\label{thrm:beat-cactus-introversion}
    For all $n\geq14$, minors of size $n(n-1)(n-2)+1$ of the tangency flattenings vanish on the $n$-th secant variety of the Segre variety in $\P(\mathbb{C}^n\otimes\mathbb{C}^n\otimes\mathbb{C}^n)$ but do not vanish on its $n$-th cactus variety.

\begin{theorem}
    \label{thrm:beat-cactus-introversion}
    For all $n\geq14$, minors of size $n(n-1)(n-2)+1$ of the tangency flattenings vanish on the $n$-th secant variety of the Segre variety in $\P(\CC^n\otimes\CC^n\otimes\CC^n)$ but do not vanish on its $n$-th cactus variety.
\end{theorem}

\begin{proposition}
    \label{prop:1de-koszul}
    Let $R$ be the apolar algebra to $e\leq \binom{d+1}2$ general quadrics in $d$ variables. If $e\geq3$, then the structure tensor $T$ of $R$ has border cactus rank at least $\floor{\frac32d}+3$. In particular, if $3\leq e\leq \floor{\frac d2}+1$, then $R$ is nonsmoothable.
\end{proposition}

\begin{proposition}
    \label{prop:quadratic-better-than-koszul}
    The space $\im\Phi_{\mathrm Q}^C(T)\subset A\otimes B$ equals the sum of all $(T(\alpha)\otimes \id_A)\left(\im\koszul^{C,B}(T)\right)$ over $\alpha\in A^*$, where we consider $T(\alpha)$ as a map $C^*\to B$. In particular, if $\dim B=\dim C$ and $T$ is $1_A$-generic, this implies $\rank \Phi_{\mathrm Q}^C(T) \geq \rank\koszul^{C,B}(T)$.
\end{proposition}

\begin{conjecture}
    \label{conj:tangency}
    Let $\CC[x_1,\dots,x_d]\surjto R$ be a surjection to an $n$-dimensional algebra $R$, let $[R]$ be the corresponding point of $\Hilb_n\CC^d$ and let $T$ be the structure tensor of $R$. Then
    \[
        \rank\tang^C(T) - n(n-1)(n-2) = dn - \dim\T_{[R]}\Hilb_n\CC^d.
    \]
\end{conjecture}

\begin{proof}
    Let us prove only the second set of equalities of \eqref{eq:koszul-quaddratic-algebra}, since the proof of the first one is very similar.

    Explicitly, the two flattenings in question map
    \begin{align*}
        \Phi_{\mathrm Q}^A(T): \bigwedge^2 R \otimes R \otimes R^* &\to R \otimes R^*,\\
        (a\wedge a')\otimes b\otimes \gamma &\mapsto (ab)\otimes \gamma(a'\cdot\bullet) - (a'b)\otimes \gamma(a\cdot\bullet)
    \end{align*}
    and
    \begin{align*}
        \koszul^{A,C}(T): \bigwedge^2 R \otimes R^* &\to R \otimes R^*,\\
        (a\wedge a')\otimes\gamma &\mapsto a\otimes \gamma(a'\cdot\bullet) - a'\otimes \gamma(a\cdot\bullet),
    \end{align*}
    where we use $\gamma(a\cdot\bullet)$ to denote the linear form $(x\mapsto \gamma(ax))\in R^*$. We claim that in fact $\Phi_{\mathrm Q}^A(T)$ and $\koszul^{A,C}(T)$ have the same images. The inclusion $\im\Phi^{A,C}(T)\subseteq \im\Phi_{\mathrm Q}^A$ is obvious by considering $b=1$ in the latter map.

