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
<br />
X+\mathfrak{a}=X'+\mathfrak{a} \quad\Leftrightarrow\quad X-X'\in\mathfrak{a}<br />
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
<br />
\mathfrak{b}/(\mathfrak{a}\cap\mathfrak{b})\simeq \mathfrak{b}/\mathfrak{a}<br />
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
<br />
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}<br />
is an isomorphism.
The map is well defined because
<br />
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}<br />
Similarly it is also injective because
<br />
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}<br />
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
<br />
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}])<br />
Chris Hillman said:
... 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.
