix. We define a character w on P by setting

w(g(A,B,C)) = u(A) v(B).

Then the induction product u \circ v is defined as the induced character {\rm Ind}_P^{G_{n+m}}(w). (Recall that for symmetric groups, we define the induction product by inducing from parabolic subgroups \mathfrak{S}_n \times \mathfrak{S}_m \subset \mathfrak{S}_{n+m}.) If we let A_n denote the complex vector space of characters of G_n, then the induction product gives a graded ring structure on A = \bigoplus_{n \ge 0} A_n. We can also put a complex inner product on A by setting the different graded components to be mutually orthogonal and using the standard inner product for characters on each component, just as in the case of the symmetric group. Now comes the important part: let \pi_{\boldsymbol{\mu}} denote the function which is 1 on the conjugacy class corresponding to \boldsymbol{\mu}, and 0 elsewhere. Then we have a characteristic map

{\rm ch} \colon A \to B, \quad {\rm ch}(\pi_{\boldsymbol{\mu}}) = \tilde{P}_{\boldsymbol{\mu}}.

Theorem. The characteristic map ch is an isometric isomorphism of graded rings.

If we continue with the analogy of the relationship between the symmetric group and symmetric functions, then the characteristic of the irreducible characters of G_n should be some kind of “Schur functions.” Unfortunately their definition will require significantly more notation. So I’ll skip that and just say that we can define functions S_{\boldsymbol{\lambda}}. One catch, though, is that the indexing set we use for these Schur functions is not the same as the indexing set for conjugacy classes. The indices {\boldsymbol{\lambda}} can be thought of as partition-valued functions, but on a different domain. But this is not such a big deal.

Theorem. The S_{\boldsymbol{\lambda}} form an orthonormal basis for B. Furthermore, their inverses under the characteristic map are the irreducible characters \chi^{\boldsymbol{\lambda}} of the groups G_n. Consequently, the values of the characters are given by the change of bases S_{\boldsymbol{\lambda}} = \sum_{\boldsymbol{\mu}} \chi^{\boldsymbol{\lambda}}(\boldsymbol{\mu}) \tilde{P}_{\boldsymbol{\mu}}.

At any rate, I think it is nice that the same kind of setup works for the finite general linear groups as does for the symmetric group, which maybe further justifies the statement that the finite general linear groups are q-analogues of the symmetric groups.

But since these symmetric functions are so horribly complicated, one doesn’t expect to have a nice combinatorial rule for changing from the S basis to the \tilde{P} basis (such as the Murnaghan–Nakayama rule for writing the Schur polynomials in terms of power sum symmetric functions in the symmetric group case). There are some nice cases though. When the conjugacy class \boldsymbol{\mu} corresponds to a unipotent conjugacy class, we can evaluate induced characters from maximal tori T of G_n at \boldsymbol{\mu} to get Green polynomials (up to a sign). And sometimes these induced characters are irreducible (precisely when the stabilizer of the character in the Weyl group of T is trivial).

Green polynomials are more manageable to think about: they arise as the change of basis coefficients when writing power sum symmetric functions as Hall–Littlewood functions (now working just in the ring \Lambda[t] from last time).

That’s basically all I want to say about the connection between symmetric functions and finite general linear groups. There is a more powerful approach to characters of these groups using \ell-adic cohomology due to Deligne and Lusztig, and it works more generally for any finite group of Lie type. Using that approach, it can be shown, for example, that the characters are integer valued.

-Steven

]]> http://concretenonsense.wordpress.com/2009/10/12/glfq-iii-characteristic-map/feed/ 0 masnevets GLFq II: Hall–Littlewood functions http://concretenonsense.wordpress.com/2009/09/28/glfq-ii-hall-littlewood-functions/ http://concretenonsense.wordpress.com/2009/09/28/glfq-ii-hall-littlewood-functions/#comments Mon, 28 Sep 2009 14:21:27 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=684 ]]>

Last time, I discussed how to parametrize the conjugacy classes of the general linear group G over a finite field. In a sense, this group is a q-analogue of the symmetric group, so we might try to imitate the constructions for the symmetric group to get information about G. There’s a nice construction that Frobenius worked out which connects the characters of the symmetric group with the combinatorics of the Schur functions. I’ll briefly summarize the statement. The conjugacy classes of the symmetric group on n letters are parametrized by partitions of n. So we can also parametrize the irreducible characters by partitions as well (though it is not clear how to do this in a “canonical” way a priori). Ignoring the indexing issue (which can be dealt with) and letting \chi^\lambda(\mu) be the irreducible character indexed by \lambda evaluated at the conjugacy class consisting of permutations whose cycle lengths are given by the parts of \mu, then one has s_\lambda(x) = \sum_\mu z_\mu^{-1} \chi^\lambda(\mu) p_\mu(x) where s_\lambda(x) is a Schur function, p_\mu(x) is a power sum (Newton) symmetric function, and z_\mu = 1^{m_1} m_1! 2^{m_2} m_2! \cdots where m_i is the number of times that i appears as a part of \mu (the meaning of z_\mu is that n! z_\mu^{-1} is the size of the conjugacy class index by \mu.)

So the question to ask might be “can we find a similar interpretation for the characters of G?” The answer is yes, but becomes a bit more involved.

Instead of Schur functions, one needs to look at another class of symmetric functions called the Hall–Littlewood functions, but we’ll actually need a much larger ring than the ring of symmetric functions. While the Schur functions are symmetric functions in a set of variables x_1, x_2, \dots usually defined over rational coefficients, the Hall–Littlewood (HL) functions are symmetric functions defined over the ring {\bf Z}[t] (so t is an additional variable which does not affect the definition of “symmetric”.) Let \Lambda[t] denote the symmetric functions in variables x_1, x_2, \dots with coefficients in {\bf Z}[t]. Like the Schur functions, the HL functions are indexed by partitions. The definition of the Hall–Littlewood function indexed by \lambda in n variables x_1, \dots, x_n is

\displaystyle P_\lambda(x_1, \dots, x_n; t) = \sum_{w \in S_n / S_n^\lambda} w\left( x_1^{\lambda_1} \cdots x_n^{\lambda_n} \prod_{\lambda_i > \lambda_j} \frac{x_i - tx_j}{x_i - x_j} \right) .

Here S_n is the symmetric group on n letters, and S_n^\lambda \subseteq S_n is the subgroup of permutations w such that \lambda_{w(i)} = \lambda_i for all i. From this, one can see that P_\lambda(x_1, \dots, x_n; 1) = m_\lambda(x_1, \dots, x_n) where m_\lambda denotes the monomial symmetric function which is the sum of all of the distinct terms x_{w(1)}^{\lambda_1} \cdots x_{w(n)}^{\lambda_n} as w ranges over all permutations of n.

Set [\lambda]_t! = [\lambda_1]_t! \cdots [\lambda_n]_t!. An equivalent definition for P_\lambda is

\displaystyle P_\lambda(x_1, \dots, x_n; t) = [\lambda]_t!^{-1} \sum_{w \in S_n} w \left( x_1^{\lambda_1} \cdots x_n^{\lambda_n} \prod_{i < j} \frac{x_i - tx_j}{x_i - x_j} \right),

from which one can deduce that P_\lambda(x_1, \dots, x_n; 0) = s_\lambda(x_1, \dots, x_n) from the Weyl character formula.

