All Questions
Tagged with ideals lie-algebras
62
questions
-1
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Nilpotent Lie-Algebra $g$: $g^{i+1} ⊆ g^i$ ideal in $g$?
Assume $g$ to be a nilpotent Lie-Algebra.
Nilpotency means that we can find an index $n$ such that:
$g^n = \{0\}$
for the series defined as:
$g^0 = g$
$g^{i+1} = \operatorname{span}\{[g,g^i]\}$
Why is ...
- 105
1
vote
1
answer
63
views
Binomial theorem for ideals
I was proving the statement that if $I$ and $J$ solvable ideals of Lie algebra $L$, then $I + J$ is a solvable ideal of $L$.
The proof is we know $$(I+J)/J\cong I/I\cap J.$$ Since $I,J$ are solvable ...
- 1,711
2
votes
2
answers
47
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What is the semidirect product we use in Levi Decomposition
So using the Levi Decomposition for any lie algebra $\mathfrak{g}$, there exists a semisimple subalgebra $\mathfrak{s}$ such that:
$Rad(\mathfrak{g})$$\ltimes$$\mathfrak{s}$=$\mathfrak{g}$
However in ...
- 69
2
votes
0
answers
81
views
Is the span of all nilpotent ideals also a nilpotent ideal?
Given a non-zero Lie algebra $\mathcal{L}$ over $\mathbb{C}$, we define $\mathcal{L}^2 = \big[\mathcal{L}, \mathcal{L} \big] = \big\{ [x, y]: x, y\in \mathcal{L} \big\}$, and for any $k\in\mathbb{N}$ ...
- 343
1
vote
1
answer
66
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Is there a proof for the existence of a complete flag of solvable Lie algebras independet on lie's theorem?
Basically title. Is there proof for the following statement, which doesn't rely on Lie's-Theorem?
For any solvable Lie-Algebra L, there exists a Flag of Ideals $0=I_0\subset \dots\subset I_n=L$ with $\...
- 364
0
votes
1
answer
101
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How to find an ideal in a Lie subalgebra?
Let's say $L$ is an Lie algebra and $H$ is an subalgebra.
It's very common to talk about the ideal generated by a subalgebra, i.e., the intersection of all ideals containing a subalgebra. That's look ...
0
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0
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38
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why is a Lie algebra not nilpotent even if an ideal is nilpotent and algebra quotient by ideal is nilpotent [duplicate]
Why can't we use the same arguments as for solvability to say that If a Lie algebra $L$ has a nilpotent ideal $I$, and $L/I$ is nilpotent, then $L$ is nilpotent. Can someone point out the fallacy in ...
- 9
1
vote
1
answer
68
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Why $\mathscr{D}^k\mathfrak{g}$ is an ideal in $\mathfrak{g}$?
We have defined the derived series $\{\mathscr{D}^k\mathfrak{g}\}$ inductively by
$$\mathscr{D}^1\mathfrak{g}=[\mathfrak{g},\mathfrak{g}],\ \mathscr{D}^k\mathfrak{g}=[\mathscr{D}^{k-1}\mathfrak{g},\...
- 430
3
votes
1
answer
119
views
Question about definition of Verma modules
I am studying Verma modules (reading Dixmier´s Enveloping Algebras) and have a question regarding the definition as a quotieng of the enveloping algebra.
Let $g$ be a Lie algebra and $h$ its Cartan ...
- 2,452
0
votes
0
answers
41
views
Corollary of Weyl's Theorem - direct sum of simple ideals
Weyl's Theorem states that if $L$ is a semisimple Lie algebra over $\mathbb{C}$, then any finite dimensional representation of $L$ is completely reducible.
I want to show that a corollary of this is ...
- 99
2
votes
1
answer
211
views
Constructing semisimple Lie algebras of dimension $n$
Suppose we want to construct a semisimple Lie algebra $L$ of dimension $n$.
We know that if we can write $L = L_1 \oplus L_2 \oplus \cdots \oplus L_s$ as a direct sum of simple ideals, then $L$ must ...
- 99
2
votes
1
answer
457
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Ideal of ideal is ideal of Lie algebra
Q: Let L be a Lie algebra, J be its ideal and I is an ideal of J, then I is an ideal of L.
My attempt:
We know that for any ideal J of Lie algebra L: [L, J] and [J, L] are in J.
So i in I, j in J and ...
- 41
2
votes
1
answer
99
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About the two sided ideal and the enveloping universal algebra
Take $L$ a $k$-Lie algebra (let $k$ be a commutative ring with unit) and $TL$ its tensor algebra. We have that $$UL=TL/I$$ is the universal enveloping algebra of $L$, where $I$ is the two sided ideal ...
- 301
2
votes
1
answer
63
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Show that $\sum_{i=0}^\infty V_i$ is an ideal of $L$.
In this paper On prime ideals of Lie algebras, the author proved this theorem:
Let $P$ be an ideal of $L$. Then the following conditions are equivalent:
(i) $P$ is a prime ideal of $L$.
(ii) If $[a,\...
- 938
1
vote
0
answers
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Show that if ${\frak h}\oplus x\Bbb F$ is solvabe and $e_0$ is a simultaneous eigenvector, then $\pi(x)^pe_0$ are all in the same eigenspace
This is a step in proving that any finite-dimensional representation of a solvable Lie algebra $\mathfrak g$ over an algebraically closed field has a simultaneous eigenvector, as discussed in these ...
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