Loopspace of an Eilenberg Maclane space K(G,n) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:42:02Z http://mathoverflow.net/feeds/question/40553 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n Loopspace of an Eilenberg Maclane space K(G,n) Sam 2010-09-29T23:30:47Z 2010-09-30T12:40:34Z <p>I've seen the fact that the loopspace $\Omega K(G,n)$ is homotopy equivalent to $K(G,n-1)$ mentioned in some places, but I have no idea why. Can anyone offer a good explanation? Also, what happens when $n=1$? Does that just mean the loopspace is a discrete space with $|G|$ points?</p> http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n/40559#40559 Answer by Andreas Blass for Loopspace of an Eilenberg Maclane space K(G,n) Andreas Blass 2010-09-30T00:10:49Z 2010-09-30T00:10:49Z <p>For any (reasonable) space, the homotopy groups of $\Omega X$ are those of $X$ shifted by 1; <code>$\pi_k(\Omega X)=\pi_{k+1}(X)$</code>. Since the homotopy groups of $K(G,n)$ are $G$ in dimension $n$ and 1 in all other dimensions, the homotopy groups of $\Omega K(G,n)$ are $G$ in dimension $n-1$ and 1 in all other dimensions. That makes $\Omega K(G,n)$ a $K(G,n-1)$.</p> http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n/40560#40560 Answer by Allan Edmonds for Loopspace of an Eilenberg Maclane space K(G,n) Allan Edmonds 2010-09-30T00:18:26Z 2010-09-30T00:18:26Z <p>In general,the map $P(X,x_0)\to X$ from the space of based paths in $X$ is a fibration with fiber $\Omega(X,x_0)$. Since $P(X,x_0)$ is contractible, by shrinking paths back toward the base point $x_0$, the homotopy long exact sequence of the fibration shows that $\pi_k(X,x_0)\cong \pi_{k-1}(\Omega(X,x_0))$. The result follows. And, yes, the loop space on a $K(G,1)$ is discrete, in one-to-one correspondence with $G$.</p> http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n/40568#40568 Answer by Jeff Strom for Loopspace of an Eilenberg Maclane space K(G,n) Jeff Strom 2010-09-30T02:20:10Z 2010-09-30T02:20:10Z <p>If $X$ is $(n-1)$-connected, and $Y$ is a space with $\pi_n(Y) \cong G$ and no other nontrivial homotopy groups, then the map $[X, Y] \to \mathrm{Hom}(\pi_n(X), G)$ is a bijection (by elementary obstruction theory). It follows that if $Z$ is a space with $\pi_n(Y) \cong H$ and no other nontrivial homotopy groups, then $[Y, Z] \cong \mathrm{Hom}(G, H)$, and consequently, that if $G\cong H$, $Y$ and $Z$ are weakly equivalent.</p> <p>Assuming they are CW complexes, they are homotopy equivalent; we call the common homotopy type $K(G,n)$.</p> <p>So, we know that the loop space operator $\Omega$ has the effect of shifting homotopy groups downward. Thus $\Omega Y$ is a space with $\pi_{n-1}(Y)\cong G$ and all other homotopy groups trivial. Thus $Y \simeq K(G, n-1)$.</p> http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n/40588#40588 Answer by Andrew Stacey for Loopspace of an Eilenberg Maclane space K(G,n) Andrew Stacey 2010-09-30T08:36:30Z 2010-09-30T08:36:30Z <p>I must confess to being a little dissatisfied with the other answers (so far). They all have a flavour of, "This is really quite simple, here's why." (I should make it clear that I do not infer anything as to the <em>intent</em> of the authors concerned). In fact, there's a lot of machinery in the background that makes these answers work. Someone working in algebraic topology has (probably) so internalised this machinery that they forget that someone coming to it afresh hasn't done so. The question, to me, reads as that of someone who hasn't yet internalised the machinery, so I'm going to give a slightly different answer.</p> <p>The key piece of machinery comes in the question that should have been asked first (except that this isn't really a good MO question ..): "What is $K(G,n)$?". The true short answer is that it is a <em>homotopy type</em>. That is, it isn't really <em>one</em> thing, but an equivalence class of things defined by a common property:</p> <blockquote> <p>A space (with the homotopy type of a CW-complex) is a $K(G,n)$ if it represents the reduced homology functor $\tilde{H}^n(-,G)$ on finite CW-complexes.</p> </blockquote> <p>In truth, to represent the homology functor we must <em>also</em> include in to the data a particular class in $\iota \in \tilde{H}^n(K(G,n),G)$ so that the representing isomorphism</p> <p>$$[X,K(G,n)] \to \tilde{H}^n(X,G)$$</p> <p>is given by $\alpha \mapsto \alpha^* \iota$.</p> <p>The first part of the Big Machine is called <strong>Brown representability</strong>. It says, in short:</p> <blockquote> <p>There is a space which is a $K(G,n)$.</p> </blockquote> <p>Now we do a bit of preparatory work. We can compute $\tilde{H}^n(S^k,G)$ quite simply to find that it is zero (by which I mean the trivial group) if $k \ne n$ and $G$ if $k = n$. By the isomorphism in Brown representability <em>and</em> the definition of homotopy groups, we deduce that if $X$ is a $K(G,n)$ then $\pi_k(X) = 1$ for $k \ne n$ and $\pi_n(X) = G$.</p> <p>The next bit of the Big Machine is the following:</p> <blockquote> <p>If $X$ and $Y$ are simply connected CW-complexes, then a <strong>weak equivalence</strong> is a <strong>homotopy equivalence</strong>.