I just want to state some of the properties that we will need later without giving too many details. For proofs, one can consult Macdonald's book Symmetric Functions and Hall Polynomials. From the first definition, one can deduce that these functions enjoy a stability property:

P_\lambda(x_1, \dots, x_n, 0; t) = P_\lambda(x_1, \dots, x_n; t),

and hence one can define P_\lambda(x;t) in infinitely many variables x_1, x_2, \dots by taking an inverse limit. Since the P_\lambda are symmetric functions, we can write

P_\lambda(x;t) = \sum_\mu w_{\lambda, \mu} (t) s_\mu(x)

for some polynomials w_{\lambda,\mu}(t). In fact, w_{\lambda, \lambda}(t) = 1 and w_{\lambda,\mu}(t) = 0 unless \lambda \ge \mu (dominance order), so the change of basis matrix from P_\lambda to s_\lambda is upper unitriangular, which implies that the P_\lambda(x;t) form a {\bf Z}[t]-basis of \Lambda[t].

The inverse of this change of basis is very interesting. In this case, write s_\lambda(x) = \sum_\mu K_{\lambda, \mu}(t) P_\mu(x;t). The K_{\lambda, \mu}(t) are the Kostka–Foulkes polynomials. Since P_\mu(x;1) = m_\mu(x), we see that K_{\lambda, \mu}(t) = K_{\lambda, \mu} are the Kostka numbers. It is a fact that the Kostka–Foulkes polynomials are in fact polynomials, and they have nonnegative integers. I hope to write a post about these at some point.

We will also need augmentations of these functions in the next post. First set

\displaystyle b_\lambda(t) = \prod_{i \ge 1} [(1-t)(1-t^2) \cdots (1-t^{m_i(\lambda)})]

where m_i(\lambda) is the multiplicity with which i appears in \lambda. Then define

Q_\lambda(x; t) = b_\lambda(t) P_\lambda(x; t).

Although we won’t use them, let me mention skew Hall–Littlewood functions. Define an inner product on \Lambda[t] by declaring that \langle P_\lambda(x;t), Q_\mu(x;t) \rangle = \delta_{\lambda, \mu}. Then we can define skew Hall–Littlewood functions for partitions \mu \subseteq \lambda via

\langle Q_{\lambda/\mu}(x;t), P_\nu(x;t) \rangle = \langle Q_\lambda(x;t), P_\mu(x;t) P_\nu(x;t) \rangle

\langle P_{\lambda/\mu}(x;t), Q_\nu(x;t) \rangle = \langle P_\lambda(x;t), Q_\mu(x;t) Q_\nu(x;t) \rangle.

From this definition, setting t=0 gives back the skew Schur functions s_{\lambda/\mu}(x) (since they are defined in a similar way). The weird thing, however, is that the skew Schur functions only depend on the shape \lambda/\mu, whereas the skew Hall–Littlewood functions remember both \lambda and \mu. One can write down a rather explicit formula for Q_{\lambda/\mu} in terms of semistandard tableaux which shows that the function depends on both \lambda and \mu (but this is only seen in the powers of t, and not the x_i), but I will omit this so that I can wrap this post up.

Let me just end with some other specializations of t that are important. When \lambda is a strict partition (i.e., the nonzero parts are distinct) then setting t=-1 gives P_\lambda(x;-1) = 2^{\ell(\lambda)} Q_\lambda where \ell(\lambda) is the number of parts, and Q_\lambda are the Schur Q-functions, which are important for the projective representation theory of the symmetric group (maybe a future topic). Also, specializations at t=q^{-1} for q a prime power are related to Hall algebras, which are used to keep track of extensions between finite Abelian groups.

In the next post, I’ll discuss the connection between characters of {\bf GL}_n({\bf F}_q) and symmetric functions.

-Steven

]]> http://concretenonsense.wordpress.com/2009/09/28/glfq-ii-hall-littlewood-functions/feed/ 0 masnevets GLFq I: Conjugacy classes of a finite general linear group http://concretenonsense.wordpress.com/2009/09/14/glfq-i-conjugacy-classes-of-a-finite-general-linear-group/ http://concretenonsense.wordpress.com/2009/09/14/glfq-i-conjugacy-classes-of-a-finite-general-linear-group/#comments Mon, 14 Sep 2009 15:26:32 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=665 ]]>

I’d like to start a series of posts, GLFq (general linear group over a finite field), in which I hope to explain the representation theory of G = {\bf GL}_n({\bf F}_q) (n and q will remain fixed). In this first post, I’ll explain how to write down the conjugacy classes of G. In later posts, I plan to introduce Hall–Littlewood polynomials and the characteristic map. I would like to also go into how to construct the actual representations, and discuss things related to Hall–Littlewood polynomials, like the q-Kostka polynomials and a lot of the interesting algebra/geometry behind them.

There are two pieces of data we would like to know. First, what is the size of G? Second, how do we parameterize the conjugacy classes? The first question is easy to answer since an invertible matrix is given by the data of n linearly independent vectors. The first one can be chosen to be any nonzero vector, so there are q^n - 1 of them. In general, the ith one can be chosen to be any vector not in the span of the last i-1 (so we are just avoiding some i-1 dimensional subspace, which has q^{i-1} elements), and hence there are q^n - q^{i-1} choices for such a vector. All together, the number of elements of G is \prod_{i=1}^n (q^n-q^{i-1}). We can rewrite this as q^{\binom{n}{2}} (q-1)^n [n]_q! to make it more analogous to the number of elements of the symmetric group.

The second question requires the rational canonical form. If we were dealing with an algebraically closed field, conjugacy classes would of course be parameterized by Jordan normal forms, so we need some kind of substitute for that. First, we need the structure theorem for finitely generated modules over a principal ideal domain R. This says that any such module is a direct sum of its torsion submodule and a free submodule. Furthermore, the torsion submodule is uniquely a direct sum of cyclic modules, which can be written in the form R/(f^m) for some irreducible element f and some positive integer m.

Given a matrix A, we’ll apply this to the case R = {\bf F}_q[t] and the module V = {\bf F}_q^n where the action of a polynomial p(t) on V is given by p(A). Irreducible elements of R are the same as irreducible polynomials, but we will never see the polynomial x show up if A is invertible, and we will only need to use the monic ones. Let \Phi be the set of all monic irreducible polynomials over {\bf F}_q which are different from the constant polynomial x. Hence, we see that the data of the decomposition is given by a partition valued function \boldsymbol{\mu} on \Phi. Explicitly, the function \boldsymbol{\mu} gives the module \bigoplus_{f \in \Phi} \bigoplus_{i \ge 0} R / (f^{\boldsymbol{\mu}(f)_i}).

