text:
\begin{document}

\begin{abstract}

Let $R$ be an $E_\infty$-ring spectrum. Given a map $\zeta$ from a space $X$ to $BGL_1R$, one can construct a Thom spectrum, $X^\zeta$, which generalises the classical notion of Thom spectrum for spherical fibrations in the case $R=S^0$, the sphere spectrum. If $X$ is a loop space ($\simeq \Omega Y$) and $\zeta$ is homotopy equivalent to $\Omega f$ for a map $f$ from $Y$ to $B^2GL_1R$, then the Thom spectrum has an $A_\infty$-ring structure. The Topological Hochschild Homology of these $A_\infty$-ring spectra is equivalent to the Thom spectrum of a map out of the free loop space of $Y$.

This paper considers the case $X=S^1$, $R=K_p^\wedge$, the p-adic $K$-theory spectrum, and $\zeta = 1-p \in \pi_1BGL_1K_p^\wedge$. The associated Thom spectrum $(S^1)^\zeta$ is equivalent to the mod p $K$-theory spectrum $K/p$. The map $\zeta$ is homotopy equivalent to a loop map, so the Thom spectrum has an $A_\infty$-ring structure. I will compute $\pi_*THH^{K_p^\wedge}(K/p)$ using its description as a Thom spectrum.
\end{abstract}
\footnotetext[1]{supported in part by ERC-Adg TMSS-228082}
\maketitle
\tableofcontents

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\parindent=0.5cm

\markboth{}{}
\thispagestyle{myheadings}

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\section{Introduction}

 The goal of this paper is to use generalised Thom spectra to calculate the Topological Hochschild Homology of $K/p$ in the category of modules over $K_p^\wedge$.

 Let $R$ be a ring spectrum and $GL_1R$ its space of units. It is the $H$-space of homotopy automorphisms of $R$ as an $R$-module. An $R$-twisting of a space $X$ is a continuous map $\zeta$ from $X$ to $BGL_1R$. Associated to $\zeta$, one can define the Thom spectrum of $\zeta$, $X^\zeta$ (see \cite{A-B-G-H-R}). This notion specialises for $R=S^0$ to the Thom spectrum of a spherical fibration. The homotopy groups of $X^\zeta$ is the group of twisted $R$ homology classes with respect to the twisting $\zeta$.

 Suppose that $R$ is an $E_\infty$-ring spectrum. Then its space of units is an infinite loop space. Given a map $f:BG \rightarrow B^2GL_1R$, let $\zeta\simeq \Omega f:G\rightarrow BGL_1R$. Then the Thom spectrum $G^\zeta$ admits an $A_\infty$ $R$-algebra structure.

\subsection{$K/p$ as a module over $K_p^\wedge$}

 Suppose that  $R=K_p^\wedge$, the spectrum of $p$-adic K-theory. Let $G$ be the group $S^1$. A twisting on $S^1$ is a map $\zeta: S^1 \rightarrow BGL_1K_p^\wedge$. This is classified by the group $\pi_1(BGL_1K_p^\wedge)\cong \pi_0(GL_1K_p^\wedge)\cong Z_p^\times$. If we choose $\zeta=1-p \in Z_p^\times$, then the Thom spectrum $(S^1)^\zeta \simeq K/p$, the $\mode$ $p$ $K$-theory spectrum. Moreover, the twisting $\zeta$ can be realised as a loop map, and so, for every way of writing $\zeta \simeq \Omega f$ we get an $A_\infty$-ring structure on $K/p$ as an $K_p^\wedge$-module.

\subsection{Topological Hochschild Homology of Thom spectra}

Given a map $f$ from $X$ to $B^2GL_1R$, let $G \simeq \Omega X$ and $\zeta \simeq \Omega f : G \simeq \Omega X \rightarrow BGL_1R$. In this case, the Thom spectrum $G^\zeta$ has an $A_\infty$-ring structure. We write $\eta^*f$ for the composite

\begindc{0}[5]
\obj(-15,15){}
\obj(0,-5){}
\obj(0,10){$LX$}
\obj(10,10){$LB^2GL_1R$}
\obj(30,10){$B^2GL_1R\times BGL_1R$}
\obj(30,0){$BGL_1R\times BGL_1R$}
\obj(50,0){$BGL_1R$}
\mor{$LX$}{$LB^2GL_1R$}{}
\mor{$LB^2GL_1R$}{$B^2GL_1R\times BGL_1R$}{$\cong$}{}
\mor{$B^2GL_1R\times BGL_1R$}{$BGL_1R\times BGL_1R$}{$\eta \times id$}
\mor{$BGL_1R\times BGL_1R$}(47,0){}
\enddc

where $\eta: \Sigma R \rightarrow R$ is induced from $S^1\stackrel{\eta}{\rightarrow}S^0$ via $S^1 \wedge R \rightarrow S^0 \wedge R \simeq R$. In the above situation, $\THH^R(G^\zeta) \simeq LX^{\eta^* f}$. The case $R=S^0$ was proved in \cite{B-C-S}. The same argument applies for general $R$.

Using this identification of $\THH$ as a Thom spectrum, we compute the Topological Hochschild Homology of $K/p$. For odd primes p,
$$ \pi_*(\THH^{K_p^\wedge}(K/p))= \left\{
\begin{array}{rl}
(Z/(p^\infty))^i & \text{if } * =2k\\
0 & \text{if } * =2k+1
\end{array} \right.$$
where $i$ is an integer between $1$ and $p-1$ depending on the choice of $f$ with $\zeta \simeq \Omega f$.

Similar results were obtained before by Angeltveit in \cite{A1}. He used the B{\"o}kstedt spectral sequence (see \cite{E-K-M-M}, chapter IX).

We can also form $\mode$ $p$ $K$-theory as a Thom spectrum by starting with $X=S^3$, $R=K_p^\wedge$ and $\zeta= p \in \pi_3(BGL_1K_p^\wedge)= \pi_2(GL_1K_p^\wedge)= Z_p$. Again, this $\zeta$ can be realised as a loop map and we can compute $\THH$ of these $A_\infty$-ring structures in an analogous way. This gives the same results.

\section{The Thom spectrum}

The notion of a generalised Thom spectrum used here is discussed in detail in \cite{A-B-G-H-R}. The construction resembles a twisted version of the group ring. Given an extension of a group $G$ by the units in a field $k$,
$$(\tau): 1\rightarrow k^* \rightarrow E \rightarrow G \rightarrow 1$$
the algebra  $k^\tau[G]= \mathbb{Z}[E]\otimes_{\mathbb{Z}[k^*]} k$ is a twisted group ring. If the extension $\tau$ is trivial, one gets the group ring $k[G]$. Imitating this definition of a twisted group ring for spectra leads to the construction of the Thom spectrum. One replaces the field $k$ by an $E_\infty$-ring spectrum $R$, and the units $k^*$ by the space of units $GL_1R$ acting on $R$.

\subsection{The space of units and the Thom spectrum}

The space of units of a ring spectrum is a generalisation of the group of units of a commutative ring, the set of invertible elements under multiplication. It is defined to be the components of $\Omega^\infty R$ that lie over the units in $\pi_0 (R)$. Following \cite{A-B-G-H-R}, we make the definition

\defn
 Let $R$ be an $E_\infty$-ring spectrum. Its space of units $GL_1R$ is defined to be the homotopy pullback,

\begindc{0}[5]
\obj(0,25){}
\obj(0,5){}
\obj(30,20){$GL_1R$}
\obj(50,20){$\Omega^\infty(R)$}
\obj(30,10){$\pi_0(R)^\times$}
\obj(50,10){$\pi_0(R)$}
\mor{$GL_1R$}{$\Omega^\infty(R)$}{}
\mor{$GL_1R$}{$\pi_0(R)^\times$}{}
\mor{$\Omega^\infty(R)$}{$\pi_0(R)$}{}
\mor{$\pi_0(R)^\times$}{$\pi_0(R)$}{}
\enddc

It follows from the definition that the homotopy classes of maps from a space $X$ to $GL_1R$ are given by
$$[X,GL_1R]= R^0(X)^\times $$
the units of the cohomology ring $R^0(X)= [X,\Omega^\infty R]$.

From the pullback diagram one can read off the homotopy groups of $GL_1R$,
$$ \pi_n(GL_1R) = \left\{
\begin{array}{rl}
\pi_n(R) & \text{if } n>0\\
\pi_0(R)^\times & \text{if } n=0
\end{array} \right.$$

We note that $GL_1R$ is an $H$-space for any ring spectrum $R$. If $R$ is $E_\infty$, then $GL_1R$ is an infinite loop space: there is a connective spectrum $gl_1R$ with $0^{th}$-space is $GL_1R$ (Theorem 3.2 in \cite{A-B-G-H-R}).

We can view $\Omega^\infty R$ as the space of endomorphisms $\End_R(R,R)$, in the topological  category of $R$-modules, and $GL_1R= \Aut_R(R,R) \subset \End_R(R,R)$ as the subset of weak equivalences. Therefore, the units $GL_1R$ is the space of homotopy automorphisms of $R$ in the category of $R$-modules. In this way, the infinite loop space $GL_1R$ acts on the spectrum $R$ by weak equivalences, and $R$ is a module over the $E_\infty$ ring spectrum $\Sigma^\infty GL_1R_+$.

\defn Given a map $\zeta : X \rightarrow BGL_1R$, let $P$ be the $GL_1R$ bundle classified by $\zeta$ described as the pullback,

 \begindc{0}[5]
\obj(0,25){}
\obj(-15,10){}
\obj(10,20){$P$}
\obj(30,20){$EGL_1(R)$}
\obj(10,10){$X$}
\obj(30,10){$BGL_1(R)$}
\mor(10,20)(25,20){}
\mor(10,20)(10,10){}
\mor(30,20)(30,10){}
\mor(10,10)(25,10){$\zeta$}
\enddc
 and define the associated Thom spectrum to be
 $$X^\zeta = \Sigma^\infty P_+\wedge{^L}_{\Sigma^\infty GL_1(R)_+}R$$

 In the above $\wedge^L$ denotes the derived smash product in the category of modules over the $E_\infty$-ring spectrum $\Sigma^\infty GL_1R_+$ as in \cite{E-K-M-M}. We note from section 7 of \cite{A-B-G-H-R}, that the Thom spectrum functor commutes with homotopy colimits, and from section 8.6 of \cite{A-B-G-H-R} that it generalises the classical Thom spectrum of a spherical fibration.

The Thom spectrum of the map $* \rightarrow BGL_1R$ is weakly equivalent to $R$, since the universal bundle associated to the inclusion of a point in $BGL_1R$ is isomorphic to $GL_1R$ and $\Sigma^\infty GL_1R_+ \wedge^L_{\Sigma^\infty GL_1R_+} R \simeq R$.

Similarly, the Thom spectrum of a map $X\rightarrow BGL_1R$ which is null homotopic is weakly equivalent to $R\wedge X_+$. Indeed, the universal bundle associated to the constant map is $X\times GL_1R$. Then the Thom spectrum is $\Sigma^\infty (X\times GL_1R)_+\wedge^L_{\Sigma^\infty GL_1R_+} R \simeq (\Sigma^\infty X_+ \wedge \Sigma^\infty GL_1R_+) \wedge^L_{\Sigma^\infty GL_1R_+} R \simeq R\wedge X_+$.

Suppose that the space $X\simeq \Sigma Y$, the reduced suspension on $Y$. Then, a map $X\stackrel{\zeta}{\rightarrow} BGL_1R$ is described by a map $Y\stackrel{\widehat{\zeta}}{\rightarrow} GL_1R$, via $[X,BGL_1R]\cong [\Sigma Y,BGL_1R]\cong [Y,GL_1R]$. Such a $\widehat{\zeta}$ is a unit in $R^0(Y)$ which induces $u_\zeta : R\wedge Y_+ \rightarrow R$.

\prop Suppose that $\zeta$ is a map from $X\simeq\Sigma Y$ to $BGL_1R$. Then, the Thom spectrum $X^\zeta$ is equivalent to the homotopy colimit of $(R\leftarrow R\wedge Y_+ \rightarrow R)$ where one of the maps is the projection $p_Y$ and the other is $u_\zeta$.

\begin{proof}

The space $X$ is the homotopy colimit of $*\leftarrow Y\rightarrow *$, and this gives a homotopy pushout square of Thom spectra,

\begindc{\commdiag}[5]
\obj(0,25){}
\obj(-15,5){}
 \obj(10,20){$Y^\zeta$}
\obj(30,20){$*^\zeta$}
\obj(10,10){$*^\zeta$}
\obj(30,10){$(\Sigma Y)^\zeta$}
\mor(10,20)(10,10){}
\mor(12,20)(29,20){}
\mor(12,10)(28,10){}
\mor(30,20)(30,10){}
\enddc
The Thom spectrum $*^\zeta$ is weakly equivalent to $R$ and $Y^\zeta \simeq R \wedge Y_+$, so the homotopy pushout can be written as

\begindc{\commdiag}[5]
\obj(0,25){}
\obj(-15,5){}
 \obj(10,20){$R \wedge Y_+$}
\obj(30,20){$R$}
\obj(10,10){$R$}
\obj(30,10){$(\Sigma Y)^\zeta$}
\mor(10,20)(10,10){}
\mor(12,20)(30,20){}
\mor(10,10)(28,10){}
\mor(30,20)(30,10){}
\enddc

From this, one obtains a  Mayer Vietoris sequence for calculating the homotopy groups
$$\ldots \rightarrow \pi_*(R \wedge Y_+) \rightarrow \pi_*(R)\oplus \pi_* (R) \rightarrow \pi_*((\Sigma Y)^\zeta) \ldots$$
To compute the maps in this sequence, one must examine the $GL_1R$-bundle over $X\simeq \Sigma Y$. This restricts to trivial bundles over the two copies of the cone of $Y$ inside $X$ and on their intersection $Y$, the bundles are identified via the map $\widehat{\zeta}:Y \rightarrow GL_1R$.

