This is exercise from Lee: Introduction to smooth manifolds.

Suppose $f \colon G \to H$ is homomorphism of Lie groups (real, finite-dimensional).

Q: Is image $Im(f) \subseteq H$ a Lie subgroup of H?

That is, is there topology and smooth structure on $Im(f)$ such that inclusion $Im(f) \hookrightarrow H$ is immersion, and such that induced operation $Im(f) \times Im(f) \hookrightarrow H \times H \to Im(f) \subseteq H$ is smooth.

The author of these notes says (last paragraph on the first page) that the answer is no, provides counterexample (dense line on torus), but the proof is ommited. Is it correct?

I need positive proof of that, to understand the following characterisation: Lie group admits faithfull finite-dimensional representation if and only if it is (isomorphic to) Lie subgroup of $GL(n,\mathbb R)$.

One more thing. It is easy to see that $Ker(f)$ is Lie subgroup of $G$ using nontrivial Closed subgroup theorem. Is there more direct proof?

Any help is very appreciated.

  • 1
    $\begingroup$ It depends on how you define Lie subgroups. I may be wrong, but if I remember correctly, Lee defines a Lie subgroup to be a sub-object in both, the manifold- and the group- sense, and that they still are compatible. If so, the given counterexample works: the dense line on the torus is not a submanifold. Another common definition doesn't require this and then, the answer to the question would be yes (almost by definition). $\endgroup$ – Ben Mar 13 '13 at 23:03
  • $\begingroup$ I am using definition which says that Lie subgroup is immersed submanifold (so not necesseraly with subspace topology). $\endgroup$ – Rafael Mrđen Mar 14 '13 at 12:07
  • $\begingroup$ So it is really not clear to me how tu put topology and smooth structure on $Im(f)$ so that everything works out. $\endgroup$ – Rafael Mrđen Mar 14 '13 at 12:08
  • 1
    $\begingroup$ Give $G/\ker(f)$ the induced Lie group structure (this requires that you know that the kernel is a closed normal subgroup) and use the natural group isomorphism from $G/\ker(f)$ to the image of $f$ to define a differentiable structure on it. W.r.t. this, the inclusion into $H$ is nothing but $f$ (modulo kernel) and hence is smooth. The multiplication should be smooth since $f$ was a morphism of lie groups. (I didn't check that!) As the inclusion then is an injective Lie group homomorphism, it automatically is an immersion. $\endgroup$ – Ben Mar 14 '13 at 16:54
  • $\begingroup$ Thank you, that is what I was looking for. At first, I thought that I must work the coordinate charts explicitely. I got the smoothness of multiplicaion myself. And for the kernel part, I can show that Lie group homomorphism must be of constant rank, and that means that level sets are embedded submanifolds of domain, and therefore closed. I still have to work out the quotienting of manifolds, but I know where to look now. $\endgroup$ – Rafael Mrđen Mar 14 '13 at 17:31

Let $f\colon G\to H$ be a homomorphism of Lie groups as in the question. Via the natural group isomorphism, we can consider $\DeclareMathOperator{\im}{im}\im(f)$ as $G/\ker(f)$. Since $\ker(f)\subset G$ is a closed subgroup, $G/\ker(f)$ has a canonical Lie group structure. Using this to impose a Lie group structure on $\im(f)$, it becomes a Lie subgroup of $H$.

We may clarify this, by showing that the given definition of Lie subgroups coincides with the following: A subset $I\subset H$ of a Lie group $H$ is an (immersed) Lie subgroup, if there exists a Lie group $I'$ and an injective homomorphism of Lie groups $I'\to H$, such that $\im(I'\to H) = I$.

Suppose this for the moment. Then the induced Lie group homomorphism $\bar{f}\colon I' := G/\ker(f)\hookrightarrow H$ has the same image as $f$ and is injective by construction. Thus $\im(f)$ is an immersed Lie subgroup of $H$.

To show the equivalence of the definitions, everything is tautological, except the fact that injective Lie group homomorphisms are automatically immersions. Actually, I didn't knew the way you are describing to see this in your comment below your question; I'm used to reduce the question to the unit and then argue via the exponential map, which is a local diffeomorphism. Indeed, it suffices to show that the derivative of $\iota\colon I'\to H$ is injective at the unit $e$. But the exponential maps $\exp_{I'}\colon T_e I'\to I'$ and $\exp_{H}\colon T_e H\to H$ are local diffeomorphisms at $0$, and since we have a commutative square $\exp_{H}\circ d\iota = \iota\circ\exp_{I'}$, the injectivity of $d\iota$ at $e$ follows from the injectivity of $\iota$.

  • $\begingroup$ Thanks. For your last paragraph, I can show that Lie group homomorphism has constant rank, and that injective map with constant rank must be an immersion (these facts are propositions in Lee's book). $\endgroup$ – Rafael Mrđen Mar 15 '13 at 20:39

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