Since the dimension of R / (f^m) is \deg(f) \cdot m, the conjugacy classes of G are given by those partition valued functions \boldsymbol{\mu} such that \| \boldsymbol{\mu} \| := \sum_{f \in \Phi} \sum_i \deg(f) \boldsymbol{\mu}(f)_i = n.

As an aside, if we look at the conjugacy classes of all matrices instead of invertible matrices, i.e., we allow the domain of the partition valued functions above to include the polynomial x, then the number of such classes is \sum_j p_j(n) q^j where p_j(n) is the number of partitions of n into j parts. I think it’s a really nice formula (though it takes some work to show). See Chapter 1, Section 10 of the second edition of Enumerative Combinatorics I for a derivation of this formula.

Let’s look at the case of n=2. The only valid partition valued functions can only have nonempty values on polynomials of degree at most 2. There are 3 types of functions:

  • There exists a single monic irreducible polynomial f of degree 1 such that \mu = \boldsymbol{\mu}(f) \ne \emptyset and we have \sum_i \mu_i = 2, so \mu \in \{(2), (1,1)\}. These correspond to matrices with a single eigenvalue, the partition (2) means that it’s a diagonal matrix, and the partition (1,1) means that it is conjugate to a size 2 Jordan block.
  • There exists two distinct monic irreducible polynomials f and g of degree 1 such that \boldsymbol{\mu} takes the value (1) on both and is empty on all other polynomials. These correspond to matrices with two distinct eigenvalues.
  • There exists a single monic irreducible polynomial f of degree 2 such that \boldsymbol{\mu}(f) = (1) and all other values are the empty partition. These are matrices without a Jordan normal form.

The only irreducible polynomials of degree 1 which are allowed are of the form x-a for a nonzero value of a. So there are 2(q-1) functions of the first kind and \binom{q-1}{2} = (q-1)(q-2)/2 functions of the second kind. For the third kind, we have q^2 monic polynomials of degree 2 over {\bf F}_q. There are q polynomials of the form (x-a)^2 and \binom{q}{2} = q(q-1)/2 of the form (x-a)(x-b) for a and b distinct, so we must have q^2 - q - q(q-1)/2 = q(q-1)/2 monic irreducible degree 2 polynomials over {\bf F}_q. Thus in total we have 2(q-1) + (q-1)(q-2)/2 + q(q-1)/2 = q^2 - 1 conjugacy classes.

Next time, I’ll say something about Hall–Littlewood polynomials. Click here for the next post.

-Steven

]]> http://concretenonsense.wordpress.com/2009/09/14/glfq-i-conjugacy-classes-of-a-finite-general-linear-group/feed/ 0 masnevets A Fock space representation http://concretenonsense.wordpress.com/2009/08/31/a-fock-space-representation/ http://concretenonsense.wordpress.com/2009/08/31/a-fock-space-representation/#comments Mon, 31 Aug 2009 13:39:23 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=623 ]]>

Lately I’ve been reading about quantum groups. I particularly enjoyed Bernard Leclerc’s paper Symmetric functions and the Fock space representation of U_q(\widehat{\mathfrak{sl}_n}), so I want to discuss a little bit of it. In particular, I want to give an example of the technique of deforming an object in order to see “hidden structure.” My notation will differ slightly from Leclerc’s since he uses the French notation for Young diagrams.

One thing that has always been really hard for me to wrap my head around is the really complicated presentations that affine Lie algebras have and how one is supposed to do anything with them. This post will be about the affine Lie algebra \widehat{\mathfrak{sl}_n}, which is defined in the paper as the Lie algebra with generators e_i, f_i, h_i (0 \le i \le n-1) and d, with 5 lines of relations. Let K be a field of characteristic 0. Another way around this is to first define the loop algebra L(\mathfrak{g}) of a simple Lie algebra \mathfrak{g} as: L(\mathfrak{g}) = \mathfrak{g} \otimes K[t,t^{-1}] with a Lie bracket given by [a \otimes t^n, b \otimes t^m] = [a,b]_{\mathfrak{g}} \otimes t^{n+m}, and then to say that \widehat{\mathfrak{g}} is its universal central extension. More precisely, we say add a central element c, and then extend the bracket above via [a \otimes t^n, b \otimes t^m] = [a,b]_{\mathfrak{g}} \otimes t^{n+m} + (a,b)_{\mathfrak{g}} n \delta_{n,-m} c where \delta is the Kronecker delta, and (,) is the Killing form of \mathfrak{g}.

In the case that \mathfrak{g} is \mathfrak{sl}_n, I want to discuss a more concrete (combinatorial) description. Just as \mathfrak{sl}_n can be thought as the traceless operators on an n-dimensional vector space (the standard representation), we can also find a standard representation for \widehat{\mathfrak{sl}_n} (the Fock space representation). For this, let {\cal F} denote the ring of symmetric functions over K in infinitely many variables. The Schur functions s_\lambda form a basis indexed by partitions, and this will be our representation. In order to describe the actions of \widehat{\mathfrak{sl}_n} on Sym, we’ll need some notation.

First, we represent partitions \lambda by their Young diagram (\lambda_i boxes drawn in the ith row, left justified). The content of a box (i,j) is the number i-j. We’ll consider contents modulo n, and say that a box is an i-node if it has content i modulo n. We define e_is_\lambda = \sum_\mu s_\mu (resp. f_is_\lambda = \sum_\mu s_\mu) where the sum is over all \mu obtained from \lambda by removing (resp. adding) an i-node, and define ds_\lambda = N_0(\lambda)s_\lambda where N_0(\lambda) is the number of 0-nodes of \lambda. Finally, set h_i = e_if_i - f_ie_i. Then \widehat{\mathfrak{sl}_n} is the Lie algebra spanned by these generators.

Unlike the \mathfrak{sl}_n case, the Fock space representation is not irreducible. Let p_\lambda = \prod_i p_{\lambda_i} be the power sum symmetric function where p_i = \sum_{j \ge 1} x_j^i. It turns out that the set \{p_{n\lambda}\} where n\lambda = (n\lambda_1, n\lambda_2, \dots) are the highest weight vectors of this representation (i.e., they are killed by the e_i, and are eigenvectors for d and the h_i). Furthermore, one has dp_{n\lambda} = |\lambda| p_{n\lambda}, so we have a natural notion of degree for our highest weight vectors. Within these graded subsets, the p_{n\lambda} give an obvious choice of basis, but there is no reason to favor them: for example, \{ p_{(2n)} + p_{(n,n)}, p_{(2n)} - p_{(n,n)} \} also forms a basis for the highest weight vectors of degree 2. The point is that the Schur functions give a “natural basis” for {\cal F} in the sense that we have defined our operators in this basis, and the basis of highest weight vectors should have “nice” properties with respect to this fixed basis, although it’s not clear what nice means right now.