In the long exact sequence, there are two maps $R_*(Y_+) \rightarrow \pi_*(R)$. One of these maps is given by the map from $Y$ to a point($p_Y$) and the other is the map $u_\zeta$ defined in the preceeding paragraph.
\end{proof}

\rem The proposition describes the homotopy groups of the Thom spectrum as twisted $R$-homology groups. An $R$-twisting on a space $X$ can be defined as a 1-cocycle in the sheaf (of groupoids) - $\lbrace \units\, \mathit{in}\, R^0(X) \rbrace$. The groupoid of units in $R^0$ is classified by the units $GL_1R$, and therefore, 1-cocycles on $X$ are equivalent to $[X,BGL_1R]$. Therefore, a twisting is given by a continuous map $\zeta$ from $X$ to $BGL_1R$.

For $X=\bigcup U_i$ a 1-cocycle defines units over $U_i\cap U_j$ satisfying a cocycle condition on further intersections. A twisted $R$ homology class is an element in each $R_*(U_i)$, two of which are identified using the values of the 1-cocycle on the intersections. The abelian group of these classes is defined to be the twisted $R$-homology of $X$ with respect to the twisting $\zeta$. This is isomorphic to the homotopy groups of the Thom spectrum $X^\zeta$. The proposition above verifies this in the case $X=\Sigma Y$, where $X$ is the union of two contractible open sets.

\subsection{Computations of some Thom spectra}

\prop Suppose that $\zeta:S^1 \rightarrow BGL_1K_p^\wedge$ represents $1-p \in \pi_1(BGL_1(K_p^\wedge))= \pi_0(GL_1(K_p^\wedge))= Z_p^\times$. Then, $(S^1)^\zeta \simeq K/p$.
 \begin{proof}
 By Proposition 2.3 with $Y=S^0$, the Thom spectrum is a homotopy pushout

\begindc{\commdiag}[5]
\obj(0,25){}
\obj(-15,5){}
 \obj(10,20){$K_p^\wedge\vee K_p^\wedge$}
\obj(30,20){$K_p^\wedge$}
\obj(10,10){$K_p^\wedge$}
\obj(30,10){$(S^1)^\zeta$}
\mor(10,20)(10,10){}
\mor(13,20)(29,20){}
\mor(10,10)(29,10){}
\mor(30,20)(30,10){}
\enddc

Therefore, there is a cofibre sequence
$$ K_p^\wedge \vee K_p^\wedge \rightarrow K_p^\wedge \vee K_p^\wedge \rightarrow (S^1)^\zeta$$
Proposition 2.3 also identifies the left map in the sequence in suitable coordinates, to be given by the matrix
\[ \left( \begin{array}{ccc}
1 & 1 \\
1 & 1-p  \end{array} \right)\]

Therefore, the cofibre sequence can be rewritten as
$$K_p^\wedge \stackrel{p}{\rightarrow} K_p^\wedge \rightarrow (S^1)^\zeta$$
so that $(S^1)^\zeta \simeq {K_p^\wedge}/p \simeq K/p$.
\end{proof}

\rem Consider the map $\zeta:S^1 \rightarrow BGL_1((S^0)_p^\wedge)$ given by $(1-p)$ as in the previous proposition. Then,  $(S^1)^\zeta \simeq (S^0)_p^\wedge /p \simeq M_p$ is the $\mode$ $p$ Moore spectrum. In fact, for any $\zeta: S^1 \rightarrow BGL_1R$, $(S^1)^\zeta \simeq \mathit{cofibre}(1-\zeta : R \rightarrow R)$. This follows from the argument above.

\prop Let $\zeta: S^3 \rightarrow BGL_1K_p^\wedge$ represent the element $p$ of
$$[S^3,BGL_1(K_p^\wedge)] = \pi_3(BGL_1(K_p^\wedge))= \pi_2(GL_1(K_p^\wedge))= \pi_2(K_p^\wedge)\cong Z_p$$
 Then $(S^3)^\zeta \simeq K/p$.

\begin{proof}

The space $S^3$ is homotopy equivalent to the suspension of $S^2$. Proposition 2.3 implies the homotopy pushout

\begindc{\commdiag}[5]
\obj(0,22){}
\obj(-15,8){}
 \obj(10,20){$ K_p^\wedge \wedge S^2_+$}
\obj(30,20){$K_p^\wedge$}
\obj(10,10){$K_p^\wedge$}
\obj(30,10){$(S^3)^\zeta$}
\mor(10,20)(10,10){}
\mor(13,20)(30,20){}
\mor(10,10)(29,10){}
\mor(30,20)(30,10){}
\enddc
and the associated Mayer Vietoris cofibre sequence
$$ K_p^\wedge \wedge (S^2) \vee K_p^\wedge \rightarrow K_p^\wedge \vee K_p^\wedge \rightarrow (S^3)^\zeta.$$
In suitable coordinates, the map in the Mayer Vietoris sequence is given by the matrix
\[ \left( \begin{array}{ccc}
1 & 0 \\
1 & p  \end{array} \right)\]
and the sequence can be rewritten as
$$\Sigma^2 K_p^\wedge \stackrel{p}{\rightarrow} K_p^\wedge \rightarrow (S^3)^\zeta $$
By Bott periodicity $\Sigma^2 K_p^\wedge \simeq K_p^\wedge$ so that $(S^3)^\zeta \simeq K_p^\wedge / p$, as claimed.
\end{proof}

\subsection{Ring Structures}

Suppose $R$ is an $E_\infty$-ring spectrum so that $GL_1R$ is an infinite loop space. Given $f:X\rightarrow B^2GL_1R$, and $\zeta:G\simeq \Omega X\stackrel{\Omega f}{\rightarrow} BGL_1R$, the Thom spectrum $G^\zeta$ has an $A_\infty$-ring structure. This follows from \cite{B1} and \cite{A-B-G-H-R}. This raises the question when a map
$$\zeta:G \rightarrow BGL_1R$$
from a monoid $G$ is homotopy equivalent to a loop map, i.e. $\zeta \simeq \Omega f$ for
$$f:BG \rightarrow B^2GL_1R.$$
We have the standard maps
$$\Sigma G \stackrel{\sigma}{\rightarrow} BG,\, \Sigma GL_1R \stackrel{\sigma}{\rightarrow} BGL_1R$$
so the question is if
$$\sigma \circ \Sigma \zeta : \Sigma G \rightarrow B^2GL_1R$$
extends over $BG$,

\begindc{\commdiag}[5]
\obj(0,25){}
\obj(-15,5){}
\obj(10,20){$\Sigma G$}
\obj(30,20){$\Sigma BGL_1(R)$}
\obj(10,10){$BG$}
\obj(30,10){$B^2GL_1(R)$}
\mor{$\Sigma G$}{$\Sigma BGL_1(R)$}{$\Sigma \zeta$}
\mor{$\Sigma BGL_1(R)$}{$B^2GL_1(R)$}{$\sigma$}
\mor{$BG$}{$B^2GL_1(R)$}{$f$}
\mor{$\Sigma G$}{$BG$}{$\sigma$}
\enddc

\prop Let $G=S^1$, $R=K_p^\wedge$ and $\zeta = 1-p$ as in Proposition 2.5, then $(S^1)^\zeta \simeq K/p$ has an $A_\infty$-ring structure.

\begin{proof}
The classifying space of $S^1$ is $CP^\infty$ so, in this case the diagram above is

\begindc{\commdiag}[5]
\obj(-23,13){}
\obj(0,-1){}
\obj(0,10){$S^2$}
\obj(20,10){$\Sigma BGL_1(K_p^\wedge)$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_p^\wedge)$}
\mor(0,10)(15,10){$\Sigma (1-p)$}
\mor(20,10)(20,0){$\sigma$}
\mor(1,0)(15,0){$f$}
\mor(0,10)(0,0){$\sigma$}
\enddc
 The space $CP^\infty$ has a CW structure made of even dimensional cells so that all the cells are attached along odd dimensional spheres. The spectrum $K_p^\wedge$ has non trivial homotopy groups only in even dimensions and hence, so does $B^2GL_1K_p^\wedge$. Thus, all the obstructions to extending the map $\Sigma 1-p$ must vanish, which implies that there is an $A_\infty$-ring structure on the Thom spectrum $K/p$.
\end{proof}

\prop Suppose that $G=S^3$, $R=K_p^\wedge$, and $\zeta =p$ as in Proposition 2.7, then the Thom spectrum has an $A_\infty$-ring structure.
\begin{proof}
 The classifying space of $S^3$ is the infinite quarternionic projective space $HP^\infty$, and $\Sigma S^3 = S^4 \rightarrow BS^3= HP^\infty$ is obtained by attaching even cells along maps of odd dimensional spheres. Therefore the extension problem can always be solved.
\end{proof}

\section{Topological Hochschild Homology of Thom spectra}

In the last section, we observed that the Thom spectrum of a loop map carries an induced $A_\infty$ structure. In this setting, there is a convenient description of the Topological Hochschild Homology as a Thom spectrum in the ideas of \cite{B-C-S} and \cite{Sch}. In the following $G$ will be a group, $X$ a space, and $G$ homotopy equivalent to $\Omega X$ as $A_\infty$-spaces. $R$ will be an $E_\infty$ ring spectrum.

 The Thom spectrum of a map $G\rightarrow BGL_1R$ is a twisted $R$-module generated by $G$. If this is a loop map, the construction is that of a twisted group ring. Recall that the Hochschild Homology of group rings over a field is given by
$$HH_*(k[G])\cong k \otimes H_*(G,G)$$
where $G$ acts on itself by conjugation. This is the homology of the Borel construction $G_{hG}\simeq EG\times_G G \simeq LBG$, the free loop space of $BG$, and so, $HH_*(k[G])\cong k\otimes H_*(LBG)$. The analogous statement for Topological Hochschild Homology is the classical result of B\"{o}kstedt and Waldhausen,
$$\THH(\Sigma^\infty \Omega X_+) \simeq \Sigma^\infty LX_+.$$

 In the category of $R$-modules, the theorem is $\THH^R( R\wedge \Omega X_+) \simeq R\wedge LX _+$, computing the Topological Hochschild Homology of the Thom spectrum of the constant map. More generally, let $f:X\rightarrow BGL_1R$ and $\zeta \simeq \Omega f :G \rightarrow BGL_1R$, the Thom spectrum has an $A_\infty$-ring structure, and the Topological Hochschild Homology is the Thom spectrum of a map from $LX$ to  $BGL_1R$.

 In the second part of the section, we apply the theorem for $R=K_p^\wedge$ and $G=S^1$, in the computation of the previous section. This implies that the Thom spectrum is homotopy equivalent to the cofibre of a certain map $K_p^\wedge \wedge CP^\infty_+ \rightarrow K_p^\wedge \wedge CP^\infty_+$.

\subsection{ Identifying Topological Hochschild Homology as a Thom spectrum}

 Recall that the free loop space $LY$ fits into a fibration
 $$\Omega Y \rightarrow LY \rightarrow Y$$
  If $Y$ is an $H$-space, then the fibration splits as $LY\simeq Y\times \Omega Y$. This is an equivalence of $H$-spaces if $Y$ is homotopy commutative.

 Let $f$ be a map from $X$ to $B^2GL_1R$ and $\eta: B^2GL_1R \rightarrow \Omega B^2GL_1R$ be induced from the Hopf map by
\begin{align}
B^2GL_1R \simeq \Maps(S^2,B^4GL_1R) & \stackrel{\eta^*}{\rightarrow} \Maps(S^3, B^4GL_1R) \nonumber \\
                              &\simeq \Maps(S^1,\Omega^2B^4GL_1R) \nonumber \\
                              &\simeq \Maps(S^1,B^2GL_1R) \nonumber \\
                              &\simeq \Omega B^2GL_1R.  \nonumber
\end{align}

Let $L^\eta f$ be the map from $LX$ to $BGL_1R$ defined by the diagram

 \begindc{0}[5]
 \obj(0,35){}
 \obj(-15,5){}
 \obj(0,30){$LX$}
\obj(15,30){$LB^2GL_1(R)$}
\obj(40,20){$\Omega B^2GL_1(R)$}
\obj(40,30){$B^2GL_1(R)\times \Omega B^2GL_1(R)$}
\obj(40,10){$BGL_1(R)$}
\mor{$LX$}{$LB^2GL_1(R)$}{$Lf$}
\mor{$LB^2GL_1(R)$}{$B^2GL_1(R)\times \Omega B^2GL_1(R)$}{$\simeq $}
\mor{$B^2GL_1(R)\times \Omega B^2GL_1(R)$}{$\Omega B^2GL_1(R)$}{$\eta \times id$}
\mor{$\Omega B^2GL_1(R)$}{$BGL_1(R)$}{$\simeq $}
\mor{$LX$}{$BGL_1(R)$}{$L^\eta f$}
\enddc
The map $\eta \times id : B^2GL_1R \times \Omega B^2GL_1R \rightarrow \Omega B^2GL_1R $ is the product of the maps $\eta$ and $\mathit{id}$ using the $H$-space structure of $\Omega B^2GL_1R$. Without proof, we state:

\thm There is a homotopy equivalence
$$\THH^R(G^\zeta)\simeq (LX)^{L^\eta f} $$

This was proved in the case of the sphere spectrum in \cite{B-C-S}, \cite{Sch}. A similar argument applies for any $E_\infty$-ring spectrum $R$. This will be accomplished in a future publication.

