# $\{\gamma'(0)\,\vert\, \gamma \in C^\infty(\Bbb R,G), \gamma(0) = I_n\}$ is a $\Bbb R$-vector space where $G$ is a closed subgroup of $GL_n(\Bbb C)$

Problem. Let $$G$$ be a closed* subgroup of $$GL_n(\mathbb C)$$. Define $$V:= \{\gamma'(0)\,\vert\, \gamma \in C^\infty(\Bbb R,G), \gamma(0) = I_n\}$$ Show that $$V$$ is a vector space over $$\mathbb R$$. Furthermore, determine $$V$$ explicitly when $$G = GL_n(\Bbb C)$$ and $$SL_n(\Bbb C)$$.

*The topology on $$GL_n(\mathbb C)$$ is due to the identification $$M_n(\mathbb C) \cong \mathbb C^{n^2}$$.

The first task is to show that $$V$$ is a real vector space.

1. Consider the constant map $$\gamma:\Bbb R\to G$$, $$t\mapsto I_n$$. Then, $$\gamma(0) = I_n$$ and $$\gamma'(0) = \mathbf{0}_n \in V$$. Note that $$I_n \in G$$ since subgroups preserve identities. So, $$\mathbf{0} := \mathbf{0}_n \in M_n(\Bbb C)$$ is the zero element of the vector space.

2. In addition, we must check the vector space axioms. After @KReiser's hint, I have shown that $$V$$ is closed under scalar multiplication. Suppose $$\gamma'(0) \in V$$ for some $$\gamma \in C^\infty(\Bbb R,G)$$ with $$\gamma(0) = I_n$$. Define $$\gamma_a \in C^\infty(\Bbb R,G)$$ for $$a\in R$$, by $$\gamma_a(t):= \gamma(at)$$. $$\gamma_a$$ satisfies $$\gamma_a(0) = \gamma(0) = I_n$$. Then, $$\gamma_a'(t) = a\gamma'(at)$$ implying $$\gamma_a'(0) = a\gamma'(0) \in V$$ for every $$a\in \Bbb R$$. For $$\gamma_1'(0),\gamma_2'(0) \in V$$, consider $$\gamma := \gamma_1\gamma_2 \in C^\infty(\Bbb R,G)$$ to conclude $$\gamma_1'(0) + \gamma_2'(0) \in V$$, i.e. $$V$$ is closed under addition. This was suggested by @José Carlos Santos.

3. For $$G = GL_n(\Bbb C)$$, I can show that $$V = M_n(\Bbb C)$$. $$V\subset M_n(\Bbb C)$$ is obvious. Take $$A\in M_n(\Bbb C)$$, and $$\gamma(t):= e^{tA}$$. Then, $$\gamma(0) = I_n$$ and $$\gamma'(0) = A \in V$$. So, $$V = M_n(\Bbb C)$$.

Thank you!

• Would would the range of $t\gamma_1+(1-t)\gamma_2$ be a subset of $G$? Jan 17, 2022 at 9:24
• Right, not necessarily - so maybe that was a bad idea. I've edited. Jan 17, 2022 at 9:26
• The vector space axioms you need to check are additivity and multiplication by scalars, and you're working with a derivative. Can you think of what you have to put in to a derivative to get a sum or a scalar multiple out? Jan 17, 2022 at 9:26
• @KReiser I'm not sure I understand your hint, but in general, we do know that $\frac{d}{dt} f(ct) = c f'(ct)$ and $\frac{d}{dt} (f(t) + g(t)) = f'(t) + g'(t)$. Did you mean something else? Jan 17, 2022 at 9:30
• @HennoBrandsma Could you elaborate? For the $\mathbf 0$ element of $V$, consider the constant map $\gamma:\Bbb R\to G$, $\gamma(t) = I_n$. Then, $\gamma(0) = I_n$ and $\gamma'(0) = \mathbf{0}_n \in V$. Jan 17, 2022 at 9:55

As I briefly said in my comment, you have to find a condition on $$\gamma'(0)$$

For example let $$G = SL_n(\mathbb C)$$ and $$\gamma : \mathbb R \to G$$ that is smooth.

What you have is $$t \mapsto \det(\gamma(t))$$ that is a constant function over $$\mathbb R$$. Its derivative is given by $$t \mapsto tr\left(\gamma'(t)\cdot {}^t Co(\gamma(t))\right)=tr\left(\gamma'(t)\cdot \gamma^{-1}(t)\right)$$ This function as the derivative of a constant function vanishes. In particular you have : $$tr \left( \gamma'(0) \right) = 0$$

Conversely for $$A$$ in $$M_n(\mathbb C)$$ and for $$t$$ in $$\mathbb R$$ : if $$tr(A) = 0$$ you have $$\det\left(e^{tA} \right) = e^{t \cdot tr(A)} = 1$$

In the end you have $$V = \{A \in M_n(\mathbb C), \quad tr (A) =0 \}$$

• Thank you! I have accepted your answer; and edited the question statement to exclude the other groups. I will try to figure those out on my own (and I will make a separate post if I need help)! Jan 18, 2022 at 16:45

If $$v_1=\gamma_1'(0)$$ and $$v_2=\gamma_2'(0)$$, then\begin{align}(\gamma_1\gamma_2)'(0)&=\gamma_1(0)\gamma_2'(0)+\gamma_1'(0)\gamma_2(0)\\&=\gamma'_1(0)+\gamma'_2(0)\\&=v_1+v_2.\end{align}And, if $$\lambda\in\Bbb R$$ and $$\eta(t)=\gamma_1(\lambda t)$$, then$$\eta'(0)=\lambda\gamma_1'(0)=\lambda v_1.$$

Note that 1 is superfluous if $$V \neq \emptyset$$, as it follows from 2. 1 can basically be seen as the proof of non-emptyness.

And for sum check explicitly by computing $$(\gamma_1\gamma_2)'(0)= \gamma_1'(0)\gamma_2(0) + \gamma_1(0)\gamma'_2(0) = \gamma_1'(0)I_n+ I_n\gamma_2'(0) = \gamma_1'(0)+\gamma_2'(0)$$ to be really complete.

• Yes, I already did this (also suggested by José Carlos Santos). I believe that completes the proof that $V$ is a vector space. Do you have any ideas for the explicit computation aspect of the problem? Jan 17, 2022 at 10:15
• @delta-divine No, I haven't thought about that part. Try $n=2$ first, as $n=1$ is too trivial..(we just get $\Bbb R$) Jan 17, 2022 at 10:16
• Alright, I'll try that! Jan 17, 2022 at 10:19