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There is a map $BG \to A(\ast)$ where $BG$ classifies stable spherical fibrations and $A(\ast)$ is Waldhausen's algebraic $K$-theory of a point. The map is induced by applying Quillen's plus construction to the inclusion $$ BGL_1(S^0) \to BGL_\infty(S^0) $$ where $BGL_1(S^0)$ is $BG$. Here $BGL_\infty(S^0)$ can be defined as the homotopy colimit over $k$ and $n$ of $BG(\vee_k S^n)$, where $G(\vee_k S^n)$ is the topological monoid of homotopy automorphisms of a $k$-fold wedge of $n$-spheres. (Note: $A(\ast) = \Bbb Z \times BGL_\infty(S^0)^+$.)

Question 1: I've heard it mentioned that there can be no retraction $A(*) \to BG$ to the map $BG \to A(\ast)$. Can someone please explain to me why there is no such splitting, and if possible, give a reference?

More generally,

Question 2: If $R$ is a structured ring spectrum, is there a reasonable set of conditions which guarantees that the map $BGL_1(R)\to K(R)$ admits a retraction?

(A related question is under what conditions does an $R$-module map $f: R^n \to R^n$ which is a weak homotopy equivalence admit a determinant $\text{det}(f) \in GL_1(R)$).

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up vote 4 down vote accepted

Question 1: There are several arguments.

In degree 2 there is a reference: the proof of corollary 3.7 of Waldhausen's "Algebraic K-theory of spaces, a manifold approach". See for a copy. Consider the maps $BG \to A(\ast) \to K(Z)$ and apply $\pi_2$. Here $\pi_2 BG = Z/2$, the composite is zero (restrict to $BSG$) and $\pi_2 A(\ast) \to K_2(Z)$ is an isomorphism. Hence the first homomorphism cannot be injective.

In degree 3 there is another argument. Right composition with the unstable map $\eta : S^3 \to S^2$ takes the generator of $\pi_2(BO)$ to zero in $\pi_3(BO) = 0$, so by naturality with respect to $BO \to BG$, composition with $\eta$ takes the generator of $\pi_2(BG) = Z/2$ to zero in $\pi_3(BG)$. On the other hand, composition with $\eta$ takes the generator of $\pi_2 Q(S^0)$ to a nonzero element in $\pi_3 Q(S^0)$, so by naturality with respect to $Q(S^0) \to A(\ast)$, composition with $\eta$ is nonzero from $\pi_2 A(*) = Z/2$ to $\pi_3 A(\ast)$. Therefore $BG$ cannot split off $A(*)$.

In degree 4 there is your argument.

Question 2: So far, I only know that $BGL_1(R) \to K(R)$ admits a retraction if $R$ is the realization of a commutative simplicial ring. There is no retraction for $R = ku$ by Corollary 2.3 of "Divisibility of the Dirac magnetic monopole as a two-vector bundle over the three-sphere" by Ch. Ausoni, B. I. Dundas and J. Rognes, Documenta Mathematica 13 (2008) 795-801, and ku is already pretty close to a commutative simplicial ring.

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I just realized how the argument for Question 1 might go (I hope this isn't self-indulgence on my part):

The composite $BO \to BG \to A(*)$ factors through $Q(S^0)$. So it suffices to show there is no map $Q(S^0) \to BG$ such that the composite $BO \to Q(S^0) \to BG$ is homotopic to the usual map. If this map existed then the composite $\pi_4(BO) \to \pi_4^{\text{st}}(S^0) \to \pi_4(BG)$ would be the map $\Bbb Z \to \Bbb Z_{24}$ which is the J-homomorphism in this dimension, so it's non-trivial.

However, we get a contradiction with this since $\pi^{\text{st}}_4(S^0)$ is trivial.

Is this the usual proof?

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I think I have heard that line of argument verbally from John Rognes. I don't know if it is written anywhere. – Neil Strickland Jan 25 '11 at 8:57

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