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It seems as though you have a few misconceptions. First, given a finite dimensinoal Lie algebra $\mathfrak{g}$, there may exist a compact, connected Lie group $G$ with Lie algebra $\mathfrak{g}$. If there does exist such a Lie group, there is no reason it must be unique. If one adds the hypothesis "simply connected" or "centerless", then existence implies ...


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The Lie functor from Lie groups to Lie algebras can be inverted on the subcategory of simply connected Lie groups. The essential surjectivity of the functor is called the third fundamental theorem of Lie, namely, every Lie algebra of finite dimension over the real or complex numbers is isomorphic to the Lie algebra of a Lie group. There are at least two ways ...


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Every supersolvable Lie algebra is solvable. For a solvable Lie algebra $L$, over an algebraically closed field of characteristic zero, Lie's theorem implies that $[L,L]$ is nilpotent - the adjoint operators $ad(x)$ can be simultaneously put into upper-triangular form, and then the adjoint operators of $[L,L]$ are strictly upper-triangular, hence nilpotent. ...


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When I google "absolutely irreducible module" the second hit I get is a passage in Lam's Exercises in classical ring theory which explains that an irreducible module $M$ over a $k$ algebra $R$ ($k$ a field) is called absolutely irreducible if $M\otimes_k K$ is irreducible over $R\otimes_k K$ for every extension field K of k. I would guess the FL part is a ...


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The following results are in this paper: Let $S$ be any set of commuting matrices. Then there is a fixed nonsingular matrix $P$ such that $P^{-1}AP$ is upper triangular for each $A$ in $S$. Let $A_j$, $1\le j\le p$, be commuting $n\times n$ matrices, and let $f=f(x_1,\dots,x_p)$ be an arbitrary polynomial in $x_1,\dots,x_p$. Then there is a fixed ...


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As you write, it is true that without using semisimplicity of $L/\mathrm{rad}(L)$, the sequence of vector spaces splits. It is not true that such a sequence of Lie algebras is necessarily split: this is where semisimplicity is crucial. Consider the example in which $L$ is $3$ by $3$ strictly upper triangular matrices (that is, with zeros on the diagonal) and ...


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Here is another argument. Note that $L$ cannot be solvable because of $[L,L]=L$. Now suppose that $I$ is a nonzero ideal, different from $L$. Hence $dim (I)=1$ or $dim(I)=2$. In the first case, $I$ is solvable and $L/I$ is of dimension $2$, hence solvable. It follows that $L$ is solvable, a contradiction. In the second case, $I$ is solvable and $L/I$ is ...



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