Winter Map Inverses

(1)

http://dx.doi.org/10.4236/apm.2014.47040

Winter Map Inverses

Thomas B. Gregory

Department of Mathematics, The Ohio State University at Mansfield, Mansfield, USA Email: [email protected]

Received 24 April 2014; revised 20 May 2014; accepted 3 June 2014

This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

Abstract

We demonstrate the functional inverse of a Winter map, which is an analog of the exponential map, for Lie algebras over fields of prime characteristic.

Keywords

Prime-Characteristic Lie Algebras, Prime-Characteristic Lie Groups

“Historically,” note Strade and Farnsteiner in [1], “Lie algebras emerged from the study of Lie groups.” In Section 1.1 of [1], they give a simple example of the close connection between Lie algebras and Lie groups. In prime characteristic, David Winter [2] has defined maps which mimic the zero-characteristic exponential maps. See also Lemma 1.2 of [3]. In this paper, we focus on the following “Winter maps”: if x is an element of a characteristic-p Lie algebra L such that

(

ad 0,_Lx

)

p = we set

(

)

(

ad ad

) (

2

)

3

(

ad

₍

)

₎

1 ad ad

2! 3! 1 !

p

L L L

L L

x x x

x I x

p

ξ

−

= + + + + +

−



where I is the identity transformation of L. Such ad-nilpotent elements of degree less than p do exist in some graded Lie algebras, as can be seen from Lemma 2.3 and Proposition 2.7 of Chapter 4 of [1], as well as from Lemma 1 of [4]; of course, it is well known that non-zero-root vectors of simple classical-type Lie algebras are ad-nilpotent of degree less than or equal to four.

We will show here that for x∈L such that

(

ad 0,_Lx

)

p = the inverse of

ξ

(

ad Lx

)

as a linear trans-

formation of L is

ξ

(

ad _L

( )

−x

)

, so that such transformations generate a group G of linear transformations of L. We will also show that

λ ξ

(

ad _Lx

)

=ad ,_Lx where, for g a linear transformation of L, and I as above, we define

( ) (

) (

2

)

3

(

₍

)

₎

1

2 3 1 !

p

g I g I g I

g g I

p

λ

−

− − −

= − − + −

−

 (1)

(2)

Lemma 1 If x and c are elements of L such that

(

ad 0,_Lx

)

p = and

(

ad 0,_Lc

)

p = then

( ) ( )

(

₍

) (

₎

)

(

) (

)

(

)

( )

1

0 0

2 2 1

1

ad ad

! !

ad ad

! !

j i j p i

L i j

j i j p p

L i p j i p

x c

j i j

x c

j i j

ξ

− −

= =

−

− −

= = − −

=

−

+

−

∑∑

∑ ∑

Proof. We group terms with respect to total degree in ad Lx and ad Lc 

Lemma 2 Let a b, ∈F, and suppose that x is an element of L such that

(

ad 0._Lx

)

p = then

( )

(

a adLx

)

(

( )

b adLx

)

(

a b

)

adLx

)

.

ξ

=

ξ

+

Proof. We have by Lemma 1 that

ξ

(

( )

a ad ad _Lx

)

ξ

(

( )

b _Lx

)

equals

(

)

(

)

1

0 0

ad 0

! !

i j i j

p i

L i j

a b x

j i j

− −

= =

+ −

∑∑

which we can write in terms of binomial coefficients as

(

)

2

0 0

ad

!

i

p i

L j i j i j

x i

a b j i

−

= =

     

∑

By the Binomial Theorem, the above expression is equal to

(

_{) ( )}

1

0

ad

! i p

i L

i

x a b i

−

=

+

∑

which we can rewrite as

(

)

(

)

1

0

ad

!

i p

L i

a b x

i

−

=

+

∑

and recognize as

ξ

(

a b+

)

ad _Lx

)

. 

Lemma 3 For any integer n≧2 and any integer j, 0< <j n, we have

( )

0

1 0

n

k j

k

n k k =

 

− _{ } =

 

∑

Proof. We proceed by induction on n and j. When n=2, we must have j=1, and we have

0 2 2− + =0. For any n>2, when j=1, we have

( )

(

)

( )

(

)

(

)

(

)

( ) ( )

₍

(

₎

)

₎

( ) ( )

( ) (

)

1

1 1

0 0

1 1

0

1

0

1

0 1

1 1 1

1

!