The next part is something that I am learning to appreciate: since there is no way to figure out a “canonical basis” for the highest weight vectors, we should introduce a new parameter to make the structure of the Fock space representation more rigid. This new parameter is made precise by replacing \widehat{\mathfrak{sl}_n} by its q-analogue U_q(\widehat{\mathfrak{sl}_n}), and similarly for {\cal F}. More precisely, we can’t deform the Lie algebra, but we can deform its universal enveloping algebra. The quantum group U_q(\widehat{\mathfrak{sl}_n}) has generators E_i, F_i, K_i, K_i^{-1}, and D, D^{-1} and even more relations than \widehat{\mathfrak{sl}_n} has, so rather than give those, I just want to mention how to change the action on {\cal F}. Let q be a transcendental element over K, and let K(q) be the function field over K. We set {\cal F}_q = {\cal F} \otimes_K K(q), and to get the actions of the E_i and F_i on {\cal F}_q, we’ll use almost the same formulas as above, but we’ll need a bit more partition notation.

Given a partition, a box is removable (resp. indent) if it can be removed (resp. added) to obtain another Young diagram. Let \lambda and \mu be two partitions such that \lambda is obtained from \mu by adding an i-node \gamma to it. Let I^r_i(\lambda, \mu) (resp. R^r_i(\lambda, \mu)) be the number of indent i-nodes of \lambda (resp. number of removable i-nodes of \lambda) which are strictly to the right of \gamma. Also set N^r_i(\lambda, \mu) = I^r_i(\lambda, \mu) - R^r_i(\lambda, \mu). Define the same numbers with the superscript r replaced by l by replacing “right” with “left.” Then we set F_is_\lambda = \sum_\mu q^{N_i^r(\lambda, \mu)} s_\mu (resp. E_is_\nu = \sum_\mu q^{-N_i^l(\mu, \nu)} s_\mu) where the sum is over all \mu such that \mu / \lambda is an i-node (resp. \mu / \nu is an i-node). We also set D^{\pm}s_\lambda = q^{\pm N_0(\lambda)}s_\lambda and define K^{\pm}_is_\lambda = q^{\pm K(i, \lambda)}s_\lambda where K(i, \lambda) is the number of removable and indent i-nodes of \lambda. And we can take the quantum group U_q(\widehat{\mathfrak{sl}_n}) to be the K(q) algebra spanned by these generators.

We’ll use a family of operators V_k to find a nice basis. To define their actions, we need some definitions about ribbons. First, an m-ribbon is a connected skew Young diagram with m boxes which does not contain a 2 \times 2 square. The most northeast box of an m-ribbon is called its origin. Its spin is the number of rows it has minus 1. A connected union of m-ribbons is a horizontal m-ribbon strip if it is a skew Young diagram, and if the origin of each ribbon does not lie below another box in the same column. The weight of a horizontal m-ribbon strip is the number of m-ribbons used to build it. Any tiling of a horizontal m-ribbon strip subject to these constraints is unique, so we can define the spin of a horizontal m-ribbon strip to be the sum of the spins of its ribbons. We define V_k s_\lambda = \sum_\mu (-q)^{{\rm spin}(\mu/\lambda)} s_\mu where the sum is over all \mu such that \mu / \lambda is a horizontal n-ribbon strip of weight k. This sort of looks like the definitions one uses to define the Murnaghan–Nakayama rule for multiplying a Schur function and power sum symmetric function. In fact, in the classical limit “q=1″, V_k reduces to multiplication by the plethysm h_n \circ p_k = h_n(x_1^k, x_2^k, \dots ).

We introduce a K-linear bar involution on K(q) via q \mapsto q^{-1}, and then extend this to a compatible K-linear involution x \mapsto \overline{x} on {\cal F}_q by requiring that it commute with the actions of F_i and V_k on {\cal F}_q, and that it fixes the basis vector s_\emptyset. Let L (resp. L^-) be the free {\bf Z}[q]-submodule (resp. {\bf Z}[q^{-1}]-submodule) of {\cal F}_q spanned by the basis \{s_\lambda\}. Then we have the following theorem.

Theorem. There exist two unique bar-invariant bases B = \{G(\lambda)\} and B^- = \{G^-(\lambda)\} of {\cal F}_q such that G(\lambda) = s_\lambda \pmod {qL} and G^-(\lambda) = s_\lambda \pmod {q^{-1}L^-}.

The two bases are called the canonical basis and dual canonical basis of {\cal F}_q. They have a lot of nice properties. Going back to highest weight vectors, it turns out that G^-(n\lambda) is a highest weight vector for all \lambda. Furthermore, “setting q=1” this basis of highest weight vectors reduces to the plethysms s_\lambda \circ p_n = s_\lambda(x_1^n, x_2^n, \dots) (this is related to the classical limit of the operators V_k). Since it comes from a more “rigid” basis, we might be satisfied with this choice for a basis of highest weight vectors in {\cal F}. Another nice property which happens with canonical bases is a nonnegativity property: write G(\mu) = \sum_\lambda d_{\lambda, \mu}(q) s_\lambda and G^-(\lambda) = \sum_\mu e_{\lambda, \mu}(-q^{-1}) s_\mu where the d and e are polynomials.

Theorem.The polynomials d and e have nonnegative coefficients as polynomials in q. Furthermore, d_{\lambda, \mu}(q) is nonzero only if \lambda \le \mu and similarly, e_{\lambda, \mu}(q) is nonzero only if \mu \le \lambda.

Here we are using the dominance order on partitions: \lambda \le \mu if \lambda - \mu can be written as a nonnegative linear combination of vectors \varepsilon_i - \varepsilon_{i+1} where \varepsilon_i is the vector with a 1 in the ith coordinate and 0s in the other coordinates.

There is a bunch of other stuff which Leclerc discusses in the paper, like connections to Kazhdan-Lusztig polynomials and Macdonald polynomials, which illustrates why these canonical bases and their change of basis matrices are important, but I’ll stop here.

-Steven

]]> http://concretenonsense.wordpress.com/2009/08/31/a-fock-space-representation/feed/ 5 masnevets Trees, The BEST Theorem, and Alexander Polynomials http://concretenonsense.wordpress.com/2009/08/20/trees-the-best-theorem-and-alexander-polynomials/ http://concretenonsense.wordpress.com/2009/08/20/trees-the-best-theorem-and-alexander-polynomials/#comments Thu, 20 Aug 2009 03:39:20 +0000 yanzhang http://concretenonsense.wordpress.com/?p=619 ]]>

Most of my “free math time” has been used to study for quals, but today I’ve made myself post to stop Steven from taking over this blog.

One of my favorite elementary algebric combinatorial results is the Matrix Tree Theorem, which states:

In a nondirected graph with vertices labelled 1, 2, \ldots n, the number of spanning trees is equal to any principal minor of the Laplacian.

This cute result gets the number of trees on n vertices (n^{n-2}) fairly quickly with some matrix manipulation, which I will leave as an exercise to the reader. I know two proofs of this theorem: the first one involves using the Cauchy-Binet formula on the Laplacian L, after making the slick observation that L = MM^t, where M is the incidence matrix. Another quick solution can be obtained by invoking the lesser-known version of the Matrix Tree Theorem for directed graphs, which is actually a bit simpler to prove:

In a directed graph with vertices labelled 1, 2, \ldots n, the number of arborescences into vertex i (that is, trees rooted at i where all the edges point towards i) is equal to the i-th minor of the Laplacian.

