4
$\begingroup$

I want to show that if $f$ is an injective/surjective Lie group homomorphism than $f_*$ is injective/surjective.

$\require{AMScd}$ $$\begin{CD}G @>f>> H\\ @AexpAA @AAexpA\\ Lie(G) @>>f_*> Lie(H) \end{CD}$$

Take open sets $U,V,U', V'$ such that $V=exp(U), V'=exp(U'), f(V)=V', f_*(U)=U'$ and $exp|_U, exp|_V$ are diffeomorphisms.

injectivity : Suppose $f_*(x)=f_*(y)$ (where $f_*(x),f_*(y)\in V$. Then take $t$ small enough such that $tx,ty\in U$. Then by the diagram, we have $exp^{-1}(f(exp(tx)))=exp^{-1}(f(exp(ty)))$ so $tx=ty$ i.e. $x=y$.

surjectivity : Consider $y\in V\subset Lie(H)$. Now chase back in the diagram a preimage of $y$ with respect to $f_*$. Then since every $\hat{y}$ can be written as $ty$ for some $t\in \mathbb{R}$ and some $y \in V$ we are done.

Is this correct

What about the converse? f_* injective/surjective $\implies$ $f$ injective/surjective?

$\endgroup$
  • $\begingroup$ The homomorphism $C_2\to C_3$ is neither injective nor surjective, but the isomorphism on Lie algebras $0\to 0$ is an isomorphism. Thus for the converse you should probably restrict to connected Lie groups (you probably meant this anyway - but I thought it worth making explicit). $\endgroup$ – tkf Jul 19 '20 at 2:06
3
$\begingroup$

To answer one of your questions: If $f_*$ is surjective and $H$ is connected then $f$ is surjective:

By commutativity of your diagram, $f$ must be surjective onto exp(Lie$(H)$), so the image of $f$ contains an open neighbourhood $U\subseteq H$ of $e\in H$. This will generate a subgroup $H'\subset H$, which is open: for any $y\in H'$, we have $yU$ a neighbourhood of $y$ contained in $H'$.

Then each coset of $H'$ is open (as multiplication by $y\in H$ is a homeomorphism), so the union of all non-trivial cosets is open, implying that $H'$ is closed.

We conclude that $H'$ is a non-empty open, closed, subset of a connected space $H$, so $H=H'$ and $H'\subseteq$im$(f)$.

As exemplified in the comments, the requirement that $H$ is connected is necessary. Otherwise, the inclusion of the connected component of $e\in H$ will not be surjective.

To answer another of your questions, even if $G,H$ are both simply connected, $f_*$ injective does not imply that $f$ is injective. Consider $f\colon \mathbb{R}\to SU(2)$ mapping: $$x\in \mathbb{R}\mapsto \left(\begin{array}{cc}e^{ix}&0\\0&e^{-ix}\end{array}\right)$$

$\endgroup$
  • $\begingroup$ I don't understand the last point. Isn't $f_*$ in this case $x\in \mathbb{R}\mapsto \begin{pmatrix}ie^{ix} & 0\\ 0& -ie^{-ix} \end{pmatrix}$ which is also not injective? $\endgroup$ – roi_saumon Jul 19 '20 at 13:30
  • 1
    $\begingroup$ Almost: $f_*$ is the linear map sending the tangent vector $\frac{\partial}{\partial x}$ at $0\in \mathbb{R}$ to $\left.\left(\begin{array}{cc}ie^{ix}&0\\0&-ie^{-ix}\end{array}\right)\right|_{x=0}=\left(\begin{array}{cc}i&0\\0&-i\end{array}\right)$. That is $$f_*\left(\lambda\frac{\partial}{\partial x}\right)=\lambda \left(\begin{array}{cc}i&0\\0&-i\end{array}\right),\qquad\forall\lambda\in \mathbb{R},$$ which is injective. $\endgroup$ – tkf Jul 19 '20 at 13:57
  • $\begingroup$ Okay, is there also a counter-example for surjectivity? $\endgroup$ – roi_saumon Jul 19 '20 at 16:09
  • $\begingroup$ As the answer shows, if $H$ is connected then there are no counterexamples for surjectivity: $f_*$ surjective implies $f$ surjective. On the other hand let $f$ be the inclusion of Lie groups $\mathbb{R}\to \mathbb{R} \times C_2$. Then $f_*$ is an isomorphism, but $f$ is not surjective. $\endgroup$ – tkf Jul 19 '20 at 17:25

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.