1 1

1 ! 1 !

1 ! 1

! 1 !

1 1

1 1 0.

n n

k k

n k

k

n k

k

n k

k n

n n

k k

n

k

k n k

n n

k n k

n n

k

n

− ₊

= =

− ₊

=

−

=

−

= −

   

− _{ } = − _ _ +

+

   

= − +

+ − +

−

= − −

− − −

 

= − − _ _

 

= − ⋅ − =

∑

Now, for any n≧3 and any positive integer j less than n, suppose that

1( 1) 0

n k i

k

n k k =

 

− _{ } =

 

∑

for all

(3)

( )

_{( ) ( )}

( ) ( )

₍

(

_{) (}

)

₎

( ) ( )

₍

_{) (}

₍

(

_{) (}

)

₎

( ) ( )

1 1

0 1 1

1 1 1 1 1 1 1 0 1 ! !

1 1 1

! ! 1 ! !

1 ! 1

1 ! 1 1 !

1 1

1

1 1

n n n

k j k j k j

k k k

n k j k n k j k n k k n n n

k k n k

k k n k k n k

n

n k

k n k

n n k k n n k k − − = = = − − = − − = − = −   − _{ } = − = − − − − −   − = − − − − − − −   = − − _ _ −   −   = − − _ _  

∑

(

)

( )

1 1 1 1

1 1 0

1 1 1

1

1 1 1

1 1

1 1 1

1 0 0 0

j j

n n

k i k

j n j

k i

i k i

n j n

n k

k i k

j n j

n k n

i k i

− − − − = = = − − − = = = +   −   −   −    = − _ − _ _ _ _ + − _ __          − − −       = − _ _ − _ _ + = − _ _⋅ =      

∑

by induction, and the fact that

( )

(

)

0 1 1 1 0 0

k n n n k n k =   − _{ }= − = =  

∑

(the “ j=0 case”). 

Lemma 4 Let x be an element of L such that

(

ad = 0_Lx

)

p . Define

(

)

2

(

₍

)

₎

1

0 ad ad = 1 ! i p L L i x x i

δ

+ − = +

∑

(2)

Then for any positive integer n less than p,

(

)

(

)

1

(

₍

)

₎

1

( ) (

)

0 0

ad

ad = 1

!

t n p n n

n L j t n

L

t j

x n

x n j

j t n δ + − − − ₊ = =    ₋ ₋      + _ _{ } _

∑

(3)

Proof. We proceed by induction on n. Since when n=1, (3) is just (2), the initial step of the induction

proof is established. Suppose (3) is true for n=k≧1. Then

(

δ

(

ad Lx

)

k 1 +

equals

(

)

(

)

( ) (

)

(

)

1

1 1 2

0 0 0

ad ad 1

! 1 !

s k i

p k k p

j s k

L L

s j i

x k x

k j j

s k i

+ + − − − ₊ − = = =  _{ }    _{ } − −    + _{ }  +  

∑



∑



We group terms with respect to total degree (t+ +k 1, in this case) in ad Lx and get that

(

)

(

)

( 1 1)

(

₍

) (

_{) (}

)

₎

1 1

( ) (

)

1

0 0 0

ad ad

ad 1

! 1 !

r k t r

p k t k

k L L j r k

L

t r j

x x k

x k j

j

r k t r

δ + − + − + − − + + = = =     = ₊ _{− +}    − −     

∑ ∑

∑

.

Rewriting the above expression using another binomial coefficient, we get that

(

δ

(

ad Lx

)

k 1 + equals

(

)

(

)

( )

(

)

1

1 1 1

0 0 0

ad 1

1 .

1 !

t k p k t k

j r k L

t r j

x k t k

k j

j r k

t k + + − + − − ₊ = = = + +     − _{ } − _ _ + + + _{ } _ _

∑

∑∑

We change the order of summation to get

(

)

(

)

1 ( 1 1)

(

₍

)

₎

1 1

( )

(

)

0 0 0

ad 1

ad 1 .

1 !

t k

p k k t

k L j r k

L

t j r

x k t k

x k j

j r k

t k

δ

+ + − + − − + + = = = + +     = − _{  } _ − + + + _{  } _

∑

We replace the index of summation r by r−k to get

(

)

(

)

₁ ( 1 1)

(

₍

)

₎

1 1

( )

(

)

0 0

ad 1

ad 1 1 !

t k

p k k t k

k L j r

L

t j r k

x k t k

x k j

(4)

Adding and subtracting terms, we get

(

)

(

)

( )

(

₍

)

₎

( )

(

)

(

)

(

)

1

1 1 1 1

1

0 0 0

1 1

0 1

ad 1

ad 1 1 !