But this is not all!

I actually recently learned of a third proof (involving Gessel-Viennot, of all things), but I will not mention it here (see the second reference of this post). The reason I mention the directed version and arborescences is to introduce a lesser-known but closely related result to the Matrix Tree Theorem, the forcibly-named BEST Theorem (‘B’ for de Bruijn, ‘S’ for Smith, ‘T’ for Tutte, and ‘E’ for… van Ardenne-Ehrenfest):

In a balanced directed graph (that is, for each vertex the out- and in-degrees equal) with vertices labelled 1, 2, \ldots n, the number of Eulerian circuits is equal to T \prod_v (d_v - 1)!, where d_v is the outdegree of vertex v and $T$ is the number of directed arborescences into any vertex.

This result is neat, for a couple of reasons. One, it shows immediately that the number of directed arborescences into any vertex in a balanced graph is equal, which is totally not obvious. Second, it implies a connection between Eulerian circuits and spanning trees, which is counterintuitive; in fact, when each vertex has in- and out- degree 2, which is a kind of graph we get quite often (especially when we consider nondirected graphs as directed graphs), we get that the number of Eulerian circuits is exactly the number of arborescences.

Sketch of Proof: pick any edge (i, j) to start with. Now, from any Eulerian circuit C, construct a directed graph T' as follows: for each vertex v (besides i), consider the last step in the circuit away from it towards a vertex j. Then add the directed edge (v, j) to T'. This is an arborescence into i. Note that besides the last exit from v (which must be to j), the other d_v - 1 exits can be done in any order. This creates a \prod_v (d_v - 1)! – fold bijection between arborescences into i and Eulerian circuits, which is exactly what we need.

Those familiar with knot theory may recall the Alexander polynomial \Delta_L(t) of a link L, which one obtains as a minorof a matrix M_D constructed from any diagram D of L (though it does not depend on the diagram). I’ll not review the construction since it is best done by example, it is a little tedious to type, and I’m too lazy to draw pictures (though there’s a functional Wikipedia page here). Now, it is well known that |\Delta_L(-1)| is well-defined (and called the determinant of L). However, it has a combinatorial meaning. Construct the following graph on the strands 1, 2, \ldots, n of L: each time the strand i is crossed above by j and comes out the other side as k, draw directed edges (i, j) and (i, k) (I think this even works if $i = j$, though I believe (I don’t know much knot theory and this is pure intuition, so correct me if I’m wrong!!!) you can always draw diagrams to avoid this). It takes a bit of thinking, but convince yourself that this creates a graph G with indegree and outdegree 2 at each vertex. Thus, by considering the construction of M_D, we have:

|\Delta_L(-1)| equals any minor of M_D(-1), which in turn equals both the number of arborescences into any vertex of G and the number of Eulerian circuits of G.

This is basically all I know about knot theory, so I’ll stop here.

References: Enumerative Combinatorics Vol. 2 (Stanley) for the Matrix Tree Theorem and the BEST Theorem, and A Course in Enumeration (Aigner) for the BEST Theorem and Alexander polynomials.

-Y

P.S. The latter reference deserves some mention, because it has a neat presentation. At the end of each chapter, Aigner gives a “highlight” section with a particularly pretty result (the BEST Theorem being one of them), which serves as fun enrichment material. I wish more math books did this (though this is not the first book I know which does this – Alon’s The Probabilistic Method also has similar sections, which were equally delightful). The exposition is quite good, coming from one of the writers of the beautiful Proofs from the BOOK.

]]> http://concretenonsense.wordpress.com/2009/08/20/trees-the-best-theorem-and-alexander-polynomials/feed/ 5 KR Knights on Tori http://concretenonsense.wordpress.com/2009/07/27/knights-on-torii/ http://concretenonsense.wordpress.com/2009/07/27/knights-on-torii/#comments Mon, 27 Jul 2009 22:15:48 +0000 yanzhang http://concretenonsense.wordpress.com/?p=550 ]]>

(guest post by Joel Lewis, another of my office-mates. -Y)

————-

I frequent the Art of Problem Solving forum (also knows as MathLinks where at some point I stumbled across the following problem (advertised (.pdf) as appearing on the 1996 Iranian Mathematical Olympiad, a claim I have no reason to doubt but no way to verify):

Consider a chessboard in which the opposite edges have been identified to yield a torus. What is the maximal number of knights that can be placed on this board so that no two attack each other?

There are several ways to attack this problem. If you’ve spent some time thinking about chess-related mathematics, you’re probably familiar with the existence of a knight’s tour on a (regular) chessboard: it’s possible to start with a knight on any square of the chessboard and make a sequence of 64 moves so that it visits every square exactly once before returning to its original position. It follows that we can place at most 32 knights on the board so that no two attack each other: no two squares visited consecutively in the knight’s tour may both be occupied, so at most half of the board may be covered. Moreover, it’s easy to find a set of 32 squares on which we can place the knights — for example, the 32 black squares do nicely, since a knight on a black square attacks only white squares. (In fact, one can extend this argument to show that this and the 32 white squares are the only sets of 32 squares on which we can place nonattacking knights.)

That was a regular chessboard — we still haven’t dealt with the torus. Note that when we identify edges, we can’t possibly increase the number of knights that fit on the board. Also, it is still true on the torus that knights on black squares attack only black squares. Thus 32 squares are still maximal, and we’ve answered our question.

To summarize what we’ve done using the terminology of graph theory, we’ve used the Hamiltonicity of the knight’s graph on an 8 \times 8 chessboard, the fact that the knight’s graph is bipartite, and the fact that adding extra edges to a graph can only decrease its independence number. This Hamiltonicity result is reasonably high-powered, though. Let’s try to avoid using it. One way to do this might be to start with a non-Hamiltonian board, for example, the 4 \times 4 board.

Consider a 4 \times 4 chessboard in which the opposite edges have been identified to yield a torus. What is the maximal number of knights that can be placed on this board so that no two attack each other?

4by4

(Showing that this board is non-Hamiltonian is perhaps nontrivial — one way to do this is by using the contrapositive of the parenthetical statement two paragraphs back.) What do we do now? One way to proceed is by a clever counting argument: on the 4 \times 4 toroidal knight’s graph, every vertex has degree four. Thus, each knight attacks four squares while each non-knight square is attacked by at most four knights. It follows that the number of knights can be no more than the number of non-knight squares and so eight squares is maximal. Once again, the toroidal board is bipartite and we can fill all the white squares with knights to achieve our bound.

(By the way, for those who are interested, the toroidal 4 \times 4 knight’s graph is isomorphic to the skeleton of the four-dimensional hypercube. I guess there just aren’t that many four-regular symmetric bipartite graphs on sixteen vertices.)

This latter argument is very elegant. It also has a lot of power: we can use it to show that we can always fit 2n^2 nonattacking knights on a toroidal 2n \times 2n board. Note, however, that if the side-length of the board is odd, bad things start to happen: the square in the upper-left corner attacks squares along the bottom and right edges of the board that are (under the usual chessboard coloring) the same color, so we can’t achieve the same maximum that we can with a regular board. In general, it seems difficult to nail down precisely what the largest number of knights is for such a board. However, some cases are still amenable to our methods. In particular, the following question allows a quite beautiful solution:

Consider a 5 \times 5 chessboard in which the opposite edges have been identified to yield a torus. What is the maximal number of knights that can be placed on this board so that no two attack each other?

