Suppose I have a Lie algebra $\mathfrak g$ which has an ideal $\mathfrak a$. Then I consider the quotient set $\mathfrak g / \mathfrak a$ which is the set of all equivalence relations of $\mathfrak g$ on $\mathfrak a$, where the appropriate equivalence relation is $v \sim u \iff v-u \in \mathfrak a$. The quotient set is also a vector space if we equip it with the following scalar multiplication and vector addition:

$$ \begin{align*} \lambda [u] &= [\lambda u] \\ [u] + [v] &= [u+v] \end{align*} $$

where $[u]$ is the equivalence class of $u$.

Now, we can also extended $\mathfrak g / \mathfrak a$ to be a Lie algebra as well if we equip it with a Lie bracket $\{\}$ defined by:

$$ \{[u],[v]\} = [\langle u,v \rangle] $$

where $\langle \rangle$ is the Lie brackets for the Lie algebra $\mathfrak g$.

My question is: is this unique? Is $\{\}$ the only Lie bracket I could possibly define for $\mathfrak g / \mathfrak a$, or is there another one?

  • 2
    $\begingroup$ This is the only structure you can put on it if you want the quotient map $\mathfrak g \rightarrow \mathfrak g/\mathfrak a$ to be a map of Lie algebras. $\endgroup$ – Nate Mar 10 '14 at 16:58

Nate's comment is correct, but here is a bit more to chew on.

Notice that if $\mathfrak{h}$ is any Lie algebra such that there is an isomorphism of vector spaces $f:\mathfrak{h}\to\mathfrak{g}/\mathfrak{a}$, then we can define a Lie algebra structure on $\mathfrak{g}/\mathfrak{a}$ by

$$\{[u],[v]\}=f(\langle a,b\rangle)$$

where $f(a)=[u]$ and $f(b)=[v]$. Notice this is well-defined because $f$ is an isomorphism. In this definition, we are simply translating structure along the isomorphism $f$, turning $f$ into a map of Lie algebras. In particular, we can always equip the vector space $\mathfrak{g}/\mathfrak{a}$ with the trivial bracket, $\{[u],[v]\}=0$ for all $[u],[v]\in\mathfrak{g}/\mathfrak{a}$. However, in this general construction, we've forgotten the particularly nice structure of the object $\mathfrak{a}$, which is a Lie ideal, and not just a vector subspace.

To take advantage of the Lie ideal structure, it turns out that the natural surjection of vector spaces $f':\mathfrak{g}\to\mathfrak{g}/\mathfrak{a}$, sending $u$ to $[u]$, induces a Lie algebra structure on $\mathfrak{g}/\mathfrak{a}$ just as defined above:

$$\{[u],[v]\}=f'(\langle u,v\rangle)=[\langle u,v\rangle]$$

In our first example, we used the fact that $f$ was an isomorphism to show that our bracket was well-defined, however, $f'$ need not be an isomorphism. Instead, to see that this bracket is well-defined, you'll need exactly what the Lie ideal structure provides. As before, we'll see then that $f'$ becomes a map of Lie algebras, not just of vector spaces. Thus, this is the most natural bracket to place on the object $\mathfrak{g}/\mathfrak{a}$, as it seems the conditions required of $\mathfrak{a}$ were predetermined specifically for this construction.


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