Mixing
Eilenberg–MacLane space $K(mathbbZ_2,n)$
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We know that the generalized classifying space / Eilenberg–MacLane space
$$
BmathbbZ_2=mathbbRP^infty
$$
$$
BU(1)=mathbbCP^infty
$$
How do one construct/derive the (infinite dimensional) space explicitly:
$$
B^n mathbbZ_2=K(mathbbZ_2,n)=?
$$How do one construct/derive, when $p$ is a prime:
$$
B^n mathbbZ_p=K(mathbbZ_p,n)=?
$$Can one explain $BmathbbZ_2$, $B^2 mathbbZ_2=K(mathbbZ_2,2)$, $B^n mathbbZ_2=K(mathbbZ_2,n)$, $B^n mathbbZ_p=K(mathbbZ_p,n)$, in a unified consistent but also intuitive way?
algebraic-topology differential-topology geometric-topology classifying-spaces eilenberg-maclane-spaces
asked Jul 14 at 20:13
wonderich
1,68021226
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up vote
2
down vote
favorite
We know that the generalized classifying space / Eilenberg–MacLane space
$$
BmathbbZ_2=mathbbRP^infty
$$
$$
BU(1)=mathbbCP^infty
$$
How do one construct/derive the (infinite dimensional) space explicitly:
$$
B^n mathbbZ_2=K(mathbbZ_2,n)=?
$$How do one construct/derive, when $p$ is a prime:
$$
B^n mathbbZ_p=K(mathbbZ_p,n)=?
$$Can one explain $BmathbbZ_2$, $B^2 mathbbZ_2=K(mathbbZ_2,2)$, $B^n mathbbZ_2=K(mathbbZ_2,n)$, $B^n mathbbZ_p=K(mathbbZ_p,n)$, in a unified consistent but also intuitive way?
algebraic-topology differential-topology geometric-topology classifying-spaces eilenberg-maclane-spaces
asked Jul 14 at 20:13
wonderich
1,68021226
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We know that the generalized classifying space / Eilenberg–MacLane space
$$
BmathbbZ_2=mathbbRP^infty
$$
$$
BU(1)=mathbbCP^infty
$$
How do one construct/derive the (infinite dimensional) space explicitly:
$$
B^n mathbbZ_2=K(mathbbZ_2,n)=?
$$How do one construct/derive, when $p$ is a prime:
$$
B^n mathbbZ_p=K(mathbbZ_p,n)=?
$$Can one explain $BmathbbZ_2$, $B^2 mathbbZ_2=K(mathbbZ_2,2)$, $B^n mathbbZ_2=K(mathbbZ_2,n)$, $B^n mathbbZ_p=K(mathbbZ_p,n)$, in a unified consistent but also intuitive way?
algebraic-topology differential-topology geometric-topology classifying-spaces eilenberg-maclane-spaces
asked Jul 14 at 20:13
wonderich
1,68021226
We know that the generalized classifying space / Eilenberg–MacLane space
$$
BmathbbZ_2=mathbbRP^infty
$$
$$
BU(1)=mathbbCP^infty
$$
How do one construct/derive the (infinite dimensional) space explicitly:
$$
B^n mathbbZ_2=K(mathbbZ_2,n)=?
$$How do one construct/derive, when $p$ is a prime:
$$
B^n mathbbZ_p=K(mathbbZ_p,n)=?
$$Can one explain $BmathbbZ_2$, $B^2 mathbbZ_2=K(mathbbZ_2,2)$, $B^n mathbbZ_2=K(mathbbZ_2,n)$, $B^n mathbbZ_p=K(mathbbZ_p,n)$, in a unified consistent but also intuitive way?
algebraic-topology differential-topology geometric-topology classifying-spaces eilenberg-maclane-spaces
asked Jul 14 at 20:13
wonderich
1,68021226
asked Jul 14 at 20:13
wonderich
1,68021226
asked Jul 14 at 20:13
wonderich
1,68021226
asked Jul 14 at 20:13
wonderich
1,68021226
1,68021226
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3 Answers
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For an abelian group $A$ with a discrete topology we have
$$K(A,0)simeq A,qquad K(A,1)simeq BA.$$
Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).
One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.
Since the projection $EBArightarrow B^2A$ is fibration with contractible total space and fibre $BAsimeq K(A,1)$ we study the long exact homotopy sequence to get
$$pi_i(B^2A)cong pi_i-1(BA)$$
In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.
Now for the same reasons as before $B^2A$ has a classifying space $B^3Asimeq K(A,3)$, and we can proceed by induction we get
$$K(A,n)=B^nA.$$
In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that
$$K(Aoplus B,n)simeq K(A,n)times K(B,n)$$
so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $Arightarrow B$ is a homomorphism then there is an induced map $K(A,n)rightarrow K(B,n)$, and in particular if $0rightarrow Axrightarrowf Bxrightarrowg Crightarrow 0$ is short exact then there is a fibration sequence
$$dotsrightarrow K(C,n-1)rightarrow K(A,n)xrightarrowKf K(B,n)xrightarrowKg K(C,n)rightarrow K(A,n+1)rightarrowdots$$
And this can be useful. For example $0rightarrow mathbbZxrightarrowtimes 2mathbbZrightarrow mathbbZ_2rightarrow 0$ gives us
$$dotsrightarrow S^1rightarrow mathbbRP^inftyrightarrow mathbbCP^inftyxrightarrowtimes 2 mathbbCP^inftyrightarrow K(mathbbZ_2,2)rightarrowdots$$
answered Jul 15 at 11:23
Tyrone
3,25611125
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
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Given a connected topological space $X$ with $pi_n(X) = G$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.
Suppose you start with $X = mathbbRP^2$ which has $pi_1(X) cong mathbbZ_2$. Then $pi_2(mathbbRP^2) = mathbbZ$ and is generated by the covering map $S^2 to mathbbRP^2$. Attaching a three-cell to $mathbbRP^2$ with attaching map given by the covering map $S^2 to mathbbRP^2$ gives a space $X'$ with $pi_1(X') = mathbbZ_2$ and $pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbRP^3$. Now note that $pi_3(mathbbRP^3) cong mathbbZ$ generated by the covering map $S^3 to mathbbRP^3$. Attaching a four-cell to $mathbbRP^3$ with attaching map given by the covering map $S^3 to mathbbRP^3$ gives a space $X''$ with $pi_1(X'') = mathbbZ_2$ and $pi_2(X'') = pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbRP^4$. Repeating this procedure ad infinitum, we see that $mathbbRP^infty$ is a $K(mathbbZ_2, 1)$.
