I need a proof of a this simple result to be convinced of a theorem:

Let $G$ be a simply connected (matrix) Lie group with Lie algebra $\mathfrak{g}$, and suppose that $\mathfrak{g} = \mathfrak{h}_1 \oplus \mathfrak{h}_2$ (so in particular $[\mathfrak{h}_1,\mathfrak{h}_2] = 0$). Now, let $\phi: \mathfrak{g} \rightarrow \mathfrak{g}$, $\phi(X+Y) = Y$ (for $X$ in $\mathfrak{h}_1$ and $Y$ in $\mathfrak{h}_2$), and let $\Phi: G \rightarrow G$ be the (unique) associated homomorphism of lie group (knowing to exist because $G$ is simply connected). $\Phi$ allows us to define a closed connected subgroup $H_1$ of $G$ by taking the identity component of its kernel. Now, I want to show that any element $A$ of $H_1$ can be written as $A = \exp(X_1) \ldots \exp(X_m)$ for all $X_i$ in $\mathfrak{h}_1$.

My intuition tells me that it is not a priori true, because the $exp$ map can fail to be both surjective and injective, thus even thought any $A$ on $G$ can be written as a product of $exp(X_i + Y_i) = exp(X_i). exp(Y_i)$, we don't know for sure why $Y_i$ should be equal to $0$ for all $i$. What did I miss ?



2 Answers 2


This is true for every connected Lie group. A Lie group is generated by a connected neighborhood of the identity and the differential of the exponential map at zero is the identity, so it's image contains a neighborhood of the identity which generates the group.

  • 1
    $\begingroup$ $H_1$ is not explicitly defined as a Lie group with Lie algebra $\mathfrak{h}_1$, so you would need to prove that first. $\endgroup$
    – Spenser
    Sep 28, 2016 at 12:54
  • $\begingroup$ I think its true because $H_1$ is the kernel of the associated homomorphism $\Phi$, thus Lie $H_1 =$ Lie ker $\Phi =$ ker $\phi$ = $\mathfrak{h}_1$. This also proves that if I define $\phi'(X+Y) = X$, with associated homomorphism $\Phi'$, then the latter acts as the identity on $H_1$ ? It's weird that the book I'm following didn't prove that any connected Lie group is generated by a neighborhood of the identity. $\endgroup$
    – sure
    Sep 28, 2016 at 12:57
  • $\begingroup$ Actually, this doesn't work because ker $\Phi$ is different than $H_1$ (can it be non connected in the case of $G$ connected/simply connected, and if yes, any example?) $\endgroup$
    – sure
    Sep 28, 2016 at 13:08

Proposition. Let $\Phi:G\to H$ be a Lie group homomorphism and $d\Phi_1:\mathfrak{g}\to\mathfrak{h}$ the induced Lie algebra homomorphism. Then, $\ker\Phi$ is a closed Lie subgroup of $G$ with Lie algebra $\ker d\Phi_1$.

Proof. The group $\ker\Phi$ is the level set at $0$ of the smooth map $\Phi$ so it is closed and hence a Lie subgroup. Moreover, its Lie algebra is the tangent space at the identity, which is $\ker d\Phi_1$ since $\ker\Phi$ is a level set of $\Phi$. $\Box$

In your case, you have homomorphism $\Phi:G\to G$ such that $d\Phi_1$ has kernel $\mathfrak{h}_1$. Thus, $\ker d\Phi$ is a closed subgroup with Lie algebra $\mathfrak{h}_1$. In particular, its identity component is also a Lie group with Lie algebra $\mathfrak{h}_1$. Then, your claim follows from the following two well-known facts:

  1. A connected Lie group is generated by any neighbourhood of the identity.
  2. The exponential map of a Lie group $G$ is a diffeomorphism from a neighborhood of $0$ in $\mathfrak{g}$ to a neighborhood of $1$ in $G$.

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