I'm working on the Poincaré Homology Sphere $P_3$ and would like to compute it's Homology $H_1$ and fundamental group. I would like to identify it's fundamental group with the binary icosahedral group $I^*$ in view of the representation of $P_3$ as the quotient space $S^3/I^*$.
Also, using the isomorphism of the abelianized fundamental group $\pi_1(P_3)/[\pi_1(P_3),\pi_1(P_3)]$ onto $H_1$ I would like to conclude that $H_1$ is trivial.
Anybody who can help me with this?
[Math] Fundamental group of the Poincaré Homology Sphere
algebraic-topologygeneral-topologygroup-theoryhomology-sphere
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I want to expand on Derek Holt's comment, because it is interesting. I also want to add a little nugget of my own.
Both ideas generalise to the torus of genus $n$, $n>1$.
Firstly, note that your group has a single defining relation. There is a paper of G. Baumslag and Taylor which gives an algorithm to determine the centre of a group with a single defining relation (The centre of groups with one defining relator). That was the nugget of my own... (EDIT: actually, you do not need to go as far as this paper - it follows immediately from the fact that $G$ is generated by more than two elements and from Theorem 5.4 on p203 of Lyndon and Schupp's fine text Combinatorial Group Theory. This theorem says that for $M=\langle a, b, \ldots, c\rangle$ where $G=\langle X; R\rangle$ is a one-relator group, $a, b, \ldots, c\subsetneq X$, then if $g\not\in M$ then $M^g\cap M$ is cyclic (actually, it is trivial, but I won't go into that here...).)
I want to expand on Derek Holt's comment, but sticking with the fact that your group has a single defining relation. So, open Lyndon and Schupp's fine text Combinatorial Group Theory. In it, we discover two things. Firstly, because the relator is not a proper power (that is, not of the form $R^m$ for some $m>1$) our group is torsion free. Secondly, we learn of a very interesting theorem: B. B. Newmann's Spelling Theorem. (A spelling theorem is a theorem which tells us the form of a word which is (non-freely) equal to the identity in the group.)
In Lyndon and Schupp, the theorem is, if a remember correctly, slightly weaker than I need...the version I need is an addition by a chap called Gurevich, and can be found in a paper of Steve Pride and Jim Howie, A spelling theorem for staggered generalized 2-complexes, with applications. I'm sure we could do this a more simple way (we can - see the addendum), but I do quite like this theorem!
In modern language, the original version of B. B. Newmann's Spelling Theorem tells us that a one-relator group with torsion (that is, the relator is a proper power) is Hyperbolic in the sense of Gromov (see, for example, Bridson and Haelfinger's book - type in their names in google and you'll find the book. Alternatively, there are plenty of good introductions to hyperbolic groups - there is one of Jim Howie I quite like). The thing to note about Hyperbolic groups is that they do not contain a copy of $\mathbb{Z}\times\mathbb{Z}$ (indeed, they do not contain any subgroup of the form $\langle a, b; a^{-1}b^ma=b^n\rangle$, a so-called Baumslag-Solitar group. This is in Bridson and Haefinger).
For our uses, the Newmann-Gurevich Spelling Theorem tells us that the group you are looking at is hyperbolic. Here it is,
Theorem: (B. B. Newman, Gurevich)
Let $G=\langle X; R^m\rangle$ where $R$ is a freely and cyclically reduced word over $X^{\pm}$. If $W=_G 1$ then either $W\equiv 1$, $W$ is a cyclic shift of $R$ or $R^{-1}$, or some cyclic shift of $W$ contains two disjoint subwords $S^{m-1}S_0$ and $T^{m-1}T_0$ where $S$ and $T$ are cyclic shifts of $R$ or $R^{-1}$ and $S=S_0S_1$ $T=T_0T_1$ with $S_0$ and $T_0$ containing all the elements of $X$ which appear in $R$.
For us $G=\langle a, b, c, d; [a, b][c, d]\rangle$, and so we can only think about the $S_0$ and $T_0$. However, this is enough. If $W=_G1$ then $W$ contains $a^{-1}b^{-1}abc^{-1}d^{-1}$, or something similar. That is, it contains "most of" the relator. So, we can replace the big slice of the relator with the smaller slice, to get a new word $W_1=_G1$. As $W_1=_G1$ we can apply the theorem again and apply this trick to get a new word $W_2$ with $|W_1|>|W_2|$, and repeat until we get a word which is either the trivial word or the relator $R=[a, b][c, d]$. This algorithm is called Dehn's algorithm, and implies that our group us hyperbolic (in fact, it is equivalent - look up one of the hyperbolic references for this).
Thus, $G$ is torsion-free and hyperbolic. As it is hyperbolic it cannot contain $\mathbb{Z}\times\mathbb{Z}$. As it is torsion-free and does not contain $\mathbb{Z}\times \mathbb{Z}$, it cannot have a centre.
Now, I wrote this all down for two reasons. Firstly, it was good procrastination. Secondly, the groups you are looking at were the "original" hyperbolic groups. You see, sometime in around 1910-1920 Max Dehn came up with a version of B.B. Newman's spelling theorem which worked for these groups, and he applied this algorithm which bears his name (B.B. Newman's spelling theorem is basically a generalisation of Dehn's observation). It wasn't until Gromov came along in the 1980s that people finally understood what was going on with Dehn's algorithm and these surface groups, and the field of Hyperbolic Groups was introduced. The study of hyperbolic groups has turned out to be a rather fruitful field and has led to many important results, most recently the proof of the Virtual Haken Conjecture and Thurston's Virtually Fibering Conjecture.
