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- Thread starter jostpuur
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- #2

Chris Hillman

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"All texts always give a theorem"?

It is never a

I suggest that you review:

1. the definition of

2. the definiton of a

3. the precise statement of

If this doesn't help, ask again.

- #3

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"All texts always give a theorem"?

It is never abadidea to cite your sources, and usually agoodidea. In fact, if you had checked your sources before posting, you might have answered your own question.

I suggest that you review:

1. the definition ofidealin Lie theory versus ring theory,

2. the definiton of amodulein ring theory,

3. the precise statement ofNoether's formula[itex](A+B)/B \simeq A/(A \cap B)[/itex].

If this doesn't help, ask again.

It could be my post was more confusing than I intended, because I don't know this "Noether's formula" if it exists also elsewhere than in theory of Lie groups. I have two sources: https://www.amazon.com/dp/0817642595/?tag=pfamazon01-20 and lecture notes of course in University of Helsinki. They both present this theorem, and Knapp puts it like this:

If [itex]\mathfrak{g}[/itex] is a Lie algebra and if [itex]\mathfrak{a}[/itex] and [itex]\mathfrak{b}[/itex] are ideals in [itex]\mathfrak{g}[/itex] such that [itex]\mathfrak{a}+\mathfrak{b}=\mathfrak{g}[/itex], then

[tex]

\mathfrak{g}/\mathfrak{a} = (\mathfrak{a}+\mathfrak{b})/\mathfrak{a} \simeq \mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}).

[/tex]

In fact, the map from left to right is [itex]A+B+\mathfrak{a}\mapsto B+(\mathfrak{a}\cap\mathfrak{b})[/itex]. The map is known from linear algebra to be a vector space homomorphism, and we easily check that it respects brackets.

I thought I understood this, until I noticed that to me it seems that also the other claim, I presented in OP, is correct. And it seems strange if they both were correct, so now I think that there is something wrong with the claim, but I cannot see what.

(Now I used the same notation as Knapp does, which is different from the notation in the OP. Sorry for the confusion)

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- #4

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[tex]

\cdots\simeq\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}) \simeq \mathfrak{b}/\mathfrak{a} ?

[/tex]

If it can, then why is it written in the intersection form, when this one is simpler?

- #5

Chris Hillman

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... but you'll learn more if you figure it out yourself.

I really think that reviewing the definitions and so on would be a very good idea! However, ask again if this really doesn't help.

Incidently, Prof. Knapp was my analysis teacher (when I was an undergraduate at Cornell; Knapp later moved to SUNY Stony Brook).

- #6

matt grime

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Chris's idea of reviewing things carefully is a good one, but I think it doesn't emphasise this point enough, and might make you try to look too hard for something that is obvious

What is A/I if A is just some subalgebra of g and I is some ideal of g?

What is A/I if A is just some subalgebra of g and I is some ideal of g?

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- #7

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For an element of a algebra, X, and a subalgebra [itex]\mathfrak{a}[/itex], the coset is defined [itex]X+\mathfrak{a}=\{X+A\;|\;A\in\mathfrak{a}\}[/itex]. An important property is

[tex]

X+\mathfrak{a}=X'+\mathfrak{a} \quad\Leftrightarrow\quad X-X'\in\mathfrak{a}

[/tex]

If [itex]\mathfrak{g}[/itex] is a Lie algebra, and [itex]\mathfrak{a}[/itex] is its subalgebra, a quotient set is defined [itex]\mathfrak{g}/\mathfrak{a}=\{X+\mathfrak{a}\;|\;X\in\mathfrak{g}\}[/itex]. For this to be a Lie algebra, [itex]\mathfrak{a}[/itex] should be an ideal. Otherwise the bracket operation is not necessarily well defined.

Comments on possible mistakes here are welcome.

The claim

[tex]

\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b})\simeq \mathfrak{b}/\mathfrak{a}

[/tex]

doesn't make fully sense, because the right side is a Lie algebra if [itex]\mathfrak{a}[/itex] is an ideal, but for the left side both [itex]\mathfrak{a}[/itex] and [itex]\mathfrak{b}[/itex] should be ideals. Am I on the right track? But if we assume them both to be ideals, I still cannot see what's wrong with the proof.

I don't see any problems arising from a possible case where we have strictly [itex]\mathfrak{b}\subset\mathfrak{a}[/itex]. [itex]\mathfrak{b}/\mathfrak{a}[/itex] merely becomes a trivial group, but it's still well defined.

