Update: As mentioned below, the answer to the original question is a strong No. However, the case of $\pi_4$ remains, and actually I think that this one would follow from Suslin's conjecture on injectivity of $H_i(\mathrm{GL}_n(F))\to H_i(\mathrm{GL}_{n+1}(F))$ up to torsion (in the case $i=4$, $n=3$) together with some results of Dupont (specifically exact sequence (15.10) in Dupont's book "Scissors congruences, group homology and characteristic classes") and Goncharov's conjecture on motivic cohomology in terms of Bloch groups in weights $2$ and $3$. Concretely:

Question. Let $F$ be any field. Consider the inclusion $F^\times\to \mathrm{PGL}_2(F)$ of the diagonal matrices. Is the induced map $$ \bigwedge^4 F^\times_{\mathbb Q} = H_4(F^\times,\mathbb Q)\to H_4(\mathrm{PGL}_2(F),\mathbb Q)$$ surjective?

(Explanation: In Dupont's (15.10), the kernel of the last map is precisely $H^2(F,\mathbb Q(3))$ under Goncharov's conjecture. Under Suslin's conjecture, $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ is injective, and the target is stable. Computing in terms of the homology of the $K$-theory space, one sees that the contributions to $H_4(\mathrm{PGL}_4(F),\mathbb Q)$ are $\mathrm{Sym}^2 K_2^M(F)_{\mathbb Q}$ and $H^2(F,\mathbb Q(3))$ and $K_4^M(F)_{\mathbb Q}$. The last term $K_4^M(F)_{\mathbb Q}$ is precisely the cokernel of the map $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ by Suslin's theorem, so $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ should have contributions from $H^2(F,\mathbb Q(3))$ and $\mathrm{Sym}^2 K_2^M(F)_\mathbb{Q}$. The latter admits a surjection from $\mathrm{Sym}^2(\bigwedge^2 F^\times)_{\mathbb{Q}} = H_4((F^\times)^2\rtimes \Sigma_3)$, which is precisely the decomposable part of $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ in Dupont's work. Looking back at (15.10), this shows that the first term should be zero, which leads to the question. There is some confusion between $\mathrm{PSL}$ and $\mathrm{PGL}$ in what I said, but at least assuming that all elements of $F$ have $6$-th roots, the argument should work.)

Here is a hopelessly naive question. Please point me to the relevant literature!

Let $F$ be any field. The Cartan subgroup of $\mathrm{PGL}_2(F)$ is $F^\times\rtimes \Sigma_2$. Let $X_2(F)$ be the homotopy cofiber of $B(F^\times\rtimes\Sigma_2)\to B\mathrm{PGL}_2(F)$, so there is a long exact sequence $$ \ldots\to H_i(F^\times\rtimes\Sigma_2)\to H_i(\mathrm{PGL}_2(F))\to H_i(X_2(F))\to \ldots $$ (I'm not sure if I should write $H_i(G)$ or $H_i(BG)$...). If I am not making a stupid mistake, then the known computations of the homology of $\mathrm{GL}_2(F)$ going up to the first unstable group $H_3(\mathrm{GL}_2(F))$ can be summarized by saying that $H_i(X_2(F))=0$ for $i=1,2$, and $H_3(X_2(F))=\mathfrak p(F)$ is the pre-Bloch group of $F$. (The pre-Bloch group is the quotient of $\mathbb Z[F^\times]$ by the ``$5$-term relations''.) (I might be off by some $2$-torsion.) By Hurewicz (and the check that $\pi_1 X_2(F)=0$), the same holds true for the homotopy groups of $X_2(F)$.

As far as I could find, little is known about the homology of $\mathrm{GL}_2(F)$ or $\mathrm{PGL}_2(F)$ in degrees $>3$. The rational structure could be determined by the following very naive question.

Question. Are the homotopy groups $\pi_i X_2(F)$ bounded torsion for $i\neq 3$?

I might even expect the implicit bound to be independent of $F$ (but of course depend on $i$; the order of magnitude should be $i!$).