1 1

t k

p k k t k

k L j r

L

t j r

k t k

r r

r r t k

x k t k

x k j

j r

t k

t k t k

k j k j

r r δ + + − + − − + + + = = = − + + = = + +  + +     = − _{ }_ _ _ − + +     + + + +       −   − −   −       

∑

Setting q= −k j, we see, as in the proof of Lemma 3, that when r ≥ 1,

( )

(

) ( )

( )

1

0 1

1

1 1 1 0

1

k k

j r k q r

j q

k k

k j k q

j q − − = = −     − _{ } − = − − _ _ = −    

∑

by that same Lemma 3. Thus,

( )

(

)

( )

(

)

1 1

0 0

1 1 1

0 0 0

1 1

1 0 0,

k k

j r

k k k

j r

r j r

k t k

k j

j r

t k k t k

k j

r j r

− − = = − − − = = = + +     − _{  } _ −     + + + +       = _ _ − _{ } − = _ _⋅ =      

∑

so from the Binomial Theorem, we get that

(

δ

(

ad _Lx

)

k+1 equals

(

)

(

)

( )

{

(

)

(

)

}

1

1 1 1

1 1

0 0

ad

1 1 1

1 !

t k

p k k

j t k t k L

t j

x k

k j k j

j t k + + − + − − _{+ +} _{+ +} = =   − _{ } − + − − − + + _{ }

∑

.

We now distribute to get that

(

δ

(

ad _Lx

)

k+1 equals

(

)

(

)

( )

( ) (

)

( )

( ) (

)

1

1 1 1 1

1 1

0 0 0

ad

1 1 1 1 .

1 !

t k

p k k k

j t k k j t k L

t j j

x k k

k j k j

j j t k + + − + − − _{+ +} − _{+ +} = = =        ₋ _{− +} _{+ −} ₋ ₋ ₋        + +      

∑

We replace the latter index of summation j by j−1 to get that

(

δ

(

ad Lx

)

k 1 + equals

(

)

(

)

( )

( ) (

)

( ) (

)

( )

1

1 1 1

0 0 1

ad

1 1 1 1 1 .

1 1 !

t k

p k k k

j t k j t k k

L

t j j

x k k

k j k j

j j t k + + − + − − _{+ +} ₋ _{+ +} = = =        ₋ _{− +} ₋ ₋ _{+ −} _{+ −}      ₋   + +      

∑

We change the order of summation and factor to get that

(

δ

(

ad Lx

)

k 1 + equals

(

)

(

)

( )

(

)

( )

( ) (

)

( )

1

1 1 1

1 1 1 1

0 1

ad

1 1 1 1 1 1 .

1 1

1 !

t k

p k k

t k j j t k k k

L

t j

x k k k

k k j

j j k

t k + + − + − _{+ +} − _{+ +} ₋ ₋ = =           ₊ ₊  ₋ ₊ ₋  _{+ −} ₋ ₋ _{− −}      ₋    ₋   + +         

∑

By binomial arithmetic

(

δ

(

ad _Lx

)

k+1 equals

(

)

(

)

( )

(

)

( ) (

)

(

)( )

1

1 1 1

0 1

ad 1

1 1 1 1 1 .

1 !

t k

p k k

t k j t k k

L

t j

x k

k k j k

j t k + + − + − _{+ +} − _{+ +} ₋ = =   +    ₊ ₊ ₋ _{+ −} ₋ ₊ ₋      + +    

∑

The above displayed formula is just (3) for n= +k 1; i.e.,

(

δ

(

ad _Lx

)

k+1 equals

(

)

(

)

( )

(

)

1 1 1 1 0 0

ad 1

1 1

1 !

t k

p k k

j t k

L t j x k k j j t k + + − + − + + = =   +    ₋ _{+ −}      + + __ _ _ __

∑

.

Thus, the induction step is complete. 

Theorem The linear transformation

ξ

(

ad Lx

)

of L has

ξ

(

ad L

( )

−x

)

as its inverse, whereas the map ξ

(5)

(a).