Let [itex]\mathfrak{a},\mathfrak{b}\subset\mathfrak{g}[/itex] be ideals of some Lie algebra. Then

[tex]

f:\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b})\to\mathfrak{b}/\mathfrak{a},\quad\quad f(B+\mathfrak{a}\cap\mathfrak{b})=B+\mathfrak{a}

[/tex]

is an isomorphism.

The map is well defined because

[tex]

B+\mathfrak{a}\cap\mathfrak{b} = B' + \mathfrak{a}\cap\mathfrak{b} \quad\implies\quad B-B'\in\mathfrak{a}\cap\mathfrak{b} \quad\implies\quad B-B'\in\mathfrak{a} \quad\implies\quad B+\mathfrak{a} = B'+\mathfrak{a}

[/tex]

Similarly it is also injective because

[tex]

B-B'\notin \mathfrak{a}\cap\mathfrak{b} \quad\big(\textrm{and}\; B-B'\in\mathfrak{b}\big) \quad\implies\quad B-B'\notin\mathfrak{a}

[/tex]

The map is surjective because for arbitrary [itex]B+\mathfrak{a}[/itex] we can choose [itex]B+\mathfrak{a}\cap\mathfrak{b}[/itex].

The map is Lie algebra homomorphism because

[tex]

f([B+\mathfrak{a}\cap\mathfrak{b},\;B'+\mathfrak{a}\cap\mathfrak{b}]) = f([B,B']+\mathfrak{a}\cap\mathfrak{b}) = [B,B']+\mathfrak{a} = [B+\mathfrak{a},\;B'+\mathfrak{a}] = [f(B+\mathfrak{a}\cap\mathfrak{b}),\; f(B'+\mathfrak{a}\cap\mathfrak{b}])

[/tex]

... but you'll learn more if you figure it out yourself.

I really think that reviewing the definitions and so on would be a very good idea! However, ask again if this really doesn't help.

I still don't know if I'm supposed to figure out that my claim is wrong or that it is right.

Incidently, Prof. Knapp was my analysis teacher (when I was an undergraduate at Cornell; Knapp later moved to SUNY Stony Brook).

Very cool.

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matt grime

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Ahaa... so it doesn't make sense to write A/I according to the standard definitions, because I is not necessarily an ideal of A. But if I still define it like this

[tex]

A/I=\{X+I\;|\;X\in A\}

[/tex]

It seems to be isomorphic to [itex]A/(A\cap I)[/itex].

- #10

matt grime

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In fact I think I proved this! I thought it was something that the book assumes, that reader must be able to figure out on his own.

So [itex]A\subset\mathfrak{g}[/itex] and [itex]I\subset\mathfrak{g}[/itex] are subalgebras of some Lie algebra, and I is an ideal. Now I want to prove that [itex]A\cap I[/itex] is an ideal in A (oh and yeah, not in [itex]\mathfrak{g}[/itex]...)

Let [itex]x\in A[/itex] and [itex]i\in A\cap I[/itex] be arbitrary.

[itex]x,i\in A[/itex], so because A is a subalgebra, we have [itex][x,i]\in A[/itex].

[itex]x\in A[/itex] and [itex]i\in I[/itex], so because I is an ideal, we have [itex][x,i]\in I[/itex].

Therefore [itex][x,i]\in A\cap I[/itex], and [itex][A, A\cap I]\subset A\cap I[/itex].

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- #12

Chris Hillman

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Jostpuur, I recommend that you carefully reread my first reply above and see if it makes more sense now. Then review all the definitions again (and the statement of Noether's formula--- make sure you know all the hypotheses). I really think you'll learn more if you figure this out yourself since your question is based upon misconceptions which are apparently due to not thinking about the issues I raised.

(I hope everyone realizes that I am trying not to give jostpuur a fish, but to teach him how to fish, as the proverb goes.)

- #13

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I thought I already understood this. The equation

[tex]

\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}) \simeq \mathfrak{b}/\mathfrak{a}

[/tex]

was correct in the sense, that if you write down both sides with expression [itex]A/B:=\{a+B\;|\;a\in A\}[/itex], they are isomorphic, but it is bad notation because in the definition of a quotient group we assume [itex]B\subset A[/itex], and hence the left side is preferred.

Now if that's wrong, then please explain. I'm believing in this now.

- #14

matt grime

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The equation

[tex]

\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b}) \simeq \mathfrak{b}/\mathfrak{a}

[/tex]

was correct in the sense, that if you write down both sides with expression [itex]A/B:=\{a+B\;|\;a\in A\}[/itex], they are isomorphic, but it is bad notation because in the definition of a quotient group we assume

ahem. assume? No, it is part of the definition.

[itex]B\subset A[/itex], and hence the left side is preferred.

the left side of what?

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