\section{Introduction}

A central problem in  three-dimensional  knot concordance theory in the smooth category consists of understanding  $\mathcal{T}$, the concordance group of topologically slice knots.  Freedman~\cites{MR1201584,MR679066} proved that the subgroup $\mathcal{T}^1$ generated by knots with Alexander polynomial one satisfies $\mathcal{T}^1 \subset \mathcal{T}$.  Early work proving that $\mathcal{T}^1$ is nontrivial includes that of Akbulut, Casson, and Cochran-Gompf~\cite{MR976591}; in~\cite{MR1334309}  it was shown that $\mathcal{T}^1$ contains an infinitely generated free subgroup.  Using  Heegaard Floer theory, in~\cite{MR3466802, MR2955197} it was shown that $\mathcal{T} / \mathcal{T}^1$ contains an infinitely generated free subgroup and infinite two-torsion.  

Recently, Jae Choon Cha~\cite{arxiv:1910.14629}  has undertaken an in-depth investigation of  {\it primary decompositions} of $\mathcal{T}$.   
One motivation for studying primary decompositions arises from Levine's work~\cite{MR0253348} in which it was shown that for all integers $n>0$ there is a homomorphism from the smooth concordance group of knotted $2n-1$ spheres in $S^{2n+1}$ to a group called the  {\it rational algebraic concordance group}:    $\psi_{2n-1} \colon\! \mathcal{C}_{2n-1} \to \mathcal{G}^{\mathbb Q}$.  It was also proved that there is a decomposition $\mathcal{G}^{\mathbb Q} \cong \oplus_{p\in \mathcal{A}}  \mathcal{G}^{\mathbb Q}_p$, where $\mathcal{A}$ is the set of irreducible Alexander polynomials.  Such a decomposition does not exist  using integer rather than  rational coefficients, but the failure was completely analyzed by Stoltzfus~\cite{MR0467764}.  In all odd higher dimensions $\psi_{2n-1}$ is injective, leading to decomposition theorems for knot concordance groups.  In the classical dimension, $n=1$, the map $\psi_1$ is not injective~\cite{MR900252}; the kernel is infinitely generated, contained the subgroup $\mathcal{T}$ of topologically slice knots.

To briefly summarize the perspective of Cha's work,  we  let  $\mathcal{Q} \subset {\mathbb Z}[t]$ be the set of irreducible  polynomials  $q(t)$ satisfying $q(1) = 1$.    Let 
$$\mathcal{P} = \{ q(t)q(t^{-1})  \in {\mathbb Z}[t,t^{-1}] \ \big| \ q(t) \in \mathcal{Q}\}.$$  According to Fox and Milnor~\cite{MR0211392}, if a knot $K$ is smoothly  slice, then its Alexander polynomial is a product of elements in $\mathcal{P}$.  The same result holds for topologically locally flat slice knots, as proved   using the existence of normal bundles for locally flat disks (see Freedman-Quinn~\cite[Section 9.3]{MR1201584}).
According to Terasaka~\cite{MR0117736}, every product of elements in $\mathcal{P}$ is the Alexander polynomial of some slice knot. 


Given any subset $\mathcal{P}_0 \subset \mathcal{P}$,  let $\mathcal{T}^{\mathcal{P}_0} \subset \mathcal{T}$ denote the subgroup generated by topologically slice knots with Alexander polynomial a product of polynomials  $p$ for  $p \in \mathcal{P}_0$.  In the case that $\mathcal{P}_0$ is a singleton  $\{p\}$, we write $\mathcal{T}^p$.  Hence, as above, $\mathcal{T}^1$ denotes the subgroup generated by knots with Alexander polynomial one.    Notice that for any pair of elements   $p, q  \in \mathcal{P}$, we have $\mathcal{T}^1 \subset \mathcal{T}^p \cap \mathcal{T}^q$.  Thus, in the following conjecture  it is necessary to consider the quotients $\mathcal{T}^p_\Delta = \mathcal{T}^p / \mathcal{T}^1$ and $\mathcal{T}_\Delta = \mathcal{T} / \mathcal{T}^1$.  

