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I am new to Lie groups and I am still getting comfortable with the interplay between the group and manifold structures. As an exercise to myself in learning how to show smoothness, let $G$ and $H$ be Lie groups. I would like to show that the constant map $$\phi: G \rightarrow H, \quad g \mapsto e$$ is a Lie group homomorphism, where $e$ is the identity element in $H$.

The map $\phi$ is clearly a group homomorphism: $$\phi(g_1 g_2) = e = ee = \phi(g_1)\phi(g_2).$$ To show smoothness, I need to show that for any chart $(U, \psi)$ containing $e \in H$ and any chart $(V, \varphi)$ of $G$, the map $$\varphi \circ \phi\circ\psi^{-1}: \mathbb{R}^n \rightarrow \mathbb{R}^m$$ is smooth. Since coordinate maps are smooth and function composition preserves smoothness, the above is smooth if and only if $\phi$ is smooth. This is where I am stuck as I've seemingly gone in a circle. What is the standard way of proving smoothness in Lie homomorphisms?

CBBAM
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    The map $\varphi \circ \phi \circ \psi^{-1}$ is literally just a constant map from $\mathbb R^n$ to $\mathbb R^m$, so is smooth because it is infinitely differentiable. Same holds for any smooth manifold. More generally, a map is smooth iff its coordinate representations are smooth, and the coordinate representation of a constant map between manifolds is simply a constant map between $\mathbb R^n$ and $\mathbb R^m$. – Brevan Ellefsen Apr 12 '23 at 22:49
  • @BrevanEllefsen Thank you. I think I was vastly overthinking this problem. – CBBAM Apr 12 '23 at 22:52

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Since coordinate maps are smooth and function composition preserves smoothness, the above is smooth if and only if $\phi$ is smooth. This is where I am stuck as I've seemingly gone in a circle.

Yeah, this is a circular reference. That's because this first sentence is misleading. $\phi$ being smooth means $\varphi \circ \phi\circ\psi^{-1}$ is smooth for all the chart maps. This is the definition of $\phi$ being smooth. But the word "smooth" for $\varphi \circ \phi\circ\psi^{-1}$ is already defined beforehand, because that is a function $\mathbb{R}^n\to\mathbb{R}^m$ between Euclidean spaces. And so the classical definition (i.e. multivariable calculus) of smoothness applies here. In other words this definition extends the term "smooth" from Euclidean case to any manifold.

So how do we show that in your case $\varphi \circ\phi\circ\psi^{-1}:\mathbb{R}^n\to\mathbb{R}^m$ is always smooth? Quite simple. Composing any function with a constant function gives us a constant function. Always. And so $\varphi \circ \phi\circ\psi^{-1}$ is a constant map between $\mathbb{R}^n$ and $\mathbb{R}^m$. And constant maps are well known to be smooth. More generally this shows that every constant map between smooth manifolds is smooth.

On the other hand, it is a more general fact that every continuous group homomorphism between Lie groups is smooth. Which might be of interest for you, see this: Continuous homomorphisms of Lie groups are smooth

freakish
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  • Thank you! In general, is smoothness rarely proved straight from the definition itself? – CBBAM Apr 13 '23 at 17:20
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    @CBBAM I suppose. We prove that a certain class of functions is smooth directly from definition. These include constants, polynomials, power series, etc. And then we use theorems like composition of smooth functions is smooth. Yeah, it is rare to do this directly. – freakish Apr 13 '23 at 17:38
  • Got it, thank you again for all your help. – CBBAM Apr 13 '23 at 17:58