Addendum: When I was using the strong version of Newman's Spelling Theorem I had a feeling that it was too much; that we didn't have to leave the pages of Lyndon and Schupp and delve into the world of research papers...and I was right! The groups we are looking at (so long as the genus $>1$) are $C^{\prime}(1/6)$ small cancellation groups. The standard reference for such groups is...Lyndon and Schupp! Basically, between Dehn and Gromov there was a class of groups called "Small Cancellation" groups which also exploited Dehn's algorithm (today they are the standard examples of Hyperbolic groups). (Okay, so that isn't quite right - there are lots of different flavours of small cancellation groups, and only the metric small cancellation ones are always hyperbolic - the $C^{\prime}(1/\lambda)$ ones, $\lambda \geq 6$.)
Small cancellation theory talks about things called pieces. Take all the relators in your presentation and their inverses, and try and lie one relator on top of another (the ends don't have to be "in line"). Anywhere where the two relators agree is called a piece. (In small cancellation theory we think of relators as "tiles" and the pieces are where the tiles can connect together.)
If every piece has length less than $1/6$ of the relator it is a part of then the presentation satisfies the $C^{\prime}(1/6)$ small cancellation condition.
You should realise that with your groups every piece has length $1$, so your group satisfies $C^{\prime}(1/7)$ (in general, $C^{\prime}(1/(4g-1))$).
Thus, $G$ is hyperbolic and we're done!
There are some conditions that guarantee that the fundamental group of a space will be abelian. For example, if the fundamental group of an H-space is abelian. In these cases, the first homology group will be isomorphic to the fundamental group (if the space is path connected).
Otherwise, if you're only given the data of $H_1(X)$, you cannot compute $\pi_1(X)$ from that. The reason is simple: if you're given the abelianization of a group $G_{ab}$, the group $G$ could be pretty much anything. It could be the product of $G_{ab}$ with a perfect group, or some other extension of $G$...
To give you an example, a knot in $\mathbb{R}^3$ is almost determined by the fundamental group of its complement (as far as I remember, you also need to specify some orientation -- a knot theorist could correct me if I remember incorrectly). But it's also a theorem that the first homology group of this complement is always $\mathbb{Z}$! Even though knot complements are very well-behaved spaces, you still get a lot with the same first homology group.
In fact the noncommutativity of $\pi_1$ can lead to "strange" situations. For example, if the fundamental group is abelian, then trivial homology ($\tilde{H}_*(X) = 0$) implies trivial homotopy ($\pi_*(X) = 0$). But when the fundamental group is not abelian, then it's not true anymore, and there are in fact tons of so-called acyclic spaces whose homology vanish but who are not contractible. Their fundamental group will be perfect because of the Hurewicz isomorphism, but after that (almost) all bets are off. See for example Acyclic spaces by Dror Farjoun. So not even the whole homology of a space is enough to determine the fundamental group if you don't know that it's abelian.
Another example of possible condition is "$X$ is a co-H-space". The fundamental group of $X$ is then free, so the rank of the abelianization is enough to find $\pi_1$ up to isomorphism. I think you can even find the generators by considering lifts of generators of $H_1$.
Best Answer
I don't have a complete answer for you, but I can say a bit.
It's a general theorem that if we have a covering space $p:\tilde{X}\rightarrow X$ with $\tilde{X}$ simply-connected and $X$ path-connected and locally path-connected, then the group of deck transformations $G$ is isomorphic to $\pi_1(X)$. (For example, see Hatcher p.71 prop. 1.39.) All the hypotheses hold for this covering $S^3\rightarrow S^3/I^*=P_3$, so this shows that $I^*=\pi_1(P_3)$.
The rest is pure group theory. First look at the icosahedral rotational group, call it $I$. $I$ is a simple group of order 60. The commutator subgroup is always normal, so $[I,I]$ is either trivial or all of $I$. $[I,I]$ is not trivial because $I$ is not abelian, so $I=[I,I]$.
Now the center $Z$ of $I^*$ is a two-element group, and $I^*/Z = I$. Clearly $[I^*,I^*]$ maps onto $[I,I]$. So $I^*/[I^*,I^*]$ is either trivial or has two elements.
The correct answer is that $[I^*,I^*]=I^*$ (wikipedia), but I'm afraid I don't know a proof. Showing that $-1\in I^*$ is a product of commutators would do the trick.
A friend referred me to Weibel's book An Introduction to Homological Algebra. Section 6.9, pp.198-199, "Universal Central Extensions", proves (Example 6.9.1 plus Lemma 6.9.2) that $I^*$ is perfect, i.e., equal to its commutator subgroup. The key fact is that $I^*$ is a universal central extension of $I$. For this fact, Weibel refers us to Suzuki, Group Theory I. I don't have that handy.
However, playing with my dodecahedral paperweight suggests a proof strategy. Let $\tilde{a}$ and $\tilde{b}$ be the two generators of $I^*$ in the presentation $\langle \tilde{a},\tilde{b} | \tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2\rangle$. Their images in the icosahedral group are $a$, a rotation of $72^\circ$, and $b$, a rotation of $120^\circ$ (and $ab$ is a rotation of $180^\circ$). In $I$, $a^5=b^3=(ab)^2=1$. Since 1 in $I$ has two preimages in $I^*$, namely $\pm 1$, we must have $\tilde{a}^5 = \tilde{b}^3 = (\tilde{a}\tilde{b})^2 = -1$. (If it equalled 1, then we'd get $I$ instead of $I^*$.)
Now $[a,b]$ is a rotation of $72^\circ$ in $I$ ("proof by paperweight"). So it's conjugate to $a$ in I, and thus $[a,b]^5=1$. That suggests that $[\tilde{a},\tilde{b}]^5$ should equal $-1$ in $I^*$. It should be possible to verify this by direct computation with $2\times 2$ matrices in $SU(2)$.
Hatcher's Example 2.38, "An Acyclic Space", p.142, is also interesting to look at.