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Let $G$ and $H$ be two connected Lie groups. By the Dold-Lashof construction the classifying space $BHom(G,H)$ is well-defined (similar to the Milnor construction).

Is there a relation between $BHom(G,H)$ and the space of pointed maps $Map_0(BG,BH)$? More precisely, could there be a homotopy equivalence or highly connected map between these spaces?

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up vote 9 down vote accepted

What is $BHom(G;H)$? Typically, $Hom(G;H)$ is not a group unless $H$ is abelian. Maybe you want to talk about the natural map

$$ Hom(G;H) \to Map_0 (BG;BH) $$

from the space of homomorphisms to the mapping space. In some cases, this is a homotopy equivalence, for example if $G$ is connected and compact and $H=U(1)$. In that case, $Hom(G,H)$ is discrete (it is $\cong Hom(\pi_1(G),\mathbb{Z})$), and $Map_0 (BG;BU(1))=Map_0 (BG;K(\mathbb{Z};2))$. The latter space has homotopy groups $\pi_i (Map_0 (BG;K(\mathbb{Z};2))) = \tilde{H}^{2-i} (BG;\mathbb{Z})$, which is zero unless $i=0$ and it is not so difficult to see that $Hom (G,H)\to Map_0 (BG;BU(1))$ is a bijection on components.

If the target $H$ is not abelian, things become much more difficult. An example is $Hom(S^3;S^3)$. Up to conjugacy, there are only two homomorphisms $S^3 \to S^3$. But there are infinitely many homotopy classes of maps $BS^3 \to BS^3$. This was proven by Sullivan. One can show that the degree of a map $BS^3 \to BS^3$ (meaning the effect on $H^4 (BS^3)=\mathbb{Z}$) is always an odd square. This is discussed in Hatchers textbook. Sullivan then proved, using profinite homotopy theory, that for each odd square, there is indeed a map with that degree.

As far as I remember, questions of this kind are also discussed in the series "maps between classifying spaces" by Adams and Mahmud, but I do not know enough to say something on it.

EDIT: if $H$ is assumed to be discrete, then the map is always a weak homotopy equivalence. Proof: In that case, $Hom(G;H)=Hom(\pi_0 (G);H)$, with the discrete topology. Moreover, for each space $X$ (which is homotopy equivalent to a CW complex), there is a homotopy equivalence $Map_0(X;BH)\cong Hom(\pi_1 (X);H)$; insert $X=BG$.

EDIT2: if $G$ is discrete, but infinite, then almost nothing can be said. Look at $H=U(1)$ and let $G$ be a perfect group. Then each homomorphism $G \to U(1)$ is trivial, but $Map(BG,BU(1))$ has the homotopy type $H^2 (G)$ (its components are contractible).

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Thank you for your answer and your example. In fact, I consider $Hom(G,H)$ as a topological monoid with the compact-open topology. Thus $Hom(G,H)$ is a H-space and the Dold-Lashof construction works for H-spaces. – stud137 Jan 9 '12 at 10:42
@stud137: How is Hom(G,H) a monoid? What is its composition (if H is not abelian) ? – HenrikRüping Jan 9 '12 at 14:43
@HenrikRüping: If $H$ is connected, then $Hom(G,H)$ is a monoid up to homotopy by pointwise multiplication, thus a H-space. – stud137 Jan 9 '12 at 15:17
$Hom(G,H)$ is not a monoid. It's not closed under multiplication in any reasonable way. The example with $S^3$ suggested by Johannes Ebert is a couterexample. Take any two homomorphisms $S^3\to S^3$ homotopic to identity. If you pointwise multiply any two representatives you get a map $S^3\to S^3$ of degree 2 which of course can not be homotopic to a homomorphism. – Vitali Kapovitch Jan 9 '12 at 16:37
In addition to Johannes' answer: Notbohm ("Maps between classifying spaces") has shown that for a torus $T$ and a compact Lie group $G$, the natural map $Rep(T,G) \to Map_0(BT,BG)$ is a bijection, where $Rep$ denotes a set of representatives of non-conjugate hom's $T \to G$. – Ralph Jan 9 '12 at 20:35

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