\begin{conjecture} The canonical homomorphisms    $\mathcal{T}^p_\Delta \to \mathcal{T}_\Delta$  induce an isomorphism  $$\Phi \colon\! \Moplus_{p \in \mathcal{P}}\mathcal{T}^p_\Delta \to \mathcal{T}_\Delta.$$ 
\end{conjecture}

In~\cite{arxiv:1910.14629}, Cha identifies and studies a specific  infinite set $\mathcal{P}_0 \subset \mathcal{P}$ with two properties:  first, for all $p \in \mathcal{P}_0$, he proves that   $\mathcal{T}^p_\Delta$ contains an infinitely generated free subgroup $\mathcal{S}^p_\Delta$; second, he proves that the  restriction of $\Phi$ is injective on $\oplus_{p \in \mathcal{P}^0} \mathcal{S}^p_\Delta$.  

The main goal of this paper is to provide a counterexample to a    splitting property related to the surjectivity of $\Phi$,  considered by Cha under the name {\it  strong existence} (see \cite[Appendix A]{arxiv:1910.14629}). 
Although this does not provide a counterexample to the conjecture, it adds to the evidence  that the conjecture is false.  More specifically, it indicates  that $\Phi$ is probably not surjective.

\begin{theorem}\label{thm:main} There exists a set of three polynomials, $\mathcal{P}_0 = \{f_1, f_2,f_3\} \subset \mathcal{P}$, such that the natural homomorphism  $$\mathcal{T}^{f_1}_{\Delta}\oplus \mathcal{T}^{f_2}_{\Delta} \oplus \mathcal{T}^{f_3}_{\Delta} \to \mathcal{T}^{\mathcal{P}_0}_\Delta$$ is not surjective.
\end{theorem}

The use of three factors is an artifact of the proof.  It will be clear that without the restriction of irreducibility for elements in $\mathcal{Q}$, we could have used two factors, as was stated in the abstract.    To be more precise, there is the following statement.

\begin{theorem}  There exist  Alexander polynomials $f_1(t)$ and $f_2(t)$ having no common factors and a topologically slice knot $K$ with $\Delta_K(t) = f_1(t)^2f_2(t)^2$ such that $K$ is not concordant to any connected sum of knots $K_1 \mathbin{\#} K_2$ where $\Delta_{K_i}(t) = f_i(t)^{n_i}$ and $n_1, n_2 \in {\mathbb Z}$.
\end{theorem} 
Notice that it follows that $K$ is a topologically slice knot that is not smoothly concordant to a knot with Alexander polynomial one.  The first  examples of  such knots were described in~\cite{MR2955197}.  The example and proof here are closely related to that earlier work. 

\vskip.05in
\noindent{\it Acknowledgments} Thanks are due to Jae Choon Cha for sharing with me    drafts of his ongoing work on primary decompositions and for his repeatedly offering thoughtful commentary on this note.  Discussions with Se-Goo Kim and Taehee Kim were also very helpful.

\section{Rational homology cobordism and the construction of an example}

The proof of Theorem~\ref{thm:main} can be  reduced to a result concerning rational homology cobordism, as we now describe.  Recall first that the Alexander polynomial of a knot determines the order of the first  homology of $M(K)$,  the 2--fold cyclic branched cover of $S^3$ branched over $K$: $\big| H_1(M(K))\big| = \big|\Delta_K(-1)\big|$. Also, 2--fold branched covers of concordant knots are rationally homology cobordant. 