    For the opposite inclusion, note that invertible elements form a dense set in $R$, so in particular, $R$ may be spanned by them, hence $\im \Phi_{\mathrm Q}^A$ is spanned by images of $(a\wedge a')\otimes b\otimes\gamma$ with $b$ invertible. These lie in $\im \koszul^{A,C}$ by the following: we set
    \[
        \tilde a := ba, \qquad \tilde a' := ba', \qquad \tilde\gamma := \gamma(b^{-1}\cdot\bullet),
    \]
    which then results in
    \begin{multline*}
        \koszul^{A,C}(T)\left(
            (\tilde a\wedge \tilde a') \otimes \tilde \gamma
        \right)
        =
        \tilde a\otimes\tilde \gamma(\tilde a'\cdot\bullet) - \tilde a'\otimes\tilde\gamma(\tilde a\cdot\bullet)
        =
        (ab)\otimes\gamma(b^{-1}ba'\cdot\bullet) - (a'b)\otimes \gamma(b^{-1}ba\cdot\bullet)
        =\\=
        (ab)\otimes \gamma(a'\cdot\bullet)-(a'b)\otimes\gamma(a\cdot\bullet)
        =
        \Phi_{\mathrm Q}^A(T)\left(
            (a\wedge a') \otimes b\otimes\gamma
        \right).
    \end{multline*}
    This justifies $\rank \Phi_{\mathrm Q}^A(T) = \rank \koszul^{A,C}(T)$. That this common rank equals $n(n-1)$ then follows from Lemma \ref{lem:easy rank Koszul for algebras}.
\end{proof}

\section{Applications}
\label{sec:applications}

\subsection{Matrix multiplication}
\label{subsec:mamu}

It is well-known that the border rank of the $2\times 2$ matrix multiplication tensor $M_2$ equals $7$ \cite{Landsberg2006, Hauenstein-Ikenmeyer-Landsberg}. However, only recently the ``first hand-checkable algebraic proof'' appeared \cite{Conner_Harper_Landsberg_2023}. Below we show that tangency flattenings provide an easy proof.
\begin{proposition}\label{prop:M2=7}
    The border rank of $M_2$ equals seven.
\end{proposition}
\begin{proof}
Strassen \cite{strassen1969gaussian} provided an explicit rank seven decomposition of $M_2$. We focus on lower-bounding the border rank.

    By Corollary \ref{cor:tan_flat} for $n=4$, $q=6$ we have to show that the tangency flattening has rank greater than $60$. We will prove that it is $64$. We fix a basis $(i,j)$ of $\CC^4$, where $i,j\in\{0,1\}$ and we use $\neg i$ to denote the negation of $i$. Recall that $M_2=\sum_{i,j,k\in \{0,1\}} (i,j)\otimes (j,k)\otimes (k,i)$.

    We will prove that the tangency flattening is a block diagonal matrix with five blocks: a $48\times 48$ identity matrix and four isomorphic, nondegenerate $4\times 4$ matrices.

The tangency flattening of $M_2$ is the restriction to $\bigwedge^3\CC^4\otimes \CC^4\otimes \CC^4$ of
\begin{align}
\label{eq:mamu-tangency}
(\CC^4)^{\otimes 3}\otimes \CC^4\otimes \CC^4 &\to \CC^4\otimes \CC^4\otimes \CC^4,\\
\nonumber
(i_1,j_1)\otimes (i_2,j_2)\otimes (i_3,j_3)\otimes (i_4,j_4)\otimes (i_5,j_5) &\mapsto
\begin{cases}
(i_1,j_1)\otimes (i_2,j_4)\otimes (i_5,j_3), & \text{if $j_2=i_4$ and $i_3 = j_5$,}\\
0, & \text{otherwise.}
\end{cases}
\end{align}
We divide the basis vectors in the domain into five groups.

The first group is formed by vectors of the form $\bigl((\neg j_5,\neg i_4)\wedge(i_2,i_4)\wedge (j_5,j_3)\bigr)\otimes (i_4,j_4)\otimes (i_5,j_5)$. By first choosing $i_4,j_4,i_5,j_5$ (16 options), and then choosing exterior products that contain $(\neg j_5,\neg i_4)$ (3 options), we see there are $48$ such (nonzero) vectors.
Note that this already forces the exterior product to be of the form above, since the two remaining exterior factors are chosen from $(\neg j_5, i_4)$, $(j_5, \neg i_4)$ and $(j_5,i_4)$.
Although the exterior product expands as a signed sum of six tensor products, only one of those -- the one ordered as in the presentation above -- maps to a nonzero vector in \eqref{eq:mamu-tangency}. Thus, the image of each such basis vector is $(\neg j_5,\neg i_4)\otimes (i_2,j_4)\otimes (i_5,j_3)$. Note that such images must satisfy $(\neg j_5,\neg i_4)\neq (\neg i_2,\neg j_3)$.