Suppose now you start with $X = S^2$ which has $pi_1(X) = 0$ and $pi_2(X) cong mathbbZ$. Then $pi_3(S^2) cong mathbbZ$ and is generated by the Hopf map $S^3 to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 to S^2$ gives a space $X'$ with $pi_1(X') = 0$, $pi_2(X') cong mathbbZ$, and $pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbCP^2$. Now note that $pi_4(mathbbCP^2) = 0$ so we don't need to glue on any five-cells, but $pi_5(mathbbCP^2) cong mathbbZ$ generated by the quotient map $S^5 to mathbbCP^2$. Attaching a six-cell to $mathbbCP^2$ with attaching map given by the quotient map $S^5 to mathbbCP^2$ gives a space $X''$ with $pi_1(X'') = 0$, $pi_2(X'') cong mathbbZ$, and $pi_3(X'') = pi_4(X'') = pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbCP^3$. Repeating this procedure ad infinitum, we see that $mathbbCP^infty$ is a $K(mathbbZ, 2)$.
The case for cyclic groups is similar to the first example. Note that $mathbbRP^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 to S^1$ is a topological space $X$ with $pi_1(X) cong mathbbZ_k$. The above procedure then shows that the infinite lens space $S^infty/mathbbZ_k$ is a $K(mathbbZ_k, 1)$. In order to obtain a $K(mathbbZ_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $pi_n(X) cong mathbbZ_k$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ_k, n)$.
As for forming $K(mathbbZ, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ, n)$.
For $n geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $pi_n(Xtimes Y) cong pi_n(X)opluspi_n(Y)$).
For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
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One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G times G to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $mathsfGrp$.
Then $K(Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(Bbb Z_2, 1) simeq BbbRP^infty$ and maps $X to BbbRP^infty$ classify line bundles on $X$ upto isomorphism. This bijection $mathcalL(X) to [X, BbbRP^infty]$ can be established by sending a line bundle $ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X times BbbR^infty$, thinking of $BbbRP^infty$ as a Grassmannian.
As Tyrone described, there is a tower of fibrations $$cdots to K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n) to K(Bbb Z, n+1) to cdots$$ induced from the short exact sequence $0 to Bbb Z to Bbb Z to Bbb Z_2 to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$cdots to H^n(X; Bbb Z) to H^n(X; Bbb Z) to H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z) to cdots$$ induced from the change-of-coefficients sequence $0 to C^*(X; Bbb Z) to C^*(X; Bbb Z) to C^*(X; Bbb Z_2) to 0$ The snake map $H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z)$ is a natural transformation $H^n(-; Bbb Z_2)! Rightarrow! H^n+1(-; Bbb Z)$ and by representability, is Yoneda-dual to the map $K(Bbb Z_2, n) to K(Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(Bbb Z_2, n) to BK(Bbb Z, n)$ of the fibration $K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n)$.
This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.
There is a sort of striking geometry to $K(Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $textTOP(n)$ the group of germs-at-origin of homeomorphisms of $Bbb R^n$ fixing the origin. Let $textPL(n) subset textTOP(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$textTOP(n)/textPL(n) to BtextPL(n) to BtextTOP(n)$$ There are natural morphisms $textTOP(n) to textTOP(n+1)$ restricting to $textPL(n) to textPL(n+1)$, given by sending a germ of a self-homeomorphism of $Bbb R^n$ to that of $Bbb R^n times Bbb R$, extended by identity on the $Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $textTOP$ and $textPL$ respectively. The aforementioned fibration should induce a fibration $$textTOP/textPL to BtextPL to BtextTOP$$ Kirby and Seibenmann proved that $textTOP/textPL$ is a $K(Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M to BtextTOP$ which lifts to a map $M to BtextPL$ if and only if $M$ admits a PL-manifold structure. The $textTOP/textPL$-bundle over $BtextTOP$ is classified by a map $BtextTOP to B(textTOP/textPL)$, which has fiber $BtextPL$. So there seems to be another bundle $BtextPL to BtextTOP to B(textTOP/textPL)$, which should be nothing but the one induced from $textPL to textTOP to textTOP/textPL$. Then the lifting condition is equivalent to demanding the pushforward $M to BtextTOP to B(textTOP/textPL)$ is nullhomotopic as it factors through the fibers. But since $textTOP/textPL cong K(Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M to BK(Bbb Z_2, 3) cong K(Bbb Z_2, 4)$, which is an element of $[M, K(Bbb Z_2, 4)] cong H^4(M; Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).
EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $textTOP/textPL$ is simply the fiber of the fibration $BtextPL to BtextTOP$, so does not have an obvious group structure, which makes $B(textTOP/textPL)$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^k+1(M; pi_k textTOP/textPL)$, and using the fact that $textTOP/textPL cong K(Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; Bbb Z_2)$.
I don't know if any other $K(Bbb Z_2, n)$ has some tractable topological model though.
answered Jul 15 at 18:54
Balarka Sen
9,57812955
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
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3 Answers
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active
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For an abelian group $A$ with a discrete topology we have
$$K(A,0)simeq A,qquad K(A,1)simeq BA.$$
Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).
One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.
Since the projection $EBArightarrow B^2A$ is fibration with contractible total space and fibre $BAsimeq K(A,1)$ we study the long exact homotopy sequence to get
$$pi_i(B^2A)cong pi_i-1(BA)$$
In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.
