Is there a notion of "cyclic element" in a simple Lie algebra? In particular, is it independent of the irreducible representation chosen?


An endomorphism A is called cyclic if there is a vector v which by the action of A generates the whole vector space (v is also called a cyclic vector for A).

Let g be a simple Lie algebra. Call an element x cyclic if its matrix in the adjoint representation is cyclic. Let r be an irreducible representation (irrep) of g and x a cyclic element of g. Is it true that r(x) is cyclic? More generally, for two irreps r_1 and r_2 of g, is it true that r_1(x) cyclic implies r_2(x) cyclic?


1) It is important to consider only irreducible representations since for a reducible one, no r(x) is cyclic. It is also important to consider simple Lie algebras since for a semi-simple one, you can easily construct counter-examples.

2) The concept of a "cyclic element" as defined above seems new. I would be happy to get references if it is not.

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    $\begingroup$ you may want to add finite dimensional irreducible since for $sl_2$, $e$ is cyclic and it does not act cyclically on a Verma module for a negative or non integer highest weight. $\endgroup$ – Reimundo Heluani Jul 3 '19 at 15:23
  • $\begingroup$ Yes, I only consider finite-dimensional representations. Thanks Reimundo. $\endgroup$ – AThomas Jul 4 '19 at 8:10
  • $\begingroup$ The title seems to suggest that you are interested in cyclic vectors (which I think led to the confusion with @JimHumphreys's answer), whereas your title suggests that you are interested in cyclic elements (those that have a cyclic vector in the adjoint representation). $\endgroup$ – LSpice Jul 4 '19 at 14:05

Summary The answer to the first question is affirmative and to the second question is negative, but for rather mundane reasons. In the simple Lie algebra case, cyclicity of ${\rm ad}\, a$ for some $a$ implies that $\frak{g}$ has rank $1$, in which case every non-zero element $x\in\frak{g}$ is cyclic in every simple module. On the other hand, for any $\frak{g}$, every $x\in\frak{g}$ acts cyclically on any one-dimensional module (e.g. the trivial representation). Thus in the higher rank case, all elements (even $x=0$) are trivial-cyclic and none are ad-cyclic. Moreover, all modules $V$ which admit a semisimple cyclic $x\in\frak{g}$ are weight multiplicity-free (WMF), which have been classified, and the corresponding cyclic elements can be explicitly described.

Details Assume that the ground field is not of characteristic $2$. It is easy to see that a simple Lie algebra $\frak{g}$ contains an ad-cyclic element only if $\frak{g}$ has rank $\ell=1$, so that ${\frak g}$ is a form of ${\frak sl}_2$. Indeed, by the standard properties of endomorphisms, the condtion "$A$ is cyclic" implies that the dimension of $\ker{A}$ is at most one, whereas for $A={\rm ad}\, x$, this dimension is at least $\ell$ by the definition of rank; thus $\ell\leq 1$. Conversely, every non-zero element $x$ of a simple rank $1$ Lie algebra acts cyclically in every finite-dimensional simple module: extend the scalars to an algebraically closed field and consider separately the cases of semisimple and nilpotent $x$.

A semisimple endomorphism (or a diagonal matrix) is cyclic iff its eigenvalues are distinct. Let $h\in{\frak h}$ be a semisimple element of a split semisimple Lie algebra $\frak{g}$ with a Cartan subalgebra ${\frak h}$ and $V$ an $N$-dimensional $\frak{g}$-module with weights $\lambda_1,\ldots,\lambda_N\in {\frak h}^*$. Then the matrix of $h$ relative to a weight basis of $V$ is diagonal with eigenvalues $\lambda_i(h)$ and $h$ acts cyclically on $V$ iff $\lambda_i(h)$ are pairwise non-equal. In particular, this is possible only if $V$ is weight multiplicity-free (WMF), and all such modules have been classified (consistent with the above, the adjoint module is WMF iff ${\frak g}$ is a direct sum of several copies of ${\frak sl}_2$). For example, in characteristic $0$, if ${\frak g}={\frak sl}_n$, the simple WMF modules are exhausted by the symmetric powers $S^k W$ of the defining module $W$, their duals, and the exterior powers $\Lambda^k W$ of $W$, and the necessary condition for $h$ to act cyclically on $V$ is that $h$ is regular (equivalently, $h$ acts cyclically on $W$, i.e. has distinct eigenvalues). Of course, this condition is not sufficient in general. For example, if $V=S^2 W$ and $h$ is diagonal with eigenvalues $a_i$ that sum to $0$, $h$ is cyclic on $V$ iff all $a_i+a_j, 1\leq i\leq j\leq n$ are pairwise distinct. For $n\geq 3$, this is clearly stronger than just all $a_i$ being pairwise distinct.

Note, however, that reducible WMF modules also exist: for example, let $\frak{g}={\frak sl}_n$ and $V=W\oplus\Lambda^2 W$, then $V$ is WMF and every diagonal $h$ with $n$ distinct eigenvalues $a_i$ satisfying appropriate additional inequalities acts cyclically on $V$. Thus contrary to OP's Remark 1, an element of ${\frak g}$ can act cyclically on a reducible module.

