Cartan's criterion for solvability(Ccs)

Cartan's criterion relates several properties (semi simplicity, solvability) of Lie algebras with properties of their invariant bilinear forms.

To prove it we need some study on bilinear forms.

DEFINITION 5.32 (in this subsection only)   For any free module $R^t$ over a ring $R$ , We denote by $\langle \rangle _R$ the ``usual inner product''. That is,

$$\langle v, w\rangle _R=\sum_i v_i w_i.$$

The first thing we do is to observe the property of this inner product when the base ring $R$ is a ``real field''. (Since we only need it for the case $R=\mathbb{Q}$ , we omit the definition of a real field and describe the following lemma only when $R=\mathbb{Q}$ .)

LEMMA 5.33   Let $W=\mathbb{Q}^t$ be a vector space. Let $b_1,b_2,\dots b_s\in W$ . Let $B$ be a $s\times s$ matrix defined by

$$B=(\langle b_i, b_j\rangle _{\mathbb{Q}}).$$

Then the determinant of $B$ is equal to 0 if and only if $\{b_j\}_{j=1}^s$ are linearly dependent over $\mathbb{Q}$ .

PROOF.. Assume $\{b_j\}_{j=1}^s$ are linearly dependent over $\mathbb{Q}$ . Then there exists a non trivial vector $(c_1,c_2,\dots c_s)\in \mathbb{Q}^s$ such that

$$\sum_{j=1}^s c_j b_j=0$$

holds. Thus

$$(c_1,c_2 \dots,c_s)B=0$$

So $B$ is a degenerate matrix which implies that $\det(B)=0$ .

Let us now prove the opposite implication. Assume $\det(B)=0$ . Then there exists a non trivial vector $(c_1,c_2,\dots,c_s)$ such that

$$(c_1,c_2 \dots,c_s)B=0$$

holds. Let us put

$$v=\sum_j c_j b_j.$$

Then we see that $\langle v,v\rangle _{\mathbb{Q}}=0$ and hence $v=0$ . (Note that for this implication we have used the fact that $\mathbb{Q}$ is a ``real field''.) Thus $\{b_j\}$ are linearly dependent over $\mathbb{Q}$ . $\qedsymbol$

The next task is to compare $\mathbb{Q}$ with other field.

DEFINITION 5.34   For any subset $S$ of a $\mathbb{Z}$ -module $W_\mathbb{Z}$ , Let us put

$$\operatorname{MG}_S=\max\{\vert\det_{l\times l}(\langle b_i,b_j\rangle _{\mathbb{Z}})\vert b_1,\dots b_l \in S\}.$$

(``The maximum modulus of Gram determinants''.) We denote by $S_k$ the subset of $W_k=W_\mathbb{Z}\otimes_\mathbb{Z}\otimes k$ defined by

$$S_k=\{x \otimes 1 \in W \otimes_\mathbb{Z}k; x \in S\}.$$

LEMMA 5.35   Let $S$ be a finite subset of a free module

$$W_{\mathbb{Z}}=\mathbb{Z}^t=\{(v_1,v_2,\dots,v_t); v_j \in \mathbb{Z}(\forall j)\}$$

over $\mathbb{Z}$ . Let $k$ be a field of characteristic $p$ . We assume either $p=0$ or $p>\operatorname{MG}_S$ holds. Then we have

$$(a\in W_k, (a^\perp \cap S_k)^{\perp\perp} \ni a) \implies a =0$$

PROOF.. Assume $a\neq 0$ . Since the inner product $\langle \bullet ,\bullet \rangle _k$ is non degenerate on $W_k$ , we see that $(a^\perp \cap S_k)^{\perp\perp}$ is equal to the $k$ -vector space spanned by $(a^\perp \cap S_k)$ . Thus there exists a set of linearly independent vectors $\{b_j\}\subset S_k$ so that we may write down $a$ as

$$a=\sum_i a_i b_i \qquad(a_i\in k).$$

Then by the assumption on $a$ , we see that

$$\sum a_i \langle b_i, b_j\rangle _k=0 \quad(\forall j)$$

Thus

$$\det(\langle b_i,b_j\rangle _k)=0$$

which is equivalent to

$$p\vert\det(\langle b_i,b_j\rangle _{\mathbb{Z}}).$$ (1)

