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Let $\mathfrak{g}$ be a complex simple Lie algebra and let $\{e,f,h\}$ be a $\mathfrak{sl}_2$-triple in $\mathfrak{g}$, i. e. we have commutation relations: $$ [e,f]=h,~[h,e]=2e,~[h,f]=-2f. $$ In particular, elements $\{e,f,h\}$ span a three-dimensional Lie subalgebra in $\mathfrak{g}$.

Question. Suppose that $\mathfrak{g}$ acts on some vector space $V$ and we know the actions of elements $e,f,h$ (so we know that $V$ is a representation of $\mathfrak{g}$). What "minimal" information we need in addition to this in order to restore the action of the whole algebra $\mathfrak{g}$? I think we can suppose that $e$, $f$, $h$ are regular elements of the algebra $\mathfrak{g}$. I am particularly interested in the case when $\mathfrak{g}=\mathfrak{sl}_n$.

Of course, this question may not have a precise answer, but any help/references would be appreciated.

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    $\begingroup$ Are you given in advance that the action does extend to $\mathfrak g$, or are you interested in determining whether it extends? (Also, are you working over $\mathbb C$?) $\endgroup$
    – LSpice
    Commented Aug 8, 2021 at 16:14
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    $\begingroup$ I think that an example from @DaveWitteMorris is relevant. $\endgroup$
    – LSpice
    Commented Aug 8, 2021 at 16:17
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    $\begingroup$ The Lie algebra $\mathfrak{g}$ could be a sum of some $\mathfrak{sl}_2$ and some arbitrary semisimple Lie algebra, with an arbitrary action on $V$. $\endgroup$
    – Ben McKay
    Commented Aug 8, 2021 at 16:28
  • $\begingroup$ @BenMcKay Indeed. I think we should restrict ourselves to the case of simple Lie algebra (actually, the initial motivation was about particular case $\mathfrak{g}=\mathfrak{sl}_{n}$). I will edit the question accordingly. $\endgroup$
    – richrow
    Commented Aug 8, 2021 at 16:37
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    $\begingroup$ @Callum The centraliser of a regular $\mathfrak{sl} _2$ in a simple Lie algebra is trivial. If $V$ is a simple ${\mathfrak g}$-module then it equals $L(\lambda)$ for some dominant weight $\lambda$, and the action of your regular $h$ tells you the height of $\lambda$. You also know the dimension of $V$, and there can't be too many possibilities for $\lambda$ given this information. However, the action of a regular $\mathfrak{sl} _2$ can't tell the difference between $V$ and $V^*$. $\endgroup$
    – Paul Levy
    Commented Aug 12, 2021 at 21:33

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