Lie algebra, ideal and isomorphism

1. Sep 1, 2007

jostpuur

Suppose $A\subset\mathfrak{g}$ and $I\subset\mathfrak{g}$ are subalgebras of some Lie algebra, and I is an ideal. Is there something wrong with an isomorphism $(A+I)/I \simeq A/I$, $a+i+I=a+I\mapsto a+I$, for $a\in A$ and $i\in I$? I cannot see what could be wrong, but all texts always give a theorem $(I+J)/J\simeq I/(I\cap J)$ instead.

2. Sep 1, 2007

Chris Hillman

"All texts always give a theorem"?

I suggest that you review:

1. the definition of ideal in Lie theory versus ring theory,
2. the definiton of a module in ring theory,
3. the precise statement of Noether's formula $(A+B)/B \simeq A/(A \cap B)$.

If this doesn't help, ask again.

3. Sep 1, 2007

jostpuur

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/Lie-Groups-Introduction-Anthony-Knapp/dp/0817642595 and lecture notes of course in University of Helsinki. They both present this theorem, and Knapp puts it like this:

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)

Last edited: Sep 1, 2007
4. Sep 1, 2007

jostpuur

Or is there anything wrong with the claim? Could the theorem continued to be

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

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

5. Sep 1, 2007

Chris Hillman

I'm not trying to be difficult...

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. Sep 2, 2007

matt grime

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?

Last edited: Sep 2, 2007
7. Sep 2, 2007

jostpuur

Basic stuff:

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

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

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

Comments on possible mistakes here are welcome.

Doubts:

The claim

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

doesn't make fully sense, because the right side is a Lie algebra if $\mathfrak{a}$ is an ideal, but for the left side both $\mathfrak{a}$ and $\mathfrak{b}$ 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 $\mathfrak{b}\subset\mathfrak{a}$. $\mathfrak{b}/\mathfrak{a}$ merely becomes a trivial group, but it's still well defined.

Proof:

Let $\mathfrak{a},\mathfrak{b}\subset\mathfrak{g}$ be ideals of some Lie algebra. Then

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

is an isomorphism.

The map is well defined because

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

Similarly it is also injective because

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

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

The map is Lie algebra homomorphism because

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

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

Very cool.

8. Sep 2, 2007

matt grime

Nope, you're looking too hard. If I have A a subalgebra of g, and I an ideal of g, how are you defining the quotient space A/I? I know how to define the quotient space g/I, for example, because I is an ideal of g, but is I an ideal of A?

9. Sep 2, 2007

jostpuur

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

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

It seems to be isomorphic to $A/(A\cap I)$.

10. Sep 2, 2007

matt grime

But is A\cap I an ideal? I.e. does it make sense to write A/(A\cap I)? As a vector space, you're certainly correct, but that isn't necessarily a Lie algebra homomorphism.

11. Sep 2, 2007

jostpuur

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

Let $x\in A$ and $i\in A\cap I$ be arbitrary.

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

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

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

Last edited: Sep 2, 2007
12. Sep 3, 2007

Chris Hillman

Back to square one, with benefit of more experience

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. Sep 13, 2007

jostpuur

This is the only context where I have seen this formula. But I can guess from Chris' comments, that there is some other theorem, not related to lie algebras, where this same formula is also present. I don't have this theorem at any sources now, and it seems difficult to find it by google and keywords "noether's formula" (OMG, the 9th google hit is my own thread! :surprised, googled 2007-09-13)

I thought I already understood this. The equation

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

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

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

14. Sep 13, 2007

matt grime

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

the left side of what?