Note on the other hand that $b_j$ are linearly independent over $\mathbb{Z}$ . Thus

$$\det(\langle b_i,b_j\rangle _\mathbb{Z})\neq 0$$

By the definition of $\operatorname{MG}_S$ , we see that

$$0<\vert\det(\langle b_i,b_j\rangle _\mathbb{Z})\vert\leq \operatorname{MG}_S$$ (2)

which contradicts to the condition (1). $\qedsymbol$

DEFINITION 5.36   For any positive integer $n$ , and for any ring $k$ , we denote by $\operatorname{Diag}_n(k)$ the set of diagonal matrices in $M_n(k)$ . For any vector $a=(a_i) \in k^n$ , we denote by $\operatorname{diag}(a)$ the diagonal matrix $\operatorname{diag}(a)=\operatorname{diag}(a_1,\dots,a_n)$ . Note that for any ring $k$ , the restriction of the Killing form of $\mathfrak{gl}_n$ coincides with the ``usual'' inner product with this identification. That is,

$$\operatorname{tr}(\operatorname{diag}(a) \operatorname{diag}(b)) =\langle a, b \rangle _k.$$

We define the following subset of $\operatorname{Diag}_n(\mathbb{Z})$ .

$$S^{[n]}=\{(\operatorname{diag}((e_i-e_j) -(e_m-e_l)); i,j,m,l\in \{1,2,3,\dots,n\}\}.$$

(where the vectors $\{e_i\}_{i=1}^n$ are elementary vectors.) We note that an obvious estimate

$$\operatorname{MG}_{S^{[n]}}\leq 4^n n!$$

holds.

LEMMA 5.37   Let $n$ be a positive integer. Let $a=(a_1,\dots,a_n), b=(b_1,\dots,b_n)\in k^n$ .

If $b$ satisfies $b\in (a^\perp\cap S^{[n]})^\perp$ , then there exist polynomials $f,g\in k[X]$ such that

$$f(\operatorname{diag}(a))=\operatorname{diag}(b) ,\quad g(\operatorname{ad}(\operatorname{diag}(a)))=\operatorname{ad}(\operatorname{diag}(b))$$

holds.

PROOF.. Let us denote by $\epsilon_{i j l m}$ the vector

$$\epsilon_{i j l m} =(e_{i} -e_{j}) -(e_{l} -e_{m}).$$

We first find a map $f_0$ from $\Lambda_a=\{a_i; i=1,2,\dots,n\}$ to $\Lambda_b=\{b_i;i=1,2,\dots,n\}$ such that

$$f_0(a_i)=b_i.$$

Such a thing exists (is ``well defined'') if and only if

$$\forall i \forall j(a_i=a_j \implies b_i=b_j)$$

holds. This condition is equivalent to the condition

$$\forall i \forall j (a \perp \epsilon_{i j 1 1} \implies b \perp \epsilon_{i j 1 1})$$

which holds by the assumption on $b$ . Thus we see that $f_0$ exists. On the other hand, by using Lagrange interpolation formula we see that there exists a polynomial $f\in k[X]$ such that $f\vert _{\Lambda_a}=f_0$ . Then we have

$$f(\operatorname{diag}(a))=\operatorname{diag}(b).$$

The adjoint action of a diagonal matrix $\operatorname{diag}(a)$ is represented by a diagonal matrix $(a_i-a_j)_{i,j}$ . Thus an argument similar to the one above proves the existence of $g$ .

$\qedsymbol$

PROPOSITION 5.38 (Cartan)   Let $V$ be an $n$ dimensional vector space over a field $k$ of characteristic $p$ . We assume that either $p=0$ or $p>\operatorname{MG}_{S^{[n]}}$ holds. Let $L$ be a Lie subalgebra of $\mathfrak{gl}(V)$ . If the Killing form of $L$ with respect to $V$ is identically equal to 0 , then $L$ is solvable.

