Def: If L is a Lie Algebra, L is said to be nilpotent if L^k = 0 for some k.  L is said to be solvable if L^(k) = 0 for some k.

Def: An endomorphism x in End(V) is called nilpotent if x^k = 0 for some k (iff all its eigenvalues are zero).

Def: The normalizer NL(K) is defined to be {y in L : [xy] is in K for all x in K}.

Note: If K is an ideal, NL(K) is L.  If K is a subalgebra then NL(K) contains K.

Def: The natural map L to gl(L) is the map from x to ad x (Where ad x (y) = [xy]).

Lemma: This natural map is a representation.

Proof: We must show that ad [xy] (z) = (ad x ad y - ad y ad x) (z).  This is true iff [[xy]z]=[x[yz]]-[y[xz]] which is true by the Jacobi identity. n

Lemma: Suppose x in End(V) and x is nilpotent, then ad x in gl(End(V)) is nilpotent.

Proof: Inside gl(V), ad x (y) = [xy] = xy-yx = L(x)y - R(x)y where L(x) is left multiplication by x and L(x) is right multiplication by x.  If x^n = 0 then L(x)^n y = x^n y = 0 and similarly R(x)^n y = y x^n = 0.  Notice L(x) and R(x) commute.  Now ad x^2n = (L(x)-R(x))^2n = 0 by the binomial theorem (Either R(x) or L(x) will be raised to a power greater or equal to n in each term). n

Lemma: Suppose we have a representation f of L on V and W is an L invariant subspace of V (Thus w in W, x in L implies xw in W).  Then the map g from L to V/W defined by g(x)=f(x)+W is a representation of L on V/W.

Proof: g([xy]) = f([xy])+W = f(x)f(y)-f(y)f(x)+W = (f(x)f(y)+f(x)W+Wf(y)+W) - (f(y)f(x)+Wf(x)+f(y)W+W) = (f(x)+W)(f(y)+W)-(f(y)+W)(f(x)+W)=g(x)g(y) - g(y)g(x).  Note we need W to be L invariant so that Wf(k) = f(k)W = 0.

Lemma: Suppose K is a subalgebra of L, then the map from K to the vector space of L/K given by f(k) = ad(k)+K is a representation.

Proof: We know ad is a representation of L on the vector space L so ad is a representation of K on the vector space L.  If K is a K invariant vector subspace of L under ad, then we are done by the previous lemma.  Let k be any element of K and k' an element of the vector space K.  Then ad(k)k'=[kk'] which is in K as K is a subalgebra. n

Lemma: If K is an ideal of L (a Lie algebra over F)  and y is in L-K then Fy+K is a subalgebra of L.

Proof: [ay+k, by+j] =[ay, by] + [ay, j] + [k, by] + [k, j] = 0 + something in K, as K is an ideal. n

Theorem: Suppose L is a Lie subalgebra of gl(V) consisting entirely of nilpotent endomorphisms, then there is some v in V so that xv=0 for all x in L.

Proof:  We proceed by induction on dim L.  The statement is obviously true for dim L in {0,1}.  Assume dim L > 1 and assume the statement is true in all cases of lesser dimension.   Let K be a maximal proper Lie subalgebra of L.  Via ad we get a representation of K on L.  Since K is a subalgebra K is ad(K) invariant, thus we have a representation of K on L/K.
             Pick x in K and remember x is nilpotent.  Thus for any x, ad_L(x) is a nilpotent endomorphism of L by lemma, and ad_L/K(x) is a nilpotent endomorphism of L/K.  Then we can apply the inductive hypothesis to ad_L/K(K) acting on L/K to get a nonzero y+K in L/K so that ad_L/K(x)y = 0 for all x in K.  If ad_L/K(x)y = 0 for all x in K then [xy] is in K for all x in K.  As y+K is nonzero in L/K, y is not in K.  Thus we have a y not in K so that [xy] is in K for all x in K.  Thus y is in NL(K) and NL(K) is strictly bigger than K.
             As NL(K) is an ideal of L and K is a maximal subalgebra, either NL(K)=K which cannot be as NL(K) is strictly bigger than K, or NL(K)=L.  Thus NL(K)=L and so K must be an ideal of L.
            Next we must show K has codimension one. By lemma, we know that as K is an ideal Fy+K is a subalgebra of L.  It clearly has dimension greater than K so it can only be equal to L.  Thus Fy+K=L and K has codimension one.
            Now let W={v in V : xv=0 for all x in K}.  As dim L > dim K we know this space is non-trivial by the inductive hypothesis.  Since K is an ideal we will show y maps W to W.  Take any w in W.  Then yw is in W as x(yw) = y(xw) + [xy]w (Since L is a subalgebra of gl(V), [xy]=xy-yx) = 0 + 0.  So yw is in W for w in W.
            As y maps W to W and y is a nilpotent endomorphism of V, y is a nilpotent endomorphism of W.  Thus there exists non-zero v in W so yv=0.  Then Lv = (Fy+K)v = (Fyv+Kv) = 0 and we are done. n

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Corollary: (Engel's Theorem) Suppose L is a Lie algebra and ad x is nilpotent for all x in L.  Then L is nilpotent.