The next four groups are indexed by $j_4,i_5\in \{0,1\}$. For fixed $j_4,i_5$ the group consists of the four basis vectors $\bigl((j_5,i_4)\wedge (\neg j_5,i_4)\wedge (j_5,\neg i_4)\bigr)\otimes (i_4,j_4)\otimes (i_5,j_5)$, where $i_4, j_5$ are arbitrary.
The image of each such vector is:
\[(j_5,i_4)\otimes (\neg j_5,j_4)\otimes (i_5,\neg i_4)-(\neg j_5,i_4)\otimes (j_5,j_4)\otimes (i_5,\neg i_4)-(j_5,\neg i_4)\otimes (\neg j_5,j_4)\otimes (i_5,i_4).\]
In particular, the images are linear combinations of basis vectors that are distinct from images of the first group as well as other groups -- in each term, the first tensor factor consist exactly of  negations of the first coordinate of the second factor  and the second coordinate of the third factor. Meanwhile, the group is determined by reading off $j_4$ and $i_5$ from the second coordinate of the second factor and first coordinate of the third factor, respectively.

Thus indeed the tangency flattening is in block-diagonal form. It remains for us to prove that the isomorphic $4\times 4$ blocks are of full rank.
Each such matrix has the following form, where we label $(j_5,i_4)\otimes (\neg j_5,j_4)\otimes (i_5,\neg i_4)$ with the first factor $(j_5,i_4)$ and subsequently order rows and columns by $(j_5,i_4)=(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$:
\[\begin{pmatrix}
    1 &  -1&-1&0\\
    -1& 1&0&-1\\
    -1& 0&1&-1\\
    0& -1&-1&1\\
\end{pmatrix}.\]
The determinant of such a matrix is $-3$, thus indeed the matrix is of full rank, and so is the tangency flattening.
\end{proof}

\label{sec:intro}
The central problem concerning tensors is that of \emph{tensor decomposition} -- presenting a tensor as a sum of simple tensors, or computing the smallest possible size of a decomposition, called the \emph{tensor rank}. While the problem is NP-hard in general \cite{RankNPcomplete, hillar2013most}, providing estimates for rank or related notions such as border rank or asymptotic rank has been a fruitful field. Perhaps the most famous application is the study of computational complexity of matrix multiplication (e.g. \cite{strassen1969gaussian, coppersmith1982asymptotic, alman2024refined, williams2024new}). Further, a recent line of research relates low asymptotic rank of certain special tensors to faster than known algorithms for NP-hard problems \cite{bjorklund2025fast, bjorklund2024asymptotic, pratt2024stronger, MaMu-SotA}.
For an introduction to tensors from an algebraic perspective, we refer the reader to \cite[Chapter 9.2]{michalek2021invitation}, or to \cite{landsberg2011tensors} for a more extensive exposition. For the connections to complexity theory, we refer to \cite{burgisser2013algebraic, landsberg2017geometry}.

The main topic of this article is obtaining lower bounds on the border rank of tensors, a longstanding question in tensor geometry. The classical approach to the problem is by providing equations for secant varieties to the Segre variety. Flattenings, or maps which produce a matrix from a given tensor, have been a bountiful source of such lower bounds (e.g. \cite{langsberg-ottaviani-equationsforsecant, landsberg2008generalizations, landsberg-ottaviani-newbounds, landsberg-michalek-mamu, landsberg-michalek-haystack}), with the equations thus obtained coming in the form of minors. In particular, \emph{Strassen's equations}, further studied by Landsberg and Manivel and  expanded by Landsberg and Ottaviani to \emph{Koszul} and \emph{Young flattenings}, became the gold standard in the field. Apart from flattenings, the techniques of \emph{border apolarity} (see \cite{Buczynscy-apolarity}) and \emph{border substitution} (see \cite{landsberg2017geometry}) have been introduced and used to prove several new lower bounds in recent years. However, all of these methods have quite restrictive limits of application.

