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I have a question about the following bit in my lecture notes:

Let $\iota_1: H_1\to G$ and $\iota_2:H_2\to G$ be injective Lie group homomorphisms with $\iota_1(H_1)=\iota_2(H_2)$. Write $\mathfrak h_i$ for the Lie algebra of $H_i$ $(i=1,2)$. Then $$ T_e(\iota_1)(\mathfrak h_1)=T_e(\iota_1(H_1))\stackrel{?}{=}T_e(\iota_2(H_2))=T_e(\iota_2)(\mathfrak h_2). $$

I don't see why the middle equality holds. I do know that $H_1$ and $H_2$ are immersed submanifolds (since $\iota_i$ is an injective Lie group morphism), but then the smooth structure on $\iota_1(H_1)$ doesn't need to coincide (a priori!) with the smooth structure of $\iota_2(H_2)$ (this is in fact what's being proved here). So why does it make sense to compare $T_e(\iota_1(H_1))$ and $T_e(\iota_2(H_2))$ when we don't know yet that the underlying manifolds have the same structure?

Edit: $T_e(\iota_1(H_1))$ and $T_e(\iota_2(H_2))$ can be compared, since $\iota_1(H_1)$ and $\iota_2(H_2)$ are immersed submanifolds of $G$, so we can consider their tangent spaces as subspaces of $T_e(G)$. However, my question remains why we have an equality.

Sha Vuklia
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  • You are comparing tangent spaces of the images, using the smooth structure of the group $G$. My suggestion is to forget about Lie groups and work out a proof for embeddings of smooth manifolds. Prove that if $f_i: N_i\to M$ are smooth embeddings with equal images then for all $x_i\in N_i$ with $f_1(x_1)=f_2(x_2)$, $Im(T_{x_1}(f_1))= Im(T_{x_2}(f_2)$. – Moishe Kohan Oct 02 '24 at 17:49
  • @MoisheKohan Thank you, I agree, I will edit my question. The proof for embeddings is fine. I'm still interested in immersions (of Lie subgroups). Btw, I know that the standard way of proving it is showing that Lie subgroups are weakly embedded, but I am curious about this specific proof. – Sha Vuklia Oct 02 '24 at 20:03
  • Doesn't $\iota_1(H_1) = \iota_2(H_2)$ imply $T_e(\iota_1(H_1)) = T_e(\iota_2(H_2))$? – Smiley1000 Oct 02 '24 at 20:15
  • Aren't they both endowed with the submanifold structure from $G$? – Smiley1000 Oct 02 '24 at 20:19
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    The trick is to note that an immersion is a local embedding. There is a little bit more to it of course, but just a bit. – Moishe Kohan Oct 02 '24 at 20:48
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    Ok, another hint: Use the Baire Category Theorem to prove first that for a closed neighborhood of identity, $B\subset H_1$, the subset $i_2^{-1}i_1(B)\subset H_2$ has nonempty interior. Then, use group action to conclude that WLOG, $1_{H_2}$ is in that interior. Incidentally, this is where you use the assumption that Lie groups are 2-nd countable; without this assumption theorem fails. :) – Moishe Kohan Oct 05 '24 at 21:57
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    For an alternative proof, see my answer here. – Moishe Kohan Oct 05 '24 at 22:11

1 Answers1

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Note: I initially constructed a false proof, as was pointed out by Jack Lee in the comments. Using Moishe's hints, I believe I've now constructed a correct proof. I've deleted the incorrect proof, but it can still be found in the version history of this post (in case someone wants to make sense of the comments below this post). Without further ado, here is the proof:

Let $B\ni e_1$ be a compact neighbourhood in $H_1$. Then $\iota_1(B)$ is compact, hence closed in $G$ (since $G$ is Hausdorff). Assume for a contradiction that $\iota_2^{-1}(\iota_1(B))$ has empty interior. There exists a countable cover of $H_1$ of the form $(h_n\cdot B)$, and $$ \iota_2^{-1}(h_n\cdot B)=\iota_2^{-1}(\iota_1(h_n))\iota_2^{-1}(B) $$ still has non-empty interior. Since $H_2$ is a Baire space (as we're assuming manifolds to be second-countable) it follows that $$ H_2=\bigcup_n \iota_2^{-1}(h_n\cdot B) $$ has empty interior, which is absurd. We conclude that $\iota_2^{-1}(\iota_1(B))$ has non-empty interior.

Let $h\in B$. Then $h^{-1}B$ is a neighbourhood of the identity of $H_1$. Note that $$ B':=\iota_2^{-1}(\iota_1(h^{-1}B))=\iota_2^{-1}(h^{-1}\cdot\iota_1(B))=\iota_2^{-1}(h^{-1})\cdot\iota_2^{-1}(\iota_1(B)) $$ is a neighbourhood of $e_2$ such that $\iota_2(B')\subset\iota_1(B)$. We therefore find that $$ T_e(\iota_1(H_1))\subset T_2(\iota_2(H_2)). $$ We are done by switching the roles of $H_1$ and $H_2$ (or noting that equality needs to hold for dimension reasons).

Sha Vuklia
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  • This argument doesn't work. Your mistake is the statement that $F_2(U_2)$ is open in $M$. To see what can go wrong, consider the case of the figure-eight in the plane (Example 4.19 in my Introduction to Smooth Manifolds). It's the image of two different injective immersions whose tangent spaces at the origin do not coincide. In the case of Lie groups, you have to use the fact that a Lie subgroup is a weakly embedded submanifold. – Jack Lee Oct 05 '24 at 18:35
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    To see an illustration of the two different immersions whose image is the figure-eight curve, see this answer. – Jack Lee Oct 05 '24 at 18:45
  • @JackLee Wow, a comment from one of my favourite authors... I'm very indebted to you, sir! Regarding my post: thank you for pointing out my mistake. The counter-example with the figure-eight curve is also very insightful. The proof I am going through is pretty elaborate, which makes me think it tries to avoid using that Lie subgroups are weakly embedded (if we can use that fact, then the proof is a one-liner!) Could be a slip-up. – Sha Vuklia Oct 05 '24 at 20:28