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Timeline for Lie algebraic Grassmannian

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Mar 17, 2014 at 6:51 comment added abx 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 $\mathfrak{gl}(2,\Bbb{R})$.
Mar 16, 2014 at 22:29 comment added Ali Taghavi So can we say that it is an algebraic variety? And what would be the singularities?
Mar 16, 2014 at 15:59 comment added abx 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 $Gr(k,n)$.
Mar 16, 2014 at 15:42 comment added Ali Taghavi 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?
Mar 16, 2014 at 15:31 comment added Ali Taghavi 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?
Mar 16, 2014 at 13:12 history edited abx CC BY-SA 3.0
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Mar 16, 2014 at 11:58 history answered abx CC BY-SA 3.0