Indeed, current lower bound methods fall far short of being able to prove a tensor in $\mathbb{C}^n\otimes\mathbb{C}^n\otimes\mathbb{C}^n$ to have the generic border rank, which is approximately $\frac{n^2}3$. Instead, known flattening methods give maximal lower bounds of up to $(2-\epsilon)n$ for small $\epsilon>0$. Combining algebraic equations with other methods, it is possible to provide examples of tensors of border rank $2.02n$ \cite{landsberg-michalek-haystack}. Still, there are well understood barriers which flattening methods cannot beat \cite{jarek_cactus, Galazka-bundles, EGOW18, garg2019more}. In contrast, the applications to asymptotic rank in computer science would require superlinear lower bounds on border rank.

The disparity between the generic border rank and available lower bounds has been attributed to the inclusion of secant varieties in \emph{cactus varieties} (see Section~\ref{sec:preliminaries}). In particular, known flattening methods have been unable to distinguish secant from cactus varieties, even when the inclusion is known to be strict  -- quoting from \cite{Conner_Harper_Landsberg_2023}:
\begin{quote}
``The geometric interpretation of the border rank lower bound barriers of \cite{EGOW18, Galazka-bundles} is that all equations obtained by taking minors, called rank methods, are actually equations for a larger variety than $\sigma_r(\Seg(\P A\times\P B\times\P C))$, called the $r$-th cactus variety \cite{Buczynscy-apolarity}.''
\end{quote}
Indeed, as proved in \cite{jarek_cactus, Galazka-bundles, galazka-thesis} all of the known methods producing equations for the $r$-th secant variety are equations for the $r$-th cactus variety. Further, it is known that the $r$-th cactus variety of the Segre variety in $(\mathbb{C}^n)^{\ot 3}$ fills the ambient space for $r$ growing linearly in $n$ \cite{ballico2019note, bernardi2018polynomials, galkazka2023multigraded}.

For these reasons, the existence of easily computable determinantal equations vanishing on secant varieties but not vanishing on cactus varieties, has long been considered unlikely or even impossible in the tensor-geometric community.
In this article, we illustrate that it is not the determinantal expression which creates the cactus barrier, but rather the linear embedding of the tensor product into a matrix space.
Thus the starting point of our approach in this article is focusing on \emph{nonlinear} maps from tensor spaces to matrix spaces.
To obtain bounds on matrix rank with linear embeddings, it is enough to investigate the embedded image of the Segre variety and appeal to linearity. With nonlinear maps on the other hand, we need to work more generally, but this does yield stronger results. We obtain a new method, which we dub \emph{Kronecker-Koszul flattenings}, or more generally \emph{Kronecker-Young flattenings}, that allows one to produce
explicit and efficiently computable determinantal equations of secant varieties that do not generally vanish on cactus varieties. In particular, Kronecker-Koszul flattenings require less computation time and memory than methods based on a more direct identification of ideals defining the secant varieties (cf.~\cite{Hauenstein-Ikenmeyer-Landsberg}).

 Broadly speaking, our method starts by tensoring together several copies of the same tensor. This, depending on the interpretation of the codomain, may be viewed as a Kronecker power or Veronese embedding \cite{kaski2025universal}, and thus is the universal map for fixed degree polynomials. Next, we further tensor with several $2$-tensors corresponding to identity maps on various vector spaces.  Finally, we group and contract some of the factors to exterior powers. This results in what we call a \emph{Kronecker-Koszul tensor}. A Kronecker-Koszul flattening is then a classical flattening of the Kronecker-Koszul tensor. Note that due to initially taking a tensor power, Kronecker-Koszul tensors are in general nonlinear as functions of the starting tensor.
See Section~\ref{sec:construction} for details of the construction. There we also comment on what we call {Kronecker-Young flattenings}, a further generalization which synthesizes the approach of Young flattenings into the method.