PROOF.. We may assume that $k$ is algebraically closed. Let us take an element $x \in [L,L]$ . Then we have

$$(\operatorname{ad}x_s) (L)=(\operatorname{ad}x)_s (L) \subset L$$

Let us now diagonalize $x_s$ and write $x_s=\operatorname{diag}(a)$ . Let us take arbitrary $b\in (a^\perp\cap S^{[n]})^\perp$ . By the lemma above we see that there exist polynomials $f,g\in k[X]$ such that

$$f(\operatorname{diag}(a))=\operatorname{diag}(b) ,\quad g(\operatorname{ad}(\operatorname{diag}(a)))=\operatorname{ad}(\operatorname{diag}(b))$$

holds. For any $w=\sum_l [y_l z_l]\in [L,L]$ , we have:

$$\operatorname{tr}(\operatorname{diag}(b)[\sum_l y_l z_l]) =\sum_l\operatorname{tr}([\operatorname{diag}(b),y_l] z_l)$$

$$= \sum_l\operatorname{tr}(\operatorname{ad}(\operatorname{diag}(b))... ...=\sum_l\operatorname{tr}(g(\operatorname{ad}(\operatorname{diag}(a))).y_l z_l)$$

$$=\sum_l \operatorname{tr}(g(\operatorname{ad}(x_s)).y_l z_l) \in \sum_l\operatorname{tr}(L L)$$

$$=0$$

That means, $\operatorname{tr}(\operatorname{diag}(b) w)=0$ . In particular, we have

$$\operatorname{tr}(\operatorname{diag}(b) x)=0.$$

Since $\operatorname{diag}(b)=f(\operatorname{diag}(a))=f(x_s)$ is a polynomial in $x$ , it commutes with $x_s$ and with $x_n$ . thus

$$\operatorname{diag}(b) x=(\operatorname{diag}(b) x_s) + (\operatorname{diag}(b) x_n)$$

gives the Jordan-Chevalley decomposition of $\operatorname{diag}(b) x$ . Therefore,

$$0=\operatorname{tr}(\operatorname{diag}(b) x) =\operatorname{tr}(... ...ame{tr}(\operatorname{diag}(b) \operatorname{diag}(a))=\langle b, a \rangle _k$$

thus $b \perp a$ .

To sum up, we have shown

$$(a^{\perp} \cap S)^\perp \ni b \implies b\in a^ \perp.$$

In other words,

$$(a^{\perp} \cap S)^{\perp\perp} \ni a$$

which is equivalent to saying that $a$ is a linear combination of elements in $(a^{\perp}\cap S)$ .

In view of Lemma 5.35, we see that $a=0$ . So $x=x_s+x_n=x_n$ is a nilpotent element.

By the theorem of Engel, we conclude that $[L,L]$ is nilpotent. Thus $L$ is solvable (since we have shown that $L/[L,L]$ and $[L,L]$ are solvable).

$\qedsymbol$

DEFINITION 5.39   We say that the Cartan's criterion for solvability (Ccs) holds for a linear Lie algebra $L\subset (\mathfrak{gl}_n(k))$ over a field $k$ if it satisfies the following condition.

(Ccs) If the Killing form on $L$ associated to $k^n$ is identically zero, $L$ is solvable.

Let $n$ be a positive integer. We denote by $\operatorname{Ccs}(n)$ the set of $p$ such that Ccs holds for any Lie algebra $L$ of dimension less than or equal to $n$ for any field $k$ of characteristic $p$ .

$$\operatorname{Ccs}(n)=\{p;$$   Ccs holds for any$$(L\subset \mathfrak{gl}_n(k))$$    provided $$\operatorname{char}(k)=p, \}$$

COROLLARY 5.40 (of Proposition)   Let $n$ be a positive integer. Then:

1. $0 \in \operatorname{Ccs}(n)$
2. For any prime $p$ which is larger than $\operatorname{MG}_{S^{[n]}}$ , we have $p \in \operatorname{Ccs}(n)$ .
3. In particular, for any prime $p$ which is larger than $4^n n!$ , we have $p \in \operatorname{Ccs}(n)$ .

Note:

The estimate given in the above corollary is presumably far from the best one.

PROPOSITION 5.41   Let $n$ be a positive integer. Let $k$ be a field of characteristic $p \in \operatorname{Ccs}(n)$ . Let $L$ be a Lie algebra over $k$ whose dimension is less than or equal to $n$ . If the usual Killing form $\kappa=\operatorname{Tr}_{\operatorname{ad},L}$ of $L$ is identically equal to zero, then $L$ is solvable. In particular, if $p>4^n n!$ or $p=0$ , then $L$ is solvable if its usual Killing form is identically equal to zero.

PROOF.. Apply the definition to

$$L/($$center of$$L) \hookrightarrow \mathfrak{gl}_n(L).$$

$\qedsymbol$