Last update: 3 July 2013
Let be a vector space of dimension over a field and let be a basis of fixed once and for all. A flag in is a sequence of subspaces of such that
with strict inclusions at each stage, so that for each In particular, if is the subspace of spanned by then is a flag in called the standard flag.
The set of flags in is called the flag manifold of
Let be the group of all automorphisims of the vector space Since we have fixed a basis of we may identify with the general linear group if and
then is identified with the matrix
The group acts on if and then is the flag Let be the subgroup of that fixes the standard flag Then if and only if is a linear combination of for that is to say if and only if whenever so that is the group of upper triangular matrices in
A basis of a flag is a sequence in such that or equivalently such that is a basis of for each Given such a basis of there is a unique such that for each and we have Hence acts transitively on the flag manifold and the mapping is a bijection of onto the coset space
For a flag let
for Then is a 'flag of sets', i.e. we have
(A.1) |
|
Proof. | |
(i) Fix and let Since it follows that or Since and there are therefore jumps in the sequence which proves (i). (ii) Suppose that so that Intersecting with we see that Hence |
From (A.1) it follows that that each determines a permutation as follows: is the unique element of for Let denote the mapping so defined.
The symmetric group acts on by permuting the basis elements
for and Hence we may regard as a subgroup of
(A.2) Let Then if and only if for some
Proof. | |
Suppose Then for we have and By virtue of (1) we can choose of the form where by 'lower terms' is meant a linear combination of and by virtue of (2). By rewriting (3) in the form we see that there exists such that for all or equivalently Hence as required. For the converse it is enough to show that (i) and (ii) for all and As to (i), is spanned by the basis vectors such that and and therefore if and only if for some Thus the set consists of which establishes (i). Finally as to (ii), we have if so that and hence as required. |
From (A2) we have immediately
(A3) (Bruhat decomposition) is the disjoint union of the double cosets
For each let
The subsets are the Schubert cells in the flag manifold By (A.3), is the disjoint union of the
Let Then if and only if has a basis such that for each We may normalize the by taking
We can then subtract from suitable multiples of the for which and so as to make the coefficient of in zero for each such Then is replaced by a vector of the form
where the sum is over such that for any i.e., such that and or equivalently the diagram of
(A.4) Let Then if and only if has a basis of the form
where the sum is over all in the row of the diagram of and the coefficients are arbitrary elements of the field Moreover, the are uniquely determined by the flag and the mapping so defined is a bijection.
Proof. | |
Clearly each "matrix" of shape determines a basis of as above, and hence a flag If determines and the same flag then each must be expressible as and from the form of and the it follows that for each and hence |
Since it follows from (A.4) that the Schubert cell is isomorphic to affine space of dimension
Let and let be any basis of Since is a basis for for each the flag determines each of the exterior products up to a nonzero scalar multiple, and hence determines the vector
up to a nonzero scalar multiple, where If denotes the projective space of (i.e. the space whose points are the lines in we have an injective mapping
(the Plücker embedding) for which is the line in generated by the vector (1).
Assume from now on that the field is the field of complex numbers. Then the embedding realizes the flag manifold as a complex projective algebraic variety, which is smooth because has a transitive group of automorphisms (namely Each Schubert cell is a locally closed subvariety of isomorphic to affine space of dimension
For each let
be the closure of in The are the Schubert varieties in and a flag lies in if and only if has a basis such that for each Each is in fact a union of Schubert cells if is a reduced word for then if and only if is of the form where is a subsequence of that is to say if and only if in the Bruhat order. In particular, is the single point At the other extreme, if is the longest element of then is the whole of and the dimension of is
Let be the cohomology ring (with integral coefficients) of the projective variety Each closed subvariety of determines an element and cup-product in corresponds, roughly speaking, to intersection of subvarieties. In particular, for each we have a cohomology class and it is a consequence of the cell decomposition (A.3) of that the form a of In particular, is the identity element.
The connection between the classes and the Schubert polynomials is given by
(A.5) There is a surjective ring homomorphism
such that
for each
Proof. | |
Let us temporarily write for Monk [Mon1959] proved that for all and where the sum on the right hand side is over all transpositions such that and as in (4.15''), Define by From (1) we deduce the counterpart of (4.16): if is the last descent of (so that then we have where are as in (4.16). Now iteration of (4.16) will ultimately express as a sum of monomiais, i.e. as a polynomial in and iteration of (2) will express as the same polynomial in Hence if we define by we have for all and the proof of (A.5) is complete. |
In fact the kernel of the homomorphism is generated by the elementary symmetric functions of the
We shall draw one consequence of (A.5) that we have not succeeded in deriving directly from the definition (4.1) of the Schubert polynomials. Since the form a of any product is uniquely a linear combination of the and it follows from intersection theory on that the coefficient of in is a non-negative integer. From this we deduce
(A.6) Let be permutations, and write as an integral linear combination of the say
Then the coefficients are non-negative.
We have only to choose sufficiently large so that and all the permutations such that lie in and then apply the homomorphism of (A,5).
Remark. The coefficients in (A.6) are zero unless
(a) | |
(b) | and |
For is homogeneous of degree which gives condition (a). Also we have
by (2.17), and the only possible nonzero term in this sum is that corresponding to Hence if we must have and by symmetry also
This is a typed excerpt of the book Notes on Schubert Polynomials by I. G. Macdonald.