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Let $\mathfrak{g}$ be a simple Lie algebra of classical type and let $V$ be the standard representation of $\mathfrak{g}$, i.e. the representation corresponding to the first fundamental weight. Given $v\in V$, is $v$ necessarily a weight vector relative to some Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$?

If the answer to the above is yes, let $v_1,\ldots,v_n\in V$ be linearly independent, where $n=\text{rank }\mathfrak{g}$. Is there a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ such that $v_1,\ldots,v_n$ are all weight vectors relative to $\mathfrak{h}$?

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    In what numeration is something "the first" fundamental weight? – Torsten Schoeneberg Feb 13 '20 at 07:07
  • Related more general question: https://math.stackexchange.com/q/3252703/96384 – Torsten Schoeneberg Jun 19 '20 at 17:23
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    For $\mathfrak{sl}_n$ and $\mathfrak{sp}_n$ the Lie group action on the set of nonzero vectors is transitive, so the answer to the first question is positive. In the case of orthogonal it should be also possible to work out the set of orbits and hence answer the question. I expect that the answer is still positive, but I haven't checked. – Blazej Jun 23 '20 at 20:51
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    Actually the answer is negative for orthogonal, since vector $v$ with nonzero $(v,v)$ can't be an eigenvector of an infinitesimal isometry to nonzero eigenvalue. – Blazej Jun 23 '20 at 20:59
  • @Blazej: Do you think that for orthogonal, all $v$ with $(v,v)=0$ do occur as eigenvectors? I think that might be the case, although it's a bit tricky to show. (The Lie algebra does not even act on that set, i.e. it can turn isotropic into anisotropic vectors, but I think the Lie group acts transitively on that set and that might be the clue, via inner automorphisms of the Lie algebra.) – Torsten Schoeneberg Jul 02 '20 at 21:02
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    I think it is true that the action of $SO(N,\mathbb C)$ on the set of vectors $v$ with $(v,v)=0$ is transitive. Thus it is sufficient to settle the question whether this $v$ is a weight vector for one such $v$. Taking $v$ to be of the simple form $v=(1,i,0,...)$ it is easy to see that the answer is affirmative. Now for vectors $v$ with $(v,v) = \lambda \neq 0$ the story is more interesting. By rescaling $v$ it is sufficient to consider the case $\lambda=1$. Those $v$ I will call normalized. The action of the orthogonal group on the set of normalized vectors is (I think) also transitive. – Blazej Jul 04 '20 at 13:26
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    Thus again it is sufficient to settle the question for one normalized $v$. I already explained above that such $v$ can't be a weight vector to a nonzero weight, but I was too hasty to conclude that $v$ can't be a weight vector at all. In fact there could exist weight vectors for the zero weight. Taking the standard Cartan subaglebra one can check that this does happen for $N$ odd and does not happen for $N$ even. Inspecting the relevant eigenvectors in the case of $N$ odd one can see that there is in fact one non-isotropic weight vector (to weight zero). – Blazej Jul 04 '20 at 13:30
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    Therefore the question is basically solved: for standard representations of the symplectic, special linear and even orthogonal groups every nonzero vector is a weight vector for some Cartan, whereas for odd orthogonal a nonzero vector is a weight vector for some Cartan if and only if it is isotropic, i.e. $(v,v)=0$. I don't make this an answer now because in the process of arguing for this I have made a number of claims which I believe to be true (and not very difficult to prove) but in fact have not checked carefully. – Blazej Jul 04 '20 at 13:33
  • @Blazej: That matches my thoughts, and it looks like a promising start even to your far more general question. By the way, do you have a quick proof, or a reference, for transitivity of the actions of $\mathfrak{sl}_n$ and $\mathfrak{sp}_n$? With google I found one reference for $n-1$-transitivity of $\mathfrak{sl}_n$ (which also makes me ask for $k$-transitivity of the others for higher $k$). – Torsten Schoeneberg Jul 07 '20 at 05:30
  • One reason I am relatively excited about this is that the focussing on isotropic vectors in the orthogonal case is not as exceptional as it seems: Also in the case of $\mathfrak{sp}_n$, the vectors which work are all for which $(v,v)=0$ for the alternating form (which just happen to be all of them precisely because it's an alternating form), and even for $\mathfrak{sl}_n$ one could say it's the "special orthogonal group with respect to the $0$-form", see the interpretation in https://math.stackexchange.com/q/3508888/96384 (whose details still confuse me though). – Torsten Schoeneberg Jul 07 '20 at 05:38
  • And even in the other case of your question where I commented on, the adjoint representation, the nilpotent elements feel as if they are another special case of "isotropic vectors" (they actually are w.r.t. the Killing form, but not all vectors isotropic w.r.t. the Killing form in general are nilpotent, so something more intricate is going on). – Torsten Schoeneberg Jul 07 '20 at 05:40
  • Regarding transitivity of $SL$, matrix from $SL(n)$ consists of $n$ linearly independent columns such that the determinant is equal to one. Thus for $n>1$ there can be anything in the first column, and in fact there can by any linearly independent vectors in first $n-1$ columns. Situation is similar for symplectic and, except that in this case also scalar products between different columns can be. This gives no restriction on any given column, but imposes a restriction on relation between distinct columns. For orthogonal also individual columns are also restricted because $(v,v)$ is fixed. – Blazej Jul 07 '20 at 20:38
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    I think both the standard module examples and the adjoint example can be be explained as follows: for an element $v$ of a representation $V$ which is a weight vector to a nonzero weight, the closure of the group orbit of $v$ contains $0$, so $f(v)=f(0)$ for any invariant function $f$. This gives $(v,v)=0$ for the standard module in the orthogonal case and nilpotence of $ad_v$ for the adjoint representation. This argument does not apply to the zero weight. In fact in the adjoint case, $v$ is a zero weight vector iff it belongs to a Cartan. Such vectors form a dense set. – Blazej Jul 07 '20 at 20:50
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    The problem for other representations is that we don't in general know how the group orbits and invariant functions look like. In this regards standard modules of classical groups and adjoint modules are very special. I am sure there are more examples that can be worked out explicitly this way, but I don't see a way to get a general answer this way. – Blazej Jul 07 '20 at 20:54
  • Re transitivity: It seems like you're talking about the group case, which as we've seen is a helpful tool here, but I really thought of the action of the Lie algebra itself, as in https://www.sciencedirect.com/science/article/pii/S0022404904002737. -- Re interpreting that as $f(v) = f(0)$ for invariant functions: This is very nice. -- Re generalisations: Well, at least one knows in principle which highest weight rep's are orthogonal in the sense that there exists a non-deg. symm. invariant BLF on them; and for these we have that only the isotropic vectors are possible, ... – Torsten Schoeneberg Jul 10 '20 at 05:05
  • ... although the adjoint case shows in general it's a proper subset of those. On the other extreme, I would find interesting to see an example without such invariant BLFs (where so far all our examples say that all vectors are possible weight vectors). What about the four-dimensional irrep of $\mathfrak{sl}_2$, for starters: Do you have any results on that? – Torsten Schoeneberg Jul 10 '20 at 05:08

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