28
$\begingroup$

Suppose I have a Lie group $G$ and a closed normal subgroup $H$, both connected. Then I can form the quotient $G/H$, which is again a Lie group. On the other hand, the derivative of the embedding $H\hookrightarrow G$ gives an embedding of Lie algebras, $\mathfrak{h}\hookrightarrow\mathfrak{g}$. Since $H$ is normal, $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, so we can form the quotient algebra $\mathfrak{g}/\mathfrak{h}$.

  1. Is it then true that the Lie algebra of $G/H$ is canonically isomorphic to $\mathfrak{g}/\mathfrak{h}$? (I guess what I really want to know is: is the functor $\phi\mapsto d_e\phi$ exact?)

  2. If so, can we always write $\mathfrak{g}$ as a direct sum of vector spaces $\mathfrak{g}=\mathfrak{h}\oplus\mathfrak{g}'$ where $\mathfrak{g}'$ is a subalgebra of $\mathfrak{g}$ isomorphic to $\mathfrak{g}/\mathfrak{h}$? (I don't mean this to be a direct sum of Lie algebras; there would certainly in general be a nontrivial bracket between $\mathfrak{g}'$ and $\mathfrak{h}$.)

Thank you.

$\endgroup$
11
$\begingroup$

The answer to your first question is yes. Let $p: G\to G/H$ denote the projection. Then $\mathfrak{h}$ is a subspace of the kernel of $p_*: \mathfrak g \to Lie(G/H)$. This is because if $X\in \mathfrak h$ then $p_* X = \frac{d}{dt}\vert_{t=0} p(\exp(tX)) = 0$ since $\exp(tX) \in H$ so $p(\exp(tX))$ is constant. By dimensionality reasons, $\mathfrak{h} = \ker p_*$. Thus $p_*$ gives a canonical isomorphism $\mathfrak g/\mathfrak h \to Lie(G/H)$.

I'm really not sure about your second question.

$\endgroup$
10
$\begingroup$

For (2), consider the Heisenberg algebra $\mathfrak g$ of dimension three, spanned by $x$, $y$ and $z$ such that $[x,y]=z$ and $z$ is central. Then $\mathfrak h=\langle z\rangle$ is an ideal, and $\mathfrak g/\mathfrak h$ is abelian of dimension $2$. But $\mathfrak g$ has no abelian subalgebra of dimension two intersecting trivially with $\mathfrak h$.

$\endgroup$
9
$\begingroup$

Contrasting with what Mariano said, for (2), in the compact case it's true. That is, given a compact Lie group $G$ with Lie algebra $\mathfrak{g}$, if $\mathfrak{h}$ is an ideal of $\mathfrak{g}$, then there is another ideal $\mathfrak{p}$ in $\mathfrak{g}$ so that $\mathfrak{g} = \mathfrak{h}\oplus \mathfrak{p}$ where this really is a sum of Lie algebras (that is, the bracket between the $\mathfrak{h}$ part and the $\mathfrak{p}$ part is 0).

For, if $G$ is compact, it has a biinvariant metric which corresponds to an $Ad(G)$ invariant inner product on $\mathfrak{g}$. One can prove that this inner product satisfies $\langle [x,y],z\rangle = \langle x, [y,z]\rangle$.

Now, given $\mathfrak{h}$, let $\mathfrak{p}$ be the orthogonal complement. I'll use $h$ with subscripts to denote arbitrary elements of $\mathfrak{h}$ and likewise for $p$.

Then $\mathfrak{p}$ is an ideal since $\langle h, [p_1 + h_1,p]\rangle = \langle [h,p_1+h_1],p\rangle = 0$ since $[h,p_1+h_1]$ is in $\mathfrak{h}$ since it's an ideal.

Finally, we must have $[h,p]\in\mathfrak{h}\cap\mathfrak{p} = \{0\}$ since both are ideals.

$\endgroup$
  • 4
    $\begingroup$ (This is related to the Principal Theorem of Wedderburn and Malcev for Lie algebras; the compact case essentially corresponds to semisimple things) $\endgroup$ – Mariano Suárez-Álvarez Aug 8 '13 at 14:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy