# Injectivity of a Lie Bracket Isomorphism

I am having an issue with a Lie bracket isomorphism.

My issue lies with a). I have to show that $$\phi$$ is a bijective linear map. As a mapping $$\Phi: \mathbb{R}^3 \to \Phi(\mathbb{R}^3)$$ is certainly surjective. In order to show injectivity I have to take $$(a_1,b_1,c_1),(a_2,b_2,c_2) \in \mathbb{R}^3$$ and suppose $$\Phi(a_1,b_1,c_1)=\Phi(a_2,b_2,c_2)$$. This implies

$$(a_1-a_2)X+(b_1-b_2)Y+(c_1-c_2)Z=0$$

If $$X,Y,Z$$ were linearly independent injectivity would follow immediately. So we have to show linear independence. Consider

$$0=\lambda_1X+\lambda_2Y+\lambda_3Z =(\lambda_3 y-\lambda_2 z) \frac{\partial}{\partial x} +(\lambda_1 z-\lambda_3 x) \frac{\partial}{\partial y} +(\lambda_2 x-\lambda_1 y) \frac{\partial}{\partial z}$$

which leads to the three equations

$$\lambda_3 y-\lambda_2 z=0 \\ \lambda_1 z-\lambda_3 x=0 \\ \lambda_2 x-\lambda_1 y=0$$

and therefore

$$\lambda_1 (z-y)+\lambda_2 (x-z)+\lambda_3 (y-x)=0$$

Unfortunately, the equation above also holds if $$\lambda_1=\lambda_2=\lambda_3=1$$, so my strategy fails. So I am confused about how to show injectivity.

The thing is that there is no equivalence between the three equations you have found out and their sum, just an implication. What you have shown is that $$(1,1,1)$$ is a potential solution, but you do not have shown that it really is one.

Clearly, $$\Phi$$ is linear. Suppose that $$\Phi(a,b,c) = 0$$: in other words, assume that $$\forall (x,y,z)\in \Bbb R^3, \quad \begin{cases} cy-bz &= 0\\ az - cx &= 0\\ bx - ay &= 0 \end{cases}$$ Alternatively setting $$(x,y,z) = (1,0,0)$$, and $$(x,y,z) = (0,1,0)$$, we find that a solution $$(a,b,c)$$ has to satisfy $$\begin{cases} -c &= 0,\\ b &= 0, \end{cases} \text{ and } \begin{cases} c &= 0,\\ -a &= 0\end{cases}.$$ Hence, it is mandatory that $$a=b=c=0$$. The converse is obviously true since $$\Phi$$ is linear, and therefore, $$\Phi(a,b,c) = 0 \iff (a,b,c) = 0$$, and $$\Phi$$ is injective.

$$\Phi$$ is linear so it is enough to show that $$\ker \Phi=0.$$ Now, $$aX+bY+cZ=0$$ leads to

$$\tag 1 cy-bz=az-cx=bx-ay=0.$$

If the claim is false, then we should be able to find a non-zero vector $$(a,b,c)$$ such that $$(1)$$ is true $$\textit{for all}$$ values of $$(x,y,z).$$

Toward a contradiction then, without loss of generality, $$a=1.$$ Then, $$y=bx$$ and $$z=cx$$.

But, choosing $$\vec x=(1,1,1)$$ implies that $$b=c=1$$ whereas choosing $$\vec x=(1,0,0)$$ gives $$b=c=0,$$ so in fact there is no such vector $$(a,b,c).$$