Note that this graph is very much nonbipartite. For example, the vertices in positions (1, 2), (3, 3) and (5, 4) form a triangle. So, we have to lower our expectations for the size of the independent set we’re looking for — maybe we can only fit eight knights on the board, instead of thirteen. But a little experimenting shows that this is overly optimistic — while it’s easy to fit five knights on the board (any row or column will do), even putting a sixth down is quite difficult.

A little more experimentation yields the following remarkable fact: not only does our graph contain triangles, but it actually contains a large number of five-cliques! For example, if we place a knight at the center of the board and choose “every other” square from those it attacks (so, the squares (1, 4), (2, 1), (3, 3), (4, 5) and (5, 2)) we have five mutually-attacking squares. This leads immediately to two possible solution methods (one found by my student K. Cordwell, the other by this blog’s proprietor): we can either note that each knight attacks eight squares while each non-knight square may be attacked by at most two knights (since the eight squares potentially attacking the given square decompose into two copies of K_4, and we may have at most one knight on each copy) so at most one fifth of the board may be occupied by knights, or we may note that we can cover the board by five disjoint shifts of the K_5 mentioned above and so choose at most one knight from each copy.

5by5

For those who are interested in this sort of thing, Noam Elkies and Richard Stanley have a nice paper titled The Mathematical Knight (.pdf) that’s worth a read. Also, there’s John J. Watkins’ extremely readable Across the Board (the first chapter is available free there) which is, I guess, a textbook for a course on graph theory viewed completely through the lens of chess problems. Also, I gather that some people actually enjoying playing chess, rather than solving math problems motivated by it. This website proposes a starting position to play toroidal chess on a conventional 8 \times 8 board.

- Joel Lewis

]]> http://concretenonsense.wordpress.com/2009/07/27/knights-on-torii/feed/ 6 KR 4by4 5by5 Lambda-rings http://concretenonsense.wordpress.com/2009/07/23/lambda-rings/ http://concretenonsense.wordpress.com/2009/07/23/lambda-rings/#comments Thu, 23 Jul 2009 22:04:02 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=534 ]]>

In this post, I want to discuss Grothendieck’s \lambda-rings and how they provide an abstract setting for Riemann–Roch formalism. The references I’ll be using are

  • Donald Knutson, Lambda-Rings and the Representation Theory of the Symmetric Group
  • William Fulton and Serge Lang, Riemann–Roch Algebra

The definition of a \lambda-ring is a bit technical, but it starts with a commutative ring R together with operations \lambda^i \colon R \to R for all nonnegative integers i such that \lambda^0(r) = 1 and \lambda^1(r) = r for all r in R together with some axioms. In particular, we should say what these lambda operations do to sums and products, and we might also want to know what compositions of them look like. To motivate these axioms, we’ll look at K-theory (where it originates).

Let X be a topological space, and consider the set of all vector bundles over X (topological, smooth, holomorphic, whatever you want). We define lambda operations using exterior powers: \lambda^i(E) = \bigwedge^i E. Of course, the set of vector bundles on X isn’t a ring, but the free Abelian group of isomorphism classes of vector bundles is a ring if we use tensor product as the multiplication. But we have to define exterior products on “negatives” of isomorphism classes of vector bundles. For actual vector bundles, we have the identity

\bigwedge^n(E \oplus F) = \bigoplus_{i=0}^n \bigwedge^i E \otimes \bigwedge^{n-i} F,

so we can use this to extend to “negatives” and we make this an axiom for a general \lambda-ring:

(L1) \lambda^i(r+s) = \sum_{j=0}^i \lambda^j(r) \lambda^{i-j}(s) for all r and s in R.

In fact, the above identity holds if we pass to the Grothendieck group K_0(X) of vector bundles over X (add the relations E'+E'' = E whenever we have a short exact sequence of the form 0 \to E' \to E \to E'' \to 0) because in general, if E’ is an extension of E and F, then \bigwedge^n E' has a filtration whose associated graded is the direct sum on the RHS above.

What about products? i.e., what should \lambda(rs) be? The exterior power of a tensor product of two vector bundles has a rather complicated expression. Nonetheless, there exist integer valued polynomials P_n in 2n variables such that

\bigwedge^n(E \otimes F) = P_n(E, \bigwedge^2 E, \dots, \bigwedge^n E, F, \bigwedge^2 F, \dots, \bigwedge^n F).

How do we get these polynomials? Let X_1, \dots, X_i, \dots, Y_1 \dots, Y_i, \dots be algebraically independent variables, and let E_i, F_i denote the ith elementary symmetric function in the Xs and Ys, respectively. Then we define the polynomials P_n via the identity

\displaystyle \sum_{n \ge 0} P_n(E_1, \dots, E_n, F_1, \dots, F_n) T^n = \prod_{i,j \ge 1} (1+X_iY_jT).

So we have some complicated family of polynomials, and the axiom

(L2) \lambda^n(rs) = P_n(\lambda^1(r), \dots, \lambda^n(r), \lambda^1(s), \dots, \lambda^n(s)) for all r and s in R.

There are also some integer-valued polynomials P_{nm} of degree nm for expressing the compositions \bigwedge^n \bigwedge^m E. I won’t get into that, but this gives the third axiom

(L3) \lambda^n(\lambda^m(r)) = P_{nm}(\lambda^1(r), \dots, \lambda^{nm}(r)) for all r in R.

A morphism of \lambda-rings is a ring homomorphism which commutes with the lambda-operations.

For a simple combinatorial example, take X to be a point. In this case, vector bundles are just vector spaces, and K_0(X) = \mathbf{Z}. Identifying vector spaces with their dimension, the lambda operations become \lambda^k(n) = \frac{n(n-1)\cdots (n-k+1)}{k!}, which is just the binomial coefficient \binom{n}{k} when n is nonnegative. A morphism \varepsilon \colon R \to \mathbf{Z} is called an augmentation. For vector bundles, this map is given by sending a vector bundle to its rank and extending linearly to virtual vector bundles. We’ll assume our \lambda-rings are augmented.

We can place some further requirements and operations on our \lambda-rings. First, in K_0(X), we naturally have a notion of what it means to be “positive”: any class which represents an actual vector bundle. The set of positive elements has the property that it’s closed under addition and multiplication, and every element of K_0(X) can be expressed as a difference of two positive elements. Furthermore, whenever x is positive, \lambda^i(x) = 0 for sufficiently large i, and all positive elements of augmentation 1 (line bundles) have multiplicative inverses. We’ll take all of these features to be an axiom system for a “positive subset” of a \lambda-ring. Motivated by the K-theory, we’ll call positive elements of augmentation 1 line elements. We’ll assume that our \lambda-rings are equipped with a positive structure.

