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I know that the real projective space $\mathbb{R}P^3$ is diffeomorphic to the Lie group $SO(3)$. But I don't know how to demonstrate that this diffeomorphism preserves the structure of Lie group.

amWhy
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Robson dos Santos Silva
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    This question is only meaningful if you already have a Lie group operation defined on $\Bbb R P^3$ in the first place, and typically if you write down such an operation you're doing so in terms of a representation of $SO(3)$. Once we have such an operation, one needs to specify what the diffeomorphism $\Bbb R P^3 \to SO(3)$ is before one can ask whether it is a Lie group isomorphism. – Travis Willse Feb 28 '20 at 23:26
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    One the other hand, if the question is, how can I construct a Lie group operation $\ast$ of $\Bbb R^3$, then you can do so by conjugating the operation $\cdot$ on $SO(3)$ by a diffeomorphism $\Phi : \Bbb R P^3 \to SO(3)$, i.e., declaring $\ell \ast \ell' := \Phi^{-1}(\Phi(\ell) \cdot \Phi(\ell'))$. It's then a matter of unwinding definitions to show that $\ast$ defines a Lie group structure. – Travis Willse Feb 28 '20 at 23:29
  • @TravisWillse would like to thanks your answer! I agree with your answer. – Robson dos Santos Silva Feb 28 '20 at 23:39
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    Does my second comment answer your question then? If so I can write up a proper answer below for the sake of resolving this question. – Travis Willse Feb 29 '20 at 02:24
  • Yes, this is my question – Robson dos Santos Silva Mar 02 '20 at 16:59

2 Answers2

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The sphere $S^3$ is a Lie group (its group structure is given by quaternion multipication). $\mathbb RP^3$ is obtained as the quotient Lie group $S^3/\{1,-1\}$.

The map $$\phi : S^3 \to SU(2), \phi(x_1,x_2,x_3,x_4) = \begin{pmatrix}x_1 +i x_2 & x_3+ix_4\\-x_3+ix_4 & x_1-ix_2 \end{pmatrix}$$ is known to be an isomorphism of Lie groups. But $SO(3)$ is isomorphic to the quotient Lie group $SU(2)/\{I,-I\}$, where $I$ denotes the identity matrix. See how to show $SU(2)/\mathbb{Z}_2\cong SO(3)$ .

Obviously $\phi(\{1,-1\}) = \{I,-I\}$, thus $\phi$ induces an isomorphism of Lie groups $$\mathbb RP^3 \to SO(3) .$$

See also Can anyone explain how the complex matrix representation of a quaternions is constructed?

Paul Frost
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As I mentioned briefly in the comments, answering the question in the title (showing that $\Bbb R P^3$ admits a Lie group structure) requires only what OP already knows: That $\Bbb R P^3$ and $SO(3)$ are diffeomorphic.

Given any diffeomorphism $$\Phi : \Bbb R P^3 \to SO(3) ,$$ we can "pull back" the group operation $\cdot$ on $SO(3)$ to a group operation $\ast$ on $\Bbb R P^3$ making $\Phi$ a Lie group isomorphism, that is, define $$\ell \ast \ell' := \Phi^{-1}(\Phi(\ell) \cdot \Phi(\ell')) .$$ With this definition, it's mostly a matter of unwinding definitions to verify that $\ast$ in fact defines a Lie group structure on $\Bbb R P^3$, that is, that the multiplication and inverse maps are smooth.

That said, constructing an explicit isomorphism $\Phi$ (and using it to write down an explicit formula for an operation $\ast$) could be an instructive exercise.

(NB this construction is very general: Given any bijection $X \to Y$ we can pull back any operation on $Y$ to an operation on $X$ making the bijection an isomorphism of sets endowed with operations.)

Travis Willse
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