Now for the same reasons as before $B^2A$ has a classifying space $B^3Asimeq K(A,3)$, and we can proceed by induction we get
$$K(A,n)=B^nA.$$
In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that
$$K(Aoplus B,n)simeq K(A,n)times K(B,n)$$
so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $Arightarrow B$ is a homomorphism then there is an induced map $K(A,n)rightarrow K(B,n)$, and in particular if $0rightarrow Axrightarrowf Bxrightarrowg Crightarrow 0$ is short exact then there is a fibration sequence
$$dotsrightarrow K(C,n-1)rightarrow K(A,n)xrightarrowKf K(B,n)xrightarrowKg K(C,n)rightarrow K(A,n+1)rightarrowdots$$
And this can be useful. For example $0rightarrow mathbbZxrightarrowtimes 2mathbbZrightarrow mathbbZ_2rightarrow 0$ gives us
$$dotsrightarrow S^1rightarrow mathbbRP^inftyrightarrow mathbbCP^inftyxrightarrowtimes 2 mathbbCP^inftyrightarrow K(mathbbZ_2,2)rightarrowdots$$
answered Jul 15 at 11:23
Tyrone
3,25611125
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
add a comment |Â
up vote
3
down vote
accepted
For an abelian group $A$ with a discrete topology we have
$$K(A,0)simeq A,qquad K(A,1)simeq BA.$$
Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).
One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.
Since the projection $EBArightarrow B^2A$ is fibration with contractible total space and fibre $BAsimeq K(A,1)$ we study the long exact homotopy sequence to get
$$pi_i(B^2A)cong pi_i-1(BA)$$
In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.
Now for the same reasons as before $B^2A$ has a classifying space $B^3Asimeq K(A,3)$, and we can proceed by induction we get
$$K(A,n)=B^nA.$$
In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that
$$K(Aoplus B,n)simeq K(A,n)times K(B,n)$$
so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $Arightarrow B$ is a homomorphism then there is an induced map $K(A,n)rightarrow K(B,n)$, and in particular if $0rightarrow Axrightarrowf Bxrightarrowg Crightarrow 0$ is short exact then there is a fibration sequence
$$dotsrightarrow K(C,n-1)rightarrow K(A,n)xrightarrowKf K(B,n)xrightarrowKg K(C,n)rightarrow K(A,n+1)rightarrowdots$$
And this can be useful. For example $0rightarrow mathbbZxrightarrowtimes 2mathbbZrightarrow mathbbZ_2rightarrow 0$ gives us
$$dotsrightarrow S^1rightarrow mathbbRP^inftyrightarrow mathbbCP^inftyxrightarrowtimes 2 mathbbCP^inftyrightarrow K(mathbbZ_2,2)rightarrowdots$$
answered Jul 15 at 11:23
Tyrone
3,25611125
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For an abelian group $A$ with a discrete topology we have
$$K(A,0)simeq A,qquad K(A,1)simeq BA.$$
Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).
One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.
Since the projection $EBArightarrow B^2A$ is fibration with contractible total space and fibre $BAsimeq K(A,1)$ we study the long exact homotopy sequence to get
$$pi_i(B^2A)cong pi_i-1(BA)$$
In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.
Now for the same reasons as before $B^2A$ has a classifying space $B^3Asimeq K(A,3)$, and we can proceed by induction we get
$$K(A,n)=B^nA.$$
In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that
$$K(Aoplus B,n)simeq K(A,n)times K(B,n)$$
so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $Arightarrow B$ is a homomorphism then there is an induced map $K(A,n)rightarrow K(B,n)$, and in particular if $0rightarrow Axrightarrowf Bxrightarrowg Crightarrow 0$ is short exact then there is a fibration sequence
$$dotsrightarrow K(C,n-1)rightarrow K(A,n)xrightarrowKf K(B,n)xrightarrowKg K(C,n)rightarrow K(A,n+1)rightarrowdots$$
And this can be useful. For example $0rightarrow mathbbZxrightarrowtimes 2mathbbZrightarrow mathbbZ_2rightarrow 0$ gives us
$$dotsrightarrow S^1rightarrow mathbbRP^inftyrightarrow mathbbCP^inftyxrightarrowtimes 2 mathbbCP^inftyrightarrow K(mathbbZ_2,2)rightarrowdots$$
answered Jul 15 at 11:23
Tyrone
3,25611125
For an abelian group $A$ with a discrete topology we have
$$K(A,0)simeq A,qquad K(A,1)simeq BA.$$
Note the homotopy equivalences rather than equalities. We can model $BA$ by using a bar construction to get a contractible free $A$-space $EA$, and then set $BA=(EA)/A$. I detailed this in an answer here Classifying space $B$SU(n).
One finds that the bar construction model for $EA$ is a topological monoid (that is, they carry an associative, unital product), and in the case that $A$ is abelian, this product descends to make $BA$ into a topological monoid. Hence the bar construction may be applied again to generate a contractible free $BA$-space $E(BA)$, whose $BA$ quotient is a classifying space $B(BA)=B^2A=E(BA)/BA$.
Since the projection $EBArightarrow B^2A$ is fibration with contractible total space and fibre $BAsimeq K(A,1)$ we study the long exact homotopy sequence to get
$$pi_i(B^2A)cong pi_i-1(BA)$$
In particular, if $BA$ is a $K(A,1)$, then $B^2A$ is a $K(A,2)$.
Now for the same reasons as before $B^2A$ has a classifying space $B^3Asimeq K(A,3)$, and we can proceed by induction we get
$$K(A,n)=B^nA.$$
In general we have no nice way to model $K(A,n)$ except for basically the few cases you outline in the question. A useful fact is that
$$K(Aoplus B,n)simeq K(A,n)times K(B,n)$$
so one can apply the theorem on the classification of abelian groups to decompose things into easier to work with factors. Another useful fact is that if $Arightarrow B$ is a homomorphism then there is an induced map $K(A,n)rightarrow K(B,n)$, and in particular if $0rightarrow Axrightarrowf Bxrightarrowg Crightarrow 0$ is short exact then there is a fibration sequence
$$dotsrightarrow K(C,n-1)rightarrow K(A,n)xrightarrowKf K(B,n)xrightarrowKg K(C,n)rightarrow K(A,n+1)rightarrowdots$$
And this can be useful. For example $0rightarrow mathbbZxrightarrowtimes 2mathbbZrightarrow mathbbZ_2rightarrow 0$ gives us
$$dotsrightarrow S^1rightarrow mathbbRP^inftyrightarrow mathbbCP^inftyxrightarrowtimes 2 mathbbCP^inftyrightarrow K(mathbbZ_2,2)rightarrowdots$$
answered Jul 15 at 11:23
Tyrone
3,25611125
answered Jul 15 at 11:23
Tyrone
3,25611125
answered Jul 15 at 11:23
Tyrone
3,25611125
answered Jul 15 at 11:23
Tyrone
3,25611125
3,25611125
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
add a comment |Â
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
thanks +1, but what is this infinite dimensional space precisely for $K(mathbbZ_2,n)$? Is that still $CP^infty$?