\subsection{The example of $G=S^1$ and $R=K_p^\wedge$}
By Proposition 2.8, we have the commutative diagram,

\begindc{\commdiag}[5]
\obj(-15,15){}
\obj(0,-5){}
\obj(0,10){$S^2$}
\obj(20,10){$\Sigma BGL_1(K_p^\wedge)$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_p^\wedge)$}
\mor(0,10)(15,10){$\Sigma 1-p$}
\mor(20,10)(20,0){$\sigma$}
\mor(1,0)(15,0){$f$}
\mor(0,10)(0,0){$\sigma$}
\enddc
and write $\THH^{K_p^\wedge}(K/p,f)$ for the Topological Hochschild Homology corresponding to this $A_\infty$-ring structure.
\prop $$\THH^{K_p^\wedge}(K/p,f)\simeq (LCP^\infty)^{\widehat{f}}$$
where $\widehat{f}$ is the composite,
$$LCP^\infty \stackrel{Lf}{\rightarrow} LB^2GL_1K_p^\wedge \simeq B^2GL_1K_p^\wedge \times BGL_1K_p^\wedge \stackrel{p_2}{\rightarrow} BGL_1K_p^\wedge$$

\begin{proof}
By Theorem 3.1, $\THH^{K_p^\wedge}(K/p,f)\simeq (LCP^\infty)^{L^\eta f}$. Since $\pi_1(K_p^\wedge)=0$, $\eta =0$ in this case. Hence, the proposition.
\end{proof}

The focus of the rest of the paper will be the calculation of $\pi_*((LCP^\infty)^{\widehat{f}})\cong \THH^{K_p^\wedge}(K/p,f)$. First of all we note that:

\prop There is a long exact sequence
$${K_p^\wedge}_* CP^\infty \rightarrow {K_p^\wedge}_* CP^\infty \rightarrow \pi_*\THH^{K_p^\wedge}(K/p,f)\rightarrow {K_p^\wedge}_{*-1} CP^\infty \ldots $$

\begin{proof}
Note that $CP^\infty$ is an infinite loop space, and hence homotopy commutative, which implies that $LCP^\infty \simeq \Omega CP^\infty \times CP^\infty\simeq S^1\times CP^\infty$. The space $S^1$ is a union of two contractible open sets whose intersection is $S^0$, so, there is a homotopy pushout,

\begindc{\commdiag}[5]
\obj(0,16){}
\obj(-25,0){}
\obj(0,13){$C P^\infty \sqcup C P^\infty$}
\obj(20,13){$C P^\infty$}
\obj(0,3){$C P^\infty$}
\obj(20,3){$LC P^\infty$}
\obj(40,8){$(*)$}
\mor(5,13)(19,13){}
\mor(0,13)(0,3){}
\mor(3,3)(18,3){}
\mor(20,13)(20,3){}
\enddc
and hence, a homotopy pushout square of Thom spectra

\begindc{\commdiag}[5]
\obj(0,15){}
\obj(-25,-5){}
\obj(0,10){$(CP^\infty \sqcup CP^\infty)^{\widehat{f}}$}
\obj(20,10){$(CP^\infty)^{\widehat{f}}$}
\obj(0,0){$(CP^\infty)^{\widehat{f}}$}
\obj(20,0){$(LCP^\infty)^{\widehat{f}}$}
\mor(6,10)(17,10){}
\mor(0,10)(0,0){}
\mor(3,0)(16,0){}
\mor(20,10)(20,0){}
\enddc

The two maps $CP^\infty \rightarrow LCP^\infty$ in (*) are the inclusion of constant loops, so, the two compositions $CP^\infty \rightarrow LCP^\infty \rightarrow LB^2GL_1K_p^\wedge \rightarrow BGL_1K_p^\wedge$ are nullhomotopic and the Thom spectra are $\simeq K_p^\wedge \wedge CP^\infty_+$. The map from $CP^\infty \sqcup CP^\infty$ to $BGL_1K_p^\wedge$ factors through $CP^\infty \rightarrow BGL_1K_p^\wedge$ so, the Thom spectrum $(CP^\infty \sqcup CP^\infty)^{f} \simeq K_p^\wedge \wedge CP^\infty_+ \vee K_p^\wedge \wedge CP^\infty_+$. Therefore, the pushout can be written as:

\begindc{\commdiag}[5]
\obj(0,15){}
\obj(-25,-5){}
\obj(0,10){$K_p^\wedge \wedge CP^\infty_+ \vee K_p^\wedge \wedge CP^\infty_+$}
\obj(25,10){$K_p^\wedge \wedge CP^\infty_+$}
\obj(0,0){$K_p^\wedge \wedge CP^\infty_+$}
\obj(25,0){$(LCP^\infty)^{Lf}$}
\mor(10,10)(21,10){}
\mor(0,10)(0,0){}
\mor(3,0)(21,0){}
\mor(25,10)(25,0){}
\enddc

This gives a Mayer Vietoris sequence on homotopy groups,
$$ \ldots \rightarrow{K_p^\wedge}_*(CP^\infty) \oplus {K_p^\wedge}_*(CP^\infty) \rightarrow {K_p^\wedge}_*(CP^\infty) \oplus {K_p^\wedge}_*(CP^\infty) \rightarrow \pi_*((LCP^\infty)^{Lf})\ldots$$
To simplify, one needs to understand the left hand map i.e., how $K_p^\wedge \wedge CP^\infty_+ \vee K_p^\infty \wedge CP^\infty_+$ maps to the two different copies of $K_p^\wedge \wedge CP^\infty_+$ in the pushout square. For that one needs to examine the structure of $P^{\widehat{f}}$, the $GL_1K_p^\wedge$-bundle over $S^1\times CP^\infty$ classified by $\widehat{f}$.

 Following the pushout square (*), we see that $P^{\widehat{f}}$ is obtained by identifying two trivial bundles over $CP^\infty$ after restricting over $CP^\infty \sqcup CP^\infty$, via a map $u:CP^\infty \sqcup CP^\infty \rightarrow GL_1K_p^\wedge$. The adjoint of $u$ is the map $\widetilde{u}$ in the diagram,

\begindc{\commdiag}[5]
\obj(0,15){}
\obj(-10,-3){}
\obj(0,10){$CP^\infty \sqcup CP^\infty$}
\obj(20,10){$CP^\infty \vee CP^\infty$}
\obj(40,10){$S^1 \times CP^\infty$}
\obj(60,10){$\Sigma CP^\infty_+ \vee \Sigma CP^\infty_+$}
\obj(40,0){$BGL_1K_p^\wedge$}
\mor(5,10)(15,10){}
\mor(25,10)(36,10){}
\mor(44,10)(53,10){}
\mor(40,10)(40,0){}
\mor(22,9)(36,1){$0$}
\mor(58,9)(44,1){$\widetilde{u}$}
\enddc

The top row is the cofibre sequence associated to the pushout (*). Since the map $S^1\times CP^\infty \rightarrow BGL_1K_p^\wedge$ is nullhomotopic on $CP^\infty\vee CP^\infty$, it factors through $\Sigma CP^\infty_+ \vee \Sigma CP^\infty_+$ as $\widetilde{u}$.

The map $u$ gives two units $u_1,\,u_2$ in the ${K_p^\wedge}^0(CP^\infty)$. In the Mayer Vietoris sequence for the Thom spectrum, these describe the map $K_p^\wedge \wedge CP^\infty_+ \vee K_p^\wedge \wedge CP^\infty_+ \rightarrow K_p^\wedge \wedge CP^\infty_+ \vee K_p^\wedge \wedge CP^\infty_+$ as the matrix,
\[ \left( \begin{array}{ccc}
1 & u_2 \\
u_1 & 1  \end{array} \right)\]

In fact, $u_1$ and $u_2$ are equal because each summand in $\Sigma CP^\infty_+$ of $\Sigma CP^\infty_+ \vee \Sigma CP^\infty_+$ is the cofibre of the map $CP^\infty \rightarrow LCP^\infty=S^1\times CP^\infty$ given by the inclusion of the constant loops and both can be defined by the same diagram,

\begindc{\commdiag}[5]
\obj(0,25){}
\obj(-25,-5){}
 \obj(0,20){$CP^\infty$}
\obj(0,10){$S^1 \times CP^\infty$}
\obj(0,0){$\Sigma CP^\infty_+$}
\obj(30,10){$BGL_1(K_p^\wedge )$}
\mor(0,20)(27,11){0}
\mor(4,10)(26,10){$\widehat{f}$}
\mor(0,20)(0,10){}
\mor(0,10)(0,0){}
\mor(3,1)(27,9){$u$}
\enddc

In terms of $u$, we can rewrite the Mayer Vietoris sequence as the long exact sequence,
$$  \ldots \rightarrow {K_p^\wedge}_*(C P^\infty) \stackrel{u-1}{\rightarrow} {K_p^\wedge}_*(C P^\infty) \rightarrow \pi_*((LC P^\infty)^{Lf})\rightarrow \ldots \ \ \ \ \ \ \  (\alpha) $$

\end{proof}

To calculate $\pi_*(THH^{K_p^\wedge}(K/p,f))$, it remains to understand the map $u$. This is done as follows:

\prop The adjoint of the map $u: \Sigma CP^\infty_+ \rightarrow BGL_1R$, is homotopy equivalent to the composite $\Sigma^2 CP^\infty_+ \stackrel{\mu}{\rightarrow}CP^\infty \stackrel{f}{\rightarrow} B^2GL_1K_p^\wedge$, where $\mu$ is the composition $\Sigma^2 CP^\infty_+ \simeq S^2 \wedge CP^\infty_+ \stackrel{\sigma \wedge id}{\rightarrow} CP^\infty \wedge CP^\infty_+ \rightarrow CP^\infty$.

\begin{proof}
The following diagram commutes:

\begindc{\commdiag}[5]
\obj(-15,15){}
\obj(0,-5){}
\obj(20,10){$S^1 \wedge LCP^\infty$}
\obj(20,0){$ CP^\infty$}
\obj(46,10){$S^1 \wedge LB^2GL_1(K_p^\wedge)$}
\obj(46,0){$B^2GL_1(K_p^\wedge )$}
\obj(0,10){$S^1 \wedge (S^1\times CP^\infty)$}
\mor{$S^1 \wedge (S^1\times CP^\infty)$}{$S^1 \wedge LCP^\infty$}{$\simeq$}
\mor{$S^1 \wedge LCP^\infty$}{$ CP^\infty$}{$\mathit{ev}$}
\mor{$S^1 \wedge LCP^\infty$}{$S^1 \wedge LB^2GL_1(K_p^\wedge)$}{$S^1 \wedge Lf$}
\mor{$ CP^\infty$}{$B^2GL_1(K_p^\wedge )$}{$f$}
\mor{$S^1 \wedge LB^2GL_1(K_p^\wedge)$}{$B^2GL_1(K_p^\wedge )$}{$\mathit{ev}$}
\enddc

 Consider the inclusion of the based loops $BGL_1K_p^\wedge \hookrightarrow LB^2GL_1K_p^\wedge$. Under the composite,
$$S^1 \times BGL_1K_p^\wedge \rightarrow S^1 \times LB^2GL_1K_p^\wedge \stackrel{\mathit{ev}}{\rightarrow} B^2GL_1K_p^\wedge,$$
 the copies $S^1\times *$ and $*\times BGL_1K_p^\wedge$ map trivially. Thus, it factors through $S^1 \wedge BGL_1K_p^\wedge$ as $\Sigma BGL_1K_p^\wedge \stackrel{\sigma}{\rightarrow} B^2GL_1K_p^\wedge$. We are trying to figure out the map
 $$S^1 \times LCP^\infty \rightarrow S^1 \times LB^2GL_1K_p^\wedge \rightarrow S^1 \times BGL_1K_p^\wedge \rightarrow B^2GL_1K_p^\wedge $$

 Then, this factors through
$$S^1 \wedge LCP^\infty \rightarrow S^1 \wedge LB^2GL_1K_p^\wedge \rightarrow S^1 \wedge BGL_1K_p^\wedge \stackrel{\sigma}{\rightarrow} B^2GL_1K_p^\wedge.$$

Also $LCP^\infty \rightarrow BGL_1 K_p^\wedge$ factors through $S^1\wedge CP^\infty_+$ as u. Putting all the remarks together, we have a commutative diagram,

\begindc{\commdiag}[5]
\obj(-5,50){}
\obj(71,-10){}
\obj(0,40){$S^2\wedge CP^\infty_+$}
\obj(66,40){$S^1\wedge BGL_1K_p^\wedge$}
\obj(20,20){$S^1 \wedge LCP^\infty$}
\obj(20,0){$ CP^\infty$}
\obj(46,20){$S^1 \wedge LB^2GL_1(K_p^\wedge)$}
\obj(66,0){$B^2GL_1(K_p^\wedge )$}
\obj(0,20){$S^1 \wedge (S^1\times CP^\infty)$}
\mor{$S^1 \wedge (S^1\times CP^\infty)$}{$S^1 \wedge LCP^\infty$}{$\simeq$}
\mor{$S^1 \wedge LCP^\infty$}{$ CP^\infty$}{$ev$}
\mor{$S^1 \wedge LCP^\infty$}{$S^1 \wedge LB^2GL_1(K_p^\wedge)$}{$S^1 \wedge Lf$}
\mor{$ CP^\infty$}{$B^2GL_1(K_p^\wedge )$}{$f$}
\mor{$S^1 \wedge LB^2GL_1(K_p^\wedge)$}{$B^2GL_1(K_p^\wedge )$}{$ev$}
\mor{$S^2\wedge CP^\infty_+$}{$S^1 \wedge (S^1\times CP^\infty)$}{}
\mor{$S^1\wedge BGL_1K_p^\wedge$}{$S^1 \wedge LB^2GL_1(K_p^\wedge)$}{}
\mor{$S^2\wedge CP^\infty_+$}{$S^1\wedge BGL_1K_p^\wedge$}{$\Sigma u$}
\mor{$S^1 \wedge LCP^\infty$}{$S^1\wedge BGL_1K_p^\wedge$}{$\Sigma \widehat{f}$}
\mor{$S^1 \wedge LCP^\infty$}{$S^2\wedge CP^\infty_+$}{}
\mor{$S^1\wedge BGL_1K_p^\wedge$}{$B^2GL_1(K_p^\wedge )$}{$\sigma$}
\enddc

The left hand vertical map from $S^2\wedge CP^\infty_+$ to $S^1\wedge (S^1\times CP^\infty)$ is the inclusion of a factor in the splitting of the suspension of $S^1\wedge (S^1\times CP^\infty)\simeq (S^2\wedge CP^\infty_+)\vee(S^1\wedge CP^\infty).$

It follows that $\widetilde{u}\simeq \sigma \circ \Sigma u \simeq f\circ g$, where,
$$g : S^2\times \Sigma CP^\infty_+ \rightarrow S^1 \wedge (S^1 \times CP^\infty) \simeq S^1 \wedge LCP^\infty \stackrel{ev}{\rightarrow} CP^\infty$$
and the composition $g\simeq \mu$.
\end{proof}

\section{The Structure of $GL_1(K_p^\wedge)$}

In this section, we prove a splitting of $GL_1K_p^\wedge$ using the logarithm $l_p:gl_1 K_p^\wedge \rightarrow K_p^\wedge$ defined by Rezk (see \cite{R2}). Throughout this section, we assume that $p$ is an odd prime.