ξ

(

ad ad _Lx

)

ξ

(

_L

( )

−x

)

=I, and (b).

λ ξ

(

ad ad _Lx

)

= _Lx.

Proof. (a) If, in Lemma 2, we let a=1 and b= −1, we see that (a) is true.

(b) Since

ξ

(

ad Lx

)

−I equals the

δ

(

ad Lx

)

of Lemma 4, we have that

λ ξ

(

ad Lx

)

equals

(

)

(

)

(

ad ad

)

2

(

)

3

(

ad

)

1 ad

2 3 1

p

L L L

L

x x x

x

p

δ δ δ

δ

−

− + − −

−



which, by Lemma 4 equals

( )

(

)

(

)

( ) (

)

1

1 1 1

1 0 0

1 ad

1 !

n t n

p p n n

j t n

L

n t j

x n

n j j

n t n

+ + − − − − ₊ = = =   −     ₋ ₋     ₊           

∑

We replace the index t by t−n to get that

(

)

(

)

1

( )

1 1

(

)

1

( ) (

)

1 0

1 ad

ad 1 !

n t

p p n

j t

L L

n t n j

x n

x n j

j n t λ ξ + − − − = = =   −      =     − −         

∑

We change the order of summation to get that

(

)

(

)

1

(

) ( )

1 1

( ) (

)

1 1 0

ad 1

ad 1 !

t n

p t n

j t L

L

t n j

x n

x n j

j t n λ ξ + − − = = = −     = _ _{ } − − _      

∑

We replace the index j by n− j to get that

(

)

(

)

1

(

) ( )

1

( )

1 1 1

ad 1

ad 1 !

t n

p t n

n j

L t

L

t n j

x n x j j t n λ ξ + − − = = = −     =    −       

∑

We cancel an n and a j and combine the −1 factors to get that

(

)

(

)

1

(

)

( )

1 1

1 1 1

ad 1

ad 1 1 !

t p t n

j

L t

L

t n j

x n

x j

j t

λ ξ

− − −

= = =   −     = _ _ _ − _ −      

∑

∑ ∑

We replace the index n by n+1 and we replace the index j by j+1, and we get that

(

)

(

)

1

(

)

1

( ) (

)

1

1 0 0

ad

ad 1 1

!

t p t n

j t L

L

t n j

x n

x j

j t

λ ξ − − −

= = =       =    − +       

∑

∑ ∑

We change the order of summation to get that

(

)

(

)

1

(

) ( ) ( )

1 1

1

1 0

ad

ad 1 1

!

t

p t t

j t L

L

t j n j

x n

x j

j t

λ ξ

− − − −

= = =

 

= − + _{ }

 

∑

We now appeal to a little more binomial arithmetic to observe that since 1 1 j j j j +     =    ₊ 

    and

1

1 1

t t t

j j j

+

     

+ =

   ₊   ₊ 

     , it follows by induction that

1 1 t n j n t j j − =     =    ₊     

∑

from which we obtain that

(

)

(

)

1

(

) ( ) ( )

1

1 0

ad

ad 1 1

1 ! t p t j t L L t j x t x j j t

λ ξ

− − −

(6)

We replace the index j by j−1 to get that

(

)

(

)

1

(

) ( )

1 1

ad

ad 1 !

t

p t

j

L t

L

t j

x t

x j

j t

λ ξ

− − −

= =

 

= − _{ }

 

∑

Finally, we use Lemma 3 to see that we are left with ad Lx 

References

[1] Strade, H. and Farnsteiner, R. (1988) Modular Lie Algebras and Their Representations. Pure and Applied Mathematics, 116, Dekker, New York.

[2] Winter, D.J. (1969) On the Toral Structure of Lie p-Algebras. Acta Mathematica, 123, 70-81.

[3] Weisfeiler, B.J. and Kac, V.G. (1971) Exponentials in Lie Algebras of Characteristic p. Mathematics of the USSR-Izvestiya, 5, 777-803.

[4] Gregory, T.B. (1990) A Characterization of the General Lie Algebras of Cartan Type. In: Benkart and Osborne, Eds.,

(7)

currently publishing more than 200 open access, online, peer-reviewed journals covering a wide range of academic disciplines. SCIRP serves the worldwide academic communities and contributes to the progress and application of science with its publication.

Other selected journals from SCIRP are listed as below. Submit your manuscript to us via either