Ideals of $\mathfrak{gl}_n(k)$

Recall that $\mathfrak{n}_n(k)$ denotes the Lie algebra of strictly upper triangular matrices. In this subsection we denote by $e_{i j}$ the elementary matrices. (as we have done so without even mentioning...)

LEMMA 5.18   Let $k$ be a field of characteristic $p$ (possibly 0 ). Let $n\in \mathbb{Z}_{>1}$ . We assume that $(p,n)\neq (2,2)$ . Then we have

$$\{x \in \mathfrak{gl}_n(k) ; [x,y]\in k.1_n \quad(\forall y\in \mathfrak{n}_n(k))\} =k.1_n+k. e_{1 n}.$$

PROOF.. Let us denote by $L$ the left hand side of the lemma. Then we trivially have $L\supset k.1_n$ . Furthermore, for all $x \in \mathfrak{n}_n(k)$ , we see easily that

$$x e_{1 n}=0, \qquad e_{1 n} x =0$$

holds. So we have

$$L\supset k.1_n +k.e_{1 n}$$

Let us prove the opposite inclusion. We take an arbitrary element $x\in L$ .

For any $(i, j)$ satisfying $i<j$ , we have $e_{i j} \in \mathfrak{n}_n(k)$ and thus

$$[e_{i,j}, x]=c 1_n \qquad (\exists c \in k).$$

The rank of the left hand side is at most $2$ . So $c$ must be equal to 0 when $n\geq 3$ . Otherwise ($n=2$ ), we compare the trace of the both hand sides. The trace of the left hand side is clearly zero. The trace of a scalar matrix $c.1_2$ is equal to $2 c$ . Thus $c=0$ by our assumption ( $(p,n)\neq (2,2)$ ). In either case, we have $[e_{i,j},x]=0$ . Then we compute some of special cases. First, let us examine the case where $i=1, j\geq 2$ . Then

$$0=[e_{1 j}, x]= \sum_{s t} [e_{1 j}, x_{s t} e_{s t}] = \sum_{t } x_{j t} e_{1 t} -\sum_s x_{s 1} e_{s j}$$

By looking at $(1,u)$ entry of the above equation, we conclude that equations in entries

$$\forall j \forall u((j\geq 2, j\neq u )\implies x_{j u}=0)$$

$$\forall j ((j\geq 2)\implies x_{j j}=x_{1 1})$$

hold. Similarly, by looking at the $(u,n)$ entry of $[e_{i n},x]$ , we conclude that equations

$$\forall i \forall u((i\leq n-1, i\neq u )\implies x_{u i}=0)$$

hold. Putting the equations all together, we conclude that $x$ is in the right hand side of the lemma. $\qedsymbol$

As an application of the Engel's theorem, we prove the following proposition.

PROPOSITION 5.19   Let $k$ be a field of characteristic $p$ (possibly 0 ). Let $n$ be a positive integer. We assume that $(p,n)\neq (2,2)$ . Then each ideal $I$ of $\mathfrak{gl}_n(k)$ is equal to the one in the following list.

1. 0 .
2. $k. 1_n$ .
3. $\mathfrak{sl}_n(k)$ .
4. $\mathfrak{gl}_n(k)$ .

PROOF.. The case $n=1$ is trivial. So let us assume $n\geq 2$ .

If $I\subset k.1_n$ , then $\dim_k(I)\leq 1$ and hence $I=0$ or $I=k.1_n$ . Assume now $I\not \subset k.1_n$ . Let us consider the Lie algebra $\mathfrak{n}_n(k)$ of strictly upper triangular matrices. Then

$$(L,V)=(\mathfrak{n}_n(k), (I+k.1_n)/k.1_n)$$

satisfies the assumption of the Engel's theorem. So there exists a non-constant element $x\in I$ such that

$$[x,y]\in k.1_n \qquad (\forall y\in \mathfrak{n}_n(k)).$$

holds. By using the previous lemma, we see that $x$ may be presented as

$$x=c_0 1_n +c_1 e_{1 n} \qquad( \exists c_0,c_1 \in k).$$

Since $x$ is non-constant, we have $c_1\neq 0$ .

$$I \ni [e_{11},x]=c_1 e_{1 n}$$

Thus $e_{1 n}$ belongs to $I$ . By changing the order of the base and repeating the above argument, we conclude that

$$\forall i \forall j ((i \neq j)\implies e_{i j} \in I.)$$

In addition we have

$$I \ni [e_{i j},e_{j i}] =e_{ii}-e_{j j}$$

This clearly proves $I\supset \mathfrak{sl}_n(k)$ . Since the codimension of $\mathfrak{sl}_n(k)$ in $\mathfrak{gl}_n(k)$ is $1$ , we have either $I=\mathfrak{sl}_n(k)$ or $I=\mathfrak{gl}_n(k)$ .

