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Let $A(X)$ denote the Waldhausen's algebraic K-theory of a space $X$, and let $n$ be odd.

  1. Are the rational homotopy groups of $A(S^n)$ known?

  2. Is the group $\pi_{2k}(A(S^n))$ finite for all positive $k\ll n$?

A reference (or proof sketch) would be appreciated.

EDIT: I found that the answers are stated (without reference or proof) on page 1 in "Homological stability of diffeomorphism groups" by Alexander Berglund and Ib Madsen, http://arxiv.org/abs/1203.4161. Namely, the answers to both questions is yes.

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Let $\tilde A(X)$ be the reduced functor, i.e., the homotopy fiber of the map $A(X) \to A(\ast)$. Since $A(*)$ is rationally a product of $K(Q,4j+1)$ for $j \ge 1$, we may as well study $\tilde A(X)$ instead.

The rational homotopy of $\Omega \tilde A(\Sigma Y)$ was studied in

G. Carlsson, R. Cohen, T. Goodwillie, and W. Hsiang, The free loop space and the algebraic K-theory of spaces. K-Theory 1 (1987), no. 1, 53–82.

(The paper has some gaps but these were later corrected.) If we assume that $Y$ is connected, then the rational homotopy type of $\Omega \tilde A(\Sigma Y)$ coincides with that of the functor: $$ \prod_{n\ge 1} Q(Y^{[n]}_{h\Bbb Z_n}) $$ where $Y^{[n]}$ is the $n$-fold smash product of $Y$, $\Bbb Z_n$ acts by cyclic permutation and ${({-})}_{h\Bbb Z_n}$ means homotopy orbits. Rationally, the homotopy groups of $Q(Y^{[n]}_{h\Bbb Z_n})$ coincide the homology groups of $Y^{[n]}_{\Bbb Z_n}$. When $Y$ is a $j$-sphere, the rational homology is not hard to compute.

Finally, if $X = S^1$, we know that $\tilde A(S^1)$ is rationally the same as $B A(*)$ by a version of Bass-Heller-Swan. But by what I mentioned above, $B\tilde A(*)$ is a rationally the product of $K(\Bbb Q,4j+2)$, $j \ge 1$.

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  • $\begingroup$ Thank you! I gather from your "not hard to compute" that there is no easy to state answer. Am I correct that $\pi_{k}(A(S^n))$ is rationally zero for even $k$ with $k<n$? I think this follows from the formula on page 1 of the linked Berglund-Madsen's paper, could you confirm this (if true)? $\endgroup$
    – Igor Belegradek
    Oct 18, 2013 at 1:53
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    $\begingroup$ No, $A(S^n)$ is not rationally $(n-1)$-connected. But the reduced space $\tilde A(S^n)$ is, by the displayed formula in my answer. The reason that $A(S^n)$ is not rationally $(n-1)$-connected is that $A(*)$ splits off of it. $\endgroup$
    – John Klein
    Oct 18, 2013 at 2:31
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    $\begingroup$ John, the description of $\Omega \tilde A(\Sigma Y)$ for connected $Y$ is valid on the nose, just rationally. I've taken the liberty of editing your answer accordingly. $\endgroup$
    – Tom Goodwillie
    Oct 18, 2013 at 4:17
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    $\begingroup$ Also, the rational answer is older than that paper. Rationally $A(X)$ is the same as the algebraic $K$-theory of the simplicial ring $\mathbb Z[G]$ where $G$ is a Kan loop group for the connected space $X$, and when $X$ is simply connected then the reduced part of this is rationally the same as cyclic homology of $\mathbb Q[G]$ by work of mine, and the latter is the same as rational homology of $S^1$ homotopy orbits of the free loopspace of $X$. $\endgroup$
    – Tom Goodwillie
    Oct 18, 2013 at 4:23
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    $\begingroup$ Tracing back the references I found all the rational homotopy groups of $A(S^n)$ listed on pp.228-229 of 1982 paper ["A Model for Computing Rational Algebraic K-Theory of Simply Connected Spaces", Hsiang-Staffeldt]. $\endgroup$
    – Igor Belegradek
    Oct 18, 2013 at 22:55

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