– wonderich
Jul 15 at 16:05
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Which infinite dimensional space do you mean? And I'm not sure I see the direct relation between $K(mathbbZ_2,n)$ and $mathbbCP^infty$ that you are referring to.
– Tyrone
Jul 15 at 16:28
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
Thanks Tyrone, $K(mathbbZ_2,1)=mathbbRP^infty$, and $K(mathbbZ_2,2)=?$
– wonderich
Jul 15 at 16:57
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
First, the correct statement is that $K(mathbbZ_2,1)simeq mathbbRP^infty$. Eilenberg-Mac Lane spaces, like group classifying spaces, are only defined up to homotopy equivalence. As I've detailed there are various ways of generating a $K(mathbbZ_2,2)$, some of which will make various features more or less accessible, but there is no "familiar" space which is a $K(mathbbZ_2,2)$. You might like to pick up a copy of May's "Simplicial Objects in Algebraic Topology" if you would like to learn a little more about the bar construction.
– Tyrone
Jul 15 at 17:17
add a comment |Â
up vote
4
down vote
Given a connected topological space $X$ with $pi_n(X) = G$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.
Suppose you start with $X = mathbbRP^2$ which has $pi_1(X) cong mathbbZ_2$. Then $pi_2(mathbbRP^2) = mathbbZ$ and is generated by the covering map $S^2 to mathbbRP^2$. Attaching a three-cell to $mathbbRP^2$ with attaching map given by the covering map $S^2 to mathbbRP^2$ gives a space $X'$ with $pi_1(X') = mathbbZ_2$ and $pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbRP^3$. Now note that $pi_3(mathbbRP^3) cong mathbbZ$ generated by the covering map $S^3 to mathbbRP^3$. Attaching a four-cell to $mathbbRP^3$ with attaching map given by the covering map $S^3 to mathbbRP^3$ gives a space $X''$ with $pi_1(X'') = mathbbZ_2$ and $pi_2(X'') = pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbRP^4$. Repeating this procedure ad infinitum, we see that $mathbbRP^infty$ is a $K(mathbbZ_2, 1)$.
Suppose now you start with $X = S^2$ which has $pi_1(X) = 0$ and $pi_2(X) cong mathbbZ$. Then $pi_3(S^2) cong mathbbZ$ and is generated by the Hopf map $S^3 to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 to S^2$ gives a space $X'$ with $pi_1(X') = 0$, $pi_2(X') cong mathbbZ$, and $pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbCP^2$. Now note that $pi_4(mathbbCP^2) = 0$ so we don't need to glue on any five-cells, but $pi_5(mathbbCP^2) cong mathbbZ$ generated by the quotient map $S^5 to mathbbCP^2$. Attaching a six-cell to $mathbbCP^2$ with attaching map given by the quotient map $S^5 to mathbbCP^2$ gives a space $X''$ with $pi_1(X'') = 0$, $pi_2(X'') cong mathbbZ$, and $pi_3(X'') = pi_4(X'') = pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbCP^3$. Repeating this procedure ad infinitum, we see that $mathbbCP^infty$ is a $K(mathbbZ, 2)$.
The case for cyclic groups is similar to the first example. Note that $mathbbRP^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 to S^1$ is a topological space $X$ with $pi_1(X) cong mathbbZ_k$. The above procedure then shows that the infinite lens space $S^infty/mathbbZ_k$ is a $K(mathbbZ_k, 1)$. In order to obtain a $K(mathbbZ_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $pi_n(X) cong mathbbZ_k$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ_k, n)$.
As for forming $K(mathbbZ, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ, n)$.
For $n geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $pi_n(Xtimes Y) cong pi_n(X)opluspi_n(Y)$).
For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
add a comment |Â
up vote
4
down vote
Given a connected topological space $X$ with $pi_n(X) = G$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.
Suppose you start with $X = mathbbRP^2$ which has $pi_1(X) cong mathbbZ_2$. Then $pi_2(mathbbRP^2) = mathbbZ$ and is generated by the covering map $S^2 to mathbbRP^2$. Attaching a three-cell to $mathbbRP^2$ with attaching map given by the covering map $S^2 to mathbbRP^2$ gives a space $X'$ with $pi_1(X') = mathbbZ_2$ and $pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbRP^3$. Now note that $pi_3(mathbbRP^3) cong mathbbZ$ generated by the covering map $S^3 to mathbbRP^3$. Attaching a four-cell to $mathbbRP^3$ with attaching map given by the covering map $S^3 to mathbbRP^3$ gives a space $X''$ with $pi_1(X'') = mathbbZ_2$ and $pi_2(X'') = pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbRP^4$. Repeating this procedure ad infinitum, we see that $mathbbRP^infty$ is a $K(mathbbZ_2, 1)$.
Suppose now you start with $X = S^2$ which has $pi_1(X) = 0$ and $pi_2(X) cong mathbbZ$. Then $pi_3(S^2) cong mathbbZ$ and is generated by the Hopf map $S^3 to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 to S^2$ gives a space $X'$ with $pi_1(X') = 0$, $pi_2(X') cong mathbbZ$, and $pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbCP^2$. Now note that $pi_4(mathbbCP^2) = 0$ so we don't need to glue on any five-cells, but $pi_5(mathbbCP^2) cong mathbbZ$ generated by the quotient map $S^5 to mathbbCP^2$. Attaching a six-cell to $mathbbCP^2$ with attaching map given by the quotient map $S^5 to mathbbCP^2$ gives a space $X''$ with $pi_1(X'') = 0$, $pi_2(X'') cong mathbbZ$, and $pi_3(X'') = pi_4(X'') = pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbCP^3$. Repeating this procedure ad infinitum, we see that $mathbbCP^infty$ is a $K(mathbbZ, 2)$.