$\qedsymbol$

For the sake of completeness, we deal with the case $(p,n)=(2,2)$ . In this case, situation is a bit different.

LEMMA 5.20   Let $k$ be a field of characteristic $2$ . Then:

1. Any two-dimensional vector subspace $V$ of $\mathfrak{sl}_2(k)$ with $V\supset k.1_2$ is equal to a vector space $L_{[b:c]}$ given by an element $[b:c]\in \mathbb{P}^1(k)$ which is defined as
   $$L_{[b:c]} =k. 1_n + k. \begin{pmatrix} 0& b \\ c & 0 \end{pmatrix}.$$

2. For any element $[b:c]\in \mathbb{P}^1(k)$ , the vector space $L_{[b:c]}$ is an ideal of $\mathfrak{gl}_2(k)$ .
3. In particular, $\mathfrak{t}_2(k)=k. 1_2 +\mathfrak{n}_2(k)$ is an ideal of $\mathfrak{gl}_2(k)$ .

PROOF.. (1) There exists a traceless non constant matrix $x \in V$ such that

$$I=k.1_n+ k.x$$

holds. By subtracting a constant matrix, one may easily replace $x$ by a matrix with zero diagonals.

(2) By a direct computation we see

$$\left [ \begin{pmatrix} x & y \\ z & w \end{pmatrix}, \begin{pma... ...trix} 0 & b \\ c & 0 \end{pmatrix}+ (b z -c y) 1_2 \qquad (x,y,z,w,b,c,\in k)$$

(Note that $\operatorname{char}(k)=0$ .)

(3) $\mathfrak{t}_2(k)=L_{[1:0]}$ . $\qedsymbol$

PROPOSITION 5.21   Let $k$ be a field of characteristic $2$ . Then each ideal $I$ of $\mathfrak{gl}_2(k)$ is equal to the one in the following list.

1. 0 .
2. $k. 1_n$ .
3. A two-dimensional Lie algebra $L_{[b:c]}$ defined as in the lemma above.
4. $\mathfrak{sl}_n(k)$ .
5. $\mathfrak{gl}_n(k)$ .

PROOF.. We divide into several cases.

(i) Case where $I\not \subset \mathfrak{sl}_2(k)$ . In this case there exists an element $x\in I$ with $\operatorname{tr}(x)\neq 0$ . Putting

$$x= \begin{pmatrix} a& b \\ c & d \end{pmatrix},$$

we compute as follows

$$I\ni [e_{1 2}, x] =\left[ \begin{pmatrix} 0& 1 \\ 0 & 0 \end{pm... ... c & d \end{pmatrix}\right] = \begin{pmatrix} c& a+d \\ 0 & c \end{pmatrix}.$$

(Note $\operatorname{char}(k)=2$ .) Then we have

$$I \ni [e_{1 1},[e_{1 2},x]] = \left[ \begin{pmatrix} 1& 0 \\ 0 &... ...\\ 0 & c \end{pmatrix}\right] = \begin{pmatrix} 0& a+d \\ 0 & 0 \end{pmatrix}$$

Thus we see that $e_{1 2}\in I$ . In a same way (by changing the order of the base), we obtain, $e_{2 1}\in I$ .

$$1_2=[e_{2 1},e_{1 2}]\in I.$$

Since $\operatorname{tr}(x)\neq 0$ , we see that $\{1_2, e_{12},e_{21},x\}$ spans the $\mathfrak{gl}_2(k)$ . thus $I=\mathfrak{gl}_2(k)$ in this case.

(ii) Case where $I\subset \mathfrak{sl}_2(k)$ and $I\cap k.1_2=0$ . Let $x$ be arbitrary element of $I$ and put

$$x= \begin{pmatrix} a& b \\ c & d \end{pmatrix}.$$

Then by computing $[e_{1 2},x]$ as in the case (i) above, we know that $c=0$ . Similarly, we know $b=0$ . Since $x$ is traceless, $a=d$ also holds. So the only possibility in this case is $I=0$ .

(iii) Case where $k.1_2 \subsetneq I\subsetneq \mathfrak{sl}_2(k)$ . By a dimension consideration, we see that $\dim I=2$ . Then we use the above lemma.

(iv) The case $I=k.1_2$ or $I=\mathfrak{sl}_2(k)$ . Excellent. There is nothing to in this case.

$\qedsymbol$