Figure~\ref{fig:knot} is a schematic diagram of a  topologically slice knot $K_n(J_n)$ (in the case of $n = 3$); we will be restricting to the case of  $n \equiv 3 \mod 4$.  These knots  bound       punctured Klein bottles in $S^3$.  In each, the left band  has framing 0,  there are   $n$ half-twists between the bands,  and the knot $J_n $ is the connected sum of $(3n-1)/4$ copies of the positive clasped untwisted double of the trefoil knot, $D(T(2,3))$.    Similar knots were used in~\cite{MR2955197} to prove the nontriviality of $\mathcal{T}_\Delta = \mathcal{T} /\mathcal{T}^1$.  Henceforth, will write $K_n$ for $K_{n}(J_{n})$.   A quick calculation in Section~\ref{sec:alex} will show   that if $n=pq$ for odd primes $p$ and $q$, then  $$\Delta_{K_n}(t) = \big(    \phi_{2p}(t) \phi_{2q}(t)\phi_{2pq}(t)  \big)^2,$$ where $\phi_k(t)$ denotes the $k$--cyclotomic polynomial.    
\begin{figure}
\fig{.6}{figure1.pdf}
\caption{Knot}\label{fig:knot}
\end{figure}

\begin{theorem} If $n = pq$, then the homology of the 2--fold branched cover of $K_n$ satisfies $\big| H_1(M(K_n))\big| = n^2$.  If $K_n$ is concordant to a connected sum $L_1 \mathbin{\#} L_2 \mathbin{\#} L_3$, with $\Delta_{L_1}(t) =  \phi_{2p}(t)^{m_1}$,    $\Delta_{L_2}(t) =  \phi_{2q}(t)^{m_2}$, and $\Delta_{L_3}(t) = \phi_{2pq}(t)^{m_3}$, then $M(K_n) $ is rationally homology cobordant to a connected sum $M_1 \mathbin{\#} M_2 \mathbin{\#} M_3$, where $\big|H_1(M_1)\big| = p^{m_1}$,  $\big|H_1(M_2) \big| = q^{m_2}$ and  $\big|H_1(M_3) \big| = 1$. 

\end{theorem}

\begin{proof}  These all follow  immediately from the facts that $  \phi_{2p}(-1) = p$,    $  \phi_{2q}(-1)  =q $, and $ \phi_{2pq}(-1) = 1$.  See Lemma~\ref{lemma:cyclotomic} for details.

\end{proof}  

The topic of~\cite{MR3259765} was the general problem of finding a primary splitting  of the rational homology cobordism group.  We will show that the techniques used there can be applied to prove that for $n=15$, a rational homology  cobordism from $M(K_n)$ to such a connected sum, $M_1 \mathbin{\#} M_2 \mathbin{\#} M_3$, does not exist. Notice that if it did exist,  we could let $N_2 =M_2 \mathbin{\#} M_3$, and reduce  our  work  to obstructing the existence of a rational homology cobordism to a connected sum of two manifolds, $M_1 \mathbin{\#} N_2$.  Our goal will be to prove that $M(K_n)$ is not rational homology cobordant to any connected sum $M_1 \mathbin{\#} M_2$, where $\big|H_1(M_1)\big| = 3^{m_{1}}$ and   $\big|H_1(M_2)\big| = 5^{m_{2}}$ for some integers $m_1$ and $m_2$.

\section{Obstructions from $d$--invariants}

We use the following notation:  for any abelian group $G$ and prime integer $p$,   let $G_{(p)}$ denote the subgroup consisting of all elements of order $p^n$ for some $n$.


All the three-manifolds $M$ we will be   working with are ${\mathbb Z}_2$--homology three-spheres.  Since $H_1(M)$ is of odd order, there are natural identifications: $H_1(M)  \cong H^2(M) \cong \text{Spin}^c(M)$.  For $g \in H_1(M)$ we denote by $d(K, g)$ the Heegaard Floer correction term associated to $g$  viewed as a Spin$^c$--structure.   Basic results concerning $\text{Spin}^c(M)$, the $d$--invariant, and its basic properties are in~\cite{MR1957829}.   Further details and examples are provided in~\cite{MR3259765}.

