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.


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...)


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