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I have seem to have two different (but maybe equivalent) notions of what a regular based covering map is. The two notions I have are:

If $p : (\tilde{X},\tilde{v}) \rightarrow (X,v)$ is a covering map. Then $p$ is regular if and only if any two points in $p^{-1}(v)$ differ by a covering transformation.

or

If $p : (\tilde{X},\tilde{v}) \rightarrow (X,v)$ is a covering map. Then $p$ is regular if for any two points $\tilde{v}_{1}, \tilde{v}_{2}$ in $p^{-1}(v)$, a loop $\ell$ based at $v$ in $X$ lifts to a loop $\tilde{\ell}$ in $\tilde{X}$ based at $\tilde{v}_{1}$ if and only if it lifts to a loop in $\tilde{X}$ based at $\tilde{v}_{2}$.

Then, we know that $(\tilde{X},\tilde{v})$ is a regular covering space of $(X,v)$ if and only if $p_{*}(\pi_{1}(\tilde{X},\tilde{v}))$ is a normal subgroup of $\pi_{1}(X,v)$.

I feel like I have an example of a normal subgroup of $F_{2}$ (the free group of two generators), whose unique corresponding regular covering space of $B_{2}$ (the bouquet of two circles) satisfies the second notion of regularity, but not the first. I've spelt out the example here, and I would appreciate it if someone could help me see where I'm going wrong.


Let $(X,v) = (B_{2},v) $ have the following cell complex:

enter image description here

Then, consider the map $\phi : F_{2} \rightarrow S_{3} $ such that $x \mapsto (1 \ 2)$ and $y\mapsto (1 \ 2 \ 3)$. Then $\phi$ is a surjective group homomorphism, so $K = \operatorname{ker}(\phi)$ is an index 6 normal subgroup of $F_{2}$.

Then let $ p : (\tilde{X},\tilde{v}) \rightarrow (X,v)$ be the unique corresponding covering map such that $p_{*}(\pi_{1}(\tilde{X},\tilde{v})) = K$. Then a loop $\ell$ based at $v$ in $X$ lifts to a loop $\tilde{\ell}$ based at $\tilde{v}$ in $\tilde{X}$ if and only if $[\ell] \in K$. Then, the pre-images of the relations of $S_{3}$ are elements of $K$, and since since $\phi$ maps the two generators of $F_{2}$ to the two generators of $S_{3}$, these pre-images are trivial to calculate.

If we equip $S_{3}$ with the presentation $\left< \tau, \sigma \mid \tau^{2}, \sigma^{3}, \tau\sigma\tau\sigma \right>$, where $\tau = (1 \ 2)$ and $\sigma = (1 \ 2 \ 3)$. Then we see that $K$ inherits the elements $x^{2}$, $y^{3}$, and $xyxy$. Hence, the loops $x^{2}, y^{3},$ and $xyxy$ all lift to $(\tilde{X},\tilde{v})$.

Then $(\tilde{X},\tilde{v})$ is a regular graph on six vertices, with 4 edges leaving every vertex, such that $x^{2}$ is a loop at every vertex, $y^{3}$ is a loop at every vertex, and $xyxy$ is a loop at every vertex. I think the following graph fits the requirements:

enter image description here

I think this graph satisfies the second notion of regularity. I cannot find a loop at one vertex that is not a loop at another vertex. However, I can't seem to describe a covering transformation that swaps any of the our vertices with any of the inner ones.


I would appreciate it if anyone could help me see my mistake.

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Let the outer vertices be listed in clockwise order starting from the upper right as $P_1, P_2, P_3$, and let the inner vertices be listed in clockwise order starting from the upper right as $Q_1,Q_2,Q_3$.

I'll describe the deck transformation $f : \tilde X \to \tilde X$ that interchanges $P_1$ and $Q_1$: $$f(P_1)=Q_1, \quad f(Q_1)=P_1 $$ Since $y$-edges go to $y$-edges, and since there are $y$-edges $P_1 \mapsto P_2$ and $Q_1 \mapsto Q_3$, I conclude that $$f(P_2)=Q_3, \quad f(Q_3)=P_2 $$ Since $x$-edges go to $x$-edges, and since there are $x$ edges $P_2 \mapsto Q_2$ and $Q_3 \mapsto P_3$, I conclude that $$f(Q_2)=P_3, \quad f(P_3)=Q_2 $$ This defines $f$ on all the vertices, and I have only left to verify that each $x$ edge or $y$-edge which I have not yet checked goes to an $x$ edge or $y$-edge respectively.

For example, I have not yet checked the $y$-edge $P_2 \mapsto P_3$: it goes to the $y$-edge $Q_3 \mapsto Q_2$. And so on.

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  • $\begingroup$ Thanks a lot for this! $\endgroup$ – Adam Higgins May 27 '18 at 7:20
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This is not what you're asking, but let me at least tell you why the two definitions are equivalent because of the lifting theorem:

$(1)\Rightarrow(2)$ Pick any $v_1,v_2\in p^{-1}(v)$, so there is a deck transformation $D$ taking $v_1$ to $v_2$, then by the lifting lemma we get $p_\#(\pi_1(\tilde X,v_1))=p_\#(\pi_1(\tilde X,v_2))$, but this equality means that any loop based at $v$ lifts to a loop based at $v_1$ iff it lifts to a loop based at $v_2$

$(2)\Rightarrow(1)$ Condition $(2)$ is equivalent to $p_\#(\pi_1(\tilde X,v_1))=p_\#(\pi_1(\tilde X,v_2))$ for all $v_1,v_2\in p^{-1}(v)$, so by the lifting theorem we get there is some deck transformation taking $v_1$ to $v_2$ for all $v_1,v_2\in p^{-1}(v)$ and thus we get $(1)$.

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  • $\begingroup$ Hey! This wasn’t the question, but it was super useful to see so thanks so much! $\endgroup$ – Adam Higgins May 27 '18 at 7:17

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