\prop (Rezk, \cite{R2}) Let $R$ be an $E_\infty$ ring spectrum. Then there is a logarithmic cohomology operation, $l_{p,n}$, from $gl_1(R)$ to $L_{K(n)}(R)$ for every $n$, and prime $p$. If $R$ is $K(n)$-local, this is a map from $gl_1(R)$ to $R$. When $n=1$, $l_p:gl_1R \rightarrow R$ is given by the formula:

$$ l_p(x) = - \frac{1}{p} log(\frac{\psi (x)}{x^p} )$$

[Recall that a $\theta$-algebra structure is described by operations $\psi$ and $\theta$ ($\psi$ is a ring homomorphism) such that
$ \psi (x)= x^p + p \theta (x) $.]

\prop Suppose that $R=K_p^\wedge$. The operation $l_p:gl_1K_p^\wedge \rightarrow K_p^\wedge$ factors through $ku_p^\wedge$, the connective cover of $K_p^\wedge$. On homotopy groups, the map is an isomorphism on $\pi_n$ for $n>2$. At $n=2$, it is 0. And for $n=0$, this is the map
$$Z_p^\times \cong Z/(p-1) \times Z_p \stackrel{p_2}{\rightarrow} Z_p$$

\begin{proof}
The spectrum $K_p^\wedge$ is $K(1)$-local, and the operation $\psi$ is the Adams operation $\psi_p$. Since $gl_1K_p^\wedge$ is connective, the map $l_p$ factors through $ku_p^\wedge$. Recall, that the homotopy groups of $gl_1K_p^\wedge$ are given by

$$ \pi_n(gl_1K_p^\wedge) = \left\{
\begin{array}{rl}
({K_p^\wedge}^0(S^n))^\times=\pi_n(K_p^\wedge) & \text{if } n>0\\
({K_p^\wedge}^0(S^0))^\times=\pi_0(K_p^\wedge)^\times & \text{if } n=0
\end{array} \right.$$

Since $\pi_nK_p^\wedge$ is nonzero only for even n, it suffices to restrict our attention to even dimensional spheres. The $K$-theory of $S^{2n}$ is generated by $\epsilon$ where $1 - \epsilon =$ the tangent bundle of $S^{2n}$.  Hence,
$$ \pi_{2n}(gl_1(K_p^\wedge))= {\widetilde{gl_1(K_p^\wedge)}}^0(S^{2n})= ({\widetilde{K_p^\wedge}}^0(S^{2n}))^\times= 1+ \epsilon \pi_{2n}(K_p^\wedge)$$
To calculate $l_p$ on $\pi_{2n}gl_1K_p^\wedge$, one needs to compute $l_p(1+k\epsilon)$ for $1+k \epsilon \in {gl_1K_p^\wedge}^0(S^{2n})=\pi_0(gl_1({K_p^\wedge}^{S^{2n}}))$. To accomplish this, we need to calculate $\psi_ p(\epsilon)$. The map $p:(S^2)^n \rightarrow S^{2n}$ which quotients out the lower cells, induces an injection in $K$-theory, and splits  $\epsilon$ as the product
$$p^*(\epsilon)= \prod(1-L_i)$$
where $L_i$ is the canonical line bundle over the $i^{th}$ copy of $S^2=CP^1$. Since the Adams operation $\psi_p$ raises line bundles to the $p^{th}$ power,
$$\psi_p(L_i)= L_i ^p$$
\begin{align}
\implies \psi_p(1-L_i) &= 1-L_i^p \nonumber \\
&= 1-(1-(1-L_i))^p \nonumber
\end{align}
 The element $1-L_i$ lies in the $K$-theory of $S^2$, so it squares to 0. Therefore,
\begin{align}
\psi_p(1-L_i) &= 1-(1-p(1-L_i)) \nonumber \\
&= p(1-L_i) \nonumber
\end{align}
$$\implies \psi_p(\epsilon) = p^n \epsilon.$$
$$\implies \psi_p(1+\epsilon)= 1+ p^n\epsilon$$
Hence,
\begin{align}
l_p(1+k\epsilon) &= - \frac{1}{p} log(\frac{\psi (1+k\epsilon)}{(1+k\epsilon)^p} ) \nonumber \\
&= - \frac{1}{p} log(\frac{1+p^n k\epsilon}{(1+k\epsilon)^p} ) \nonumber \\
&\equiv  - \frac{1}{p}log( 1+(p^n -p) k \epsilon)\,(\mode\ p) \nonumber
\end{align}
which becomes multiplication by $1- p^{n-1}$ ($\mode\ p$) if $n > 0$. Since the homotopy group $\pi_{2n}(gl_1K_p^\wedge)=Z_p$ for $n>0$, this is an isomorphism for $n >1$. For $n=1$, this map is $0$. For $n=0$, the map $l_p : Z_p^\times \cong \mu_{p-1} \times Z_p\rightarrow Z_p$ is given by
$$- \frac{1}{p} log(x^{1-p} )$$
This map has kernel $\nu_{p-1}$, the group of $(p-1)^{st}$ roots of unity, as it takes $p$-adic integers of the form $1+pk$ to
\begin{align}
l_p(1+pk)&=-\frac{1}{p} log((1+pk)^{1-p}) \nonumber \\
         &= -\frac{1}{p}log(1+p(1-p)k)\nonumber \\
         &= -(1-p)k + O(p) \nonumber \\
         &\equiv -k (\mode\ p) \nonumber
\end{align}
 Therefore, the map $l_p$ on $Z_p^\times = \nu_{p-1} \times Z_p$, has kernel $\nu_{p-1}$ and is an isomorphism onto $Z_p$ .
\end{proof}

Recall that the spectrum $ku_p^\wedge$ splits into Adams summands,
$$ku_p^\wedge \simeq B \vee \Sigma^2 B \ldots \Sigma ^{2p-4} B $$
where $B$ is the $p$-adic Adams summand($\pi_*(B)= Z_p[v_1]$). Using this, we identify the image of the logarithmic cohomology operation. We construct $K_p(\widehat{2})$ from the spectrum $ku_p^\wedge$ by killing the $2^{nd}$ homotopy group:

\defn Let $B_2$ be the 2-connective cover of $B$. Define
$$K_p(\widehat{2})=B \vee \Sigma^2 B_2 \ldots \vee \Sigma ^{2p-4}  B $$

\prop There is a split cofibre sequence,
$$ H\nu_{p-1} \vee \Sigma^2 HZ_p \rightarrow gl_1(K_p^\wedge) \rightarrow K_p(\widehat{2}) $$

\begin{proof}
From the definition above, note that $gl_1K_p^\wedge \stackrel{l_1}{\rightarrow}ku_p^\wedge \rightarrow K_p(\widehat{2})$ is surjective on homotopy groups. The fibre $F$ has homotopy only in dimensions 0 and 2. The Postnikov tower of $F$ then is a cofibre sequence,
$$\Sigma^2 HZ_p \rightarrow F \rightarrow H\nu_{p-1} \rightarrow \Sigma^3 HZ_p$$

 Since the group $H^3(H\nu_{p-1};Z_p)=0$, the sequence splits and one obtains
$$F\simeq H\nu_{p-1}\vee \Sigma^2 HZ_p$$
Therefore, there is a cofibre sequence
$$ H\nu_{p-1} \vee \Sigma^2 HZ_p \rightarrow gl_1(K_p^\wedge) \rightarrow K_p(\widehat{2}) $$
The next term in this sequence is
$$\Sigma (H\nu_{p-1} \vee \Sigma^2 HZ_p)\simeq \Sigma H\nu_{p-1} \vee \Sigma^3 HZ_p$$
and the next map is $K_p(\widehat{2})\rightarrow \Sigma H\nu_{p-1} \vee \Sigma^3 HZ_p$. Since the spaces in the Adams summands are retracts of $bu_p^\wedge$, their homology concentrated in even dimensions. Therefore,
$$[\Sigma^{2k} B, \Sigma H\nu_{p-1} \vee \Sigma^3 HZ_p] \cong H^1(B;\nu_{p-1})\oplus H^3(B;Z_p) \cong 0$$
Since the spectrum $B_2$ is 3-connected,
$$[\Sigma^2B_2,\Sigma H\nu_{p-1} \vee \Sigma^3 HZ_p] \cong H^{-1}(B_2;\nu_{p-1})\oplus H^1(B_2;Z_p)\cong 0$$
$$\implies [K_p(\widehat{2}),H^1(B;\nu_{p-1})\oplus H^3(B;Z_p)]=0$$
Hence, the cofibre sequence splits and
$$gl_1(K_p^\wedge) \simeq K_p(\widehat{2})\vee H\nu_{p-1} \vee \Sigma^2 HZ_p.$$
\end{proof}

We will use this decomposition later to calculate  homotopy classes of extensions. For that, we also have to understand how the splitting looks like when we map a space $X$ to $GL_1(K_p^\wedge)$. Recall, $[X,GL_1(K_p^\wedge)]= {K_p^\wedge}^0(X)^\times$. The map $l_p$ gives the way to map this to $[X,K_p(\widehat{2})]$. The map $ {K_p^\wedge}^0(X)^\times \rightarrow H^0(X;\nu_{p-1})$ is the composite
$$X\rightarrow GL_1(K_p^\wedge)\rightarrow \pi_0GL_1(K_p^\wedge) \cong Z_p^\times \cong \nu_{p-1}\times Z_p \rightarrow \nu_{p-1}\simeq K(\nu_{p-1},0)$$

 The third factor is $\Sigma^2 HZ_p$, and we have to understand the map from $H^2(X;Z_p)$ to ${K_p^\wedge}^0(X)^\times$. Now, $H^2(X;Z_p)= [X,K(Z_p,2)]=[X,{CP^\infty}_p^\wedge]$. The space $CP^\infty$ classifies line bundles which are invertible elements in $K$-theory.

\prop The map $H^2(X;Z_p) \rightarrow {K_p^\wedge}^0(X)^\times$ is given by $f\in [X,{CP^\infty}_p^\wedge]\mapsto L^f$ where $L^f$ is the line bundle classified by $f$.

 \begin{proof}
  The formula in the statement of the proposition defines a map of infinite loop spaces ${CP^\infty_p}^\wedge \rightarrow GL_1K_p^\wedge$, and hence, a map of spectra $\Sigma^2 HZ_p \rightarrow gl_1K_p^\wedge$. Composing it with $l_p$, we get

  \begin{align}
  l_p(L^f)&= - \frac{1}{p} log(\frac{\psi_p (L^f)}{(L^f)^p} ) \nonumber \\
          &= -\frac{1}{p} log(\frac{(L^f)^p}{(L^f)^p}) \nonumber \\
          &= -\frac{1}{p}log(1) \nonumber \\
          &= 0 \nonumber
 \end{align}

 The computation above shows that the composition $\Sigma^2 HZ_p \rightarrow gl_1(K_p^\wedge) \rightarrow K_p(\widehat{2})$ equals 0. Therefore, it  factors through $\nu_{p-1}\times \Sigma^2 HZ_p$ in the diagram,

\begindc{\commdiag}[5]
\obj(-15,20){}
\obj(0,-5){}
\obj(20,13){$\Sigma^2 HZ_p$}
\obj(20,0){$gl_1(K_p^\wedge)$}
\obj(40,0){$ K_p(\widehat{2})$}
\obj(0,0){$H\nu_{p-1} \times \Sigma^2 HZ_p$}
\mor(6,0)(17,0){}
\mor(23,0)(39,0){}
\mor(19,12)(1,1){}
\mor(20,13)(20,0){}
\enddc

To complete this proof, we need to show that the map $\Sigma^2 HZ_p \rightarrow H\nu_{p-1}\vee \Sigma^2 HZ_p \rightarrow \Sigma^2 HZ_p$ is an equivalence. The only non zero homotopy group of $\Sigma^2 HZ_p$ is $\pi_2$, so it suffices to check that the map $[S^2,CP^\infty] \rightarrow H^2(S^2; Z_p)$ as described by the statement is an isomorphism. The left group is isomorphic to $Z_p$, via $k\mapsto L^k$, $L=$ the tangent bundle of $S^2$. The right group is $H^2(S^2; Z_p) \cong Z_p$ inside $K_p^\wedge(S^2)^\times$ as elements $1+k\epsilon$, $\epsilon = 1- L$. The map between the two is $L^k \mapsto (1-\epsilon)^k= 1-k\epsilon$ because $\epsilon^2=0$, and is evidently an isomorphism.
 \end{proof}

\section{Calculation of $\THH$}

In this section, we complete the computation of $\THH$ for odd primes $p$. We first parameterise the homotopy classes of extensions $f$,

\begindc{\commdiag}[5]
\obj(-23,13){}
\obj(0,10){$S^2$}
\obj(20,10){$\Sigma BGL_1(K_p^\wedge)$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_p^\wedge)$}
\mor(0,10)(15,10){$\Sigma (1-p)$}
\mor(20,10)(20,0){$\sigma$}
\mor(1,0)(15,0){$f$}
\mor(0,10)(0,0){$\sigma$}
\enddc
using the results of the previous section.

