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Assume that $L$ is a Lie algebra structure on $\mathbb{R}^{n}$, and $1<k<n$ is given.

We define $Gr(k,n)_{L}$, the space of all $k$ dimensional Lie subalgebra of $(\mathbb{R}^{n}, L)$.

For what type of Lia algebra structures $L$, $Gr(k,n)_{L}$ is a compact submanifold of ordinary Grassmannian $Gr(k,n)$?

Can the Lie algebra structure of $L$ be determined by topology of $Gr(k,n)_{L}$?That is: Are there two non isomorphic Lie algebra structures $L$ and $L'$ on $\mathbb{R}^{n}$ such that $Gr(k,n)_{L}$ is homemorphic to $Gr(k,n)_{L'}$?

Is there any relation between characteristic classes of canonical $k$- plane bundle restricted to $Gr(k,n)_{L}$, and Lie algebraic invariants of $L$?

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    $\begingroup$ You have asked a few questions like this, and it feels a bit like you are on a »fishing expedition»... Have you tried to compute an example of this $GLr(k,n)_L$, do you have an example where this is a manifold? why would one expect a sensible description of the Lie algebras where this works?, why do you expect any relation whatsoever between characteristic classes? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Mar 16, 2014 at 11:40
  • $\begingroup$ What are "Lie algebraic invariants"? $\endgroup$
    – Alex Degtyarev
    Commented Mar 16, 2014 at 11:45
  • $\begingroup$ @AlexDegtyarev every quantity or object which is invariant under Lie automorphisms. for example Killing form $\endgroup$
    – Ali Taghavi
    Commented Mar 16, 2014 at 13:57
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    $\begingroup$ @DeaneYang thank you for your suggestion. I try to compute it in a low dimensional case. I would like to know your opinion about this question. do you think that this leads to triviality? Do you think that there are examples of "manifold" case and also examples of singular case?(thanks again for you comment) $\endgroup$
    – Ali Taghavi
    Commented Mar 16, 2014 at 22:53
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    $\begingroup$ @MarianoSuárez-Alvarez Thank you very much for your comment. I am not on "fishing expedition". In MO I just try to learn some thing from specialist and positive mathematician who encourage even an $\epsilon$ new idea, and positively help to develop this $\epsilon$ new idea. $\endgroup$
    – Ali Taghavi
    Commented Mar 17, 2014 at 17:21

1 Answer 1

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Let me call $L$ your Lie algebra $\mathbb{R}^n$. The condition for a subspace $V\subset L$ to be a subalgebra is a closed condition: you want the map $V\otimes V\rightarrow L/V$ deduced from the bracket to be zero. In other words, let $\mathcal{V}$ be the universal $k$-plane bundle on $Gr(k,n)$, and $\mathcal{Q}$ the universal quotient; $Gr(k,n)_{L}$ is the zero locus of the map $\mathcal{V}\otimes \mathcal{V}\rightarrow \mathcal{Q} $ induced by the bracket. Thus $Gr(k,n)_{L}$ is a closed subset of $Gr(k,n)$, hence it is always compact.

As for the second question, I am not sure I understand it : I suppose you are talking about the restriction of $\mathcal{V}$ to $Gr(k,n)_{L}$; then its characteristic classes are just the restrictions to $Gr(k,n)_{L}$ of the characteristic classes of $\mathcal{V}$.What else?

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  • $\begingroup$ thanks for the answer. I do not underestand something in your answer. To prove closedness of $G(k,n)_{L}$ are you defining a continous map from $G(k,n)$ to $\mathcal{Q}$, wh which send $V$ to $[V,V]$? the dimension of $[V,V]$ may vary by choosing different $V$'s. Is your map well defined? what is the exact definition of $\mathcal{Q}$? Morover, as I asked in my question, we want to know wether it is a compact submanifold. Is not possible that the zero locus would be a singular level set? $\endgroup$
    – Ali Taghavi
    Commented Mar 16, 2014 at 15:31
  • $\begingroup$ concerning the second part of my question, as you said, the characteristic classes of the restricted bundle =the restriction of characteristic classles. Yes it is a general fact. But it would be interesting to compute the ring cohomology of $Gr(k,n)_{L}$, then we restrict the generators of cohomology of ordinary grassmannian to $Gr(k,n)_{L}$. I explain what I am meaning, with the following particular question: assume that $1<k<n$ are given. Is there a Lie structure $L$ such that the canonical bundle is the trivial on $G(k,n)_{L}$ but this spacehas nontrivial cohomolgy? $\endgroup$
    – Ali Taghavi
    Commented Mar 16, 2014 at 15:42
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    $\begingroup$ I doubt very much that $Gr(k,n)_L$ is a submanifold. All I said is that it is compact, because it is the zero locus of a homomorphism of vector bundles on $Gr(k,n)$. $\endgroup$
    – abx
    Commented Mar 16, 2014 at 15:59
  • $\begingroup$ So can we say that it is an algebraic variety? And what would be the singularities? $\endgroup$
    – Ali Taghavi
    Commented Mar 16, 2014 at 22:29
  • $\begingroup$ Yes, it is an algebraic subvariety (real if you are over $\Bbb{R}$). I would guess that for instance abelian Lie subalgebras would give singularities. I think you have to check examples, e.g. 2-dimensional subalgebras in $\mathfrak{gl}(2,\Bbb{R})$. $\endgroup$
    – abx
    Commented Mar 17, 2014 at 6:51

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