# Show that the Lie algebra homomorphism $f$ cannot be induced by a smooth Lie group homomorphism.

Consider the Lie algebras $$\mathfrak {so} (3)$$ and $$\mathfrak {su} (2)$$ and let $$f : \mathfrak {so} (3) \longrightarrow \mathfrak {su} (2)$$ be the Lie algebra homomorphism given by $$f \left (\begin{pmatrix} 0 & x & y \\ -x & 0 & z \\ -y & -z & 0 \end{pmatrix} \right ) = \frac {1} {2} \begin{pmatrix} iz & y - ix \\ -y - ix & -iz \end{pmatrix}$$ $$x,y,z \in \mathbb R.$$ Show that there does not exist any smooth Lie group homomorphism $$F : SO (3) \longrightarrow SU(2)$$ such that $$F(\exp (X)) = \exp (f(X)),$$ for all $$X \in \mathfrak {so} (3).$$

If we can show that then by Lie's theorem it yields an alternative way of showing that the matrix Lie group $$SO(3)$$ is not simply connected. But I am having hard time showing the non-existence of such a smooth $$F.$$ Could anyone help me out here?

Suppose there exists such a homomorphism $$F$$. Then take $$X=2\pi \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$

Note that $$\text{Exp}(X)=I$$ and since $$F$$ is a homomorphism $$F(I)=I'$$ (identity in $$SU(2))$$.

On the other hand $$f(X)=\pi\begin{pmatrix} 0 & -i \\ -i & 0 \\ \end{pmatrix}$$

And $$\text{Exp}(f(X))=-I'$$

• Why $\exp (X) = I$? Note that $\exp$ is locally diffeomorphic near $0$ and hence so is $\log$ near $I.$ So if you take log in both sides it follows that $X = 0.$ So doesn't it imply that no non-zero matrix exponentiates to $I.$ Aug 22, 2022 at 7:38
• @AnilBagchi But this isn't near $0$. Consider the usual complex exponential/logarithm: $\exp(2\pi i) = 1$. All you've found is that the only answer near to $0$ is $0$ but $\exp$ is distinctly not injective so there are other answers. Indeed you could argue the reason the answer here works is the lack of injectivity of $\exp$ Aug 22, 2022 at 7:48