I am solving a multiple part problem on Lie algebra representations. I have done the first three parts, but am stuck on part (iv) as follows:

Lie algebra

Define a linear map $\phi : \mathbb{g} \rightarrow End(V)$ by specifying it on the basis as $\phi(X)=\hat{x}$, $\phi(Y)=\hat{p}$ and $\phi(Z)=\hat{A}$ where $A$ is an operator of $V$

I have already shown that the basis satisfies the lie bracket, $Z$ is in the centre of $\mathbb{g}$, $V$ is invariant under $\hat{x}$ and $\hat{p}$ and computed the lie bracket of $\hat{x}$ and $\hat{p}$

I am really stumped by this question. I know that an endomorphism maps a space back onto itself, here the space is $V$. I was thinking of taking the differential $\mathbb{g'}$ which would be linear but am unsure how to construct it

There is a lot of notation here so I thought I would provide the picture of the problem - I hope this is clear

What should $A$ be so that the map $\phi$ is a lie algebra representation?

I know that a finite dimensional representation $(\pi, V)$ of a lie algebra $\mathbb{g}$ is a lie algebra homomorphism $\pi : \mathbb{g} \rightarrow \mathbb{g}(V)$

So we must have that $\phi (XY)=\phi(X)\phi(Y)$

Thank you for your time!

  • $\begingroup$ Check the definition of a Lie algebra homomorphism again on wikipedia. What you have written at the end of your question is unfortunately not correct. $\endgroup$ Commented Nov 13, 2016 at 10:26

1 Answer 1


$A$ should be $[\hat x,\hat p]=\hat x\circ\hat p- \hat p\circ \hat x$. To complete the proof, show that $[\hat x,\hat p]$ commutes with $\hat x$ and $\hat p$.


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