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We know that the classifying space $$ BO(1)=B\mathbb{Z}_2=\mathbb{RP}^{\infty} $$ $$ BU(1)=\mathbb{CP}^{\infty} $$ $$ BSU(2)=\mathbb{HP}^{\infty} $$

  • How do one construct/derive $$ BSU(n)=? $$

  • Can one explain $B\mathbb{Z}_2$, $BU(1)$, $BSU(n)$ in a unified consistent but also intuitive way?

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In general for a group $G$, there exists a contractible free $G$-space $EG$ whose quotient $BG=EG/G$ is a classifying space for $G$.

The construction of $EG$ is not unique and may be carried out functorially using, say, a bar construction or Milnor infinite join construction. Whatever the method adopted, the space $EG$ exists and is unique up to a $G$-equivariant homeomorphism. The resulting classifying space $BG=EG/G$ is also unique up to homotopy equivalence.

An important point is that the equalities you have written in your question are not correct, and should be replaced with homotopy equivalences since these spaces are only unique up to homotopy.

The construction can be made functorial in $G$, so that a homomorphism $\phi:G\rightarrow H$ induces a continuous $G$-map $E\phi:EG\rightarrow EH$ and hence a continuous quotient $B\phi:BG\rightarrow BH$. In particular, if $\phi$ is a group isomorphism, then $B\phi$ is a homotopy equivalence, and the isomorphic groups $G$, $H$ have the same classifying space.

For the cases of $G=\mathbb{Z}_2$, $G=U(1)\cong S^1$ and $G=SU(2)\cong S^3$, one finds that the infinite dimensional sphere $S^\infty=colim\; S^n$ is contractible (it is weakly contractible by definition and carries CW structure) and models $EG$. Indeed $\mathbb{Z}_2$ acts freely on each $S^n$ with quotient $\mathbb{R}P^{n-1}$. Since the equatorial inclusions $S^n\hookrightarrow S^{n+1}$ defining the colimit are $\mathbb{Z}_2$-equivariant, $\mathbb{Z}_2$ also acts freely on $S^\infty$. Hence

$$B\mathbb{Z}_2\simeq S^\infty/\mathbb{Z}_2\cong colim\;(S^n/\mathbb{Z}_2)=colim\;\mathbb{R}P^{n-1}=\mathbb{R}P^\infty.$$

Similarly $S^1\subseteq\mathbb{C}$ acts freely on each $S^{2n-1}\subseteq \mathbb{C}^n\cong\mathbb{R}^{2n}$ with quotient $\mathbb{C}P^{n-1}$. Since the sequence $S^{2n-1}$ is cofinal, $S^1$ acts freely on $S^\infty$. Hence

$$BU(1)\simeq BS^1\simeq S^\infty/S^1\cong colim\;(S^{2n-1}/S^1)=colim\;\mathbb{C}P^{n-1}=\mathbb{C}P^\infty.$$

Finally $S^3\subseteq\mathbb{H}$ acts freely on each $S^{4n-1}\subseteq \mathbb{H}^n\cong\mathbb{R}^{4n}$ with quotient $\mathbb{H}P^{n-1}$, and by the same reasoning we get

$$BSU(2)\simeq BS^3\simeq S^\infty/S^3\cong colim\;(S^{4n-1}/S^3)=colim\;\mathbb{H}P^{n-1}=\mathbb{H}P^\infty.$$

This nice little run stops here. The 8-dimensional division algebra of the octonians is not a group, since it is not associative. You can form the Cayley plane but cannot go all the way to a classifying space.

In general there will be no nice space representing the homotopy type of the classifying space of a given group $G$. Of course, there are homotopy equivalences like $E(G\times H)\cong EG\times EH$, giving $B(G\times H)\simeq BG\times BH$ which can be useful. Also, if $H\leq G$ is a subgroup then $H$ acts freely on $EG$. Hence $BH\simeq (EG)/H$ (which is another way to think about the previous results for $\mathbb{Z}_2\leq S^1\leq S^3$). In this case the homotopy class $BH\rightarrow BG$ induced by the subgroup inclusion has a canonical representative, namely the coset projection $(EG)/H\rightarrow (EG)/G$, $eH\mapsto eG$, which makes it clear that the (homotopy) fibre of this map is the orbit space $G/H$.

Since the classical groups are obviously useful, and feature prominently in the classification of Lie groups, their classifiny spaces have been much studied. Indeed, for $\mathbb{K}=\mathbb{R},\mathbb{C},\mathbb{H}$ the general linear group $Gl_n(\mathbb{K})$ acts freely on the Stiefel manifold $V_{n}(\mathbb{K}^{n+k})$ of $n$-frames in $\mathbb{K}^{n+k}$. Forming the colimit $V_n(\mathbb{K}^{\infty})=colim_k\; V_n(\mathbb{K}^{n+k})$ one finds that it is contractible and carries a free $Gl_n(\mathbb{K})$-action. Hence

$$EGl_n(\mathbb{K})\cong V_n(\mathbb{K}^{\infty}),\qquad BGl_n(\mathbb{K})\simeq V_n(\mathbb{K}^{\infty})/Gl_n(\mathbb{K}^\infty)=Gr_n(\mathbb{K}^\infty)$$