Lemma: If L/Z(L) is nilpotent then L is nilpotent.

Proof: As (L/Z(L))^k = 0 for some k, L^k must be inside Z(L).  Then L^k+1 is in [L, Z(L)] = 0.

Proof: (of Engel's Theorem) Again proceed by induction on the dimension of L.  Consider ad(L) as a subalgebra of gl(L).  By the previous result there is some z in L such that ad(x)z = 0 for all x in L.  Thus [xz]=0 for all x in L and Z(L), the center of L, is nontrivial.  The elements of L/Z(L) are ad nilpotent as well.  As L/Z(L) has dimension less than L, we can say L/Z(L) is nilpotent by induction.  Then by lemma, L is nilpotent. n

Note: The converse of Engel's Theorem is trivial.  If [x1[x2[...[xny]]...] = 0 for any x1,...xn in L then [x[x[..[xy]]..] = 0 for any one particular x in L.  Thus L is nilpotent iff L is ad-nilpotent.

Def: Suppose V is a vector space.  A flag in V is a sequence of subspaces V0<V1<...<Vn=V with dim Vi = i.

Corollary: Suppose L is a nilpotent subalgebra of gl(V).  Then L strictly stabilizes some flag V0<V1<...<Vn=V (Thus xVi<Vi+1).  Hence if we choose a basis v1,v2,..,vn with vi in Vi then the elements of L are all strictly triangular matrices.

Proof: Let L be a nilpotent Lie algebra.  Let v1 be a common nullvector for x in L.  Then x(cv1) = 0 for all x in L, c in F so if we let V1={cv1 : c in F} then LV1<V1.  Thus V1 is L invariant and we get a representation of L on V/V1.  The image of L over the projection map p from L to V/V1 is again nilpotent.  Thus we can find v2 in V so x(v2 + V1) = 0 + V1 so xv2 is in V1.  Let V2 equal {span of v1,v2} and continue by letting Vi={span of v1,v2,...,vi} and vi+1 be a null vector for the image of L of V/Vi until we are done. n

Theorem: Suppose L is a solvable subalgebra of gl(V), then there exists a nonzero v in V which is a common eigenvector.

Note: We know commuting matrices have a common eigenvector.  Commuting matrices correspond to an abelian Lie algebra.  We must prove this for solvable Lie algebras, which will imply the other case.

Proof: We proceed by induction on dim L.  The statement is obvious if dim L = 1.  Then all endomorphisms in L are multiples of one endomorphism, thus they all have the same eigenvectors.  If L is solvable [L L] is not equal to L.  We also know L/[LL] is abelian and thus any subspace of L/[LL] is an ideal.  Let K be any ideal of codimension one containing [LL].  Then if x is in L, y in K, [xy] is in [LL] a subset of K.  Thus K is an ideal.
            By our inductive hypothesis and the fact that any ideal of a solvable algebra is solvable, we can choose nonzero v in V to be a common eigenvector for K.  Thus we can define l to be the eigenvalue map from K to C by xv = l(x)v for all x in K.  Next we can define W={w in V : xw= l(x)w for all x in K}={The set of eigenvectors for all of K which share the same eigenvalues}. We want to prove W is invariant under L.
            To prove this suppose w is in W, x in K, y in L/K and we want to show x(yw)=l(x)yw.  Well x(yw) = y(xw)+[xy]w (as xy - yx = [xy]) = l(x)yw + l([xy])w.  We thus need to show l([xy]) = 0 for all x in K and y in L.
            Fix w in W and consider the vectors {w, yw, yyw, ..., y^n-1w} where n is the smallest integer so that yⁿ is a linear combination of the previous vectors in the list.  Let Wi=span{w, yw, yyw, ..., y^i-1w} for each i in {1,..,n}.  First notice yWn is in Wn.  Next by induction we can see that xyⁱw is in yⁱxw + Wi = l(x)yⁱw + Wi.  Thus the matrix of x on Wn has all diagonal entries equal to l(x) with respect to basis {w, yw, yyw, ..., y^n-1w}.  Note also that x was an arbitrary element of K.  We have shown Wn is closed under K and y, and as L=Cy+K we have shown Wi is invariant under L.
            If we pick the element [xy] in k in place of our x, we get the matrix of [xy] on Wn has all diagonal entries equal to l([xy]).  Thus the trace of [xy] on Wn is nl([xy]).  However [xy]=xy-yx so the trace([xy])=trace(xy)-trace(yx)=0.  Thus l([xy])=0 and we have shown W to be invariant under L.
            Now we need only write L=Cy+K and consider it's action on W.  There must be some eigenvector for y in W.  This eigenvector is already an eigenvector for K, being in W, so it is a common eigenvector for the entire space.  We are done. n
 

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