# Redundance of the Smoothness of the Inversion Map in the Definition of a Lie Group.

$$\DeclareMathOperator{\inv}{inv}$$ I am trying to understand the proof of the following from this document:

Let $$M$$ be a smooth manifold which admits a group structure such that the multiplication map $$m:G\times G\to G$$ defined as $$m(g, h)=gh$$ for all $$g, h\in G$$ is a smooth map. Then the inversion map $$\inv:G\to G$$ defined as $$\inv(g)=g^{-1}$$ for all $$g\in G$$ is a smooth map.

The proof in the document proceeds as follows: First note that the multiplication map is a constant rank surjective smooth map, and is therefore a submersion. Therefore, the level set $$\Delta=m^{-1}(e)$$ of $$e\in G$$, $$e$$ being the identity of the group structure, is an embedded submanifold of $$G\times G$$ of dimension $$n$$.

Now let $$\pi_1:G\times G\to G$$ be the projection on the first coordinate. Consider the map $$\pi_1\circ i:\Delta\to G$$, where $$i:\Delta\to G\times G$$ is the inclusion map, which we know is a smooth embedding.

It is claimed in the document that $$\pi_1\circ i$$ is a diffeomorphism. The reasoning given seems to be this: The map $$\pi_1\circ i$$ is smooth and is a homeomorphism and hence by the inverse function theorem, it is also a diffeomorphism.

This seems to suggest that any smooth homeomorphism is a diffeomorphism. But this is not true since the map $$x\mapsto x^3:\mathbf R\to \mathbf R$$ is a smooth topological embedding but not an immersion at $$x=0$$.

How do I show that that composition $$\pi_1\circ i$$ is a diffeomorphism?

• The author of the note is not suggesting that every smooth homeomorphism is a diffeomorphism. The phrase "by the inverse function theorem" is shorthand for the following argument: Since $\pi_1\circ\iota$ is a smooth bijection whose differential is everywhere nonsingular, it has an inverse (by bijectivity), and its inverse is smooth in a neighborhood of each point (by the inverse function theorem), so the inverse is smooth. The note didn't give details of the proof that the differential is nonsingular -- can you fill that in? Commented May 27, 2015 at 19:29
• Thanks. I am having trouble seeing how the differeintial of $\pi_1\circ i$ is surjective at each point of $\Delta$. Can you please give me a hint? Commented May 27, 2015 at 19:32
• Since the domain and codomain have the same dimension, you only need to show it's injective. Commented May 27, 2015 at 19:40
• I wasn't easily able to see that the differential of $\pi_1\circ i$ at each point of $\Delta$ is injective. Oliver Begassat has provided a full argument below. Do you have a shorter argument in mind? Thanks. Commented May 27, 2015 at 20:09
• Not really shorter. Another way to go about it is to figure out what $T_x\Delta$ is at each point, and then just compute the action of $d\pi_1|_{\iota(x)}\circ d\iota_x$ on such a vector. Either way, there's some work to be done. Commented May 27, 2015 at 20:20

You are right, the argument isn't conclusive, it's missing a piece, but it's fixable. You need to note that the map $p_1=\pi_1\circ\iota:\Delta'\to G$ has constant rank equal to $\dim G=\dim\Delta'$.

Indeed, let $m=(g,g^{-1})\in\Delta'$ be a point in $\Delta'$, then \begin{array}{cccc} \theta_g&:&G\times G&\longrightarrow & G\times G\\ &&(x,y) & \longmapsto &(gx,yg^{-1}) \end{array} is a diffeomorphism of $G\times G$ and sends $\Delta'$ diffeomorphically to itself (let's call $\widetilde{\theta_g}$ the induced diffeomorphism of $\Delta'$). Then $$p_1\circ\widetilde{\theta_g}=\pi_1\circ\iota\circ\widetilde{\theta_g}=\pi_1\circ\theta_g\circ\iota=L_g\circ\pi_1\circ\iota=L_g\circ p_1$$ (where $L_g$ is left multiplication by $g$). If we calculate the differentials at $m_0=(e,e)\in\Delta'$, we get $$d_{m}p_1\circ d_{m_0}\widetilde{\theta_g}=d_eL_g\circ d_{m_0}p_1$$ Since $\widetilde{\theta_g}$ and $L_g$ are diffeomorphisms, $p_1$ has the same rank at $m\in\Delta'$ as at $m_0\in\Delta'$, so that it has constant rank ($\leq\dim G$). It follows from the constant rank theorem (see the first theorem in these notes) and the bijectivity of $p_1$, that $p_1$ has constant rank equal to $\dim G$, and now you can apply the global inversion theorem.

• Thanks. I am unable to see that $\theta_g:G\times G\to G\times G$ is a diffeomorphism. This is because we do not a priori know that the inverse map is smooth. Can you please address this? Commented May 27, 2015 at 19:38
• You don't need smoothness of the inverse map, you only need smoothness of the maps "left multiplication by some fixed $g$" and "right multiplication by some fixed $h$" (of course you'll take $h=g^{-1}$), which follow from the assumption that multiplication is a smooth map. Commented May 27, 2015 at 19:48
• Ah! My blunder. Reading the proof again. Thanks. Commented May 27, 2015 at 19:49
• I get it now. Great job! Commented May 27, 2015 at 20:07

Here is perhaps a more straightforward proof of your statement. We can realize the inversion as the composition $$\mathrm{inv}=\mathrm{pr_2}\circ F^{-1}\circ \iota$$, where we have defined $$\iota\colon G\rightarrow G\times G$$ as $$\iota(g)=(g,e)$$ and

$$F\colon G\times G\rightarrow G\times G,\quad F(g,h)=(g,gh).$$

Note that the map $$F$$ is a smooth bijection with inverse $$(g,k)\mapsto(g,g^{-1}k)$$, so we would like to show it's a diffeomorphism. By dimensionality and inverse map theorem, it's enough to show it's an immersion at all points.

Suppose $$\mathrm d F_{(g,h)}(v,w)=0$$ for some $$(v,w)\in T_gG\oplus T_hG$$. Since $$\mathrm{pr}_1=\mathrm{pr}_1\circ F$$, we get $$\mathrm d(\mathrm{pr}_1)_{(g,h)}(v,w)=0$$ and hence $$v=0$$. To prove that $$w=0$$, we consider the equality $$m=\mathrm{pr}_2\circ F$$, where $$m$$ denotes the multiplication map; differentiating it yields $$\mathrm d m_{(g,h)}(v,w)=0$$. On the other hand, $$\mathrm d m_{(g,h)}(v,w)=\mathrm d(L_g)_h(w)+\mathrm d(R_h)_g(v)=\mathrm d(L_g)_h(w),$$ hence $$\mathrm d(L_g)_h(w)=0$$, implying $$w=0$$ since $$L_g$$ is a diffeomorphism. So $$F$$ is an immersion at $$(g,h)\in G\times G$$.