If $M$ bounds a rational homology four-ball, then there is a subgroup $\mathcal{M} \subset H_1(M)$ such that: $\big|\mathcal{M}\big| ^2= \big|H_1(M)\big|$;  the nonsingular linking form vanishes on $\mathcal{M}$; and $d(M, g) = 0$ for all $g \in \mathcal{M}$.    We can equivalently view $\mathcal{M} \subset H^2(M)$. 


\begin{theorem}\label{thm:main:d}  Let $p$ and $q$ be distinct odd primes and let $M$ be a three-manifold satisfying $H_1(M) \cong {\mathbb Z}_{p^2} \oplus {\mathbb Z}_{q^2}$,  generated by elements $a$ and $b$ of order $p^2$ and $q^2$, respectively.  If  $M$ is rationally homology cobordant  to a connected sum $ M_1 \mathbin{\#} M_2$ where $H_1(M_1)_{(q)} = 0$ and $H_1(M_2)_{(p)} = 0$, then the value of    $$d(K,ipa +jqb) - d(K,ipa) - d(K,jqb)$$ is independent of $i$ and $j$. 
\end{theorem}

\begin{proof}  Since $M$ and $M_1 \mathbin{\#}  M_2$  are rationally homology  concordant, $M \mathbin{\#}  -M_1 \mathbin{\#}  -M_2$ bounds a           rational homology $4$--ball $W$.  The image of $H^2(W)$   in $H\!:= H^2(M \mathbin{\#} -M_1 \mathbin{\#} -M_2)$ is the desired  subgroup $\mathcal{M}$ satisfying $\big|\mathcal{M}\big|^2 = \big|H\big|$.  Furthermore, $\mathcal{M}$ is self-annihilating with respect to the nonsingular linking form.  The order of a self-annihilating subgroup of a group $G$ is of order at most $\sqrt{|G|}$; it follows that  the subgroup $\mathcal{M}_{(p)} \subset H(M)_{(p)} \oplus H(-M_1)_{(p)}$ cannot be contained in $H(-M_1)_{(p)}$.  In particular, some element of the form $(x_p, y_p) \in  H(M)_{(p)} \oplus H(-M_1)_{(p)}$ with $x_p \ne 0 \in Z_{p^2}$ is contained in $\mathcal{M}$.  By taking a multiple, we can assume $x_p = pa$.  Similarly, there is an element $(qb, y_q) \in  H(M) \oplus H(-M_2)$ in $\mathcal{M}_q$.

Notice that we are viewing $H_1(-M_1) \subset H_1(-M_1) \oplus H_1(-M_2)$, so in this sense  $y_p$ can be interpreted as an ordered pair $(y_p, 0) \in H_1(-M_1) \oplus H(-M_2)$; similarly, $y_q$ represents an ordered pair   $(0,y_q) \in H_1(-M_1) \oplus H_1(-M_2)$. The correction term is additive under connected sum and vanishes for elements in $\mathcal{M}$.  Thus for any $i$ and $j$:
$$d(M, ipa) - d(M_1, iy_p) - d(M_2,0)= 0,$$
$$d(M, jqb)- d(M_1,0) - d(M_2, jy_q) = 0,$$
and
$$d(M, ipa + jqb ) - d(M_1, iy_p)  - d(M_2, jy_q) = 0.$$
Subtracting the first two equations from the third shows that for all $i$ and $j$,
$$d(M, ipa + jqb ) - d(M, ipa)  - d(M, jqb) = -d(M_1, 0)   -d(M_2, 0).$$ The right hand side is independent of $i$ and $j$.

\end{proof}


\section{The knots $K_n(J_n)$  and their Alexander polynomials}\label{sec:alex}

As described in the introduction, we are considering the knots illustrated schematically in  Figure~\ref{fig:knot}.    

\begin{theorem}  Let $K_n = K_n(J_n)$. For  $n$   odd, the Alexander polynomial is given by $$\Delta_{K_n}(t) = \left( \frac{ t^{n}+1}{t+ 1}\right)^2.$$ \end{theorem}

\begin{proof}
This knot is a winding number two companion of $J_n$.  Since $J_n$ has Alexander polynomial one, a  standard formula for the Alexander polynomial of a satellite knot~\cite{MR0035436} implies  that the Alexander polynomial of $K_n(J_n)$ is the same as that for $K_n(U)$, where $U$ is the unknot.  In this case, a simple manipulation shows that  $K_n(U) = P(n, -n, n-1)$,   a three-stranded pretzel knot, illustrated in the case of $n=3$ in Figure~\ref{fig:pretzel}.  To compute its Alexander polynomial, we consider instead the Conway polynomial $\nabla_{K_n}(z)$.  (Recall that for an arbitrary oriented knot $K$, $\Delta_K(t) = \nabla_K(t^{1/2} - t^{-1/2})$.)

