Singular components of Springer fibers

# Singular components of Springer fibers in the two-column case

Lucas Fresse Department of Mathematics, the Weizmann Institute of Science, Rehovot 76100, Israel
###### Abstract.

We consider the Springer fiber corresponding to a nilpotent endomorphism of nilpotent order . As a first result, we give a description of the elements of a given component of which are fixed by the action of the standard torus relative to some Jordan basis of . By using this result, we establish a necessary and sufficient condition of singularity for the components of .

###### Key words and phrases:
flag varieties, Springer fibers, singularity criteria, Young tableaux
###### 1991 Mathematics Subject Classification:
14M15, 14B05

Let be a -vector space of dimension and let be a nilpotent endomorphism. We denote by the set of -stable complete flags, i.e. flags such that for any . The set is a projective subvariety of the variety of complete flags. The variety is called Springer fiber since it can be seen as the fiber over of the Springer resolution of singularities of the cone of nilpotent endomorphisms of (see for example [5]). The Springer fiber is not irreducible in general and the geometry of its irreducible components has been an important topic of study for more than thirty years. Many problems remain unsolved, among them the problem to determine the singular components of .

The geometry of depends on the Jordan form of , which can be represented by a Young diagram: let be the sizes of the Jordan blocks of , and denote by the Young diagram of rows of lengths . The problem to determine singular components has a complete answer in two cases: when the diagram is of hook type or has two rows, every component of is nonsingular (see [1] and [6]). When has two columns, singular components can arise (see [4] or [6]). In this article we give a necessary and sufficient condition of singularity for the components of in the two-column case.

In section 1, the diagram is general. Following [4], we recall the classical parameterization of the components of by the standard tableaux of shape . Let be the component corresponding to the standard tableau . We fix a Jordan basis of and we denote by the torus of diagonal automorphisms in this basis. We show that the elements of which are fixed by are parameterized by the so-called row-standard tableaux, i.e. row-increasing numberings of by . Set to be the flag corresponding to the row-standard tableau .

From section 2, we suppose that the diagram has two columns of lengths . Let be the tableau numbered from top to bottom by in the first column and by in the second column. We show that the flag belongs to every component of , and that a component is nonsingular if and only if is a nonsingular point of . In section 2.2, we give some relations satisfied by the elements of . For and standard and row-standard tableaux of shape , in section 2.3, we establish a necessary and sufficient condition for to be in . This question can be related to the description of orbital varieties in term of -orbits given by A. Melnikov (see [2]), and the result that we prove here can be connected to [3, Theorem 3.15].

Let be the set of row-standard tableaux which are obtained from by switching two entries , with . Let denote the number of elements in a set . Our main result, proved in section 3, is the following

###### Theorem.

Suppose that has two columns. Let be the length of the first column of . The component is singular if and only if

 #{T′∈X(Y):FT′∈KT}>r(r−1)/2.
###### Notation.

Fix some conventional notation. Let denote the field of complex numbers. In fact all our constructions and results hold for any algebraically closed field. Let be the number of elements in the set . In what follows flags will be denoted by or or , Young diagrams will be denoted by standard tableaux by and row-standard tableaux by Other pieces of notation will be introduced in what follows.

## 1. The geometry of Bu and the combinatorics of Young

### 1.1. Young diagram Y(u)

Recall that a Young diagram is a collection of boxes displayed along left-adjusted rows of decreasing lengths. For example

is a Young diagram with boxes. Let be the sizes of the Jordan blocks of the endomorphism , and let be the Young diagram of rows of lengths . This diagram has boxes. The dimension of the variety depends on the diagram as shown by the following theorem (see [4, §II.5.5]).

###### Theorem.

The variety is equidimensional. Moreover, denoting by the lengths of the columns of , we have

 dimBu=s∑q=1nq(nq−1)2.

Two-column case. If , then the diagram has (at most) two columns. When has two columns of lengths and , we get

 dimBu=r(r−1)2+s(s−1)2.

### 1.2. Components of Bu parameterized by standard tableaux

From now on, set for simplicity .

A standard tableau of shape is a numbering of the boxes of from to such that numbers increase in each row from left to right and in each column from top to bottom. We denote by the set of standard tableaux of shape .
For example

For , let for be the subtableau obtained by deleting boxes with numbers . The shape of the subtableau is a subdiagram with boxes. The standard tableau can be regarded as the maximal chain of subdiagrams .

Let be a -stable flag. For the restriction map is a nilpotent endomorphism. Let be the Young diagram representing the Jordan form of in the sense of section 1.1. The diagrams form an increasing sequence of subdiagrams of . We set

 BTu={F∈Bu:Yi(F)=YTi ∀i}.

The subsets , for running over , form a partition of . Setting , we have (see [4, §II.5.4–5]):

###### Theorem.

Let be standard. The subset is an irreducible component of . Every component of is obtained in that way.

### 1.3. Elements FT′∈Bu parameterized by row-standard tableaux

In this subsection, we fix a Jordan basis of . Since the lengths of the rows of coincide with the sizes of the Jordan blocks of , it is possible to index the basis on the boxes of so that the following is true. Writing the vector of the basis associated to the box , we have if is in the first column of , and where is the box just on the left of otherwise.