Recall that,
$$GL_1(K_p^\wedge)= \nu_{p-1} \times K(Z_p,2) \times \Omega^\infty K_p(\widehat{2})$$
$$\implies B^2GL_1(K_p^\wedge)= B^2 \nu_{p-1} \times K(Z_p,4)\times \Omega^\infty \Sigma^2 K_p(\widehat{2})$$

The condition on the map $f$ is that its restriction to $S^2$ is $1-p$. The homotopy classes of maps from $S^2$ to $B^2GL_1(K_p^\wedge)$ is split into  three factors,

1. $[S^2,B^2\nu_{p-1}]=H^2(S^2;\nu_{p-1})\cong \nu_{p-1}$

2. $[S^2,K(Z_p,4)]= H^4(S^2;Z_p)=0$

3. $[S^2,\Omega^\infty \Sigma^2 K_p(\widehat{2})]= [S^2,\Omega^\infty (\Sigma^2B \vee \Sigma^4 B_2 \vee \Sigma^6 B \ldots \vee \Sigma^{2p-4}B]
                                             = [S^2,\Omega^\infty \Sigma^2 B]
                                             =B^2(S^2)\cong Z_p $

In the splitting,
$$[S^2,B^2GL_1(K_p^\wedge)] = \nu_{p-1} \oplus B^2(S^2)\oplus H^4(S^2; Z_p)= \nu_{p-1} \oplus Z_p \oplus 0,$$
$1-p$ is in the factor $Z_p$, where it equals $l_p(1-p)= \alpha_p$ and,
\begin{align}
\alpha_p        &= - \frac{1}{p} log((1-p)^{1-p} )\nonumber \\
                &\cong -\frac{1}{p} log ( 1- (1-p)p) \nonumber \\
                &\cong -1(\mode \ p) \nonumber
\end{align}

\subsection{Calculation at the prime 3}

Let us begin the calculation at the prime $3$. The cofibre sequence for $gl_1K_3^\wedge$ is
$$HZ/2 \vee \Sigma^2 HZ_3 \rightarrow gl_1(K_3^\wedge) \rightarrow  K_3(\widehat{2})$$
and
$$K_3(\widehat{2}) = B\vee \Sigma ^2 B_2.$$
Therefore, $$GL_1K_p^\wedge = Z/2 \times K(Z_p,2)\times \Omega^\infty B \times \Omega^\infty B_2.$$

We will study the extension to $CP^\infty$ of the map $1-p$, to the four factors $Z/2,\, K(Z_3,2)$, $\Omega^\infty B,\, \Omega^\infty B_2$ one by one. Let us start with the factor $B$. The Adams summands are the eigenspaces of the action of the $(p-1)^{st}$ roots of unity by Adams operations. The spectrum $B$ is fixed by all the Adams operations. The projection from ${K_p^\wedge}^*(X)$ to $B^*(X)$ is given by
$$\pi = \frac{1}{p-1}(1+ \psi_\zeta + \psi_{\zeta^2} +\ldots +\psi_{\zeta^{p-2}}),$$
where $\zeta \in \nu_{p-1} \subset Z_p^\times$.

For the prime $3$, we can take $\zeta = -1$ and then the projection operator is
$$\pi =  \frac{1+\psi_{-1}}{2}$$
Let us start by working out an example.

\ex
Consider the element $\beta L \in {K_3^\wedge} ^2(CP^\infty)$ where $\beta$ is the Bott element. Applying the projection, we get
\begin{align}
\pi (\beta L) &= \frac{\beta (L-L^{-1})}{2} \nonumber
\end{align}

Restricting to $S^2$, using $L=1-\epsilon$ and $\epsilon^2= 0$, we obtain
\begin{align}
\frac{\beta ((1-\epsilon) -(1-\epsilon)^{-1})}{2} &= \frac{\beta ((1-\epsilon)-(1+\epsilon))}{2} \nonumber \\
                                                  &= -\beta \epsilon \nonumber \\
                                                  &= -1 \nonumber
\end{align}

In order for it to be an extension of the kind required, this restriction must be $\alpha_3$, so we multiply by $-\alpha_3$. This defines,
$$f= -\alpha_3 \frac{\beta (L-L^{-1})}{2}.$$

Recall that, $\THH^{K_3^\wedge}(K/3,f)$ is the cofibre of
$$ K_3^\wedge \wedge CP^\infty \stackrel{u-1}{\rightarrow} K_3^\wedge \wedge CP^\infty  \ \ \ \ \ \  \ \ (\alpha)$$
where $u\in {K_3^\wedge}^0(CP^\infty) ^\times= [CP^\infty_+,GL_1(K_3^\wedge)]$ is the adjoint of,

\begindc{\commdiag}[5]
\obj(-25,-3){}
\obj(0,13){}
 \obj(0,10){$S^2 \wedge CP^\infty_+$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_3^\wedge )$}
\mor(3,0)(16,0){$f$}
\mor(0,10)(0,0){$\mu$}
\mor(2,9)(18,1){$\widetilde{u}$}
\enddc

 The group structure of $CP^\infty$ classifies tensor product of line bundles so, $\mu^*L=L\otimes L$. This implies,
 \begin{align}
\mu^*(f)&=-\alpha_3 \frac{\beta(L\otimes L - L^{-1}\otimes L^{-1})}{2} \nonumber
\end{align}

 The $K$ theory of $S^2$ is generated by $\epsilon = 1-L$ with $\epsilon^2 =0$. We can rewrite the equation using the generator
\begin{align}
\mu^*(f) &=-\alpha_3 \frac{\beta((1-\epsilon)\otimes L - (1+\epsilon) \otimes L^{-1})}{2} \nonumber \\
                       &= -\alpha_3 \frac{\beta \epsilon \otimes (L + L^{-1})}{2} \nonumber
\end{align}

Using the suspension isomorphism (given by $\beta \epsilon =1$) we get,
$$\mu^*(f)= -\alpha_3 \frac{ L + L^{-1}}{2}$$

To get $u$ we need to invert the logarithmic cohomology operation. Suppose that $u=h(x)\in {K_3^\wedge}^0(CP^\infty)^\times$. Then, we have to solve,
$$- \frac{1}{3} log(\frac{\psi_3 (h(x))}{h(x)^3})= -\alpha_3 \frac{ L +L^{-1}}{2}$$
 $$\implies \frac{\psi_3 (h(x))}{h(x)^3}= exp( 3\alpha_3 \frac{ L + L^{-1}}{2})\ \ \  \ \   \ \ (*) $$

Note that $\psi_3(x)= 1- (1-x)^3$ and hence,
$$\frac{ h(1-(1-x)^3)}{h(x)^3}= exp( 3\alpha_3 \frac{ L + L^{-1}}{2})$$

Let us look at the equation $(\mode \ 3^2,x^3)$. The right side of the equation can be written in terms of $x$ using $L= 1-x$, and then, $L^{-1}= 1+x+x^2 \, (\mode \ 3^2,\, x^3)$. Therefore, the right side simplifies to
\begin{align}
exp(3\alpha_3 \frac{ L + L^{-1}}{2})&= exp( 3 \alpha_3 \frac{2 + x^2}{2}) \nonumber \\
                                    &=1 + 3\alpha_3 + 3\frac{\alpha_3 x^2}{2} \nonumber
\end{align}

Now we will simplify the left side of (*). Suppose that $h(x)= a + bx+ cx^2$. In order to solve the equation, we have to invert $l_3$. We know that $l_3$ has a kernel $Z/2\vee K(Z_3,2)$, so the equation can be solved once we know the restriction to these.

In the part of $HZ/2$,  $\sigma:S^2 \rightarrow CP^\infty$ induces an isomorphism in $H^2(-; Z/2)$. Therefore, the extension is $0$ here. The map ${K_3^\wedge}^0(CP^\infty)^\times \rightarrow H^0(CP^\infty;Z/2)$ sends $a \mapsto a\,(\mode\ 3)$ (identifying $Z/2$ with the group of units in $\mathbb{F}_3$). Therefore, since $\mu^*(0)=0$, we get the equation
$$a\equiv 1\,(\mode\ 3)$$

In the factor $K(Z_3,2)$, there is no restriction on $f$. Assume that it is trivial, so $\mu^*(0)=0$.  This maps into $GL_1(K_3^\wedge)$ by taking a line bundle over $CP^\infty$ to the corresponding unit in $K$-theory. If we look at $k\in Z_p= H^2(CP^\infty; Z_3)=[CP^\infty,K(Z_3,2)]$, this is the line bundle $L^k= (1-x)^k= 1-kx + \frac{k(k-1)}{2}x^2 \,(\mode\, x^3)$. This is the only factor that gives a non zero coefficient of $x$ so, we get that $b=0$.

Therefore, $h(x)= a + c x^2(mode\ 3^2,\,x^3)$ and $a\equiv 1(\mode \ 3)$. The left side of (*) is
\begin{align}
\frac{\psi (h(1-(1-x)^3))}{h(x)^3}&\equiv \frac{h(3x-3x^2+x^3)}{h(x)^3} \nonumber \\
                                  &\equiv \frac{a}{a^3 + 3ca^2x^2} \nonumber \\
                                  &\equiv a^{-2}( 1- 3\frac{c}{a}x^2)\,(\mode\ 3^2,\,x^3) \nonumber
\end{align}

Working $(\mode\ 3^2,\, x^3)$, we have
$$a^{-2}( 1- 3\frac{c}{a}x^2)= 1 + 3\alpha_3 + 3\frac{\alpha_3 x^2}{2}$$
$$\implies a\equiv 1+3\alpha_3 (\mode\ 3^2) \ \ and \ \ c\equiv \alpha_3 (\mode\ 3)$$
Thus $a= 1+ 3.(unit)$ and $c$ is a unit (since $\alpha_3$ is a unit).

Therefore, $u-1$ looks like $3.(\mathit{unit})+ x^2(\mathit{unit})$. We can choose a different parameterisation for $K$-theory  of $CP^\infty$ to assume that $u-1= 3+ x^2$.

Now ${K_3^\wedge}_*(CP^\infty)= {K_3^\wedge}_*\left\{\beta_0,\beta_1,\ldots \right\}$ where $\beta_i$ is dual to $x^i$. Therefore,
$$ < (u-1)(\beta_i),x^j>= <\beta_i, x^j(3+ x^2)>= \left\{
\begin{array}{rl}
3 & \text{if } j=i\\
1 & \text{if } j=i-2\\
0 & \text{} otherwise
\end{array} \right.$$
Therefore, the map $u-1$ on ${K_p^\wedge}_*(CP^\infty)$ is given by
$$(u-1)(\beta_i)=\left\{
 \begin{array}{rl}
3\beta_i & \text{if } i=0,1\\
3\beta_i + \beta_{i-2} & \text{if } i>1
\end{array}\right.$$

Following the cofibre $(\alpha)$, we understand that $u-1$ is injective, and its cokernel has two copies of $Z/(3^\infty)$ in even dimensions. Thus,
$$\pi_k(\THH^{K_3^\wedge}(K/3),f)=\left\{
 \begin{array}{rl}
0  & \text{if $k$ is odd }\\
Z/(3^\infty) \oplus Z/(3^\infty) & \text{if $k$ is even}
\end{array}\right.$$
completing the calculation in this example.

\begindc{\commdiag}[5]
\obj(0,0){}
\obj(0,5){}
\enddc

 Now we perform the calculation at the prime 3 for all extensions that are non trivial only on the factor $\Omega^\infty B$ of $GL_1(K_3^\wedge)$. The extension in the example was of this kind. So, we are looking at elements in $B^2(CP^\infty)$ which restrict to $\alpha_3$ in $S^2$.

An element in ${K_3^\wedge}^2(CP^\infty)$ is given by $\beta g(x)$. Therefore, an element in $B^2(CP^\infty)$ is
$$\pi(\beta g(x))= \frac{\beta(g(x)-g(1-\frac{1}{1-x}))}{2}$$

Suppose that $g(x)= a'+b'x+c'x^2 \,(\mode\ 3^2,\, x^3)$. Restricting to $S^2$ (using $x=\epsilon$ and $\epsilon^2= 0$) we get $b'$. We need to get $\alpha_3$. Thus, to get an extension we must have  $b'= \alpha_3$. This gives us all possible extensions $f$ on the factor $B$. Let us work as before $(\mode \ 3^2,\,x^3)$. Then,

\begin{align}
f &= \frac{\beta(g(x)-g(1-\frac{1}{1-x}))}{2} \nonumber \\
                &= \frac{\beta(a'+ b'x+c'x^2 - g(-x -x^2))}{2} \nonumber \\
                &= \frac{\beta(a'+ b'x+ c'x^2-(a'-b'x-b'x^2+c'x^2))}{2}\nonumber \\
                &=\frac{\beta(2b'x +b'x^2)}{2} \nonumber \\
                &=\beta b'x + \frac{\beta b'}{2} x^2\nonumber
\end{align}

We have to calculate $u$ using

\begindc{\commdiag}[5]
\obj(-25,-3){}
\obj(0,13){}
 \obj(0,10){$S^2 \times CP^\infty$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_3^\wedge )$}
\mor(3,0)(16,0){$f$}
\mor(0,10)(0,0){$\mu$}
\mor(2,9)(18,1){$u$}
\enddc

By definition, the multiplication map takes $x$ to the formal group, which for $K$-theory is the multiplicative group. Therefore,
$$x\mapsto \epsilon \otimes 1 + 1\otimes x -\epsilon \otimes x$$
\begin{align}
\implies x^2 &\mapsto (\epsilon \otimes 1 + 1\otimes x -\epsilon \otimes x)^2 \nonumber \\
     &= 1\otimes x^2 + 2 \epsilon \otimes x -2\epsilon \otimes x^2 \nonumber
\end{align}
To get $\mu^*$ we must project onto the factor $S^2\wedge CP^\infty_+$. Thus, we obtain
$$\mu^*(x)= \epsilon \otimes 1 - \epsilon \otimes x  \ , \   \mu^*(x^2)= 2\epsilon \otimes x - 2\epsilon \otimes x^2.$$

Using these formulae and the suspension isomorphism $\beta \epsilon = 1$ we calculate $\mu^*(f)$.

\begin{align}
\mu^*(f) &= \beta b'(\epsilon \otimes 1- \epsilon \otimes x)+ \frac{\beta b'}{2}(2 \epsilon \otimes x-2\epsilon \otimes x^2) \nonumber \\
                       &= b'(1-x) + \frac{b'}{2}(2x-2x^2) \nonumber \\
                       &= b' - b'x^2 \nonumber
\end{align}

To get $u$, we have to invert the logarithmic cohomology operation $l_3$, as in the example. Suppose that $u= h(x)$. Then, we need to solve
$$l_3(u)= \frac{\psi_3 (h(x))}{h(x)^3}= exp(-3b'(1-x^2))$$

We have the formula $\psi_3(x)= 1-(1-x)^3$. Similar to the example, we assume that in our extension the contribution from $HZ/2$ is $1$ and $HZ_3$ is $0$. In the same way, this implies that if $h(x)= a + bx +c x^2$ ,
$$a\equiv 1(\mode \ 3) \ , \ b=0$$
Then, the equation becomes,
\begin{align}
a^{-2}( 1- 3\frac{c}{a}x^2) &= exp(-3b'(1-x^2)) \nonumber \\
                            &= 1 - 3b' + 3b'x^2 \nonumber
\end{align}

In the same way, we understand that the unit $u= 1+ 3.\mathit{unit} + x^2.\mathit{unit}$, and so, we obtain the same computation

$$\pi_k(\THH^{K_3^\wedge}(K/3),f)=\left\{
 \begin{array}{rl}
0  & \text{if $k$ is odd }\\
Z/(3^\infty) \oplus Z/(3^\infty) & \text{if $k$ is even}
\end{array}\right.$$

 Now we want to see what happens if we allow extensions with non trivial contributions from the other 3 factors of $GL_1(K_3^\wedge)= Z/2 \times K(Z_3,2) \times \Omega^\infty B \times \Omega^\infty B_2$. In the part $Z/2$, the restriction $H^2(CP^\infty;Z/2)\rightarrow H^2 (S^2;Z/2)$ is an isomorphism. So, this factor always contributes trivially.