</p> </blockquote> <p>(I want to attribute this to Whitehead, but my reference doesn't make any attribution so I shan't in case I'm wrong).</p> <p>One very important thing to note here is that to apply this, one has to have a map. So saying $\pi_k(X) \cong \pi_k(Y)$ for all $k$ <em>is not enough</em>. These isomorphisms have to be generated by a map. But in our case, we do have such a map. If $X$ and $Y$ are $K(G,n)$s then $[X,X] \cong H^n(X,G) \cong [X,Y]$ so there is a map $X \to Y$ corresponding to the identity on $X$. As it corresponds to the identity on $X$, it's easy to show that it induces an isomorphism on homotopy groups. Hence (as $X$ and $Y$ are <strong>assumed</strong> to have the homotopy type of CW-complexes), $X \simeq Y$.</p> <p>To apply all of this to the case in question, we need to prove two things:</p> <ol> <li>If $X$ is a $K(G,n)$ then $\Omega X$ represents $H^{n-1}(X,G)$ (on finite CW-complexes).</li> <li>If $X$ has the homotopy type of a CW-complex, then $\Omega X$ has the homotopy type of a CW-complex.</li> </ol> <p>The second is a result due to Milnor. The first follows from the <em>suspension isomorphism</em>:</p> <p>$$H^n(\Sigma Z, G) \cong H^{n-1}(Z, G)$$</p> <p>If $X$ is a $K(G,n)$, then we get a natural isomorphism:</p> <p>$$H^{n-1}(Z,G) \cong H^n(\Sigma Z, G) \cong [\Sigma Z, X] \cong [Z, \Omega X]$$</p> <p>and hence $\Omega Z$ represents $H^{n-1}(-,G)$ on finite CW-complexes. Thus, by all the above, it is a $K(G,n-1)$ space and so is homotopy equivalent to any other $K(G,n-1)$ space.</p> <p>Finally, I'd like to underline Kevin Walker's comment. If $G$ is discrete, $\Omega K(G,1)$ is <strong>homotopy equivalent</strong> to $G$, but may not be it exactly. A simple example is the case $G = \mathbb{Z}$. The circle $S^1$ is a $K(\mathbb{Z},1)$. $\Omega S^1$ is the space of all continuous maps from the circle to itself which fix the basepoint. That is most certainly not a discrete space! But it has components corresponding to $\mathbb{Z}$ (given by the winding number) and each component is contractible. Hence $\Omega S^1 \simeq \mathbb{Z}$. In fact, by interpreting $\Omega$ a little differently, you <em>can</em> get a situation in which $\Omega S^1 = \mathbb{Z}$, but to do this you need to interpret $\Omega$ as being <em>polynomial</em> loops.</p> http://mathoverflow.net/questions/40553/loopspace-of-an-eilenberg-maclane-space-kg-n/40611#40611 Answer by Allen Hatcher for Loopspace of an Eilenberg Maclane space K(G,n) Allen Hatcher 2010-09-30T12:40:34Z 2010-09-30T12:40:34Z <p>When I first saw this question I expected to find a number of votes to close it since the question asks about something that is surely covered in any algebraic topology textbook that includes basic homotopy theory. However, the question was not closed, and to my further surprise there have been some rather elaborate answers to so elementary a question. </p> <p>Whether a simple answer is possible depends on which definition of a $K(G,n)$ one uses. Surely the simplest definition, and the one I thought was standard, given for example in Spanier, is that a $K(G,n)$ is a space whose homotopy groups are all trivial except for $\pi_n$ which is isomorphic to $G$. (When $n=0$ the definition needs to be adjusted slightly, or this case can be excluded entirely, as in Spanier.) With this definition the answer given by Allan Edmonds is perfectly satisfactory, using the path fibration, and so is the answer by Andreas Blass which is even more elementary, using just the definitions of the objects involved (and basic properties of function spaces).</p> <p>Now, in practice one wants $K(G,n)$ spaces to be unique up to homotopy equivalence, and this requires restricting to spaces having the homotopy type of CW complexes. Uniqueness can then be proved by elementary arguments not using cohomology or obstruction theory. If one makes the requirement of having the homotopy type of a CW complex part of the definition of a $K(G,n)$ then to answer the original question one would have to quote Milnor's theorem that the loopspace of a CW complex has the homotopy type of a CW complex. This is not a trivial theorem, and it involves digressions into point-set topology. Milnor's theorem is surely a reassuring fact to know, but it can often be avoided by using CW approximations. Overall, it seems best not to require $K(G,n)$'s to have the homotopy type of CW complexes, although this may be a matter of personal preference.</p> <p>As Andrew Stacey's answer shows, quite different definitions of $K(G,n)$'s are possible. These may have their virtues in certain contexts, but I would argue that when it comes to definitions, simplicity and minimality of prerequisites should be the highest priorities. With the simple definition of $K(G,n)$'s in terms of homotopy groups, the fact that they represent cohomology groups $H^n(X;G)$ of CW complexes $X$ as $[X,K(G,n)]$ then comes as a miraculous surprise, rather than something that is true by definition.</p>