We call row-standard tableau of shape a numbering of the boxes of from to such that numbers increase in each row from left to right. We denote by the set of row-standard tableaux of shape .
For example

For , the boxes of the subtableau form a subset . Let be the flag defined by for every . It is easy to see that belongs to .

Moreover, let be standard and suppose that each belongs to the same column of and . Then we can see that the flag belongs to the set . Hence belongs to the component . In particular, considering as a row-standard tableau, we get , thus .

We prove the following

###### Lemma 1.1.

A component is nonsingular if and only if every flag of the form with which belongs to is nonsingular.

###### Proof..

The implication is immediate. To prove the other implication, suppose there is a singular . Let be the subgroup of diagonal automorphisms with respect to the basis. Observe that the flags for running over are the elements of fixed by for its natural action on flags. However, this action does not leave invariant. Let us construct a subtorus with the same fixed points and leaving invariant. To this end, we choose pairwise distinct numbers associated to the boxes , so that if is the box on the left of . For , let be defined by for any . Then is a subtorus of . Since we have for any , the natural action of on flags leaves invariant the Springer fiber and all its components. As the ’s are pairwise distinct, the flags for row-standard are the -fixed points of .
The curve admits a limit at the infinity and this limit is necessarily a fixed point of , hence it is some with . Since is singular, the point is singular for any , and finally is also a singular point of the component. ∎

## 2. Fixed points of the components in the 2-column case

From now on, we suppose that the diagram has two columns of lengths and , with . As in section 1.3 we fix a Jordan basis indexed on the boxes of . For convenience we write the vectors of the basis associated to the boxes of the first column of from top to bottom, and the vectors associated to the boxes of the second column. Thus we have for and for .

### 2.1. Property of the flag F¯¯¯T associated to the tableau ¯¯¯¯T

Let be the tableau of shape numbered from top to bottom by in the first column and by in the second column:

Thus for every , the -th subspace of the flag is generated by the vectors of the basis. The flag satisfies the following properties.

###### Proposition 2.1.

(a) The flag belongs to every irreducible component of .
(b) A component is nonsingular if and only if is nonsingular in .

###### Proof..

Let be the centralizer of . The natural action of on flags leaves invariant. Moreover, the group is connected, since it is an open subset of the vector subspace of formed by the endomorphisms which commute with . Hence leaves each component of invariant. Applying Lemma 1.1, and using the observation made just before Lemma 1.1 that a component always contains some , it is sufficient to prove that the flag belongs to the closure of the -orbit of for any . We reason by induction on the first entry which has not the same place in and . If , then there is nothing to prove. Suppose , and take minimal which has not the same place in and . Write . We have thus and for some . For let be the automorphism such that and for with , and
-  if , -  if and , -  if .
We have . Moreover, the curve admits a limit at the infinity which is the flag associated to some tableau such that the entries have the same place in and . By induction, we have . Since , we get . The proof is complete. ∎

### 2.2. Some relations satisfied by F∈KT

Let . Let . The subspaces and are both invariant by , hence the quotient map can be considered, and it is still a nilpotent map. First, we show the following formula.

We have: .

###### Proof..

By applying the rank formula for the map , we get:

 ranku|Vj/Vi=j−dimu−1(Vi)∩Vj.

The map is surjective and its kernel is , hence

 dimu−1(Vi)∩Vj=dimVj∩keru+dimVi∩u(Vj).

On one hand, we have . On the other hand, the rank formula gives . The lemma follows. ∎

Let be standard. We associate to a row-standard tableau . Let (resp. ) be the entries of the first (resp. second) column of . We renumerate the entries of the first column from to :
- Set .
- If have been constructed for , then let be the maximal element among the such that .
- For , set .
Then let be the tableau numbered from top to bottom by in the first column, and in the second column.
For example

As observed in section 1.3, since the content of the columns of and coincides, the flag associated to belongs to the component .
Let . Set

 sTj/i=#{p:1≤p≤s and i

We prove the following

###### Lemma 2.2.

Let . The set

 {F=(V0,...,Vn)∈KT:ranku|Vj/Vi=sTj/i}

is a nonempty open subset of .

###### Proof..

Recall that denotes the subgroup of elements which commute with . Observe that for any in the -orbit of the flag . Thus, to prove the lemma, it is sufficient to prove that the -orbit of the flag is an open subset of . Let be the subgroup of elements which fix the flag . It is sufficient to prove that .
Recall we have fixed a Jordan basis with for . Any is thus determined by the images , which lie in , and by . Then we see that . By Theorem 1.1 we have . Thus, we have to prove

 dimZ(T∗)=r(r+1)2+s(s+1)2.