\begin{figure}[h]
\fig{.5}{figure2.pdf}
\caption{The pretzel knot $P(3,-3,2)$}\label{fig:pretzel}
\end{figure}

The standard crossing change formula  for the Conway polynomial is $$\nabla_+(z) - \nabla_-(z) = -z \nabla_s(z),$$    where $\nabla_\pm$ denotes the Conway polynomial of  an oriented link with a specified crossing made positive or negative  and $\nabla_s$ is the Conway polynomial of the    link formed by smoothing that same crossing.  This can be applied to a crossing on the right-most band of the pretzel knot, indicated by the dot in Figure~\ref{fig:pretzel}.  Smoothing the crossing yields an unlink, which has  0 Conway polynomial.  Changing the crossing removes two half-twists.  Thus, $\nabla_{P(n, -n, n-1)}(z) =\nabla_{P(n, -n, n-3)}(z)$.  Since $n$ is odd, continuing in this way removes the right-most crossings, ultimately yielding the connected sum $-T(2,n) \mathbin{\#} T(2,n)$.   The Alexander polynomials of torus knots is well-known; in this case it is $$\Delta_{T(2,n)}(t) = \frac{(t^{2n} -1)(t-1)}{(t^2 - 1)(t^n -1)} = \frac{t^n +1}{t+1}.$$
\end{proof}

\begin{lemma}\label{lemma:cyclotomic} For distinct odd primes $p$ and $q$, there is the following identity, where the $\phi_i$ are cyclotomic polynomials.
$$ \frac{t^{pq} +1}{t+1} =  \phi_{2p}(t) \phi_{2q}(t) \phi_{2pq}(t).$$ Furthermore, $\phi_{2p}(-1) = p$, $\phi_{2q}(-1) = q$, and $\phi_{2pq}(-1) = 1$.
\end{lemma}

\begin{proof}  The polynomial $t^n -1$ has factors $\phi_d(t)$ for all divisors $d$ of $n$.  Thus 

\begin{equation}{t^{2pq} -1} = \phi_{2pq}(t)\phi_{2p}(t)\phi_{2q}(t) \phi_{p}(t)\phi_{q}(t)\phi_2(t)\phi_1(t)
\end{equation}
and $${t^{ pq} -1} = \phi_{ pq}(t)\phi_{ p}(t)\phi_{ q}(t)\phi_1(t).$$ Dividing the first equation by the second, and then dividing by $\phi_2(t) = t +1$ yields   $$ \frac{t^{pq} +1}{t+1} =  \phi_{2pq}(t) \phi_{2p}(t) \phi_{2q}(t) .$$  

L'Hopital's rule can be used to determine that the left hand side evaluated at $-1$ is $pq$.  Thus, if we show $\phi_{2p}(-1) = p$, and, similarly, $\phi_{q}(-1) = q$, we are done.  Proceeding as before, $$t^{2p} -1 = \phi_{2p}(t) \phi_p(t) \phi_2(t) \phi_1(t) $$ and $$t^{p} -1 =  \phi_p(t)  \phi_1(t). $$  Dividing yields   $$\frac{ {t^p +1}}{t+1}  =  \phi_{2p}(t). $$  In this case, L'Hopital's rule shows that $\phi_{2p}(-1) = p$.

\end{proof}

\begin{corollary} If $n = pq$, where $p$ and $q$ are distinct odd primes, then $\Delta_{K_n(J_n)}(t) =\big( \phi_{2p}(t)\phi_{2q}(t)\phi_{2pq}(t)\big)^2$, where the $\phi_k(t)$ are cyclotomic polynomials.  For $r$ an  odd prime,  $\phi_{2r}(-1) = r$ and for  a product of two distinct odd primes, $\phi_{2pq}(-1) = 1$.
\end{corollary}

\section{Computing $d(M(K), i)$; the completion of the proof of Theorem~\ref{thm:main}}

We will now restrict to the  case of $n = 15$.     The methods of~\cite{MR593626} apply to show that the  2--fold branched cover of $S^3$ branched over $K_n(J_n)$ is given by $15^{2}$--surgery on $T_{14,15} \mathbin{\#} 22D(T_{2,3})$.  We continue to denote $K_{15}(J_{15})$ by $K_{15}$  and denote its 2--fold  branched cover simply by $M$.  Note that $H_1(M) \cong {\mathbb Z}_{15^2}$.  

