$\DeclareMathOperator\SL{SL}$The stable real cohomology of $\SL_n(\mathbb Z)$ was computed by Borel: it is given by $\mathbb R[z_i\mid i=5,9,13,\dotsc]$ with $z_i$ in degree $i$. One may wonder whether the pull back of the stable class $z_i$ on $\SL(\mathbb Z)$ to $\SL_n (\mathbb Z)$ for some finite $n$ is non-zero. That is, whether $\iota_n^* z_i = 0 \in H^i(\SL_n(\mathbb Z),\mathbb R)$ for $\iota_n \colon \SL_n(\mathbb Z) \to \SL(\mathbb Z)$.

Borel's proof also gives a real cohomological stability result and hence gives a range of $n$'s for which this is true, but in 1978 Ronnie Lee announced a proof in On unstable cohomology classes of $\SL_n(\mathbb Z)$ that $z_i$ is non-zero as long as $i<2n-3$, a much larger range of $n$'s than what can be deduced from Borel's proof. His paper says

Because the proofs of Theorems 1 and 2 require careful geometric construction, it is planned to present the detailed proof later, and we will indicate here some of the ideas involved in the case when $n$ is odd.

but as far as I know no detailed proof has appeared. My questions are: Has any proof of Ronnie Lee's theorem appeared in the literature? If not, what is the best known range for non-vanishing of the stable Borel classes?

  • 1
    $\begingroup$ There is a paper by Jens Franke on " a topological model for some summand of Eisenstein cohomology" (see the math review ams.org/mathscinet-getitem?mr=2402680 ) . The section 7.5 deals with the case of $SL_n(\mathbb Z)$ ; Franke's result is very general and it is an exercise to read the special case of SL(n) from his paper. (I hope the following link to the chapter works:) rdcu.be/xQhw $\endgroup$ Oct 29 '17 at 3:19

You can indeed read this off from the work of Franke, as was done in Section 4.3 of Characteristic classes of bundles of K3 manifolds and the Nielsen realization problem by Jeffrey Giansiracusa, myself, and Bena Tshishiku. In particular, Lee's result is true.


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.