We reason by induction on , with immediate initialization for . Observe that has the same dimension as the vector space of endomorphisms of commuting with and leaving stable each subspace in the flag . Then, using the definition of the flag associated to a tableau , it is straightforward to establish the formula

 dimZ(T∗) = #{(p,q):1≤p,q≤r, s

We distinguish two cases.
A) Suppose that for some . Let be the tableau obtained from by removing the row containing , this tableau is row-standard up to moving if and if . By the above formula, we get . By induction hypothesis, . Thus follows.
B) Suppose that for . Let be the tableau obtained from by removing , this tableau is row-standard up to moving , . Likewise, the above formula implies . By induction hypothesis, we have . Again follows. ∎

We get now the following

Let . If , then
(a) for any .
(b) for any .

###### Proof..

By Lemma 2.1, (a) and (b) are equivalent. By Lemma 2.1 and Lemma 2.2, the set is a nonempty open subset of . Claim (b) follows from the lower semicontinuity of the map . ∎

### 2.3. A necessary and sufficient condition for FT′∈KT

In this section, we characterize the row-standard tableaux such that the flag belongs to the component .

Let be row-standard. Let (resp. ) be the entries from top to bottom in its first (resp. second) column. For , set

 sj/i(T′)=#{p:1≤p≤s and i

Then, writing , we have . By Proposition 2.2, we have

 FT′∈KT ⇒ sj/i(T′)≤sTj/i for any 0≤i

The following theorem shows that this implication is in fact an equivalence.

###### Theorem 2.1.

Let be row-standard. The following conditions are equivalent.
(a) .
(b) for any .

###### Proof of Theorem 2.1..

It remains to prove that claim (b) implies (a). First, we introduce some notation and convention. For in the first column of , we denote by its right neighbor entry, and put if has no entry on its right in . We set for any integer , so that the set is totally ordered. Recall that denotes the subgroup of elements which commute with . The natural action of on flags leaves and its components invariant.

We know that the flag belongs to the component when the columns of and have the same entries. Now, let be the minimal entry which lies in different columns in and , and reason by induction on . Since , the number belongs to the first column of and to the second column of . Thus contains strictly more entries in its second column than . Since , there is in the second column of whose entry on its left is some . Take minimal.
We distinguish two cases:

A) Suppose that there is in the first column of such that . Take minimal, and let denote the tableau obtained by switching and in . This tableau is row-standard: indeed, by minimality of we always have , hence .

B) Suppose that for any in the first column of . Then let denote the tableau obtained by switching and in . Let us prove that this tableau is row-standard. Assume the contrary: . Set . By the hypothesis every in the first column of has a right neighbor entry among . By definition of , every in the second column of has a left neighbor entry among . Thus is even and . On the other hand, since is in the second column of , we have . We get , which contradicts the hypothesis. Hence , and the tableau is row-standard.

In both cases, we have defined row-standard.

Claim. The flag lies to the closure of the -orbit of the flag .

###### Proof of the claim..

Recall that we have fixed a Jordan basis of . We renumber the vectors of the basis from to so that generate the -th subspace of . For we define an automorphism . As above we distinguish two cases: Suppose we are in case A above. For different from and , we set . Next, set . If , then set in addition . Suppose we are in case B above. For different from , set . In addition set . In both cases , and the flag is the limit at the infinity of the curve . The claim follows. ∎

By the claim, it is sufficient to prove that the flag belongs to , and by induction hypothesis, it is sufficient to prove that we have for any . As above, we distinguish two cases.

A) Suppose we are in the case A above. Then is the tableau obtained by switching and in the tableau . By the minimality of , we have , hence . We have to show that . If , then it is immediate. We have unless we are in the following subcase (A.1).

(A.1) We suppose that .
Then , hence it is sufficient to prove . As and as is in the second column of , we have . Moreover we have by hypothesis. As and by minimality of , the entry on the left of any in the second column of also belongs to . As , the entry on the right of any in the first column of also belongs to . Thus , so that . We have , since is the number of rows of the subtableau of of entries . The desired inequality follows.

B) Suppose we are in the case B above. Then is the tableau obtained by switching and in the tableau . Let denote the entry on the left of in . As already observed, we have . Moreover . We have to show that . If , then it is immediate. We have unless we are in one of the following subcases.

(B.1) We suppose that .
The proof is exactly the same as in the subcase (A.1) with instead of .

(B.2) We suppose that .
Then , hence it is sufficient to prove . By minimality of , for any in the first column of , we have . It follows . Let be the entry on the left of in the tableau . If , then we have and the desired inequality ensues. Suppose now that . By definition of , for any in the first column of , we have . Hence is equal to the number of elements in the first column of . The entry belongs to and to the first column of . As , the number is strictly lower that the number of elements in the first column of . Moreover, by minimality of , every is in the same column in and . Thus . Therefore we get . This completes the proof. ∎

## 3. Characterization of singular components

### 3.1. Statement of the result

The diagram is always supposed to have two columns of length . Let be the tableau introduced in section 2.1. We define as the set of row-standard tableaux which are obtained from by switching two entries such that and .  For example, if and , then the elements of are the tableaux:

Observe that is exactly the set of tableaux which are obtained from by switching two entries such that and .

Our main result is the following

###### Theorem 3.1.

Suppose has two columns. Let be the length of the first column of