Lecture 9  -  October 7,  2004

Prof .  Victor Kač
Scribe :  Yaim Cooper

Definition .   A Polynomial map f : (𝔽^m) ->  (𝔽^(n )) is a map of the following form, f(({{(Subscript[x, 1])}, {.}, {.}, {.}, {(Subscript[x, m])}})) -> (({{(Subscript[P, 1]) ((Subscript[x, 1.  .. ]) (Subscript[x, m]))}, {.}, {.}, {.}, {(Subscript[P, n]) ((Subscript[x, 1.  .. ]) (Subscript[x, m]))}}))  where (Subscript[P, i ])are polynomials.

Exercise 9.1  Let A be a nilpotent operator in a finite dimensional vector space V over a field 𝔽 of charachteristic 0.  Let Exp(A)=(∑_(j = 1)^∞) ((A^j)/(j!)) 
   Show that Exp (A) : V -> 
        V is an invertible linear map with inverse Exp (-A).

Solution .   A is nilpotent so (A^m) = 0 for some m ∈ ℕ . 
Let n be the smallest natural number for which (A^n) = 0.  Then 
Exp (A) =(∑_(j = 1)^(n - 1))((A^j)/(j!)).

(1)  Exp (A) is linear:

(Exp (A)) ((Subscript[c, 1]) (Subscript[v, 1]) + (Subscript[c, 2]) (Subscript[v, 2]) )=((∑_(j = 1)^(n - 1))((A^j)/(j!)))((Subscript[c, 1]) (Subscript[v, 1]) + (Subscript[c, 2]) (Subscript[v, 2]) )
=(∑_(j = 1)^(n - 1))(((A^j) ((Subscript[c, 1]) (Subscript[v, 1]) + (Subscript[c, 2]) (Subscript[v, 2])))/(j!)) = (∑_(j = 1)^(n - 1))((((Subscript[c, 1]) (A^j) (Subscript[v, 1]) + (Subscript[c, 2]) (A^j) (Subscript[v, 2])))/(j!))
=(Subscript[c, 1]) (∑_(j = 1)^(n - 1))(((A^j) (Subscript[v, 1]))/(j!)) + (Subscript[c, 2]) (∑_(j = 1)^(n - 1))(((A^j) (Subscript[v, 2]))/(j!)) = (Subscript[c, 1]) Exp (A) (Subscript[v, 1]) + (Subscript[c, 2]) 
                Exp (A) (Subscript[v, 2]).

(2)  Exp (-A) 
        is the inverse of Exp (A) .   Note that if we show Exp (-A) Exp (A) = 
      I,  
it follows that Exp (A) 
        Exp (-A) = I  since  - (-A) = A

Consider Exp (xA) Exp (cA) and Exp ((x + c) A),  
    where x and c are real
numbers .    Since differentiation with respect to x gives the same result on both sides,  
    and the expressions are equal when x  =  0,  
    these two expressions are equal  
Using x  = 1 and a = -1  gives the desired result.

Exercise 9.2:  If in addition, V is an algebra and A is a nilpotent derivation of V,
then Exp (A) is an automorphism of the algebra.

Solution .   We' ve shown in 9.1  
      that Exp (A) is an invertivble linear map so we only have 
left to show that Exp (A) preserves multiplication in V.

First we show by induction that (A^j) ((Subscript[v, 1]) 
            (Subscript[v, 2]))=(∑_(k = 0)^j)((j/k))(A^k)(Subscript[v, 1])(A^(j - 
                  k))(Subscript[v, 2]) .   
Base case :  A ((Subscript[v, 1]) (Subscript[v, 2]))=((Subscript[Av, 1])) 
            (Subscript[v, 2]) + (Subscript[v, 1]) ((Subscript[Av, 2])) by the definition of a derivation
Inductive step :   (A^j) ((Subscript[v, 1]) (Subscript[v, 2]))=A((∑_(k = 0)^(j - 1))(((j - 1)/k))(A^k)(Subscript[v, 1])(A^(j - k - 1))(Subscript[v, 2]))
=(∑_(k = 0)^(j - 1))(((j - 1)/k))((A^(k + 1)) (Subscript[v, 1]) (A^(j - k - 1)) 
                  (Subscript[v, 2]) + (A^k) (Subscript[v, 1]) (A^(j - k)) (Subscript[v, 2]))=(∑_(k = 0)^j)((j/k))(A^k)(Subscript[v, 1])(A^(j - 
                  k))(Subscript[v, 2])

So Exp (A) ((Subscript[v, 1]) (Subscript[v, 2]))=((∑_(j = 1)^(n - 1))((A^j)/(j!))) ((Subscript[v, 1]) 
              (Subscript[v, 2]))=(∑_(j = 1)^(n - 1))(((∑_(k = 0)^j)((j/k))(A^k)(Subscript[v, 1])(A^(j - 
                      k))(Subscript[v, 2]))/(j!))
=(∑_(j = 1)^(n - 1))((1/(j!)))(∑_(i = 0)^(n - 
                  1))((1/(i!))) (A^j) (Subscript[v, 1]) (A^i) 
                (Subscript[v, 2])=((∑_(j = 1)^(n - 1))((1/(j!))) (A^j) 
                (Subscript[v, 1]))((∑_(i = 0)^(n - 
                  1))((1/(i!))) (A^i) (Subscript[v, 2]))
=(Exp (A) 
            (Subscript[v, 1])) (Exp (A) (Subscript[v, 2])).