K-theory also has a nice involution: send a vector bundle to its dual bundle. In general, we’ll say that x \mapsto x^\vee is an involution of our \lambda-ring if it satisfies

  • (x^\vee)^\vee = x for all x,
  • \varepsilon(x^\vee) = \varepsilon(x) for all x,
  • u^\vee = u^{-1} for all line elements u.

We’ll further assume that our \lambda-rings are equipped with an involution.

Another example of \lambda-rings comes from representation rings. Given a group G and a representation V, we define \lambda^i(V) to be the representation \bigwedge^i V with the diagonal action of G. The augmentation here sends a representation to its dimension (over the ground field), the positive elements are the representations, and the involution sends a representation to its dual. One particular example is when G is the general linear group \mathbf{GL}_n and we consider only rational representations, so that the representation ring is the ring of symmetric functions in n variables (together with a multiplicative inverse for the product of the n variables). In this case, the lambda operations are plethysm: \lambda^n(p) = e_n \circ p where e_n is the nth elementary symmetric function, and positive means Schur positive. [I first saw \lambda-rings in context of symmetric functions, so the definitions seemed a bit mysterious to me.]

In the next post, I’ll discuss abstract Chern classes in the context of \lambda-rings and Riemann–Roch formalism, and say how this relates to Grothendieck–Riemann–Roch for proper maps between nonsingular varieties.

-Steven

]]> http://concretenonsense.wordpress.com/2009/07/23/lambda-rings/feed/ 8 masnevets P-partitions and Gorenstein algebras http://concretenonsense.wordpress.com/2009/06/01/p-partitions-and-gorenstein-algebras/ http://concretenonsense.wordpress.com/2009/06/01/p-partitions-and-gorenstein-algebras/#comments Mon, 01 Jun 2009 23:19:33 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=486 ]]>

In this post I’d like to say what a P-partition is, what a Gorenstein ring is, and plan to discuss a chain of topics which will lead from one to the other. Roughly the first half of this post can be found in Section 4.5 of Richard Stanley’s Enumerative Combinatorics, Vol 1.

First, let’s start with P-partitions. Here P is a finite poset with p elements. The standard definition of partition is of course a way of decomposing a positive number into a sum of positive numbers, i.e., 5 = 1 + 3 + 1. Since we don’t care about the order, we just list them in descending order, so the example becomes (3,1,1). Another way to interpret this is in terms of posets: let [m] be the usual ordering on the set {1, 2, …, m}. Then a partition of n using at most m parts is the same as an order-reversing map \sigma \colon \left[m\right] \to {\bf N}, where {\bf N} is the natural numbers (including 0) under the usual order, such that \sum_{i=1}^m \sigma(i) = n. Now replace [m] with an arbitrary partition P and we can talk about P-partitions. Also, we’ll say that a P-partition is strict if \sigma from before is strictly order-reversing.

Given a poset P, let a(n), resp. \overline{a}(n), be the number of P-partitions, resp. strict P-partitions of n, and define the generating functions G_P(x) = \sum_{n \ge 0} a(n) x^n and \overline{G}_P(x) = \sum_{n \ge 0} \overline{a}(n) x^n. I’ll note two facts: as rational functions, we have

\displaystyle G_P(x) = \frac{W_P(x)}{(1-x)(1-x^2) \cdots (1-x^p)}

where W_P(x) is a polynomial of degree strictly less than \binom{p}{2}, so that a(n) is a quasi-polynomial. Second, we have a reciprocity law:

x^p \overline{G}_P(x) = (-1)^p G_P(1/x),

and this implies that a(-n) = \overline{a}(n-p), where a(-n) makes sense since it is a quasi-polynomial. There is a refinement of this reciprocity which I won’t mention.

Here’s another way to think of P-partitions: think of them as points inside of {\bf Z}^P. But not just any points: the order-reversing requirement gives some linear inequalities, so they live in some rational cone. Then a(n) is the number of lattice points of this cone with the intersection of the hyperplane \sum x_i = n. These intersections will be polytopes, but in general not ones with integer vertices.

But there is a way to get integral polytopes. Instead of P-partitions of n, let \Omega(P,n) be the number of P-partitions \sigma such that \sigma(x) \le n for all x in P. This is the order polynomial of P (it’s not that bad to show that it is indeed a polynomial. Similarly, let \overline{\Omega}(P,n) be the number of strict P-partitions \sigma such that \sigma(x) \le n for all x in P. Then we can interpret \Omega(P,n) as the number of lattice points in {\bf Z}^P with inequalities 0 \le x_i \le n-1 and x_i \le x_j for all inequalities in P. This is a polytope O(P,n) (called the order polytope of P), and in fact, it will have integral vertices. We can even describe what they are: define a filter F of P to be subset such that if x \ge y and y is in F, then so is x. Then each vertex is $\sum_{x \in F} x$ where F is a filter. Then the integer points in O(P,n) are in bijection with order-preserving maps P \to \left[ n \right], which are in turn in bijection with order-reversing maps. In this case, the order polynomial is a special case of the Ehrhart polynomial of a polytope O(P) = O(P,2). Since \overline{\Omega}(P,n) counts the interior points of O(P,n+1), reciprocity for Ehrhart polynomials implies that \Omega(P,-n) = (-1)^p \overline{\Omega}(P, n).

Well now that we have an integral polytope, we also get a toric variety. This is the projective variety corresponding to the Ehrhart ring, which is defined as follows. Given an integral polytope Q \subset {\bf R}^n and a field K, let K[Q] be the K-vector space with basis elements x_1^{d_1} \cdots x_n^{d_n} z^d corresponding to integer points (d_1, \dots, d_n) = dQ, where dP is the dth dilation of Q. These are multiplied just like monomials are multiplied, and we grade K[Q] by the degree of z. Then by definition, the Hilbert function of K[Q] is the Ehrhart polynomial of Q.

Now let’s give some properties of the order polynomial of P which might tell us some information about the Ehrhart ring K[O(P)] (and its toric variety). First, let L be the length of the longest chain of P. Then \Omega(P, -i) = (-1)^p\overline{\Omega}(P, i) = 0 for 1 \le i \le L. We say that P is graded if for any given elements x and y, all maximal chains between x and y have the same length. Then P is graded if and only if \Omega(P, -L-m) = (-1)^p \Omega(P,m) for all m. We can see this as saying that the number of interior integer points of O(P, L+m) is the same as the number of integer points of O(P,m) for all m. Combined with the vanishing statements, this is also equivalent to saying that when we write the generating function \sum_{n \ge 0} \Omega(P, n+1) x^n as a rational function, then the numerator polynomial has palindromic coefficients. Since this is the Hilbert series of K[O(P,2)], we now mention

Theorem (Stanley). If R is a Cohen–Macaulay domain over a field K, then R is Gorenstein if and only if the numerator polynomial of its Hilbert series has palindromic coefficients.

A proof can be found in Stanley’s paper Hilbert functions of graded algebras, along with other interplay between numerical conditions of Hilbert series and properties of the ring.