The case for cyclic groups is similar to the first example. Note that $mathbbRP^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 to S^1$ is a topological space $X$ with $pi_1(X) cong mathbbZ_k$. The above procedure then shows that the infinite lens space $S^infty/mathbbZ_k$ is a $K(mathbbZ_k, 1)$. In order to obtain a $K(mathbbZ_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $pi_n(X) cong mathbbZ_k$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ_k, n)$.
As for forming $K(mathbbZ, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ, n)$.
For $n geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $pi_n(Xtimes Y) cong pi_n(X)opluspi_n(Y)$).
For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Given a connected topological space $X$ with $pi_n(X) = G$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.
Suppose you start with $X = mathbbRP^2$ which has $pi_1(X) cong mathbbZ_2$. Then $pi_2(mathbbRP^2) = mathbbZ$ and is generated by the covering map $S^2 to mathbbRP^2$. Attaching a three-cell to $mathbbRP^2$ with attaching map given by the covering map $S^2 to mathbbRP^2$ gives a space $X'$ with $pi_1(X') = mathbbZ_2$ and $pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbRP^3$. Now note that $pi_3(mathbbRP^3) cong mathbbZ$ generated by the covering map $S^3 to mathbbRP^3$. Attaching a four-cell to $mathbbRP^3$ with attaching map given by the covering map $S^3 to mathbbRP^3$ gives a space $X''$ with $pi_1(X'') = mathbbZ_2$ and $pi_2(X'') = pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbRP^4$. Repeating this procedure ad infinitum, we see that $mathbbRP^infty$ is a $K(mathbbZ_2, 1)$.
Suppose now you start with $X = S^2$ which has $pi_1(X) = 0$ and $pi_2(X) cong mathbbZ$. Then $pi_3(S^2) cong mathbbZ$ and is generated by the Hopf map $S^3 to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 to S^2$ gives a space $X'$ with $pi_1(X') = 0$, $pi_2(X') cong mathbbZ$, and $pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbCP^2$. Now note that $pi_4(mathbbCP^2) = 0$ so we don't need to glue on any five-cells, but $pi_5(mathbbCP^2) cong mathbbZ$ generated by the quotient map $S^5 to mathbbCP^2$. Attaching a six-cell to $mathbbCP^2$ with attaching map given by the quotient map $S^5 to mathbbCP^2$ gives a space $X''$ with $pi_1(X'') = 0$, $pi_2(X'') cong mathbbZ$, and $pi_3(X'') = pi_4(X'') = pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbCP^3$. Repeating this procedure ad infinitum, we see that $mathbbCP^infty$ is a $K(mathbbZ, 2)$.
The case for cyclic groups is similar to the first example. Note that $mathbbRP^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 to S^1$ is a topological space $X$ with $pi_1(X) cong mathbbZ_k$. The above procedure then shows that the infinite lens space $S^infty/mathbbZ_k$ is a $K(mathbbZ_k, 1)$. In order to obtain a $K(mathbbZ_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $pi_n(X) cong mathbbZ_k$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ_k, n)$.
As for forming $K(mathbbZ, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ, n)$.
For $n geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $pi_n(Xtimes Y) cong pi_n(X)opluspi_n(Y)$).
For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
Given a connected topological space $X$ with $pi_n(X) = G$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$, you can build a $K(G, n)$ by attaching cells of dimension at least $n + 2$ to kill the higher homotopy groups.
Suppose you start with $X = mathbbRP^2$ which has $pi_1(X) cong mathbbZ_2$. Then $pi_2(mathbbRP^2) = mathbbZ$ and is generated by the covering map $S^2 to mathbbRP^2$. Attaching a three-cell to $mathbbRP^2$ with attaching map given by the covering map $S^2 to mathbbRP^2$ gives a space $X'$ with $pi_1(X') = mathbbZ_2$ and $pi_2(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbRP^3$. Now note that $pi_3(mathbbRP^3) cong mathbbZ$ generated by the covering map $S^3 to mathbbRP^3$. Attaching a four-cell to $mathbbRP^3$ with attaching map given by the covering map $S^3 to mathbbRP^3$ gives a space $X''$ with $pi_1(X'') = mathbbZ_2$ and $pi_2(X'') = pi_3(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbRP^4$. Repeating this procedure ad infinitum, we see that $mathbbRP^infty$ is a $K(mathbbZ_2, 1)$.
Suppose now you start with $X = S^2$ which has $pi_1(X) = 0$ and $pi_2(X) cong mathbbZ$. Then $pi_3(S^2) cong mathbbZ$ and is generated by the Hopf map $S^3 to S^2$. Attaching a four-cell to $S^2$ with attaching map given by the Hopf map $S^3 to S^2$ gives a space $X'$ with $pi_1(X') = 0$, $pi_2(X') cong mathbbZ$, and $pi_3(X') = 0$; in fact, $X'$ is homotopy equivalent to $mathbbCP^2$. Now note that $pi_4(mathbbCP^2) = 0$ so we don't need to glue on any five-cells, but $pi_5(mathbbCP^2) cong mathbbZ$ generated by the quotient map $S^5 to mathbbCP^2$. Attaching a six-cell to $mathbbCP^2$ with attaching map given by the quotient map $S^5 to mathbbCP^2$ gives a space $X''$ with $pi_1(X'') = 0$, $pi_2(X'') cong mathbbZ$, and $pi_3(X'') = pi_4(X'') = pi_5(X'') = 0$; in fact, $X''$ is homotopy equivalent to $mathbbCP^3$. Repeating this procedure ad infinitum, we see that $mathbbCP^infty$ is a $K(mathbbZ, 2)$.