 For the factor $K(Z_3,2)$, the group $[S^2,B^2K(Z_3,2)]=[S^2,K(Z_3,4)]=H^4(S^2;Z_3) =0$. Therefore, there is no condition on $f$ here. The group $H^4(CP^\infty;Z_3)$ is generated by $x^2$ and $f$ is given by $ax^2$ for some $a \in Z_p$. To compute $u$, consider:

\begindc{\commdiag}[5]
\obj(-25,-3){}
\obj(0,13){}
 \obj(0,10){$S^2 \times CP^\infty$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$K(Z_3,4)$}
\mor(3,0)(16,0){$f$}
\mor(0,10)(0,0){$\mu$}
\mor(2,9)(18,1){$u$}
\enddc

Note that in this case, $\mu^*(x)= \epsilon \otimes 1 + 1\otimes x$, which implies
\begin{align}
 \mu^*(x^2) &=(\epsilon \otimes 1 + 1\otimes x)^2 \nonumber \\
                    &= 2\epsilon \otimes x + 1 \otimes x^2 \nonumber
\end{align}

To get $u$ we have to project to $S^2\wedge CP^\infty_+$ and apply the suspension isomorphism. Then, we get $2a x \in H^2(CP^\infty; Z_3)$.
Recall from the previous section that, from this we get the unit by taking $L^{2a}$, where $L= (1-x)$ is the canonical line bundle. Therefore, the contribution to $u$ from this factor is $(1-x)^{2a}$.

Now if $a$ is divisible by $3$ then, we still get that our $u= 1 + 3.\mathit{unit} + x^2.\mathit{unit}$ which results in the same calculation for $\pi_*(\THH^{K_3^\wedge}( K/3,f))$. If $a$ is not divisible by $3$ then, it is a unit, so that $u= 1+ 3.\unit + x.\unit$. Therefore, by reparameterising we can write $u-1= 3 + x$.
 $$\implies < (u-1)(\beta_i),x^j>= <\beta_i, x^j(3 +x )>= \left\{
\begin{array}{rl}
3 & \text{if } j=i\\
1 & \text{if } j=i-1\\
0 & \text{} otherwise
\end{array} \right.$$
$$\implies (u-1)(\beta_i)=\left\{
 \begin{array}{rl}
3\beta_i & \text{if } i=0\\
3\beta_i + \beta_{i-1} & \text{if } i>0
\end{array}\right.$$

Therefore, in this case,

$$\pi_k(THH^{K_3^\wedge}(K/3),f)=\left\{
 \begin{array}{rl}
0  & \text{if $k$ is odd }\\
Z/(3^\infty) & \text{if $k$ is even}
\end{array}\right.$$

Now consider the factor $\Sigma ^2 B_2$. We know that this is $5$-connected. So, if we look at extensions we know that they always restrict to $0\in K_3^\wedge (CP^2)$. Since we are working $(\mode\ x^3)$, this means that these extensions always give $0$.

Therefore, we get that, depending on $f$ either $ \pi_*(\THH^{K_3^\wedge}( K/3,f))= (Z/(3^\infty))^i$ in even degrees where $i=1$ or $2$ depending on $f$.

This finishes our calculation at the prime 3.

\subsection{Calculation at primes $\geq 5$}
Let us now look at the other odd primes and work $(\mode \ x^p,\,p^2)$. Recall that there is a splitting
$$GL_1(K_p^\wedge)= \nu_{p-1} \times K(Z_p,2)\times \Omega^\infty K_p(\widehat{2})$$
$$K_p(\widehat{2})=B \vee \Sigma^2 B_2\vee \ldots \Sigma ^{2p-4} B $$

We start by working in the factor $B$ of $K_p(\widehat{2})$. The projection operator from ${K_p^\wedge} ^*(X)$ to $B^*(X)$ is given by
$$ \pi = \frac{1+ \psi_\zeta + \psi_{\zeta^2} + \ldots \psi_{\zeta^{p-2}}}{p-1}$$

Define $\kappa$ to be the composite,
$${K_p^\wedge} ^*(CP^\infty) \stackrel{\mu^*}{\rightarrow} {K_p^\wedge} ^*(S^2 \wedge CP_+^\infty) \stackrel{\cong}{\rightarrow} {K_p^\wedge} ^{*-2} (CP^\infty)$$

First observe that the following diagram commutes:

\begindc{\commdiag}[5]
\obj(-20,15){}
\obj(0,-5){}
\obj(20,10){${K_p^\wedge}^2(CP^\infty)$}
\obj(0,10){${K_p^\wedge}^2(CP^\infty)$}
\obj(0,0){${K_p^\wedge}^0(CP^\infty)$}
\obj(20,0){${K_p^\wedge}^0(CP^\infty)$}
\mor(5,10)(15,10){$\psi_a$}
\mor(0,10)(0,0){$\kappa$}
\mor(5,0)(15,0){$\psi_a$}
\mor(20,10)(20,0){$\kappa$}
\enddc

This implies all Adams operations hence $\pi$, commutes with $\kappa$.

Write $x=1-L$ for the generator in ${K_p^\wedge}^2(CP^\infty)$ amd $\epsilon$ its restriction to $S^2$. We have to look for $f$ as in the diagram,

\begindc{\commdiag}[5]
\obj(-23,13){}
\obj(0,10){$S^2$}
\obj(20,10){$\Sigma BGL_1(K_p^\wedge)$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$B^2GL_1(K_p^\wedge)$}
\mor(0,10)(15,10){$\Sigma (1-p)$}
\mor(20,10)(20,0){$\sigma$}
\mor(1,0)(15,0){$f$}
\mor(0,10)(0,0){$\sigma$}
\enddc

Suppose that  $f$ is given by $\pi (\beta g(x))$ where $g(x)= a_0 +a_1 x+\ldots +a_{p-1}x^{p-1}\,(\mode\ x^p,\,p^2)$.

\textbf{Claim:}
$$\kappa(\beta g(x))= g'(x)(1-x)$$

\begin{proof}
It is enough to check this on the generators $x^n$. The multiplication takes $x$ to the formal group of $K$-theory, which is the multiplicative formal group.
$$\mu ^*(x)= \epsilon \otimes 1 - \epsilon \otimes x$$
Therefore,
\begin{align}
\mu ^*(\beta x^n)&= \beta(\epsilon \otimes 1 + 1\otimes x - \epsilon \otimes x)^n \nonumber \\
                 &= \beta(1 \otimes x^n + n \epsilon \otimes x^{n-1} - n \epsilon \otimes x^n) \nonumber
\end{align}

$\kappa$ is obtained by projecting this onto the factor $S^2\wedge CP^\infty_+$ of the product, and then applying the suspension isomorphism ($\beta\epsilon =1$). Therefore, we obtain

\begin{align}
\kappa(\beta x^n)&= nx^{n-1} - n x^n \nonumber \\
                 &= nx^{n-1}(1-x) \nonumber \\
                 &= (x^n)'(1-x) \nonumber
\end{align}

\end{proof}

If we restrict $f$ to $S^2$, we get
\begin{align}
\pi (\beta g(\epsilon)) &= \pi (\beta( a_0 +a_1 \epsilon))\nonumber \\
                        &= (\frac{1+ \psi_\zeta + \psi_{\zeta^2} + \ldots \psi_{\zeta^{p-2}}}{p-1})(\beta(a_0 + a_1 \epsilon ))\nonumber
\end{align}
The action of the Adams operations on the Bott element and $\epsilon$ are given by
$$\psi_a(\beta)=\frac{\beta}{a} \ , \ \psi_a(\epsilon)=1-(1-\epsilon)^a= a\epsilon$$
Therefore,
\begin{align}
\pi (\beta g(\epsilon)) &= (\frac{1+ \psi_\zeta + \psi_{\zeta^2} + \ldots \psi_{\zeta^{p-2}}}{p-1})(\beta(a_0 + a_1 \epsilon ))\nonumber \\
                        &= \frac{\beta( a_0(1+ \zeta^{-1} +\zeta^{-2}+\ldots +\zeta^{2-p}) +(p-1)a_1 \epsilon) }{p-1} \nonumber \\
                        &= \frac{\beta( a_0(1+ \zeta^{-1} +\zeta^{-2}+\ldots +\zeta^{2-p}))}{p-1} +a_1 \epsilon \nonumber
\end{align}
Since $\zeta$ is a $(p-1)^{st}$ root of unity, we get
\begin{align}
1+ \zeta^{-1} +\zeta^{-2}+\ldots +\zeta^{2-p} &= \zeta^{p-1} + \zeta^{p-2} +\ldots +\zeta \nonumber \\
                                              &= 0 \nonumber
\end{align}
This shows that $\pi(\beta g(x))$ restricts to $a_1 \in B^2(S^2)$. Thus, we have that $a_1= l_p(1-p)= \alpha_p$.

We need to calculate $u$ from the extension $\pi(\beta g(x))$ by solving,

\begin{align}
l_p(u) &= \kappa \pi(\beta g(x))\nonumber \\
       &= \pi \kappa(\beta g(x))\nonumber \\
       &= \pi(g'(x)(1-x)) \nonumber
\end{align}

Suppose that $h(x)= g'(x)(1-x)= c_0+c_1 x +\ldots + c_{p-1}x^{p-1} (\mode \ x^p,\,p^2)$. Then,

\begin{align}
\pi (h(x)) &= \frac{1+ \psi_\zeta + \psi_{\zeta^2} + \ldots \psi_{\zeta^{p-2}}}{p-1}(h(x))\nonumber \\
           &= \sum_{i=0}^{p-2} \frac{ h(1-(1-x)^{\zeta^i})}{p-1} \nonumber
\end{align}

Let us look at the coefficient of $x^a$ in the above equation.

\begin{align}
[\pi (x^n)]_a &= [\sum_{i=0}^{p-2} \frac{ (1-(1-x)^{\zeta^i})^n}{p-1}]_a \nonumber \\
              &=\sum_{i=0}^{p-2} \frac{[(1-(1-x)^{\zeta^i})^n]_a}{p-1} \nonumber \\
              &=\sum_{i=0}^{p-2} \frac{[ \sum {{n \choose l} (-1)^l(1-x)^{l\zeta ^i}}]_a}{p-1} \nonumber \\
              &=\sum_{i=0}^{p-2} \frac{\sum { {n\choose l} (-1^l) {{l\zeta^i}\choose a}}}{p-1} \nonumber
\end{align}

 Since $\zeta$ is a $(p-1) ^{st}$ root of unity,
$$(\zeta) ^i + (\zeta^2)^i +\ldots + (\zeta^{p-1})^i =  \left\{
 \begin{array}{rl}
0 & \text{if } i=1,2,\ldots ,p-2\\
p-1  & \text{if } i=0,p-1
\end{array}\right.$$
The binomial coefficient ${y \choose a}$ is a polynomial in $y$ of degree $a$ with the constant term 0  and the top coefficient $1/a!$. Therefore,
$$  {{l(\zeta)}\choose a} + {{l(\zeta^2)}\choose a} +\ldots + {{l(\zeta^{p-1})}\choose a} =  \left\{
 \begin{array}{rl}
0 & \text{if } a=1,2,\ldots ,p-2\\
\frac{p-1}{(p-1)!}l^{p-1}  & \text{if } a=p-1\\
p-1 &\text{if } a=0
\end{array}\right.$$
Therefore, we get
$$[\pi (x^n)]_a = \left\{
\begin{array}{rl}
0 & \text{if } a=1,2,\ldots ,p-2\\
\frac{1}{(p-1)!}\sum{{n\choose l}(-1)^l l^{p-1}}  & \text{if } a=p-1\\
\sum{{n\choose l}(-1)^l} & \text{if } a=0
\end{array}\right.$$
$$\implies [\pi (x^n)]_0 = \sum{{n\choose l}(-1)^l} =\left\{
 \begin{array}{rl}
(1-1)^n = 0  & \text{if } n > 0 \\
1  & \text{if } n=0
\end{array}\right.$$
The other possible non zero coefficient is $[\pi(x^n)]_{p-1}$. If $n=0$, this must be 0. If $n>0$ this gives
 \begin{align}
 [\pi (x^n)]_{p-1} &\equiv \frac{1}{(p-1)!}\sum{{n\choose l}(-1)^l l^{p-1}}\nonumber \\
                   &\equiv \frac{1}{(p-1)!}\sum{{n\choose l}(-1)^l l^{p-1}}\nonumber \\
                   &\equiv - \sum{{n\choose l}(-1)^l} \nonumber \\
                   &\equiv -(1-1)^n +1 \nonumber \\
                   &\equiv 1\, (\mode\ p) \nonumber
 \end{align}
 Summarising the calculation $(\mode\ p)$, we get
$$[\pi (x^n)]_a =  \left\{
\begin{array}{rl}
1 & \text{if } a=0,\,n=0\\
1  & \text{if } a=p-1,\,n>0\\
0 & \text{otherwise }
\end{array}\right.$$
 Now we are in a position to calculate $\pi(h(x))\,(\mode\ p)$
 \begin{align}
 \pi(h(x)) &=\pi(c_0 + c_1 x +\ldots +c_{p-1}x^{p-1}) \nonumber \\
           &=c_0\pi(1)+c_1\pi(x)+\ldots +c_{p-1}\pi(x^{p-1}) \nonumber \\
           &=c_0+ c_1 x^{p-1} + \ldots +c_{p-1}x^{p-1} \nonumber \\
           &=c_0+bx^{p-1} \nonumber
 \end{align}
 where $c_0=a_1$ and
 \begin{align}
 b&= c_1+ \ldots c_{p-1} \nonumber \\
  &=a_1 -2a_2 + 2a_2 -3a_3 \ldots -(p-1)a_{p-1}+(p-1) a_{p-1} -p a_p\nonumber \\
  &\equiv a_1 \,( \mode \ p) \nonumber
 \end{align}
Thus the equation for $u \, (\mode\ p)$ reduces to
$$l_p (u)= a_1+ b x^{p-1}\ (\mode\ p)$$
$$\implies -\frac{1}{p}log(\frac{\psi_p(u(x))}{u^p})= a_1 + bx^{p-1}\ (\mode\ p)$$
$$\implies \frac{\psi_p(u(x))}{u^p}= exp( -p(a_1 + bx^{p-1}))= 1 -p a_1 + pb x^{p-1}(\mode\ p^2)$$
We are looking at extensions which are non trivial only on the factor $B$. This implies $u(x) \in B^0(CP^\infty)$ which implies $u$ is in the image of $\pi$. By the calculations above, this implies that $u(x)= d_0 + d_1 x^{p-1} \,(\mode\, x^p)$. Then,