So I’ve fulfilled my promise of connecting the two topics in the title, but I need to say what a Gorenstein ring is! The significance comes from Serre duality. We’ll give a geometric definition first. For a variety X of dimension n, X is Gorenstein if the canonical bundle \omega_X = \bigwedge^n \Omega_{X/k} is a line bundle, where \Omega_{X/k} is the cotangent bundle of X. Of course, X nonsingular implies that X is Gorenstein. For a finitely generated domain over K, we say it is Gorenstein if its corresponding affine variety is Gorenstein. Chasing through all of the above, we have conclude that if P is a graded poset, then the affine cone of the toric variety associated to its order polytope (this means the affine variety of K[O(P)] forgetting that its graded) is Gorenstein (what a mouthful!). In other words, the toric variety is arithmetically Gorenstein (this implies that it’s Gorenstein also).

Now an algebraic definition says that a local K-algebra R of dimension n is Gorenstein if and only if \mathrm{Ext}^i_R(K, R) is 0 for i \ne n, and is K for i=n. And then a ring is Gorenstein if all of its localizations are Gorenstein. There’s a lot more equivalent definitions, which you can find on the wiki article.

Finally, here’s something which connects to my previous posts on Boij–Söderberg theory. If K[x_1, \dots, x_n] / I is Gorenstein where I is a homogeneous ideal, then the graded Betti table exhibits symmetry (we may have to assume that I is prime, but I’m not sure). To be more precise, if we write the table with the convention that \beta_{i,j} is the rank of the ith syzygy module in degree -i-j, and the table has c+1 columns and r+1 rows, then \beta_{i,j} = \beta_{d-i,r-j} (assuming that the top left corner is entry (0,0)). In fact, this is another characterization.

And now we have a lot of examples of symmetric Betti tables!

-Steven

]]> http://concretenonsense.wordpress.com/2009/06/01/p-partitions-and-gorenstein-algebras/feed/ 0 masnevets q-analogues and homogeneous spaces http://concretenonsense.wordpress.com/2009/04/28/q-analogues-and-homogeneous-spaces/ http://concretenonsense.wordpress.com/2009/04/28/q-analogues-and-homogeneous-spaces/#comments Tue, 28 Apr 2009 22:13:10 +0000 Steven Sam http://concretenonsense.wordpress.com/?p=452 ]]>

While I was working on my final paper for the course on quivers that I’m taking this semester, I came across the following result (the notation \mathbf{F}_{p^r} means the finite field with p^r elements, and \mathbf{C} is the field of complex numbers):

Theorem. Let X be a variety defined over Z and assume for some prime p, and all r>0, that the function \#X({\bf F}_{p^r}) is obtained by plugging in p^r into some polynomial P(t). Then P(t) has integral coefficients, and P(1) is the Euler characteristic of X({\bf C}).

If you’re not comfortable with varieties, just think of the solution set of some collection of polynomials in multiple variables, and if you don’t know what the rest of the words mean, that won’t matter much for the stuff starting with the next paragraph. Here I’m using the notation X(F) to denote the solutions for X over F, and I’m thinking of X({\bf C}) as a complex analytic space. The proof uses \ell-adic cohomology (with compact support) and the Grothendieck–Lefschetz trace formula. I won’t get into that, but I’d like to use this as an excuse to talk about the q-analogues of the natural numbers.

Let’s fix a field with q elements, and work over this field. First let’s use the theorem on projective space \mathbf{P}^{n-1}, whose points are the one-dimensional subspaces (lines) in an n-dimensional vector space V. The number of lines of an n-dimensional vector space is (q^n-1)/(q-1) = q^{n-1} + q^{n-2} + \cdots + q + 1 because any nonzero vector defines a line, and each line is spanned by q-1 different vectors. Let’s denote this number [n] because substituting q=1 gives n, which is the Euler characteristic of n-1 dimensional complex projective space. Now let’s move onto complete flag varieties Flag(V) for V n-dimensional: the elements are just increasing chains V_1 \subset V_2 \subset \cdots \subset V_n where \dim V_i = i. The number of flags is [ n ] [ n - 1 ] \cdots [ 2 ] [1] because to define a flag, we first pick v_1 nonzero to get V_1 = \langle v_1 \rangle, and in general, after we pick v_1, \dots, v_{i-1}, we can pick any lift v_i of a nonzero vector \overline{v}_i \in V / \langle v_1, \dots, v_{i-1} \rangle, and the space V_i = \langle v_1, \dots, v_i \rangle is independent of the choice of lift. So let’s call this number [ n ] ! because specializing q=1 gives n!, which we now know is the Euler characteristic of the complex complete flag variety.

Let’s take a look at the Grassmannian Gr(k,V), whose points are the k-dimensional subspaces of an n-dimensional vector space V. To count its number of points, notice that we have a map Flag(n) to Gr(k,n) which just sends a flag V_1 \subset \cdots \subset V_n to the subspace V_k. This map is obviously surjective, but what are the fibers? It follows from the definitions that the preimage of a k-dimensional subspace W is Flag(W) x Flag(V/W), so the number of points is just \displaystyle \frac{[ n ] ! }{ [ k ] ! [ n - k ] !}. And we’ll call this number \left[ \begin{matrix} n \\ k \end{matrix} \right] because it specializes to \binom{n}{k} upon setting q=1, and as before, this is the Euler characteristic of the complex Grassmannian Gr(k,V) where now V is a complex n-dimensional vector space.

Now we can even do partial flag varieties for tuples (d_1, \dots, d_r), whose points consists of partial flags V_1 \subset \cdots \subset V_r where \dim V_i = d_i. The number of points is going to be \left[ \begin{matrix} n \\ d_1 \end{matrix} \right] \left[ \begin{matrix} n-d_1 \\ d_2 - d_1 \end{matrix} \right] \cdots \left[ \begin{matrix} n-d_1 - \cdots - d_{r-1} \\ d_r - d_{r-1} - \cdots - d_1 \end{matrix} \right], and setting q=1 gives the analogous product of binomial coefficients for the Euler characteristic.

In all of the cases, it turned out that each of the polynomials in q had positive coefficients. So we might ask what these numbers are counting, if anything. It turns out that setting q=t^2 gives the generating function \sum_{i \ge 0} \dim_{\mathbf{C}} \mathrm{H}^i(X; \mathbf{Z}) t^i where X is whichever variety we are talking about, and H refers to plain old singular cohomology. I can give a better answer though: in each of the cases of interest, there is an explicit cellular decomposition of variety in question, and the number of i-dimensional cells is exactly the coefficient of q^i in the corresponding polynomial. We could discuss this at length, but for now, I’ll just refer the interested reader to William Fulton’s book Young Tableaux.

And for those who don’t know what a homogeneous space is: a homogeneous space is a manifold (resp. algebraic variety) with a transitive smooth (resp. algebraic) action of a Lie (resp. algebraic) group. And one can verify that all of the examples I’ve discussed are homogeneous spaces for the group \mathbf{GL}(n) of n \times n invertible matrices. For fun, the reader can work out the counts for the symplectic and orthogonal groups and/or figure out what the right analogues of Grassmannians and flag varieties are.

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