# Homotopy Groups of Connected Sums

This was sparked because I wanted to compute $\pi_2(Sym^2(\Sigma_2))$ via $Sym^2(\Sigma_2)\approx \mathbb{T}^4$# $\bar{\mathbb{C}P}^2$.
We know how to compute $\pi_1$ of $M$ # $N$ via van-Kampen's theorem. But what about higher homotopy groups? I looked in the literature and google without luck, and so I am wondering if no such procedure exists. Are there any results for calculating $\pi_n$ of connected sums?

There was mention of "higher van Kampen theorem"... has this actually been used to do such computations? I'd be interested in references if not just examples.

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–  Qiaochu Yuan Apr 6 '12 at 6:47
Higher homotopy groups, higher van Kampen theorems. –  Fernando Muro Apr 6 '12 at 13:39
Sorry can this even be used in a connected-sum example? It seems just like "abstract theory". –  Chris Gerig Apr 6 '12 at 16:26
I don't know of any examples of people using these higher homotopy van kampen theorems to compute anything new. But if people can point out examples that would be great. –  Ryan Budney Apr 6 '12 at 19:31
"Abstract theory", hmm... are there non-abstract mathematics? Think of some elementary mathematics you think of as non-absrtract, pick an arbitrary individual in the street and ask him/her about his/her opinion... –  Fernando Muro Apr 10 '12 at 22:30

The 2nd homotopy group of a connect sum is fairly reasonable to compute. $\pi_i X$ is isomorphic to $\pi_i \tilde X$ provided $i \geq 2$ and $\tilde X$ indicates any covering space of $X$, so we might as well take the universal cover. By the Hurewicz theorem, $\pi_2 \tilde X$ is isomorphic to $H_2 \tilde X$. In the case of a connect sum, the universal cover has a very nice description (take disjoint unions of the universal covers of the punctured manifolds and glue them together appropriately).

Since $\mathbb CP^2$ is simply connected this is a fairly easy thing to compute. The universal cover looks like $\mathbb R^4$ with a $\mathbb CP^2$ summand at every integer lattice point. So,

$$\pi_2 ((S^1)^4 \# \mathbb CP^2) \simeq \bigoplus_{\pi_1 T^4} \pi_2 \mathbb CP^2$$

i.e. a direct sum over $\mathbb Z^4$ of copies of the integers, i.e. $\mathbb Z[t_1^\pm, t_2^\pm, t_3^\pm, t_4^\pm]$ a laurent polynomial ring in four variables. $\pi_1$ acts by multiplication by units in the Laurent polynomial ring.

Higher homotopy groups in general can be fairly painful to compute but $\pi_2$ is usually quite reasonable, like this case.

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Ah perfect for $\pi_2$! Is there a deep reason why we can't do similar things for higher homotopy? –  Chris Gerig Apr 6 '12 at 16:41
The anthropic principal response would be "homotopy groups are really useful, so if you could compute them easily, your life would become too easy". :) I don't have a deep reason to offer you, but if you take as an axiom that the homotopy groups of spheres are very complicated and there's mysterious patterns relating them, then since $S^n=\Sigma S^{n−1}$, there can't be a simple way to compute the homotopy groups of a CW-complex since that would force too obvious a symmetry into the homotopy-groups of spheres. Rational homotopy groups are much more manageable, but torsion is tricky. –  Ryan Budney Apr 6 '12 at 19:29
The general framework I'm using starts running out of steam but it is usable. $\pi_2$ we compute by going to the universal cover and computing $H_2$. Okay, so what about $\pi_3$? You'd kill $\pi_2$ via a fibration and compute $H_3$ of the total space of that fibration. Basically we're using the Whitehead tower (as described in Hatcher's algtop book) where you kill the low-dimensional homotopy groups successsively. If $\pi_2$ is free abelian you can kill it using a fibre bundle with fibre a product of circles (a generalized Hopf fibration). –  Ryan Budney Apr 6 '12 at 20:03
This is, in principle, the old Serre method for computing homotopy groups, isn't it? –  Lennart Meier Apr 9 '12 at 10:09
I thought Serre did a similar thing but usually the other way around, using a Postnikov system rather than the Whitehead tower. They're very similar ideas, though. I would have known the answer when I was a grad student! –  Ryan Budney Apr 9 '12 at 19:15

Here is something that's valid in the stable range.

If $M$ and $N$ are closed $n$-manifolds, there is a cofibration sequence $$S^{n-1} \to M_0 \vee N_0 \to M\sharp N$$ where $M_0$ denotes the effect of deleting a point from $M$.