The only evidence I have is that a back-of-the-envelope calculation seems to suggest that this holds true for finite fields. In that case the order of $\mathrm{PGL}_2(\mathbb F_q)$ is the product of $q-1$, $q$ and $q+1$. Localized at $q-1$, the homology agrees with the homology of the Cartan; localized at $q$, it is only in degrees larger than $q$ (which is OK as the bound on torsion may depend on $i$), and localized at $q+1$, the homology seems to agree with the homology of a $K(\mathbb Z/(q+1),3)$ (up to bounded torsion in each degree), which is roughly $\mathbb Z/(q+1)$ in all degrees $\equiv 3\mod 4$ (and is bounded torsion in other degrees). The pre-Bloch group $\mathfrak p(\mathbb F_q)$ is also $\mathbb Z/(q+1)$ up to $2$-torsion, so things add up.

The case of $F=\mathbb Q$ might be amenable to computations, but I was unable to do those. Are there any relevant results about $H_4(\mathrm{GL}_2(F))$ that might shed light on $\pi_4 X_2(F)$?

A final remark: I expect that it is critical that $F$ is a field. The similar statement should be false already for discrete valuation rings, or $\mathbb Z[\tfrac 1n]$ (which is why the case of $F=\mathbb Q$ is not so easy to compute; I presume one would try to first compute for all $\mathbb Z[\tfrac 1n]$ and then pass to a colimit, and the desired structure should only appear in the colimit).

Update: As mentioned by Matthias Wendt in the comments, a special case of the work of Borel-Yang says that $H_i(\mathrm{PGL}_2(\mathbb Q),\mathbb Q)=0$ for $i>0$. Inserting this into the above long exact sequence, one immediately sees that $X_2(\mathbb Q)$ is in fact rather complicated, and for example has nontorsion $\pi_5$. Thus, the answer to the question is a strong No. Thanks for the help in sorting this out!

  • 6
    $\begingroup$ @Will Sawin: This is basically Quillen's classical computation. $\endgroup$ – Peter Scholze Apr 21 '18 at 15:19
  • 3
    $\begingroup$ @Matthias Wendt: "Bounded torsion" means "bounded exponent". I would expect something like "$\pi_i X_2(F)$ is killed by $i!$ for $i\neq 3$". I would imagine that the torsion is of the same cardinality as $F$. $\endgroup$ – Peter Scholze Apr 21 '18 at 15:38
  • 3
    $\begingroup$ @S. carmeli: I don't think the cohomological dimension of $F$ should be relevant. In fact, there is no "Galois descent" (even approximate, like maybe in degrees bigger than the cohomological dimension or so (like for $K$-theory)) for $X_2(F)$. In fact, for $F=\overline{\mathbb F}_p$, the Bloch group is $0$, and all homotopy groups of $X_2(F)$ are bounded torsion. Thus, there is no hope to recover $X_2(\mathbb F_p)$ from $X_2(\overline{\mathbb F}_p)$. $\endgroup$ – Peter Scholze Apr 22 '18 at 16:23
  • 5
    $\begingroup$ According to Borel and Yang, the rational cohomology of $SL_2(F)$ with $F$ a number field is the tensor product: for every real place an exterior algebra generated by an Euler class in degree 2, for every complex place an exterior algebra generated by a class in the Bloch group. So it's reasonable to assume that in this case $X_2(F)$ is rationally a product of 3-spheres. I haven't checked the details, but that would be partial support for assuming that the higher homotopy groups (above degree 3) of $X_2(F)$ are torsion. $\endgroup$ – Matthias Wendt Apr 23 '18 at 10:42
  • 4
    $\begingroup$ For the function field case, Harder's results show that rational cohomology of $SL_2(F)$ is torsion. I don't know of any results bounding torsion there. The homology of $X_2(F)$ is the equivariant homology of a complex of configurations of points on $\mathbb{P}^1$. The exponential bounds for torsion (in cohomology) could follow if we knew that all torsion comes from symmetry groups of configurations. But I guess we don't know that. Finally: why should we assume that it's easier to study homotopy of $X_2(F)$ instead of its homology? $\endgroup$ – Matthias Wendt Apr 23 '18 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.