\begin{align}
\frac{\psi_p(u(x))}{u^p}&= \frac{ d_0}{ {d_0}^p + p d_0^{p-1} d_1 x^{p-1}} \nonumber \\
                       &= (d_0)^{1-p}(1- p \frac{d_1}{d_0} x^{p-1})\nonumber
\end{align}

Therefore, we obtain

$$d_0 ^{1-p}= 1- pa_1$$
 \begin{align}
 \implies d_0 &= (1-pa_1)^{\frac{1}{1-p}} \nonumber \\
              &= 1- \frac{p}{1-p} a_1 \nonumber \\
              &= 1- p a_1 \,(\, \mode\, p^2) \nonumber
 \end{align}
\begin{align}
\implies d_1 &= - {d_0}^p b \nonumber \\
             &\equiv -1\, (\mode\ p) \nonumber
\end{align}

Therefore, $d_0 = 1+ p.\unit$ and $d_1=\unit$. Thus, $u= 1+ p.\unit + \unit. x^{p-1}$. We can reparameterise so that $u= 1+ p + x^{p-1}$.
\begin{align}
< (u-1)(\beta_i),x^j>&= <\beta_i, x^j(p +x^{p-1} )> \nonumber \\
                     &= \left\{
\begin{array}{rl}
p & \text{if } j=i\\
1 & \text{if } j=i-(p-1)\\
0 & \text{} otherwise
\end{array} \right. \nonumber
\end{align}
$$\implies (u-1)(\beta_i)=\left\{
 \begin{array}{rl}
p\beta_i & \text{if } i=0\\
p\beta_i + \beta_{i-(p-1)} & \text{if } i>0
\end{array}\right.$$

Inputting this in the long exact sequence $(\alpha)$, we get
$$ \pi_k(\THH^{K_p^\wedge}(K/p,f))=\left\{
 \begin{array}{rl}
0  & \text{if $k$ is odd }\\
(Z/(p^\infty))^{p-1} & \text{if $k$ is even}
\end{array}\right.$$
   Now let us look at what happens if we allow non trivial extensions in the other factors. Under restriction to $S^2$, $H\nu_{p-1}^2(S^2) \cong H\nu_{p-1}^2(CP^\infty) = \nu_{p-1}$. The element $1-p$ gives $1 \in \nu_{p-1}$. So, this part always contributes trivially.

The factor $\Sigma^2 B_2$ is $(2p-1)$-connected. So, $[ CP^{p-1} , \Sigma ^2 B_2 ] = 0$. Thus $(\mode \  x^p)$ this factor is always trivial.

Next lets look at the factor $\Sigma^2 HZ_p$. Since $[S^2,\Sigma^4 HZ_p] = HZ_p^4(S^2)= 0$, we have no condition on $f$ from this factor. The group $HZ_p ^4(CP^\infty)$ is generated by $x^2$. Suppose that $f$ is given by $ax^2 \in HZ_p ^4(CP^\infty)$. To compute the contribution to  $u$, we have the diagram,

\begindc{\commdiag}[5]
\obj(-25,-3){}
\obj(0,13){}
 \obj(0,10){$S^2 \times CP^\infty$}
\obj(0,0){$CP^\infty$}
\obj(20,0){$K(Z_p,4)$}
\mor(1,0)(18,0){$f$}
\mor(0,10)(0,0){$\mu$}
\mor(2,9)(18,1){$u$}
\enddc

Under $\mu$, $x$ pulls back to the formal group and thus,
\begin{align}
\mu^*(x^2)&=(\epsilon \otimes 1 + 1\otimes x)^2 \nonumber \\
          &= 2\epsilon \otimes x + 1 \otimes x^2\nonumber
\end{align}
Projecting this to the factor $S^2\wedge CP^\infty_+$, and applying the suspension isomorphism we get $2a x \in HZ_p^2(CP^\infty)$. The map from $HZ_p^2(CP^\infty) \rightarrow [CP^\infty, GL_1(K_p^\wedge)] = {K_p^\wedge} ^0(CP^\infty) ^\times $ is given by $\alpha x \rightarrow (1+x)^\alpha $.

Therefore, if $a$ is divisible by $p$ then we still get that $u= 1 + p.\unit + x^{p-1}.\unit$. This does not change the calculation of $\THH^{K_p^\wedge}( K/p,f)$. If $a$ is not divisible by $p$, then it is a unit. Then, $u= 1+ p.\unit + x.\unit$. This can be reparameterised to $u= 1+p+x$. Then

\begin{align}
< (u-1)(\beta_i),x^j>&= <\beta_i, x^j(p+ x)> \nonumber \\
                     &= \left\{
\begin{array}{rl}
p & \text{if } j=i\\
1 & \text{if } j=i-1\\
0 & \text{} otherwise
\end{array} \right. \nonumber
\end{align}

$$\implies (u-1)(\beta_i)=\left\{
 \begin{array}{rl}
p\beta_i & \text{if } i=0\\
p\beta_i + \beta_{i-1} & \text{if } i>0
\end{array}\right.$$

Therefore, we obtain

$$ \pi_k(\THH^{K_p^\wedge}(K/p,f))=\left\{
 \begin{array}{rl}
0  & \text{if $k$ is odd }\\
Z/(p^\infty) & \text{if $k$ is even}
\end{array}\right.$$

The other factors are $\Sigma^{2k} B$ for $k=2,3,\ldots p-2$. These correspond to the eigenspaces of the action of the Adams operations where $\psi_{\zeta^i}$ acts as $\zeta^{ki}$. The projection operator is given by
$$\pi_k= \frac{1+\zeta^{-k} \psi_\zeta + \zeta^{-2k} \psi_{\zeta^2} +\ldots + \zeta^{-k(p-2)} \psi_{\zeta^{p-2}}}{p-1}$$

The group $[S^2, \Omega^\infty \Sigma^2\Sigma^{2k} B]= B^{2k+2}(S^2)=0$, so, there is no condition on restriction to $S^2$. Then, we may choose any $\pi_k(\beta h(x))$ for $f$, and $u$ must satisfy,
\begin{align}
l_p(u)&= \kappa (\pi_k(\beta h(x)) ) \nonumber \\
      &=\pi_k(\kappa (\beta h(x))) \nonumber \\
      &=\pi_k(f'(x)(1-x)) \nonumber
\end{align}

Now assume $g(x)= h'(x)(1-x)= c_0+c_1 x +\ldots + c_{p-1}x^{p-1}$. Then
\begin{align}
\pi_k (g(x))&= \frac{1+ \zeta^{-k}\psi_\zeta + \zeta^{-2k}\psi_{\zeta^2} + \ldots +\zeta^{-k(p-2)}\psi_{\zeta^{p-2}}}{p-1}(g(x)) \nonumber \\
            &= \frac{1+ \zeta^{-k}g(1-(1-x)^\zeta) + \zeta^{-2k}g(1-(1-x)^{\zeta^2}) + \ldots \zeta^{-k(p-2)}g(1-(1-x)^{\zeta^{p-2}})}{p-1}\nonumber
\end{align}

The following proposition is useful to complete the calculation
\prop
There is a polynomial $f_k(x)= x^k + a_{k+1} x^{k+1} + \ldots $ such that, $Im(\pi_k)$ has polynomials that are multiples of $f_k\,(\mode\ x^p)$ .

\begin{proof}
These polynomials are in the $p$-adic $K$-theory of $CP^\infty$. By looking $(\mode\ x^p)$, we are restricting to the $K$-theory of $CP^{p-1}$. It splits into eigenspaces
$${K_p^\wedge}^0(CP^{p-1})= \oplus_{k=0}^{p-2} \Lambda_k$$
  where $\Lambda_k=[CP^{p-1}, \Omega^\infty \Sigma^{2k} B]=B^{2k}(CP^{p-1})$ is the eigenspace on which the Adams operations $\psi_\zeta$ act as multiplication by $\zeta^k$. $\pi_k$ is the projection on to the eigenspace $\Lambda_k$. In this decomposition, $dim(\Lambda_0)=2$ and $dim(\Lambda_k)=1$ for all $k\geq 1$. Therefore, $\Lambda_k = span(f_k)$ for some polynomial $f_k$. To see how the polynomial $f_k$ looks like we compute $\pi_k(x)$ (note $\pi_k(1)=0$, so we don't get any information out of it).
\begin{align}
\pi_k (x)&= \frac{1+ \zeta^{-k}\psi_\zeta + \zeta^{-2k}\psi_{\zeta^2} + \ldots +\zeta^{-k(p-2)}\psi_{\zeta^{p-2}}}{p-1}(x) \nonumber \\
            &= \frac{1}{p-1}\sum_{i=0}^{p-2}\zeta^{-ik}(1-(1-x)^{\zeta^i}) \nonumber \\
            &= \frac{1}{p-1}\sum_{i=0}^{p-2}\zeta^{-ik} \sum_{n=1}^{\infty} (-1)^{i-1}{\zeta^i \choose n} x^n \nonumber \\
            &= \frac{1}{p-1}\sum_{i=0}^{p-2}\sum_{n=1}^{\infty}   (-1)^{i-1}\zeta^{-ik}{\zeta^i \choose n} x^n \nonumber
\end{align}

Lets look at the coefficient of $x^n$ in the above formula. ${y \choose n}$ is a polynomial of degree $n$ in $y$, and therefore, $-1)^{i-1}y^{-k}{y \choose n}$ has terms of degree $-k$ to $-k+n$. So, if we sum the series, it is 0 if $k< n$. Thus, the first possible non zero coefficient of $x$ is in degree $k$. The coefficient of $x^k$ in $\pi_k(x)$ is given by
\begin{align}
[\pi_k(x)]_k &\equiv \frac{1}{p-1}\sum_{i=0}^{p-2} (-1)^{i-1}\zeta^{-ik}{\zeta^i \choose k}  \nonumber \\
             &\equiv \frac{1}{p-1}\sum_{i=0}^{p-2} (-1)^{i-1}\frac{1}{k!} \nonumber \\
             &\equiv \frac{1}{(p-1) k!} \nonumber \\
             &\neq 0 \,(\mode\ p) \nonumber
\end{align}

So, this is a unit in $Z_p$. Therefore, $Im(\pi_k)= Span(f_k)$ where $f_k$ looks like $x^k+ O(x^{k+1})$.

\end{proof}

Therefore, $\pi_k(g(x))= cf_k(x)$ for some constant $c$. The equation for $u$ is
$$l_p(u)= \pi_k(g(x))= c f_k(x)$$
$$\implies \frac{\psi_p(u)}{u^p} = exp(-p c f_k(x) )$$

If $c$ is divisible by $p$, then $(\mode\ p^2)$ the above equation is 0. If $c$ is not divisible by $p$, then the coefficient of $x^k$ in the right side is $p$ times an unit. We can solve for $u$ as in the cases before. From here, we get a contribution = $\unit.x^k$. Therefore, the unit becomes $u= 1+p.\unit + x^k.\unit$. As before, we have the long exact sequence $(\alpha)$
$$  {K_p^\wedge}_*(C P^\infty) \stackrel{u-1}{\rightarrow} {K_p^\wedge}_*(C P^\infty) \rightarrow \pi_*(\THH^{K_p^\wedge}(K/p,f)) $$
and,
\begin{align}
< (u-1)(\beta_i),x^j>&= <\beta_i, x^j(p+ x^k)> \nonumber \\
                     &= \left\{
\begin{array}{rl}
p & \text{if } j=i\\
1 & \text{if } j=i-k\\
0 & \text{} otherwise
\end{array} \right. \nonumber
\end{align}
$$\implies (u-1)(\beta_i)=\left\{
 \begin{array}{rl}
p\beta_i & \text{if } i=0\\
p\beta_i + \beta_{i-k} & \text{if } i>0
\end{array}\right.$$

Therefore, we obtain that
$$\implies \pi_n(\THH^{K_p^\wedge}(K/p,f))=\left\{
 \begin{array}{rl}
0  & \text{if $n$ is odd }\\
(Z/(p^\infty))^k & \text{if $n$ is even}
\end{array}\right.$$

This ends the calculation for all odd primes. The homotopy groups of $\THH^{K_p^\wedge}(K/p)$ are 0 in odd degrees and $(Z/(p^\infty))^k$ in even degrees, where $k$ is a number between 1 and $p-1$ depending on the $A_\infty$ structure on $K/p$. This result was proved before by Angeltveit (\cite{A1}). He used the B\"okstedt spectral sequence to calculate Topological Hochschild Homology.

