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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?

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  • $\begingroup$ 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, 2020 at 2:06

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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: $$x\in \mathbb{R}\mapsto \left(\begin{array}{cc}e^{ix}&0\\0&e^{-ix}\end{array}\right)$$

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  • $\begingroup$ 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, 2020 at 13:30
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    $\begingroup$ 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, 2020 at 13:57
  • $\begingroup$ Okay, is there also a counter-example for surjectivity? $\endgroup$
    – roi_saumon
    Jul 19, 2020 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, 2020 at 17:25

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