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Let $\mathfrak{g}$ denote a Lie algebra. A proposition of a course of mine on Lie algebra claims that any one-dimensional representaiton of a simple Lie algebra is $0$. However, doesn't this extends to any any Lie algebra $\mathfrak{g}$?

Indeed, a one-dimensional Lie algebra representation is a Lie algebra homomorphism $\rho:\mathfrak{g}\rightarrow \mathbb{K}$, and thus in particular one has that $\rho([x,y])=[\rho(x),\rho(y)]=0\,\forall x,y\in\mathfrak{g}$ where I equated the last term to $0$ by commutativity of the field $\mathbb{K}$, as $\rho(x),\rho(y)\in\mathbb{K}$.

QuantizedObject
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1 Answers1

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No, it doesn't apply to any Lie algebra. Take, say, $\mathfrak g=\Bbb R$, with, of course, $(\forall x,y\in\mathfrak g):[x,y]=0$. Then you have the $1$-dimensional representation $\rho$ of $\mathfrak g$ on $\Bbb R$ defined by $\rho(x)(y)=x\cdot y$. Clearly, you don't have $\rho=0$.

In your attempt to prove that it does apply to any Lie algebra, all that you managed to prove was that you always have $\rho([x,y])=0$. But, in general, $\mathfrak g$ is not generated by$$\{[x,y]\mid x,y\in\mathfrak g\}.\label{a}\tag1$$By the way, the subalgebra of $\mathfrak g$ generated by \eqref{a} is the derived algebra of $\mathfrak g$ and it is usually denoted by $\mathfrak g'$.

José Carlos Santos
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  • I see, indeed my proof doesn't prove that $\rho=0$ identically. I am then not sure how to prove that $\rho(x)=0,\forall x\in\mathfrak{g}$ for $\mathfrak{g}$ simple. I am thinking of using that the subreps of the adjoint representation are ideals, and that a simple Lie algebra has no poper ideals, but I am not sure this is the way to go. Could you provide some hints? – QuantizedObject May 05 '25 at 09:04
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    It's simple: $\mathfrak g'$ is an ideal. So, since $\mathfrak g$ is simple and $\mathfrak g'\ne{0}$, $\mathfrak g'=g$. So, since $\rho|_{\mathfrak g'}=0$, $\rho=0$. – José Carlos Santos May 05 '25 at 09:09
  • Sorry if this might be trivial, but why $\rho|_{\mathfrak{g}'}=0$ holds for $\mathfrak{g}'$ an ideal of $\mathfrak{g}$? – QuantizedObject May 05 '25 at 09:12
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    Because $\mathfrak g'$ is not just any ideal. It's the ideal generated by the elements of $\mathfrak g$ of the form $[x,y]$. And you proved that $\rho([x,y])$ is always equal to $0$, right?! – José Carlos Santos May 05 '25 at 09:15
  • Thank you very much. Indeed, $\mathfrak{g}'=\mathrm{span}{[x,y]; x,y\in\mathfrak{g}}$ is equal to $[\mathfrak{g},\mathfrak{g}]$ the dervied subalgebra, as you state in your answer, and it must be equal to $\mathfrak{g}$ for it to be simple. – QuantizedObject May 05 '25 at 09:18