\rem : This is the calculation identifying $K/p$ as the Thom spectrum of $S^1$. A similar calculation can be carried out for the Thom spectrum of $S^3$ to get the same results.

config: arxiv


TITLE: Asymptotic Curves and Lines of Curvature of Helicoid
QUESTION [5 upvotes]: I have a question that asks me to find the asymptotic curves and lines of curvature of the helicoid given by: $x = v \cos u$, $y = v \sin u$, $z = cu$, for some fixed real $c$. Can you show me how best to do this every step of the way, from finding the coefficients of the fundamental forms to solving the differential equations for $u$ and $v$?
$X = (x,y,z)$.
$X_{u} = (-v \sin u, v \cos u, c)$,
$X_{v} = (\cos u, \sin u, 0)$.
First fundamental form:
$E = X_{u} • X_{u} = v^{2} + c^{2},$
$F = X_{u} • X_{v} = 0$,
$G = X_{v} • X_{v} = 1$.
Information that follows swiftly from it:
$X_{uu} = (-v \cos u, - v \sin u, 0)$
$X_{uv} = (-\sin u, \cos u, 0)$,
$X_{vv} = (0,0,0)$.
$N = \frac{1}{\sqrt{(EG-F^2)}}\cdot(-c \sin u, c \cos u, -v).$
Second Fundamental Form:
$e = \langle N, X_{uu}\rangle = \frac{2c v \sin u\cos u}{\sqrt{EG-F^2}},$
$f = \langle N, X_{uv}\rangle = \frac{c \sin^2 u - c \cos^2 u}{\sqrt{EG-F^2}},$
$g =\langle N, X_{vv}\rangle = 0.$
These can be simplified somewhat.
Asymptotic curves must satisfy:
$e(u')^{2} - 2f u' v' + g (v')^{2} = 0.$
I can plug these values in, but cannot really proceed.
The differential equation for the lines of curvature is given by:
$$\begin{vmatrix}(v')^2& -u'v'& (u')^2\\ E& F& G\\ e& f& g\end{vmatrix}= 0.$$
Again, I can plug them in, but how to proceed eludes me.

REPLY [3 votes]: We try to generalize all the cases in between catenoid and helicoid.
Let $\, \boldsymbol {x} (u,v)=
\cos t \left(
     \begin{array}{c}
       \cos u \cosh v \\
       \sin u \cosh v \\
       v
     \end{array}
   \right)+
\sin t \left(
     \begin{array}{c}
       \sin u \sinh v \\
      -\cos u \sinh v \\
       u
     \end{array}
   \right)=
\Re \left[ e^{it}\left(
     \begin{array}{c}
       \cos (u+vi) \\
       \sin (u+vi) \\
       -i(u+vi)
     \end{array}
   \right) \right]
$
It's a catenoid for $t=0, \pi$ and helicoid for $t=\frac{\pi}{2}, \frac{3\pi}{2}$.
Now $\,
\left|
  \begin{array}{ccc}
    E & F & G \\
    e & f & g \\
    dv^{2} & -du \, dv & du^{2} \\
  \end{array}
\right| =
\left|
  \begin{array}{ccc}
    \cosh^{2} v & 0 & \cosh^{2} v \\
   -\cos t & \sin t & \cos t \\
    dv^{2} & -du \, dv & du^{2} \\
       \end{array}
\right| = 0
$.
We have $\left(\sin t \, du^{2}+2\cos t \, du \, dv-\sin t \, dv^{2}\right)
\cosh^{2} v=0, \,$ thus
$
\left \{
  \begin{array}{rcl}
    \cos \frac{t}{2} du-\sin \frac{t}{2} dv & = & 0 \\
    \sin \frac{t}{2} du+\cos \frac{t}{2} dv & = & 0 \\
  \end{array}
\right. \implies
\left \{
  \begin{array}{rcl}
    u\cos \frac{t}{2}-v\sin \frac{t}{2} & = & \alpha \\
    u\sin \frac{t}{2}+v\cos \frac{t}{2} & = & \beta \\
  \end{array}
\right. \;
$ or equivalently,
$
u+vi=\exp\left( \frac{1}{2}it \right) (\alpha+\beta i) \,$ where $\alpha, \beta$ are integration constants and hence the parameters for the lines of curvature.
Note that $t=0$ is trivial for catenoid.


TITLE: Vector ratios on parallelogram
QUESTION [0 upvotes]: There is a parallelogram ABCD.
On length CD there is point E with statement |CE|=2|ED|
On length AD there is point F with statement |AF|=1/4|AD|
Point S is intersection of length BE and CF.
In which ratio does point S divide length BE and FC?

The answer to this problem is:
|FS| : |SC| = 5 : 4
|ES| : |SB| = 1 : 2
I have been stuck on this problem for some time now and any guidance would be helpful.

REPLY [0 votes]: Here's a solution using vectors . . .

Place the origin at $A$.

Letting $\mathbf{u}=\vec{D}$, and $\mathbf{v}=\vec{B}$, we get
\begin{align*}
\vec{C}&=\vec{D}+\vec{B}=\mathbf{u}+\mathbf{v}\\[4pt]
\vec{F}&={\small{\frac{1}{4}}}\vec{D}={\small{\frac{1}{4}}}\mathbf{u}\\[4pt]
\vec{E}&=\vec{D}+{\small{\frac{1}{3}}}(\vec{C}-\vec{D})=\mathbf{u}+{\small{\frac{1}{3}}}\mathbf{v}\\[4pt]
\end{align*}
Also, since $S$ is on the line segments $FC$ and $EB$, we get
\begin{align*}
\vec{S}&=\vec{F}+x(\vec{C}-\vec{F})=\Bigl(\frac{1+3x}{4}\Bigr)\mathbf{u}+x\mathbf{v}\\[4pt]
\vec{S}&=\vec{E}+y(\vec{B}-\vec{E})=(1-y)\mathbf{u}+\Bigl(\frac{1+2y}{3}\Bigr)\mathbf{v}\\[4pt]
\end{align*}
for some $x,y\in\mathbb{R}$, with $0 < x < 1$ and $0 < y < 1$.

Since the vectors $\mathbf{u},\mathbf{v}$ are linearly independent, we must have
\begin{align*}
\frac{1+3x}{4}&=1-y\\[4pt]
x&=\frac{1+2y}{3}\\[4pt]
\end{align*}
which yields $x={\large{\frac{5}{9}}}$ and $y={\large{\frac{1}{3}}}$.

It follows that

$S$ is ${\large{\frac{5}{9}}}$ of the way from $F$ to $C$, so $|FS|{\;:\,}|SC|=5{\,:\,}4$.$\\[10pt]$
$S$ is ${\large{\frac{1}{3}}}$ of the way from $E$ to $B$, so $|ES|{\;:\,}|SB|=1{\,:\,}2$.


TITLE: What are the invertible elements of the $K[X]$ ring where $K$ is a field?
QUESTION [1 upvotes]: What are the invertible elements of the $K[X]$ ring where $K$ is  a field?
I know that in a field all the elements are invertible but I'm having trouble linking $K$ and $K[X]$

REPLY [4 votes]: If we have two nonzero polynomials $f$ and $g$ of degrees $n$ and $m$ then consider the leading terms, say $aX^n$ and $bX^m$. The product $fg$ must have leading term $abX^{n+m}$. Since $K$ is a field we can't have $ab=0$ and so $fg$ has degree $n+m$. Thus $fg$ can't be $1$ unless both $f$ and $g$ have degree $0$.
Conversely it's clear that every nonzero polynomial of degree $0$ is invertible.


TITLE: Poisson Process - Finding conditional expectation and conditional variance
QUESTION [0 upvotes]: Consider a Poisson process $X={X(t);t\ge0}$ of rate $λ=5$. Here $X(t)$ is the number of customers arrived up to the time$=t$. Suppose that $X(1)=5$ (so 5 customers arrived by the end of the first hour). Find the conditional expectation and conditional variance for (total waiting time) W=W1+W2+...+W5, given that $X(1)=5.$
I'm lost on how to solve this problem. Any help is appreciated!

REPLY [1 votes]: Hint: conditioned on $X(1)=5$, the interarrival times $W_1,\ldots,W_5$ are distributed as the order statistics of 5 independent and uniformly distributed random variables over $[0,1]$.


TITLE: If two balls are chosen at random and one of them is found to be of standard quality. What is the probability that the other one is defective?
QUESTION [1 upvotes]: A football factory has two machines, A and B. Machine A manufactures 70% of the balls and machine B manufactures 30%. 80% of the balls from machine A and 90% of the balls from machine B are rated of standard quality.
Q1. A ball is chosen at random and is found to be of standard quality. What is the probability that it has come from machine A? The answer here is 0.67 and can be found using Bayes' rule.
Q2. If two balls manufactured using the same machine are chosen at random and one of them is found to be of standard quality, what is the probability that the other ball is defective?
My attempt:
I have used Bayes theorem to answer the question.
Here, D = event of having defective item, S = event of having  Standard quality item.
Pr(D|S) = Pr(D).Pr(S|D)/ Pr(S)
Pr(S) = 0.7*0.8 + 0.3*0.9 = 0.83
Pr(D) = 0.7*0.2 + 0.3*0.1 = 0.17
Since they are independent events. They have no common intersection between them, Pr(S|D) = Pr(S)
So, In my opinion, the answer is 0.17.
But, when I discussed this with my friends, one of them highlighted a point saying that "how do you know the exact first pick will be standard or defective?" In his opinion, I need to dig more. So it can happen in 4 ways. They are given below:
Event 1 could be One Standard, other is defective;
Event 2 could be One Defective, other is Standard;
Event 3 could be Both Defective;
Event 4 could be both Standard.
I have used the conditional probability approach to solve that.
Now let's say X = event of other one is defective, Y = event of one of them is Standard
Pr(one of them being Standard) = 0.83*0.83 + 0.83*0.17 = 0.9711 (sum of event 1,2,4) ---> Pr(Y)
Pr(other one defective | one of them Standard) = Pr(X intersets Y) / P(Y)
Pr(X intersets Y) = 0.83*0.17 + 0.17*0.83 = 0.2822 (Sum of event 1,2)
therefore the answer is 0.2822/0.9711 = 0.2906
I have applied Bayes, got the same answer in this case.
which one was the correct approach? and if you could contrast between them, it will help me a lot to understand it better. TYIA

REPLY [2 votes]: The second part of the question is ill-posed in two ways.  First, it is not clear whether the two balls were drawn from the same randomly selected machine.  Second, it is not clear whether the standard quality ball was determined by random selection and the second ball's quality is unknown, or if the quality of both balls is known to another party and you are told that at least one ball is of standard quality.  To illustrate why these things matter, consider the following scenarios:

The machines that make the balls dump them into a large bin, mixing up all of the balls; however, balls are stamped with a serial number that identifies which machine that made them.  You randomly select one ball from the bin, inspect it, and find it is of standard quality.  You then draw a second ball from the bin.  What is the probability it is defective?  The answer is $$1 - ((0.8)(0.7) + (0.9)(0.3)) = 0.17.$$
Same as Scenario 1, except this time, a factory worker randomly select two balls from the bin and inspects them.  Then he tells you that one of the balls is standard quality.  What is the probability the other ball is defective?  The answer is $$\frac{2(0.17)(0.83)}{1 - (0.17)^2} = \frac{34}{117} \approx 0.290598.$$
Now instead suppose the machines put the balls into separate bins.  The factory worker, with equal probability, selects one of the two bins, and randomly draws one ball for you to inspect; it is standard.  If he draws a second ball from the same bin (thus the same machine), what is the probability it is defective?  The answer is computed as follows.  Let $X_i = 1$ if ball $i$ is standard, and $0$ if defective.  Then the posterior probability of the worker having selected Bin $A$ given the $X_1 = 1$ is $$\Pr[A \mid X_1 = 1] = \frac{\Pr[X_1 = 1 \mid A]\Pr[A]}{\Pr[X_1 = 1]} = \frac{(0.8)(0.5)}{(0.8)(0.5)+(0.9)(0.5)} = \frac{8}{17} \approx 0.470588.$$  Consequently, the probability the second ball drawn from the same bin is defective is $$\begin{align*}
\Pr[X_2 = 0 \mid X_1 = 1] &= \Pr[X_2 = 0 \mid A]\Pr[A \mid X_1 = 1] + \Pr[X_2 = 0 \mid B]\Pr[B \mid X_1 = 1] \\
&= (0.2)(8/17) + (0.1)(9/17) = \frac{5}{34} \approx 0.147059.
\end{align*}$$
As in Scenario 3, we have separate bins, but now this time, the factory worker selects a bin with probability proportional to the number of balls it contains; that is to say, because machine $A$ makes $70\%$ of the balls, he chooses bin $A$ with $70\%$ probability rather than $50\%$.  Everything else is the same.  Then the answer becomes $$(0.2)(56/83) + (0.1)(27/83) = \frac{139}{830} \approx 0.16747.$$
Now let's consider Scenario 3 except the factory worker has drawn two balls from the same bin, inspects them, and informs you that one of them is standard.  What is the probability the other is defective?  By now you should clearly realize that this probability will not be the same as any of the ones we have calculated so far.  I will not do the computation; I leave it as a straightforward exercise for the reader.
Consider Scenario 5 except now the probability of selecting a bin is as in Scenario 4.  Again, the computation will not be performed here and is left as an exercise.

As you can see, we must be absolutely clear about how the balls are drawn in order to perform the requisite computation.  The way the question is posed does not adequately describe the sampling scheme.

Now that the question has been rephrased, we can reasonably interpret it to match Scenario 3 or Scenario 4.  But which one is again not clear because it is not explained how two balls are randomly selected from the same machine.