# Is it possible for the covering space of a topological space to have a fundamental group isomorphic to the fundamental group of the base?

Let $M$ and $N$ be two connected manifolds, such that $M$ is a covering space of $N$. Is it possible for $\pi_1(M)$ to be isomorphic to $\pi_1(N)$ in some non-canoncial way?

EDIT: Of course, I'm excluding the trivial case where $M=N$ and the covering map is the identity.

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When you say $M = N$ do you mean the case that $M \to N$ is the trivial cover or do you mean $M \cong N$? (I don't consider the latter case trivial in the sense that the existence of a connected manifold that can cover itself nontrivially isn't a completely trivial fact.) – Qiaochu Yuan Feb 7 '12 at 21:08
A more interesting question is finding an example where $M\not\cong N$. – user641 Feb 7 '12 at 23:12
@QiaochuYuan - Yes, sorry. I meant the trivial cover. – Braindead Feb 8 '12 at 3:37
So, can we have a space with fundamental group $\mathbb Q$ and a covering space with fundamental group $\mathbb Q \oplus \mathbb Q$, which is isomorphic to $\mathbb Q$ but in many different ways...??? – GEdgar May 9 '12 at 13:41

Take a double cover $z \mapsto z^2$ of the circle $S^1 \to S^1$.

The underlying group-theoretic question is whether there exist groups with subgroups isomorphic to themselves, and of course there are; above I used the subgroup $2\mathbb{Z}$ of $\mathbb{Z}$, but there are also less trivial examples (for example many subgroups of $F_2$ are isomorphic to $F_2$; in fact $[F_2, F_2] \cong F_{\infty}$). So it remains to find manifolds with appropriate fundamental groups, and as it turns out every finitely presented group is the fundamental group of a $4$-manifold (see Wikipedia).

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Edit As pointed out by Steve D, step (1) of the following is wrong, but easily fixable. First, to fix it, merely change all occurrences of $S^3$ to $S^2$. The key point is that the antipodal map of $S^2$ is orientation reversing, while that of $S^3$ is orientation preserving.

The problem with my calculation where I "prove" $N$ is nonorientable is that I didn't compute in a chart, but rather in the ambient Euclidean space. It's easy to see it's wrong if one simply replaces $S^3$ by $S^1$. Then one can visually see that the corresponding $N$ I construct is not the Klein bottle, but it just $S^1\times S^1$.

End Edit

For a slighty less trivial example, consider $M = S^3\times S^1$ and $N = S^3 \hat{\times} S^1$, the unique nontrivial $S^3$ bundle over $S^1$. (If you want, $N$ can be thought of as the unit sphere bundle in Möbius+rank 3 trivial bundle over $S^1$. Alternatively, I'm thinking of $N$ as $S^3\times [0,\pi]/$~ where we identify $((x,y,z,w),0)$ with $((-x,-y,-z,-w),\pi)$.)

I claim that (1) $N$ is nonorientable (and therefore not even homotopy equivalent to $M$), (2) $M$ double covers $N$, and (3) $\pi_1(M)$ is isomorphic to $\pi_1(N)$.

To see (1), consider the curve $\gamma(t) = ((\cos(t),\sin(t),0,0), t)$ on $S^3\times [0,\pi]$. When projected to $N$, $\gamma$ is a closed curve. The claim is the a basis chosen at $\gamma(0)$ changes orientation coming back to $\gamma(\pi)$. To see this, notice the vector $e_1(t) = ((-\sin(t), \cos(t),0,0),0)$ is always in the tangent space of $S^3\times[0,1]$ at $\gamma(t)$ and likewise so are $e_2(t) = ((0,0,1,0),0)$, $e_3(t) = ((0,0,0,1),0)$ and $e_4(t) = ((0,0,0,0),1)$.

The point is the the differential of the gluing map sends $e_i(\pi)$ to $-e_i(0)$ for $i=1,2,3$ and sends $e_4(\pi)$ to itself. This is a negative determinant transformation, so the orientation has reversed.

Now on to (2). The motivation comes from the picture of the torus double covering the Klein bottle.

Thinking of $M$ as $S^3\times [0,2\pi]/$~ where we identify $((x,y,z,w),0)$ with $((x,y,z,w),2\pi)$, define the map $f:M\rightarrow N$ by identifying $((x,y,z,w),t)$ with $((-x,-y,-z,-w),t+\pi)$ for $t\leq \pi$. Check that $f$ is well defined, really maps onto $N$, is continuous, and really does double cover it. (I'll admit I haven't checked all the details myself).

Finally, (3). The long exact sequence for the homotopy groups of a fibration immediately imply (since $\pi_1(S^3) = 0$), that $\pi_1(N)\rightarrow\pi_1(S^1)$ is an isomorphism (and likewise for $\pi_1(M)$). Thus, $\pi_1(M)\cong \pi_1(N)\cong\mathbb{Z}$.

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This is similar to $S^2\times S^1$ covering $S^2 \hat{\times} S^1$, which is slightly easier to handle since these are the only two closed 3-manifolds with infinite cyclic $\pi_1$. – user641 May 9 '12 at 2:08
@SteveD: Agreed, but I wasn't sure how to show $S^2 \hat{\times} S^1$ and $S^2\times S^1$ were not homotopy equivalent. I think a similar argument to the above shows $S^2\hat{\times} S^1$ is orientable, so that won't work. – Jason DeVito May 9 '12 at 3:42
It is not orientable; it comes from the antipodal map on $S^2$. Explicitly, take $S^2\times I$ and identify the boundary spheres via the antipodal map. – user641 May 9 '12 at 4:51
In fact, I think this case is easier to handle, because $S^2$ is even-dimensional. In particular, if $S^2\hat{\times} S^1$ was orientable, it would be bundle-isomorphic to $S^2\times S^1$. But with $S^2\hat{\times} S^1$, the generator of $\pi_1(S^1)$ acts via inversion on $H_2(S^2)$ (this is by design). – user641 May 9 '12 at 5:11
I got it - the problem is that I didn't use a chart for my computations. Thanks for catching my error - now comes a big edit. – Jason DeVito May 9 '12 at 13:29