Generalised geometric weak conjecture on spherical classes and non-factorisation of Kervaire invariant one elements

Abstract

This paper is on the Curtis conjecture. We show that the image of the Hurewicz homomorhism $h:{\pi }_{\ast }{Q}_0{S}^0\to H_{\ast }(Q_0{S}^0;\mathbb{Z})$, when restricted to product of positive dimensional elements, is determined by $\mathbb{Z}\{h({\eta }^2),h({\nu }^2),h({\sigma }^2)\}$. Localised at $p=2$, this proves a geometric version of a result of Hung and Peterson for the Lannes-Zarati homomorphism. We apply this to show that, for $p=2$ and $G=O(1)$ or any prime $p$ and $G$ any compact Lie group with Lie algebra $\mathfrak{g}$ so that $\dim \mathfrak{g}>0$, the composition $${}_p{\pi }_{\ast }Q{\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n}\to {}_p{\pi }_{\ast }{Q}_0{S}^0\stackrel{h}{\to }{H}_{\ast }(Q_0{S}^0;\mathbb{Z}/p)$$ where ${\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n}\to S^0$ is the $n$-fold transfer, is trivial if $n>2$. Moreover, we show that for $n=2$, the image of the above composition vanishes on all elements of Adams filtration at least $1$, i.e. those elements of ${}_2{\pi }_{\ast }^s{\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n}$ represented by a permanent cycle ${\mathrm{Ext}}_{A_p}^{s,t}({\tilde{H}}^{\ast }{\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n},\mathbb{Z}/p)$ with $s>0$, map trivially under the above composition. The case of $n>2$ of the above observation proves and generalises a geometric variant of the weak conjecture on spherical classes due to Hung, later on verified by Hung and Nam. We also show that, for a compact Lie group $G$, Curtis conjecture holds if we restrict to the image of the $n$-fold transfer ${\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n}\to S^0$ with $n>1$. Finally, we show that the Kervaire invariant one elements ${\theta }_j\in {}_2{\pi }_{2^{j+1}-2}^s$ with $j>3$ do not factorise through the $n$-fold transfer ${\Sigma }^{n\dim \mathfrak{g}}B{G}_{+}^{\wedge n}\to S^0$ with $n>1$ for $G=O(1)$ or any compact Lie group as above.

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