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Let $G$ be $l$-dimensional compact Lie group and consider any smooth $F : \mathbb{R}^n \to G$.

Then, the first-order derivative of $F$ at each $x \in \mathbb{R}^n$ can be regarded as a linear mapping $D_x F : \mathbb{R}^n \to \mathbb{R}^l$. Let us fix a matrix norm $\lVert \cdot \rVert$ and assume that \begin{equation} x \mapsto \lVert D_x F \rVert \text{ is polynomially bounded} \end{equation} We can extend the above notion of polynomial boundedness to higher order derivatives as well.

Now, let us define a set $\mathfrak{G}$ as the set of $F : \mathbb{R}^n \to G$ such that $F$ is smooth and all its derivatives are polynomially bounded in the above sense.

Now, my question is that

Is this $\mathfrak{G}$ a group under pointwise multiplication and inverse?

That is,

  1. For any $F_1, F_2 \in \mathfrak{G}$ and $x \in \mathbb{R}^n$, let $(F_1 F_2)(x) := F_1(x) F_2(x)$. Then, do we have $F_1 F_2 \in \mathfrak{G}$?

  2. For any $F \in \mathfrak{G}$ and $x \in \mathbb{R}^n$, let $F^{-1}(x) := [F(x)]^{-1}$. Then, do we have $F^{-1} \in \mathfrak{G}$.

I think the item 1 must be true, but not sure about item 2. For example, can a derivative of $F$ decay exponentially at infinity, making a derivative of $F^{-1}$ "grow exponentially" at infinity?

Could anyone please clarify for me? If I am wrong about $\mathfrak{G}$ being a group, what additional conditions should I impose to make it into a group?

Add) I think this is indeed the case for $\operatorname{SU}(N)$ with $N \geq 2$ or any direct product of them. Still, I haven't figured out for general $G$.

Add 2) This ME post shows that my guess is also true for $\operatorname{U}(1)$.

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1 Answer 1

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Note that the maps $$\iota: G\to G,\ x\mapsto x^{-1}\qquad \mu:G\times G\to G,\ (x,y)\mapsto x\cdot y$$ have bounded derivatives, since you are assuming $G$ to be compact.

Then for $F_1, F_2\in\mathfrak G$ you have that the derivatives of $$F_1^{-1}=\iota\circ F_1,\qquad F_1\cdot F_2 = \mu\circ (F_1,F_2)$$ will have polynomial growth, being the composition of a map whose derivative has polynomial growth and a map with bounded derivative.

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  • $\begingroup$ I do not clearly see how we can use compactness of $G$ to show that $\iota: G\to G$ has bounded derivatives. Could you provide more details? $\endgroup$
    – Isaac
    Commented Jul 16 at 22:44
  • $\begingroup$ $d\iota: G \to \Bbb R^\ell$ is a continuous function with compact domain. $\endgroup$
    – mme
    Commented Jul 17 at 0:18
  • $\begingroup$ @mme I am confused about the domain and range of your differential. Are you thinking pointwise on $G$ or the whole tangent bundle? $\endgroup$
    – Isaac
    Commented Jul 17 at 9:48
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    $\begingroup$ @Isaac the norms $\|d\iota\|$ and $\|d\mu\|$ are continuous $\Bbb R$-valued functions on $G$ and $G\times G$. Compactness of $G$ guarantees that their image is bounded. $\endgroup$
    – o r
    Commented Jul 17 at 15:30
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    $\begingroup$ Yes, I mean that function. The argument can be applied to show the boundedness of the norms of the higher order derivatives of $\iota$ and $\mu$ as well (higher derivatives are usually described via "jets" in differential geometry). $\endgroup$
    – o r
    Commented Jul 17 at 17:36

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