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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

$\displaystyle \{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

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

holds. So we have

$\displaystyle 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

$\displaystyle [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

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

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

      $\displaystyle \forall j \forall u((j\geq 2, j\neq u )\implies x_{j u}=0)$
      $\displaystyle \forall j ((j\geq 2)\implies x_{j j}=x_{1 1})$

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

$\displaystyle \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

$\displaystyle (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

$\displaystyle [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

$\displaystyle 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$ .

$\displaystyle 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

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

In addition we have

$\displaystyle 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

    $\displaystyle 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

$\displaystyle 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

$\displaystyle \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

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

we compute as follows

$\displaystyle 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

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

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$ .

$\displaystyle 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

$\displaystyle 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$


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Next: Ideals of . Up: generalities in finite dimensional Previous: Theorem of Engel
2007-12-19