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Let $\mathfrak{h}$ be an 1-dimensional ideal of the lie algebra $\mathfrak{gl}(n,\mathbb{F})$. Does the equality $\mathfrak{h}=Z(\mathfrak{gl}(n,\mathbb{F}))$ holds for arbitrary field $\mathbb{F}$? According to the book "Representations of Lie Algebras" (written by Anthony Hederson) that equality holds if $\mathbb{F}=\mathbb{C}$.

I was able to prove that $Z(\mathfrak{gl}(n,\mathbb{F}))=\text{Span}(\{I_n\})$ in which $I_n$ is the identity matrix. I tried to use this fact to prove that equality however I didn't make any progress.

Thank you for your attention!

Torsten Schoeneberg
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rfloc
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1 Answers1

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As said in the comments, much stronger statements are possible in the case that $char(\mathbb F)$ does not divide $n$. Because in that case, the Lie algebra of traceless matrices $\mathfrak{sl}_n(\mathbb F)$ ($=$ the derived subalgebra of $\mathfrak{gl}_n(\mathbb F)$) is semisimple, $\mathfrak{sl}_n(\mathbb F) \cap \mathbb F \cdot id = \{0\}$, and the map

$$\mathfrak{gl}_n(\mathbb F) \rightarrow \mathfrak{sl}_n(\mathbb F) \oplus \mathbb F \cdot I_n \\ A \mapsto (A-\frac1n tr(A)\cdot id, \frac1n tr(A) \cdot id)$$ is an isomorphism of Lie algebras, showing that $\mathfrak{gl}_n(\mathbb F)$ is reductive with centre spanned by the identity matrix, and semisimple part $\mathfrak{sl}_n (\mathbb F)$. Then your assertion follows trivially from the fact that semisimple Lie algebras do not have one-dimensional ideals.

But here is a proof for it in general, without that assumption on the characteristic, which just generalizes the proof from the accepted answer by user NebulousReveal to A linear operator commuting with all such operators is a scalar multiple of the identity.

Let $\mathfrak h$ be a one-dimensional ideal in $\mathfrak g := \mathfrak{gl}_n(\mathbb F)$, and choose $0 \neq T \in \mathfrak h$. We will show that $T$ is a scalar multiple of $id$, i.e. in the one-dimensional centre of $\mathfrak g$.

Namely, that $\mathfrak h$ is an ideal means there is a linear functional

$$\alpha : \mathfrak g \rightarrow \mathrm{End}_{\mathbb F}(\mathfrak h) \simeq \mathbb F$$

given by $\alpha(S) \cdot T = ad_S(T) = ST-TS$ for all $S \in \mathfrak g$.

If for any $v \in V := \mathbb F^n$, the vectors $v$ and $Tv$ were linearly independent, then we could take as $S$ any linear map that maps both $v$ and $Tv$ to $v$, and would get $(ST-TS)v = v-Tv$ but also $=\alpha(S) \cdot Tv$ i.e. $v$ and $Tv$ would be linearly dependent, contradiction. So $T$ acts on each vector $v$ via a scalar $t_v$.

But now assume $v,w$ were vectors such that $t_v \neq t_w$. They would necessarily be independent, and we could take a linear map $S$ that interchanges $v$ and $w$. This would make $(ST-TS)v= (t_v - t_w) w$ on the one hand, but on the other hand $=\alpha(S) \cdot Tv \in \mathbb F\cdot v$. Contradiction to $v,w$ being independent.

So $T$ is just scalar multiplication with the same $t$ for all vectors $v$. QED.

Torsten Schoeneberg
  • 26,051