Thus Exp (A) is an automorphism of V.

\(Lemma\ 3. \ \ If\ \(f : 𝔽\^\(\(n\)\(\ \)\)\) -> \ \(\(\(\(𝔽\^\(\(n\)\(\ \)\)\) 
              is\ a\ polynomial\ map\ such\ that\ df\)\( | \_\(\(a\)\(\ \)\)\)\) : \ 𝔽\^n -> \ 𝔽\^\(\(n\)\(\ \)\)\ is\ a\)\)
nonsingular linear operator for some a ∈ (𝔽^n) 
        then f ((𝔽^n)) contains a nonempty Zariski
 open subset containing f (a) . 
 (Note :  this is like the implicit function theorem with Zariski open sets replacing open sets)

Lemma 4.   Take 𝔥 to be a Cartan subalgebra of 𝔤 with a ∈ 𝔥 a regular element .   Then 𝔥 ∈ ((Subscript[𝔤, 0])^a).

\(Proof . \ \ 𝔥\ is\ a\ nilpotent\ subalgebra\ so\ \ ad \((a)\)\( | \_𝔥\)\ 
      is\ nilpotent\ \ so\ 𝔥\  ⊂ \ 𝔤\_0\^a\ \ But\ 𝔤\_0\^a\ is\ a\)
 nilpotent Lie Algebra and 𝔥 is a maximal nilpotent subalgebra .   Thus 𝔥  = ((Subscript[𝔤, 0])^a).

Proof of Theorem .   Take 𝔥 a Cartan subalgebra of 𝔤 .   Then the root space
\(decomposition\ is\ given\ by\ 𝔤 = \(⊕\_\(α ∈ \(𝔥\^*\)\)\ \ 𝔤\_α\), 
where\ 𝔤\_α\  = \ {a ∈ 𝔤 | \(\((ad \((h)\) - α \((h)\))\)\^n\) a = 
          0\ for\ n\ sufficiently\ large}\)
Moreover,  [(Subscript[𝔤, α]),  (Subscript[𝔤, β])] ⊂ (Subscript[𝔤, α + β])  and (Subscript[𝔤, 0]) = 𝔥.

Note that if a ∈ (Subscript[𝔤, α]) and α ≠ 0,  
    then ad (a) 
      is a nilpotent operator
(since (((ad (a)))^N) (Subscript[𝔤, β]) ⊂ (Subscript[𝔤, β + Nα]) and (Subscript[𝔤, α]) ≠ 
          0  for a finite number of 
distinct α' 
            s since the dimension of 𝔤 is finite)
Let (Subscript[b, 1]) ...  
        (Subscript[b, n]) be the union of  bases of (Subscript[𝔤, α]), (Subscript[𝔤, α]) ≠ 0.   Define the polynomial map 
f : 𝔤 -> 𝔤 by
f ((∑_(i = 1)^m) (Subscript[x, i]) (Subscript[b, i]) + h) = (Exp ((Subscript[x, 1]) (ad ((Subscript[b, 1]))) ... Exp((Subscript[x, m]) (ad ((Subscript[b, m]))))(h)
where (Subscript[x, i])(Subscript[b, i])∈\(\(⊕\_\(α ∈ \(𝔥\^*\)\)\ \ 𝔤\_α\ for\ α\) ≠ \ 
                0\ \ and\ h\) ∈𝔥  Note that
 (Exp ((Subscript[x, 1]) (ad ((Subscript[b, 1]))) ... Exp((Subscript[x, m]) (ad ((Subscript[b, m]))))is an automorphism of 𝔤, by Lemma 2.   Moreover, f is a polynomial map in the entries h and (Subscript[x, i]).

Now,  
    apply Lemma 3.   Take a  ∈  𝔥 such that α (a) ≠ 
        0 for all nonzero α for which (Subscript[𝔤, α]) is 
nonzero.

\(Compute\ df\( | \_a\)\((b + h)\) = 
      d\/dt\( | \_\(t = 
              0\)\)\((f \((t \((∑\+\(i = 1\)\%m\( x\_i\) b\_i + h)\) + 
                    a)\))\) . \ \ Taylor\ expanding\ we\ get\)

=((I + (Subscript[tx, 1]) (Subscript[adb, 1]) + 
                    o ((t^2))) ...  (I  + (Subscript[tx, m]) (Subscript[adb, m])) + 
            o ((t^2))) (a + th)

\(\(\(=\)\(d\/dt\( | \_\(t = 0\)\)\ \ \ t[b, a] + 
          Ith\)\)\)
\(\(\(=\)\(d\/dt\( | \_\(t = 0\)\)\ \ \ t \((\([b, a]\) + 
              h)\)\)\)\)
=[b, a] + h

\(So\ df\( | \_a\)\((b + h)\) = \([b, a]\) + 
        h\ which\ is\ nonsingular\ since\ it\ is\ the\ identity\ on\ \(𝔤\_0\) and\ on\ 𝔤\_α\)
for all nonzero α it is  - ad a which is invertible because ad a has the form 
α (a)*
        Identity  +  nilpotent part  and α (a) is nonzero