The case for cyclic groups is similar to the first example. Note that $mathbbRP^2$ can be viewed as a circle with a two-cell attached via the double covering map $S^1 to S^1$. Likewise, a circle with a two-cell attached via the $k$-sheeted covering map $S^1 to S^1$ is a topological space $X$ with $pi_1(X) cong mathbbZ_k$. The above procedure then shows that the infinite lens space $S^infty/mathbbZ_k$ is a $K(mathbbZ_k, 1)$. In order to obtain a $K(mathbbZ_k, n)$, one can form the space $X$ obtained by attaching an $(n + 1)$-cell to $S^n$ with attaching map given by a degree $k$ map $S^n to S^n$; equivalently, $X$ is the previous space suspended $n - 1$ times. This space has $pi_n(X) cong mathbbZ_k$ and $pi_i(X) = 0$ for $i = 1, dots, n - 1$. I don't know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ_k, n)$.
As for forming $K(mathbbZ, n)$'s, there is a simple starting space, namely $X = S^n$ (which is $S^1$ suspended $n - 1$ times). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(mathbbZ, n)$.
For $n geq 2$, a $K(G, n)$ only exists if $G$ is abelian. If in addition $G$ is finitely generated, then a $K(G, n)$ can be formed by taking products of the examples above (because $pi_n(Xtimes Y) cong pi_n(X)opluspi_n(Y)$).
For $n = 1$, if $G$ is abelian and finitely generated, then as before, you can just take a product of spaces constructed above to obtain a $K(G, 1)$. However, $G$ can be non-abelian, and taking products of the spaces above cannot produce a $K(G, 1)$ for $G$ not abelian. However, there is a simple way to construct a space $X$ with $pi_1(X) = G$ which you can then apply the above procedure to in order obtain a $K(G, 1)$. Take a presentation for $G$. First form a bouquet of circles, one for each generator, and now attaching two-cells, one for each relation, with attaching map determined by the word used in the relation; this is the desired $X$ (which is sometimes called a presentation complex for $G$). Again, I do not know of an alternate description of the space you obtain from the above procedure applied to $X$, other than it is a $K(G, 1)$.
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
edited Aug 2 at 2:45
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
answered Jul 15 at 22:06
Michael Albanese
61.3k1591290
61.3k1591290
add a comment |Â
add a comment |Â
up vote
2
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One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G times G to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $mathsfGrp$.
Then $K(Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(Bbb Z_2, 1) simeq BbbRP^infty$ and maps $X to BbbRP^infty$ classify line bundles on $X$ upto isomorphism. This bijection $mathcalL(X) to [X, BbbRP^infty]$ can be established by sending a line bundle $ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X times BbbR^infty$, thinking of $BbbRP^infty$ as a Grassmannian.
As Tyrone described, there is a tower of fibrations $$cdots to K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n) to K(Bbb Z, n+1) to cdots$$ induced from the short exact sequence $0 to Bbb Z to Bbb Z to Bbb Z_2 to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$cdots to H^n(X; Bbb Z) to H^n(X; Bbb Z) to H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z) to cdots$$ induced from the change-of-coefficients sequence $0 to C^*(X; Bbb Z) to C^*(X; Bbb Z) to C^*(X; Bbb Z_2) to 0$ The snake map $H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z)$ is a natural transformation $H^n(-; Bbb Z_2)! Rightarrow! H^n+1(-; Bbb Z)$ and by representability, is Yoneda-dual to the map $K(Bbb Z_2, n) to K(Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(Bbb Z_2, n) to BK(Bbb Z, n)$ of the fibration $K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n)$.
This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.
There is a sort of striking geometry to $K(Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $textTOP(n)$ the group of germs-at-origin of homeomorphisms of $Bbb R^n$ fixing the origin. Let $textPL(n) subset textTOP(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$textTOP(n)/textPL(n) to BtextPL(n) to BtextTOP(n)$$ There are natural morphisms $textTOP(n) to textTOP(n+1)$ restricting to $textPL(n) to textPL(n+1)$, given by sending a germ of a self-homeomorphism of $Bbb R^n$ to that of $Bbb R^n times Bbb R$, extended by identity on the $Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $textTOP$ and $textPL$ respectively. The aforementioned fibration should induce a fibration $$textTOP/textPL to BtextPL to BtextTOP$$ Kirby and Seibenmann proved that $textTOP/textPL$ is a $K(Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M to BtextTOP$ which lifts to a map $M to BtextPL$ if and only if $M$ admits a PL-manifold structure. The $textTOP/textPL$-bundle over $BtextTOP$ is classified by a map $BtextTOP to B(textTOP/textPL)$, which has fiber $BtextPL$. So there seems to be another bundle $BtextPL to BtextTOP to B(textTOP/textPL)$, which should be nothing but the one induced from $textPL to textTOP to textTOP/textPL$. Then the lifting condition is equivalent to demanding the pushforward $M to BtextTOP to B(textTOP/textPL)$ is nullhomotopic as it factors through the fibers. But since $textTOP/textPL cong K(Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M to BK(Bbb Z_2, 3) cong K(Bbb Z_2, 4)$, which is an element of $[M, K(Bbb Z_2, 4)] cong H^4(M; Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).
EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $textTOP/textPL$ is simply the fiber of the fibration $BtextPL to BtextTOP$, so does not have an obvious group structure, which makes $B(textTOP/textPL)$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^k+1(M; pi_k textTOP/textPL)$, and using the fact that $textTOP/textPL cong K(Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; Bbb Z_2)$.
I don't know if any other $K(Bbb Z_2, n)$ has some tractable topological model though.
answered Jul 15 at 18:54
Balarka Sen
9,57812955
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
add a comment |Â
up vote
2
down vote
One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G times G to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $mathsfGrp$.
Then $K(Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(Bbb Z_2, 1) simeq BbbRP^infty$ and maps $X to BbbRP^infty$ classify line bundles on $X$ upto isomorphism. This bijection $mathcalL(X) to [X, BbbRP^infty]$ can be established by sending a line bundle $ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X times BbbR^infty$, thinking of $BbbRP^infty$ as a Grassmannian.
As Tyrone described, there is a tower of fibrations $$cdots to K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n) to K(Bbb Z, n+1) to cdots$$ induced from the short exact sequence $0 to Bbb Z to Bbb Z to Bbb Z_2 to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$cdots to H^n(X; Bbb Z) to H^n(X; Bbb Z) to H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z) to cdots$$ induced from the change-of-coefficients sequence $0 to C^*(X; Bbb Z) to C^*(X; Bbb Z) to C^*(X; Bbb Z_2) to 0$ The snake map $H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z)$ is a natural transformation $H^n(-; Bbb Z_2)! Rightarrow! H^n+1(-; Bbb Z)$ and by representability, is Yoneda-dual to the map $K(Bbb Z_2, n) to K(Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(Bbb Z_2, n) to BK(Bbb Z, n)$ of the fibration $K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n)$.
