I am trying to get some more intuitions about the statement:

Killing the $\pi_3$ homotopy group in Spin(n), one obtains the infinite-dimensional string group String(n).

Formally: I know that here is a short exact sequence of topological groups

$$ {0\rightarrow K(\mathbb{Z},2)\rightarrow {\text{String}}(n)\rightarrow {\text{Spin}}(n)\rightarrow 0} $$ where $K(\mathbb{Z}, 2)=K(U(1), 1)=P^{\infty}(C)$ is an Eilenberg–MacLane space and Spin(n) is a spin group.

Since $P^{\infty}(C)$ is infinite dimensional space, ${\text{String}}(n)$ is also infinite dimensional space.


  • How do we see that killing the $\pi_3$ homotopy group of Spin(n), gets this large infinite-dimensional space String(n)?

Expected answer for other examples: In contrast, killing the $\pi_1$ homotopy group of SO(n), we only get the double cover as Spin(n) which is a finite-dimensional space.

  • Question: How do we "visualize" this String(n) space?

Expected answer for other examples: In contrast, we can visualize the Spin(n) easily which is just a double cover of SO(n).

  • $\begingroup$ Personally I visualise $String(n)$ by writing down the short exact sequence in your question. That's about as visual as I like to get. I don't entirely agree that it is easy to 'visualise' $Spin(n)$, or in fact $SO(n)$ when $n>2$. It's easy to write down a set of matrices and give algebraic conditions that describe it as $SO(n)$, but this is not the same as visualising it. For example $SO(3)\cong\mathbb{R}P^3$, and I have trouble visualising projective 3-space. $\endgroup$ – Tyrone Aug 22 '18 at 11:45
  • $\begingroup$ @tyrone: Look at this: math.stackexchange.com/questions/60522/… $\endgroup$ – Thomas Rot Aug 22 '18 at 13:15
  • 2
    $\begingroup$ p.s. I have trouble with many things, including getting out of bed and understanding hip hop lyrics. Please don't think my comment is meant to be discouraging! It certainly was not intended in that way. $\endgroup$ – Tyrone Aug 22 '18 at 14:58
  • $\begingroup$ "How do we see that killing the π3 homotopy group and its relation to K(ℤ,2)?" This may be an easier question? I also meant to say that how do we go from Spin(n) to String(n) by the construction? $\endgroup$ – wonderich Aug 22 '18 at 21:50

This is a partial answer designed to help you understand the relation between killing $\pi_3$ and the appearance of a $K(\mathbb{Z},2)$. It was way too long for a comment, but I thought it might be useful to you.

For $n=1$, $Spin(1)\cong\mathbb{Z}_2$ and for $n=2$ $Spin(2)\cong S^1$. In these cases the string groups as you desribe them don't really exist, so we'll assume $n\geq 3$ from now on.

Then $Spin(3)\cong S^3\cong SU(2)$ so $String(3)\simeq S^3\langle 3\rangle$ is just the $3$-connected cover of $S^3$. All the homotopy groups of this space are torsion, by a Theorem of Serre, and it has non-trivial homotopy groups in infinitely many degrees.

On the other hand $Spin(4)\cong S^3\times S^3$ is not simple, so again $String(4)$ is a special case. To avoid these difficulties we'll assume that $n>4$ in the following. Note that what we say will still apply to $Spin(3)$, but if care is taken with statements (importantly, for example, the inclusion $Spin(3)\hookrightarrow Spin(n)$ does not induce an isomorphism on $\pi_3$).

Now for $n>4$, $Spin(n)$ is a simply connected, simple compact Lie group. Therefore it holds that $\pi_1Spin(n)=0=\pi_2Spin(n)$ and $\pi_3Spin(n)\cong\mathbb{Z}$. The Hurewicz theorem tells us that $H_2Spin(n)=0$ and $H_3Spin(n)\cong \pi_3Spin(n)\cong\mathbb{Z}$, and the universal coefficient theorem then tells us that $H^3Spin(n)\cong Hom(H_3Spin(n),\mathbb{Z})\cong \mathbb{Z}$. Let $x_3\in H^3Spin(n)$ be a generator. Then this cohomology class is represented by a map

$$x_3:Spin(n)\rightarrow K(\mathbb{Z},3)$$

into the integral Eilenberg-Mac Lane space in degree $3$. This cohomology class is primitive for degree reasons so its homotopy fibre is an H-space. It will actually turn out that $x_3$ is the cohomology suspension of a universal class $c^s_2\in H^4BSpin(4)$. Therefore $x_3$ is a loop map and its homotopy fibre has the homotopy type of a topological group, which is homotopy equivalent to $String(n)$. In any case we have a homotopy fibration sequence

$$String(n)\rightarrow Spin(n)\xrightarrow{x_3} K(\mathbb{Z},3)$$

which represents a delooping of the sequence written in your question. This is how $K(\mathbb{Z},2)\simeq \Omega K(\mathbb{Z},3)$ appears in your sequence.

To understand what is going on it may help to first note that the complexification map $SU(2)\rightarrow SO(n)$ may be lifted to the universal cover as a homomorphism $i:SU(2)\cong S^3\rightarrow Spin(n)$ whose homotopy class generates $\pi_3Spin(n)$, and this homomorphism classifies to a map $Bi:BSU(2)\rightarrow BSpin(n)$. We know $\pi_4BSpin(n)\cong H^4BSpin(n)\cong \mathbb{Z}$ are the first nontrivial homotopy/cohomolgy groups, so for an appropriate choice of generators $c^s_2\in H^4BSpin(4)$ it holds that $Bi^*c_2^s=c_2$, where $c_2\in H^4BSU(2)$ is the 2nd Chern class. This follows since $Bi$ induces an isomorphism on $\pi_4$.

The point is that $x_3\simeq \Omega c_2^s$ so there is another homotopy fibration sequence

$$BString(n)\rightarrow BSpin(n)\xrightarrow{c_2^s} K(\mathbb{Z},4)$$

which returns the previous sequence I wrote above upon applying $\Omega$.

The main point to understand is that this a general procedure. For any simply connected space $X$ and any spherical homology class $a\in H_n(X;A)$ in the image of the Hurewicz map $\pi_nX\rightarrow H_nX$. Choose $\alpha\in \pi_n$ representing $a$ in this sense. Then $a$ generates a subgroup $A\leq H_nX$ and so corresponds to a class in $H^n(X;A)$. Then there is a homotopy fibre sequence

$$X\langle\alpha\rangle\rightarrow X\xrightarrow{x}K(A,n)$$

defining the space $X\langle\alpha\rangle$ which is obtained by killing the class $\alpha$. This is obviously clearest when $X$ is, say, $n$-connected.

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  • $\begingroup$ thank you +1, great! $\endgroup$ – wonderich Aug 23 '18 at 15:06
  • $\begingroup$ How does one visualize the "3-connected cover of $S^3$," what is it? $\endgroup$ – annie marie heart Sep 1 '18 at 3:34
  • $\begingroup$ @annie It is the homotopy fiber of the map $S^3 \to K(Z,3)$ corresponding to the homotopy class $1 \in \pi_3(K(Z,3)) = Z$. In general, one can obtain these “$k$-connected covers” of a space $X$ by taking homotopy fibers of the maps $X \to X[n]$ in the Postnikov tower of $X$. Unfortunately Postnikov towers are a very “non-visual” decomposition of a space (as opposed to, say, cellular decomposition). $\endgroup$ – Aleksandar Milivojevic Sep 2 '18 at 19:15

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