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- Thread starter Silviu
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- #2

fresh_42

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Full stop.Hello! I am a bit confused by some definitions. We have that a Lie algebra is abelian if ##[a,b]=0## for all ##a,b \in L##

Do you mean an ideal by invariant subalgebra? If not, what means invariant here? Invariant under what? ##L'## is always an ideal of ##L##. Further conditions are not necessary.... and ##L'## is an invariant subalgebra of ##L##

If ##[L',L]=0## then of course is ##[L',L']=0## for the subset, i.e. ##L'## is Abelian.... if ##[a,b]=0## for all ##a \in L'## and ##b \in L##. From here I understand that ##L'## is abelian.

This is not the usual definition and I doubt it's true, although I don't have a counterexample in mind. But for fields which are not algebraically closed, things are not self-evident. A Lie algebra is defined semisimple if its radical is ##\{0\}##. A consequence is, that it is semisimple if and only if its Killing form is non-degenerate or if and only if it is a direct sum of simple Lie algebras.Then they define a Semi-simple Lie algebra as an algebra which is not abelian and has no Abelian invariant Lie subalgebra.

As mentioned above, I assume they mean ideals by invariant subalgebras. As the radical of a Lie algebra is a solvable ideal, it contains an Abelian ideal, if it's not zero. However, I can't see, why this should be an ideal of the Lie algebra itself.I am not sure I understand this definition. What do they mean by "no Abelian invariant Lie subalgebra".

No, at least not in my understanding. However, without knowing the definition of "invariant" I can only guess.Aren't all invariant subalgebras abelian?

It could be a direct sum of simple ideals. But once again: What is invariant? Which is the operation that defines invariance?This would mean it has no invariant subalgebra at all, but this would be the definition of Simple Lie algebra. What am I missing here?

- #3

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Thank you for your reply. Invariant is given as a definition: "A subalgebra ##L'## of a Lie algebra ##L## is said to be invariant if ##[a,b]=0## for all ##a \in L'## and all ##b \in L##". To my understanding it is not invariant "under something", it is invariant if that condition holds.Full stop.

Do you mean an ideal by invariant subalgebra? If not, what means invariant here? Invariant under what? ##L'## is always an ideal of ##L##. Further conditions are not necessary.

If ##[L',L]=0## then of course is ##[L',L']=0## for the subset, i.e. ##L'## is Abelian.

This is not the usual definition and I doubt it's true, although I don't have a counterexample in mind. But for fields which are not algebraically closed, things are not self-evident. A Lie algebra is defined semisimple if its radical is ##\{0\}##. A consequence is, that it is semisimple if and only if its Killing form is non-degenerate or if and only if it is a direct sum of simple Lie algebras.

As mentioned above, I assume they mean ideals by invariant subalgebras. As the radical of a Lie algebra is a solvable ideal, it contains an Abelian ideal, if it's not zero. However, I can't see, why this should be an ideal of the Lie algebra itself.

No, at least not in my understanding. However, without knowing the definition of "invariant" I can only guess.

It could be a direct sum of simple ideals. But once again: What is invariant? Which is the operation that defines invariance?

- #4

fresh_42

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That's another unusual notation: ##L'## normally symbolizes the product subalgebra ##[L,L]##. It is a bit disturbing to name a subalgebra this way. Anyway. If ##[L',L]=0 ## then ##L'## is necessarily an Abelian ideal of ##L##. Semisimple Lie algebras don't have Abelian ideals, so they cannot have such an invariant subalgebra, i.e. ##L' \neq \{0\} \Longrightarrow L \text{ not semisimple } \Longleftrightarrow \operatorname{rad}L \neq \{0\}##.Thank you for your reply. Invariant is given as a definition: "A subalgebra ##L'## of a Lie algebra ##L## is said to be invariant if ##[a,b]=0## for all ##a \in L'## and all ##b \in L##". To my understanding it is not invariant "under something", it is invariant if that condition holds.

Now in the other direction if ##L## is not semisimple, then ##\operatorname{rad}L \neq \{0\}## is a non trivial solvable ideal of ##L##. Thus ##\operatorname{rad} L## contains an Abelian ideal ##J \trianglelefteq \operatorname{rad}L## which would be a candidate for ##L'##, however, I can't see why ##[J,L]=\{0\}## should hold. I don't even see, why it should be an ideal of ##L##.

The world outside semisimplicity is pretty big and there are really many kinds of Lie algebras. Maybe it is true, which I'm not convinced of, but then it has to be proven.

##[L',L]=\{0\}## is equivalent to ##L' \subseteq C(L)## the center of ##L##. But not semisimple doesn't imply the existence of a non trivial center. E.g. ##\mathbb{F}\cdot H + \mathbb{F}\cdot X## with the multiplication ##[H,X]=X## is solvable and has no center elements, hence not semisimple. It does contain an Abelian ideal ##L'=\mathbb{F}\cdot X## but ##[L',L] \neq \{0\}##.

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George Jones

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Then they define a Semi-simple Lie algebra as ... What do they mean by "no Abelian invariant Lie subalgebra".

Who is/are "they"?

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