5
$\begingroup$

By the work of Hill-Yarnall, for the group $G=C_p,$ all the slices for any spectrum, in particular, for $S^V \wedge H\underline{\mathbb{Z}}$, are classified. Here $V$ is a representation of $C_p.$

Again, following Yarnall's work, we know the spectrum $S^n \wedge H\underline{\mathbb{Z}}$ has the $n$-slice of the form $S^{W(n)}\wedge H\underline{\mathbb{Z}}$. Here $W(n)$ is a certain representation defined in Definition 3.5 of Yarnall's work "The slices of $S^n \wedge H\underline{\mathbb{Z}}$ for cyclic $p$-groups.

$\mathbf{Question:}$ If we take any $C_p$-representation $V=m+n\xi$. Is it true the $\dim(V)$-slice of the spectrum $S^V \wedge H\underline{\mathbb{Z}}$ is of the form $S^{U(m,n)}\wedge H\underline{\mathbb{Z}}$? If so, can we write explicitly what this representation $U(m,n)$ is (may be in terms of $W(m)$ and $n$)?

Thank you so much in advance. Any help will be appreciated.

$\endgroup$
3
$\begingroup$

This follows from the Hill-Yarnall formula for slices (I guess they do regular slices, so you have to deal with a shift if you want the classical ones). The reason is as follows: the slice of $X$ in dimension $n$ is given by first applying some algebraic procedure to $\pi_WX$ where $W$ is a certain representation of dimension $n$, and then suspending that Mackey functor by $W$. It turns out that, in our case, this algebraic procedure will always split out a Mackey functor which is equivalent to a suspension of $\underline{\mathbb{Z}}$.

Specifically, we have $S^V \wedge \underline{\mathbb{Z}}$ and you want the $\mathrm{dim}(V)$-slice. So we'll need to compute some $\pi_{W-V}\underline{\mathbb{Z}}$ where $W-V$ has dimension zero. You either get $\underline{\mathbb{Z}}$ or you get the transferred version $\underline{\mathbb{Z}}_{\mathrm{tr}}$ (when $p=2$ there is one further possibility, which is the Mackey functor that has $\mathbb{Z}$ on underlying with the sign representation and $0$ on fixed points). Now you do one of three possible algebraic procedures: (i) nothing, (ii) mod out the kernel of the restriction, or (iii) take the submackey functor generated by the transfer. The possible results of these algebraic procedures are again either $\underline{\mathbb{Z}}$ or $\underline{\mathbb{Z}}_{\mathrm{tr}}$ (or that extra possibility at $p=2$). So the $\mathrm{dim}(V)$-slice is given by the $W$-suspension of either $\underline{\mathbb{Z}}$ or $\underline{\mathbb{Z}}_{\mathrm{tr}}$. But now $\underline{\mathbb{Z}}_{\mathrm{tr}} \simeq \Sigma^{2-\lambda}\underline{\mathbb{Z}}$, so that's secretly also a suspension of $\underline{\mathbb{Z}}$. (Similarly, when $p=2$ you have the additional equivalence that $\Sigma^{1-\sigma}\underline{\mathbb{Z}}$ is the Mackey functor with $0$ fixed point part and $\mathbb{Z}$ with the sign representation on underlying).

You can work out what $W$ is explicitly (there's just gonna be several cases depending on writing the dimension of $V$ in the form $rp+2k+\varepsilon$ and splitting up into subcases depending on whether some quantity is positive or negative and so on...)

$\endgroup$

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.