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The universal cover of $SO(2)$ is $\mathbb{R}$, whilst the fundamental group is $\mathbb{Z}$. That is $$ SO(2) \cong \mathrm{universal\ cover}/\pi_1 $$ Likewise, I believe that the universal cover of $SO(n)$ for $n \geq 3$ satisfies $$ SO(n) \cong \mathrm{universal\ cover}/\mathbb{Z}_2$$ and indeed $\pi_1(SO(n)) = \mathbb{Z}_2$. So $SO(n)$ is the quotient of its covering group by its fundamental group for all $n \geq 2$. I was wondering if this was part of a more general result? I have very little knowledge of algebraic topology so a basic answer would be appreciated. Thanks.

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Yes it is. Let $X$ be a topological space with universal cover $\widetilde{X}$ and covering map $p : \widetilde{X} \to X$. A homeomorphism $f : \widetilde{X} \to \widetilde{X}$ is called a Deck transformation of $p$ if $p\circ f = p$; that is, $f$ preserves the fibres of $p$ so if $y \in p^{-1}(x)$, $f(y) \in p^{-1}(x)$. The set of all Deck transformations of $p$ is denoted $\operatorname{Deck}(p)$ and forms a group under composition. The quotient of $\widetilde{X}$ by $\operatorname{Deck}(p)$ is $X$. Moreover, $\operatorname{Deck}(p) \cong \pi_1(X)$.

A good reference for this material is Hatcher's Algebraic Topology.

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  • $\begingroup$ There's a bit of a subtlety here that I'm curious about...can the group of deck transformations be realized as a subgroup of the covering space? $\endgroup$ – Matt Samuel Jan 19 '16 at 23:40
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    $\begingroup$ @MattSamuel Yes it is isomorphic to the kernel of the covering map and the action of the deck group is the action of (left or right) multiplication. The subgroup is actually necessairly central (hence both left and right multiplication gives the same action). $\endgroup$ – PVAL-inactive Jan 20 '16 at 6:14
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If $H$ is a topological group which is both path-connected and locally path-connected (i.e. a connected Lie group such as $SO(n)$), then any path-connected cover of $H$ inherits a unique group structure making the covering map a group homomorphism. In fact for any such cover $p:G \to H$,we have $ker(p) \cong \pi_1(H)/p_*(\pi_1(G))$. This generalizes the statements in the question (when $\pi_1(G)=0$). (Here $p_*$ denotes the induced map on fundamental groups coming from $p$.)

See https://en.wikipedia.org/wiki/Covering_group for an outline of the construction.

For $SO(n)$, the universal covering group is called the spin group and denoted as $\operatorname{Spin}(n)$. It shows up throughout topology, geometry and physics.

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