If $e$ is a nilpotent element acting cyclically on a $\frak{g}$-module $V$ then $V$ has to be simple with respect to the corresponding ${\frak sl}_2$-subalgebra, so that the weights of $V$ belong to a single line. This places severe constraints on the possible data. For a simple classical Lie algebra ${\frak g}\ne{\frak sl}_2$ and a non-trivial module $V$, this leaves only the cases of ${\frak g}={\frak sl}_n$ and $V$ the defining module or its dual, and a symplectic or odd orthogonal ${\frak g}$ and the defining ("vector") module $V$, with the element $e$ being a principal nilpotent in all cases.

Finally, if $x=s+n$ is the Jordan decomposition of a general element acting cyclically on a $\frak{g}$-module $V$ then the centralizer $Z(s)$ of the semisimple part $s$ is a reductive subalebra containing $x$, so that $V$ is a $Z(s)$-module with a cyclic action of $x$. This provides an approach to classifying general cyclic pairs $(V,x)$.

  • $\begingroup$ Thank you very much Victor for your answer! I think what I really need is the notion of WMF module. Can you give a reference for that? $\endgroup$ – AThomas Jul 5 '19 at 9:11
  • $\begingroup$ You are welcome! For a comprehensive survey of multiplicity-freeness, including WMF modules, I recommend Roger Howe's Schur Lectures. They contain various classification results and further references. $\endgroup$ – Victor Protsak Jul 5 '19 at 14:22
  • $\begingroup$ MO discussion of where to get the Schur lectures: mathoverflow.net/questions/85562/… . It's really a shame no electronic version is available; I seem to remember Roger sayng he was working on TeXing and revising at some point. $\endgroup$ – LSpice Jul 5 '19 at 18:23
  • $\begingroup$ Does "$\operatorname{ad}$-simple element" mean "element $x$ such that $\mathfrak g$ is simple as an $\operatorname{ad}(x)$-module"? $\endgroup$ – LSpice Jul 5 '19 at 18:24
  • $\begingroup$ Thanks, Loren. That was a typo, I meant "ad-cyclic". $\endgroup$ – Victor Protsak Jul 5 '19 at 21:13

In an irreducible representation (finite or infinite dimensional), every nonzero vector is cyclic. This has nothing to do with Lie algebras as such, as it is true over an arbitrary ring.

Thus for example any nonzero vector in a simple Lie algebra is by definition "cyclic", since it generates the entire Lie algebra under the adjoint representation.

[Way back in 1972, I proposed the label "standard cyclic module" for what is now referred to as a "Verma module", but my label never caught on. Since these modules are rarely irreducible, this is a reasonable distinction.)

ADDED: I noticed in today's arXiv posting under the tag 'math.RT' a paper by Elashvili, Kac, and a co-author referrring back to a definition of "cyclic element" in a semisimple Lie algebra by Kostant, which is different from the one proposed here. See for example the earlier published paper here, which is probably also available on arXiv as a preprint. This reinforces my earlier point that it is better to call the newer elements "special" or whatever.

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    $\begingroup$ If I understand correctly, the question is not which vector are cyclic (which is a trivial question), but which elements of the Lie algebra act cyclically. $\endgroup$ – YCor Jul 4 '19 at 13:24
  • $\begingroup$ @YCor: I'm not sure what is exactly the question being asked, but it's clear in a finite dimensional simpe Lie algebra that each nonzero vectror in the Lie algebra generates the entire Lie algebra under the (irreduible) adjoint representaton. $\endgroup$ – Jim Humphreys Jul 4 '19 at 13:32
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    $\begingroup$ Well, what I mean is that if you take any diagonal $x\in\mathfrak{g}=\mathfrak{sl}_d$ for $d\ge 3$, then $\mathfrak{ad}(x)$ does not make $\mathfrak{g}$ a cyclic module (over the single endomorphism $\mathfrak{ad}(x)$); actually for no $x$ at all by a simple additional argument. $\endgroup$ – YCor Jul 4 '19 at 13:35
  • $\begingroup$ @YCor: I don't understand your comment. Why does it matter that $x$ is diagonal (any element will do provided it's nonzero, to generate the simple Lie algebra under the irreducible adjoint rep)? $\endgroup$ – Jim Humphreys Jul 4 '19 at 23:04
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    $\begingroup$ I repeat, the question is whether an irreducible $\mathfrak{g}$-module $V$ is cyclic as module of the 1-dimensional Lie algebra generated by $\mathrm{ad}(x)$ for given $x\in\mathfrak{g}$ (and not whether it's cyclic as $\mathfrak{g}$-module). For instance when $V$ is the adjoint rep of $\mathfrak{sl}_d$, the answer is different for $d=2$ and $d\ge 3$. For $d=2$ it holds for every $x\neq 0$; for $d\ge 3$ it holds for no $x$. $\endgroup$ – YCor Jul 4 '19 at 23:12

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