# Slowly increasing smooth mappings with values in a Lie group?

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 $$$$x \mapsto \lVert D_x F \rVert \text{ is polynomially bounded}$$$$ 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 $$\operatorname{U}(1)$$.

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.

• 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? Commented Jul 16 at 22:44
• $d\iota: G \to \Bbb R^\ell$ is a continuous function with compact domain.
– mme
Commented Jul 17 at 0:18
• @mme I am confused about the domain and range of your differential. Are you thinking pointwise on $G$ or the whole tangent bundle? Commented Jul 17 at 9:48
• @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.
– o r
Commented Jul 17 at 15:30
• 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).
– o r
Commented Jul 17 at 17:36