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Let $F$ be a field that we want to compute its rational algebraic $K$-theory using the Quillen's $Q$-construction. Let $QF$ be the $Q$ construction of the category of finite dimensional vector spaces over $F$. There is a rank filtration on $QF$. Because of a theorem of Solomon-Tits that says the homotopy type of the Tits building of a vector space is bouquet of spheres, this rank filtration gives the following spectral sequence for computing the rational homology of $QF$:

$E^1_{p,q}= H_q(GL_p(F^p),st(F^p))\otimes \mathbb{Q}=H_q((F^{\times})^p,\mathbb{Q})\implies H_{p+q}(QF)\otimes \mathbb{Q}$

For details of the spectral sequence you can see Theorem 4.3.3 here. It is kind of saying that the rational homology of $QF$ which contains the rational homotopy groups as primitive elements is determined by only multiplicative group of $F$.

Any integer $n$ acts on these homologies by raising the elements of $F^{\times}$ to the power of $n$. Raising to the power of $n$ acts on $H_q((F^{\times})^p,\mathbb{Q})$ by multiplication by $n^q$. So for each $n$ you can split the $H_i(QF,\mathbb{Q})$ into subspace that eigenvalue of the action of $n$ on those subspaces are $1,n,n^2,\cdots , n^{i-1}$ so subsequently these integers $n$ are acting on the rational algebraic $K$-theory such that $K_{i-1}(F)\otimes \mathbb{Q}$ splits to eigenspaces corresponding to the eigenvalues $1,n,n^2,\cdots , n^{i-1}$. As you can see these eigenspaces do not depend on the choice of $n$.

Does the action of $n$ defined above coincide with the action of the Adams operator $\psi^n$? Since they satisfy very similar properties but I cannot see how they are the same.

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    $\begingroup$ I'm not sure about the identification $H_q(GL_p(F^p),st(F^p))=H_q((F^\times)^p)$. This doesn't seem to be in the linked paper of Kahn. It also seems to contradict the homological vanishing of Ash-Putnam-Sam arxiv.org/abs/1704.08344 An identification as claimed would follow via Shapiro's lemma if the Steinberg representation was induced from the trivial rep of the Borel, but the character of the Steinberg is the alternating sum of induced reps from all parabolics. $\endgroup$
    – Matthias Wendt
    Commented Jul 15, 2019 at 12:57
  • $\begingroup$ My idea was that the generators of $st(F^p)$ are given by a choice of $p$ linearly independent lines. $GL_p(F^p)$ is acting transitively on these and I assumed that the stabilizer of $p$ linearly independent lines are just the monomial matrices which their rational homology coincides with $H_q((F^{\times})^p)$. But I think the problem is that the stabilizer is not quite the monomials i.e. those with even permutation is fixing but odd permutations are just multiplying by -1 and that prevents us to use Shapiro. (If you think this was incorrect as well please let me know!) $\endgroup$
    – user127776
    Commented Jul 15, 2019 at 15:34
  • $\begingroup$ If what I wrote above is correct then using the Shapiro lemma you can deduce that $H_q(GL_p(F^p),st(F^p))$ is isomorphic to the group homology of monomial matrices i.e. the product of permutation matrices and $(F^\times)^p$ with coefficients in $\mathbb{Z}$ such that action on $\mathbb{Z}$ is defined by the parity of the permutation. If it is odd it multiplies by -1 otherwise it is identity. $\endgroup$
    – user127776
    Commented Jul 15, 2019 at 16:12
  • $\begingroup$ @MatthiasWendt The more I read the paper you have linked the more confused I get. They claim that $H_0(GL_p(K^p),st(K^p))$ for a field $K$ is zero for $p\geq 2$. But the zeroth homology is the co-invariants since $GL_p(K^p)$ is acting transitively on the spheres and sends each sphere to another sphere or fixes it or multiplies it by -1 the co-invariants will be always $\mathbb{Z}/2$. I'm definitely missing something ... $\endgroup$
    – user127776
    Commented Jul 15, 2019 at 17:57
  • $\begingroup$ If we write the Steinberg representation as homology of the Tits building, then generators can be described as follows: any apartment of the building gives a sphere. Any basis of the vector space gives an apartment by considering all the possible flags built from the basis. But to get a basis of the representation, we restrict to strictly upper triangular matrices and the bases given by their column vectors. So for $GL_n(\mathbb{F}_q)$ the Steinberg representation has dimension $q^m$ with $m=n(n-1)/2$ the number of positive roots. $\endgroup$
    – Matthias Wendt
    Commented Jul 16, 2019 at 8:29

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