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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?

Thanks for your time.

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

Leading to a contradiction.

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  • $\begingroup$ 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 $I.$ $\endgroup$
    – Anil Bagchi.
    Aug 22, 2022 at 7:38
  • $\begingroup$ @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 $\exp$ $\endgroup$
    – Callum
    Aug 22, 2022 at 7:48
  • $\begingroup$ @AnilBagchi: See the above answer. Also you could calculate it explicitly to confirm it, it should be straightforward (You could also check the Rodrigues formula for a quick route, which is what I did). More generally, the Exp map will not be injective (even onto the connected component) if the Lie group is compact (like in the case of SO(3)); This is because the domain of the Exp map is a vector space (Lie algebra). $\endgroup$
    – Leonid
    Aug 22, 2022 at 12:18

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