# 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: http://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?