If $M$ and $N$ are $r$-connected, then so is the connected sum. The Blakers-Massey excision theorem then implies an exact sequence $$\pi_k(S^{n-1}) \to \pi_k(M_0 \vee N_0) \to \pi_k(M\sharp N) \to \pi_{k-1}(S^{n-1}) \to \cdots$$ as long as $k \le n-2+r$.

Furthermore the map $M_0 \vee N_0 \to M_0 \times N_0$ is $(2r+1)$-connected, so if $k \le 2r$ we get $\pi_k(M_0 \vee N_0) = \pi_k(M) \oplus \pi_k(N)$.

Assembling this, we have an exact sequence $$\pi_k(S^{n-1}) \to \pi_k(M) \oplus \pi_k(N) \to \pi_k(M\sharp N) \to \pi_{k-1}(S^{n-1}) \to \cdots$$ which is valid for $k \le 2r$, $r \le n-2$.

I just realized one could simply note that the cofiber sequence gives a long exact sequence on stable homotopy $$\pi_k^{st}(S^{n-1}) \to \pi_k^{st}(M_0) \oplus \pi_k^{st}(N_0) \to \pi_k^{st}(M\sharp N) \to \pi_{k-1}^{st}(S^{n-1}) \to \cdots$$ and then if $M$ and $N$ are $r$-connected with $k \le 2r$ and $r\le n-2$ we can use the Freudenthal suspension theorem to identify the stable groups with the corresponding unstable ones. This gives a more elementary argument.

Here's a special case: when $M$ and $N$ are framed, so is $M\sharp N$ and the connecting map in the exact sequence splits to give a splitting $$\pi_k(M\sharp N) = \pi_k(M) \oplus \pi_k(N) \oplus \pi_{k-1}(S^{n-1})$$ (assuming the constraints on $k,r$ and $n$).

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Nice! I like this better than the abstract "higher van-Kampen". –  Chris Gerig Apr 8 '12 at 0:14
@ John Klein: The higher van Kampen theorems can sometimes give more information on $K \cup L$ when the intersection $K \cap L$ is not simply connected. For the connected sum we essentially have $P- K \cap L$ is a solid ball. What about other constructions, e.g. $P= S^1 \times D^2$ ? or more complicated? Are those of interest? –  Ronnie Brown Aug 9 '13 at 17:01
That should have been $P=K \cap L$. One example which worked well was when $P= K(G,1)$ for any group $G$ and $K,L$ are cones on $P$, so their union is the suspension of $P$. –  Ronnie Brown Aug 9 '13 at 20:20

Just a comment on the role of the Higher Homotopy Seifert-van Kampen Theorems:

they should be regarded as an extra tool in algebraic topology. There are quite severe conditions on their applicability but when they apply they compute quite a lot. Just as the 1-dimensional theorem, in its groupoid formulation, is about calculating 1-types, so the 2-dim theorem is about computing 2-types, in the form of crossed modules (over groupoids). However computing the second homotopy group from this 2-type may not be straightforward. But then the situation is the same for the 1-dim theorem, as is evidenced by the complications of theorems such as the Kurosh subgroup theorem, which can be seen to be about the fundamental group(s) of a cover of a wedge of $K(G_i,1)$'s.

As a taster, based on the 2-d theorem, work with Chris Wensley enabled the computation of the crossed module representing the 2-type of the mapping cone of a map $Bf: BG \to BH$ induced by a morphism $f: G \to H$. Of course. the second homotopy group, even as a module over the fundamental group, is but a pale shadow of the 2-type. You can see some of this in our book (pdf available from my web page on the book).

R. Brown, P.J. Higgins, R. Sivera, Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids, EMS Tracts in Mathematics Vol. 15, 703 pages. (August 2011).

November 8, 2013: As a taster, let $X$ be the homotopy pushout of the classifying spaces of the two maps of groups $P \to P/M, P \to P/N$ where $M,N$ are normal subgroups of the group $P$. The the homotopy 2-type of $X$ is determined by the crossed module $M \circ N \to P$, the coproduct of the two crossed $P$-modules, which is given by the pushout of crossed modules

$$\begin{matrix} (1 \to P) & \to & (N \to P) \cr \downarrow && \downarrow \cr (M \to P)& \to & (M \circ N \to P) \end{matrix}$$. It follows that $$\pi_2(X) \cong (M \cap N)/ [M,N].$$ (Of course we know $\pi_1 X$ by the 1-dimensional van Kampen Theorem.) This result is applied in Bardakov, Valery G; Mikhailov, Roman; Vershinin, Vladimir V.; Wu, Jie, "Brunnian braids on surfaces". Algebr. Geom. Topol. 12 (2012), no. 3, 1607–1648.

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