This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.
There is a sort of striking geometry to $K(Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $textTOP(n)$ the group of germs-at-origin of homeomorphisms of $Bbb R^n$ fixing the origin. Let $textPL(n) subset textTOP(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$textTOP(n)/textPL(n) to BtextPL(n) to BtextTOP(n)$$ There are natural morphisms $textTOP(n) to textTOP(n+1)$ restricting to $textPL(n) to textPL(n+1)$, given by sending a germ of a self-homeomorphism of $Bbb R^n$ to that of $Bbb R^n times Bbb R$, extended by identity on the $Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $textTOP$ and $textPL$ respectively. The aforementioned fibration should induce a fibration $$textTOP/textPL to BtextPL to BtextTOP$$ Kirby and Seibenmann proved that $textTOP/textPL$ is a $K(Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M to BtextTOP$ which lifts to a map $M to BtextPL$ if and only if $M$ admits a PL-manifold structure. The $textTOP/textPL$-bundle over $BtextTOP$ is classified by a map $BtextTOP to B(textTOP/textPL)$, which has fiber $BtextPL$. So there seems to be another bundle $BtextPL to BtextTOP to B(textTOP/textPL)$, which should be nothing but the one induced from $textPL to textTOP to textTOP/textPL$. Then the lifting condition is equivalent to demanding the pushforward $M to BtextTOP to B(textTOP/textPL)$ is nullhomotopic as it factors through the fibers. But since $textTOP/textPL cong K(Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M to BK(Bbb Z_2, 3) cong K(Bbb Z_2, 4)$, which is an element of $[M, K(Bbb Z_2, 4)] cong H^4(M; Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).
EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $textTOP/textPL$ is simply the fiber of the fibration $BtextPL to BtextTOP$, so does not have an obvious group structure, which makes $B(textTOP/textPL)$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^k+1(M; pi_k textTOP/textPL)$, and using the fact that $textTOP/textPL cong K(Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; Bbb Z_2)$.
I don't know if any other $K(Bbb Z_2, n)$ has some tractable topological model though.
answered Jul 15 at 18:54
Balarka Sen
9,57812955
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
add a comment |Â
up vote
2
down vote
up vote
2
down vote
One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G times G to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $mathsfGrp$.
Then $K(Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(Bbb Z_2, 1) simeq BbbRP^infty$ and maps $X to BbbRP^infty$ classify line bundles on $X$ upto isomorphism. This bijection $mathcalL(X) to [X, BbbRP^infty]$ can be established by sending a line bundle $ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X times BbbR^infty$, thinking of $BbbRP^infty$ as a Grassmannian.
As Tyrone described, there is a tower of fibrations $$cdots to K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n) to K(Bbb Z, n+1) to cdots$$ induced from the short exact sequence $0 to Bbb Z to Bbb Z to Bbb Z_2 to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$cdots to H^n(X; Bbb Z) to H^n(X; Bbb Z) to H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z) to cdots$$ induced from the change-of-coefficients sequence $0 to C^*(X; Bbb Z) to C^*(X; Bbb Z) to C^*(X; Bbb Z_2) to 0$ The snake map $H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z)$ is a natural transformation $H^n(-; Bbb Z_2)! Rightarrow! H^n+1(-; Bbb Z)$ and by representability, is Yoneda-dual to the map $K(Bbb Z_2, n) to K(Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(Bbb Z_2, n) to BK(Bbb Z, n)$ of the fibration $K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n)$.
This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.
There is a sort of striking geometry to $K(Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $textTOP(n)$ the group of germs-at-origin of homeomorphisms of $Bbb R^n$ fixing the origin. Let $textPL(n) subset textTOP(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$textTOP(n)/textPL(n) to BtextPL(n) to BtextTOP(n)$$ There are natural morphisms $textTOP(n) to textTOP(n+1)$ restricting to $textPL(n) to textPL(n+1)$, given by sending a germ of a self-homeomorphism of $Bbb R^n$ to that of $Bbb R^n times Bbb R$, extended by identity on the $Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $textTOP$ and $textPL$ respectively. The aforementioned fibration should induce a fibration $$textTOP/textPL to BtextPL to BtextTOP$$ Kirby and Seibenmann proved that $textTOP/textPL$ is a $K(Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M to BtextTOP$ which lifts to a map $M to BtextPL$ if and only if $M$ admits a PL-manifold structure. The $textTOP/textPL$-bundle over $BtextTOP$ is classified by a map $BtextTOP to B(textTOP/textPL)$, which has fiber $BtextPL$. So there seems to be another bundle $BtextPL to BtextTOP to B(textTOP/textPL)$, which should be nothing but the one induced from $textPL to textTOP to textTOP/textPL$. Then the lifting condition is equivalent to demanding the pushforward $M to BtextTOP to B(textTOP/textPL)$ is nullhomotopic as it factors through the fibers. But since $textTOP/textPL cong K(Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M to BK(Bbb Z_2, 3) cong K(Bbb Z_2, 4)$, which is an element of $[M, K(Bbb Z_2, 4)] cong H^4(M; Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).
EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $textTOP/textPL$ is simply the fiber of the fibration $BtextPL to BtextTOP$, so does not have an obvious group structure, which makes $B(textTOP/textPL)$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^k+1(M; pi_k textTOP/textPL)$, and using the fact that $textTOP/textPL cong K(Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; Bbb Z_2)$.
I don't know if any other $K(Bbb Z_2, n)$ has some tractable topological model though.
answered Jul 15 at 18:54
Balarka Sen
9,57812955
One way to understand Eilenberg-Maclane spaces is using the fact that they represent cohomology. Namely, over the category of CW-complexes, the functors $H^n(-; G)$ and $[-, K(G, n)]$ are naturally isomorphic. If $G$ is moreover an abelian group, then the multiplication map $G times G to G$ is a group homomorphism, therefore induces a multiplication map on $K(G, n)$ that makes it an $H$-space; hence $[X, K(G, n)]$ gets a group structure, and these are isomorphic as functors to $mathsfGrp$.