Suppose now that $K_{15}$ were concordant to a connected sum of three knots, one with Alexander polynomial $ \phi_{6}(t) ^{m_1} $, one with Alexander polynomial   $ \phi_{10}(t) ^{m_2}$, and one with Alexander polynomial $   \phi_{30}(t) ^{m_3}$.   Then, as described in the introduction, $M$ would by rationally homology cobordant to $M_1 \mathbin{\#} M_2$, where $H_1(M_1)_{(5)} = 0$ and $H_1(M_2)_{(3)} = 0$.  Thus,  Theorem~\ref{thm:main:d} would apply. 

In~\cite[Section 6]{MR2955197}, an algorithm is presented for computing the values of the $d$--invariants of $M(K_n(J_n))$.  Here is   the results of the computation; readers are referred to~\cite{MR2955197} for general background.  In Appendix~\ref{app:dcomp} we provide a summary of the details of this specific computation.  Since $H_1(M) \cong Z_{225}$, the order 9 subgroup is generated by $a = 25$ and the order 25 subgroup is generated by $b = 9$.   The values that result from the computation  of $d(M, ipa + jqb ) $ are as shown in Table~\ref{table:d}. 

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | } 
\hline
&     $i = 0  $        & $i=1$        &    $ i = 2$   \\ 
\hline
$j=0$         &          		22             &     		14       & 	14   \\ 
\hline
$j=1$         &              	18          &    		6       &  			20  \\ 
\hline
$j=2$         &           		10              &   		16      & 			22  \\ 
\hline
$j=3$         &          		10               &    	 22     & 		16   \\ 
\hline
$j=4$         &         		18                &     	20       	&	 	6   \\ 
\hline
\end{tabular}
\vskip.1in
\end{center}
\caption{Values of  $d(M, ipa + jqb )  $}\label{table:d}
\end{table}

The values of  $d(M, ipa + jqb ) - d(M, ipa)  - d(M, jqb)$ are as shown (with sign reversed for readability) in Table~\ref{table:d2};  they are not all equal.

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | } 
\hline
&     $i = 0  $        & $i=1$        &    $ i = 2$   \\ 
\hline
$j=0$         &          		22             &     		22      & 	22  \\ 
\hline
$j=1$         &              	22          &    		26       &  			12  \\ 
\hline
$j=2$         &           		22             &   		8      & 			2  \\ 
\hline
$j=3$         &          		22              &    	 2     & 		8  \\ 
\hline
$j=4$         &         		22               &     	12       	&	 	26   \\ 
\hline
\end{tabular}
\vskip.1in
\end{center}
\caption{Values of  $-\big(d(M, ipa + jqb ) - d(M, ipa)  - d(M, jqb)\big)$}\label{table:d2}
\end{table}

\subsection{Infinite families}  Let $\{p_i\}$ be an infinite increasing sequence of  primes for which $p_i \equiv 3 \mod 4$ if and only if $i$ is odd.  Let $\mathcal{P}_0 = \cup_{i =1  }^\infty  \{\phi_{2p_i -1 }(t), \phi_{2p_i  }(t), \phi_{2 p_{2i-1} p_{2i}}(t)\}$.  Most of the previous argument is easily generalized.  The only step that we have not been able to complete in general is the computation of the $d$--invariants.   Our expectation is that this would lead to the conclusion that $$ \oplus_{p \in \mathcal{P}_0} \mathcal{T}^p_\Delta \to  \mathcal{T}^{\mathcal{P}_{0}}_\Delta$$ is not surjective.