where $Gr_n(\mathbb{K}^\infty)$ is the Grassmannian of $n$-planes in $\mathbb{K}^\infty$. Since the $\mathbb{K}$-orthogonal and special orthogonal groups $O(\mathbb{K}^n)$ and $SO(\mathbb{K}^n)$ are subgroups of $Gl_n(\mathbb{K})$, we get classifying spaces from the above construction. Firstly we have $BO(\mathbb{K}^n)\simeq V_n(\mathbb{K}^{\infty})/O(\mathbb{K}^n)$, which fibres over $BGl_n(\mathbb{K}^\infty)$ with contractible fibre $Gl_n(\mathbb{K}^\infty)/O(\mathbb{K}^n)\simeq\mathbb{K}^N$ (this follows using Gram-Schmidt). Hence the map $BO(\mathbb{K}^n)\rightarrow BGl(\mathbb{K}^n)$ induced by the subgroup inclusion is a homotopy equivalence, and in particular

$$BO(n)\simeq Gr_n(\mathbb{R}^\infty),\qquad BU(n)\simeq Gr_n(\mathbb{C}^\infty),\qquad BSp(n)\simeq Gr_n(\mathbb{H}^\infty).$$

In fact we can model this a little better by replacing $V_n(\mathbb{K}^\infty)$ with the Steifel manifold $\widetilde V_n(\mathbb{K}^\infty)$ of orthogonal $n$-frames in $\mathbb{K}^\infty$ and proceeding as before.

For $\mathbb{K}=\mathbb{R},\mathbb{C}$ we have also the special orthogonal groups and fibration sequences

$$O(\mathbb{K}^n)/SO(\mathbb{K^n})\rightarrow BSO(\mathbb{K}^n)\rightarrow Gr_n(\mathbb{K}^\infty)$$

where $O(n)/SO(n)\cong\mathbb{Z}_2$ and $U(n)/SU(n)\cong S^1$. We can model the middle classifying space explicitly by replacing $V_n(\mathbb{K}^\infty)$ with the Stiefel manifold $\widehat V_n(\mathbb{K}^\infty)$ of oriented orthonormal $n$-frames in $\mathbb{K}^\infty$. In particular

$$BSO(n)\simeq \widehat{Gr}_n(\mathbb{R}^\infty)$$

is the Grassmannian manifold of oriented $n$-planes in $\mathbb{R}^\infty$. The complex case isn't so simple, since any complex plane has a canonical orientation. We are left with a principal $S^1$-bundle

$$S^1\rightarrow BSU(n)\rightarrow\widehat{Gr}_n(\mathbb{C}^\infty)$$

which is classified by the map $B\text{det}:BU(n)\rightarrow BU(1)\simeq BS^1\simeq K(\mathbb{Z},2)$ which classifies the determinant homomorphism and corresponds to the first Chern class. Thus

$$BSU(n)\rightarrow BU(n)\xrightarrow{c_1}K(\mathbb{Z},2)$$

is a homotopy fibration sequence, and we can identify $BSU(n)$ with the $3$-connected cover of $BU(n)$ (recall that if $G$ is a compact Lie group then $\pi_2G=\pi_3BG=0$, so $\pi_3BSU(n)$ is automatically $0$).

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    $\begingroup$ That is an amazing answer. +1. $\endgroup$ – wonderich Jul 14 '18 at 16:24
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    $\begingroup$ Isn't $\operatorname{Gr}_n(\mathbb{R}^{\infty})$ actually a model of $BO(n)$ rather than $BSO(n)$? $\endgroup$ – Michael Albanese Jul 15 '18 at 9:25
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    $\begingroup$ It is, you just need to liberally add in the adjectives oriented and orthonormal to get things straight. Thanks for pointing out the error, its been corrected. $\endgroup$ – Tyrone Jul 15 '18 at 11:02
  • $\begingroup$ I think $\operatorname{Gr}_n(\mathbb{C}^{\infty})$ is a model of $BU(n)$, not $BSU(n)$. The latter should be $3$-connected (because $SU(n)$ is a simply connected Lie group), but that is not true of $\operatorname{Gr}_n(\mathbb{C}^{\infty})$, e.g. $\operatorname{Gr}_1(\mathbb{C}^{\infty}) = \mathbb{CP}^{\infty}$. $\endgroup$ – Michael Albanese Jul 15 '18 at 23:04
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    $\begingroup$ @Tyrone I think the quotient of $\hat V_n(\Bbb C^\infty)$ by $SU(n)$ removes the dependence on the unitary frame, but doesn't change $\det(e_1, \cdots, e_n) \in S^1$. Then $BSU(n)$ is the space of $n$-planes with a fixed unitary isomorphism $\Lambda^{\text{rk}}_{\Bbb C}(E) \cong \Bbb C$ (given by the determinant). There are $S^1$ such isomorphisms/determinants, which explains your fiber sequence. $\endgroup$ – user98602 Jul 16 '18 at 12:30

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