Then $K(Bbb Z_2, n)$ can be thought as the (unique upto homotopy equivalence) space which represents $H^n(X; Bbb Z_2)$. For $n = 1$ there is an explicit way to describe this correspondence, since $K(Bbb Z_2, 1) simeq BbbRP^infty$ and maps $X to BbbRP^infty$ classify line bundles on $X$ upto isomorphism. This bijection $mathcalL(X) to [X, BbbRP^infty]$ can be established by sending a line bundle $ell/X$ to the Gauss map of it's fiberwise embedding in some Euclidean bundle $X times BbbR^infty$, thinking of $BbbRP^infty$ as a Grassmannian.
As Tyrone described, there is a tower of fibrations $$cdots to K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n) to K(Bbb Z, n+1) to cdots$$ induced from the short exact sequence $0 to Bbb Z to Bbb Z to Bbb Z_2 to 0$, but the dual to this tower (or what this tower of fibrations represent) is the Bockstein long exact sequence $$cdots to H^n(X; Bbb Z) to H^n(X; Bbb Z) to H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z) to cdots$$ induced from the change-of-coefficients sequence $0 to C^*(X; Bbb Z) to C^*(X; Bbb Z) to C^*(X; Bbb Z_2) to 0$ The snake map $H^n(X; Bbb Z_2) to H^n+1(X; Bbb Z)$ is a natural transformation $H^n(-; Bbb Z_2)! Rightarrow! H^n+1(-; Bbb Z)$ and by representability, is Yoneda-dual to the map $K(Bbb Z_2, n) to K(Bbb Z, n+1)$ that is the "connecting morphism" in the fiber sequence. An alternative description of this map would be as classifying map $K(Bbb Z_2, n) to BK(Bbb Z, n)$ of the fibration $K(Bbb Z, n) to K(Bbb Z, n) to K(Bbb Z_2, n)$.
This is mostly how I think about higher Eilenberg-Maclane spaces, as usually they don't have an easy topological model.
There is a sort of striking geometry to $K(Bbb Z_2, 3)$, and I'll try to explain my limited understanding of the story (if there are one or several misunderstandings here I'd really appreciate if someone could comment on that): Let's call $textTOP(n)$ the group of germs-at-origin of homeomorphisms of $Bbb R^n$ fixing the origin. Let $textPL(n) subset textTOP(n)$ be the subgroup of germs of piecewise linear such homeomorphisms. There is a fibration $$textTOP(n)/textPL(n) to BtextPL(n) to BtextTOP(n)$$ There are natural morphisms $textTOP(n) to textTOP(n+1)$ restricting to $textPL(n) to textPL(n+1)$, given by sending a germ of a self-homeomorphism of $Bbb R^n$ to that of $Bbb R^n times Bbb R$, extended by identity on the $Bbb R$-factor. Call the direct limit of the directed system given by these morphisms to be $textTOP$ and $textPL$ respectively. The aforementioned fibration should induce a fibration $$textTOP/textPL to BtextPL to BtextTOP$$ Kirby and Seibenmann proved that $textTOP/textPL$ is a $K(Bbb Z_2, 3)$-space. A striking consequence of this seems to be the following: if $M$ is a topological manifold, there is a classifying map $M to BtextTOP$ which lifts to a map $M to BtextPL$ if and only if $M$ admits a PL-manifold structure. The $textTOP/textPL$-bundle over $BtextTOP$ is classified by a map $BtextTOP to B(textTOP/textPL)$, which has fiber $BtextPL$. So there seems to be another bundle $BtextPL to BtextTOP to B(textTOP/textPL)$, which should be nothing but the one induced from $textPL to textTOP to textTOP/textPL$. Then the lifting condition is equivalent to demanding the pushforward $M to BtextTOP to B(textTOP/textPL)$ is nullhomotopic as it factors through the fibers. But since $textTOP/textPL cong K(Bbb Z_2, 3)$, this is equivalent to saying the obstruction to $M$ having a PL-structure is homotopical nontriviality of a map $M to BK(Bbb Z_2, 3) cong K(Bbb Z_2, 4)$, which is an element of $[M, K(Bbb Z_2, 4)] cong H^4(M; Bbb Z_2)$. If I understand correctly this is the Kirby-Seibenmann class. A consequence of this is that any manifold with $H^4 = 0$ has a PL structure (so for example, topological 3-manifolds automatically always have PL structures).
EDIT: As @MikeMiller points out in the comment below, the argument for this conclusion is going to be more subtle that I described. $textTOP/textPL$ is simply the fiber of the fibration $BtextPL to BtextTOP$, so does not have an obvious group structure, which makes $B(textTOP/textPL)$ nonsensical. Apparently if the obstruction theory is done, the obstruction classes for the existence of that lifting would belong to $H^k+1(M; pi_k textTOP/textPL)$, and using the fact that $textTOP/textPL cong K(Bbb Z_2, 3)$, one arrives at the fact that the only obstruction lies in $H^4(M; Bbb Z_2)$.
I don't know if any other $K(Bbb Z_2, n)$ has some tractable topological model though.
answered Jul 15 at 18:54
Balarka Sen
9,57812955
edited Jul 15 at 21:24
answered Jul 15 at 18:54
Balarka Sen
9,57812955
answered Jul 15 at 18:54
Balarka Sen
9,57812955
answered Jul 15 at 18:54
Balarka Sen
9,57812955
9,57812955
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
add a comment |Â
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
1
1
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
This is nice, but note 1) The KS class is rather obtained by obstruction theory on the fibration TOP/PL -> BPL -> BTOP, as opposed to constructing a map to some B(TOP/PL); 2) The obstruction theory in Kirby-Seibenman only works in dimension at least 5 - TOP = PL in dimension 3 is proved in different ways, and there are topological 4-manifolds with ks=0 that do not have a smooth structure; see Freedman's classification.
– Mike Miller
Jul 15 at 20:07
add a comment |Â
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