Our main result uses the \emph{tangency flattening}, a very particular Kronecker-Koszul flattening which depends quadratically on a tensor. See the paragraph preceding Corollary~\ref{cor:tan_flat} for the definition.

The main topic of this article is obtaining lower bounds on the border rank of tensors, a longstanding question in tensor geometry. The classical approach to the problem is by providing equations for secant varieties to the Segre variety. Flattenings, or maps which produce a matrix from a given tensor, have been a bountiful source of such lower bounds (e.g. \cite{langsberg-ottaviani-equationsforsecant, landsberg2008generalizations, landsberg-ottaviani-newbounds, landsberg-michalek-mamu, landsberg-michalek-haystack}), with the equations thus obtained coming in the form of minors. In particular, \emph{Strassen's equations}, further studied by Landsberg and Manivel and expanded by Landsberg and Ottaviani to \emph{Koszul} and \emph{Young flattenings}, became the gold standard in the field. Apart from flattenings, the techniques of \emph{border apolarity} (see \cite{Buczynscy-apolarity}) and \emph{border substitution} (see \cite{landsberg2017geometry}) have been introduced and used to prove several new lower bounds in recent years. However, all of these methods have quite restrictive limits of application.

Indeed, current lower bound methods fall far short of being able to prove a tensor in $\mathbb{C}^n\otimes\mathbb{C}^n\otimes\mathbb{C}^n$ to have the generic border rank, which is approximately $\frac{n^2}3$. Instead, known flattening methods give maximal lower bounds of up to $(2-\epsilon)n$ for small $\epsilon>0$. Combining algebraic equations with other methods, it is possible to provide examples of tensors of border rank $2.02n$ \cite{landsberg-michalek-haystack}. Still, there are well understood barriers which flattening methods cannot beat \cite{jarek_cactus, Galazka-bundles, EGOW18, garg2019more}. In contrast, the applications to asymptotic rank in computer science would require superlinear lower bounds on border rank.

The disparity between the generic border rank and available lower bounds has been attributed to the inclusion of secant varieties in \emph{cactus varieties} (see Section~\ref{sec:preliminaries}). In particular, known flattening methods have been unable to distinguish secant from cactus varieties, even when the inclusion is known to be strict -- quoting from \cite{Conner_Harper_Landsberg_2023}:
\begin{quote}
``The geometric interpretation of the border rank lower bound barriers of \cite{EGOW18, Galazka-bundles} is that all equations obtained by taking minors, called rank methods, are actually equations for a larger variety than $\sigma_r(\Seg(\P A\times\P B\times\P C))$, called the $r$-th cactus variety \cite{Buczynscy-apolarity}.''
\end{quote}
Indeed, as proved in \cite{jarek_cactus, Galazka-bundles, galazka-thesis} all of the known methods producing equations for the $r$-th secant variety are equations for the $r$-th cactus variety. Further, it is known that the $r$-th cactus variety of the Segre variety in $(\mathbb{C}^n)^{\ot 3}$ fills the ambient space for $r$ growing linearly in $n$ \cite{ballico2019note, bernardi2018polynomials, galkazka2023multigraded}.

For these reasons, the existence of easily computable determinantal equations vanishing on secant varieties but not vanishing on cactus varieties, has long been considered unlikely or even impossible in the tensor-geometric community.
In this article, we illustrate that it is not the determinantal expression which creates the cactus barrier, but rather the linear embedding of the tensor product into a matrix space.
Thus the starting point of our approach in this article is focusing on \emph{nonlinear} maps from tensor spaces to matrix spaces.
To obtain bounds on matrix rank with linear embeddings, it is enough to investigate the embedded image of the Segre variety and appeal to linearity. With nonlinear maps on the other hand, we need to work more generally, but this does yield stronger results. We obtain a new method, which we dub \emph{Kronecker-Koszul flattenings}, or more generally \emph{Kronecker-Young flattenings}, that allows one to produce
explicit and efficiently computable determinantal equations of secant varieties that do not generally vanish on cactus varieties. In particular, Kronecker-Koszul flattenings require less computation time and memory than methods based on a more direct identification of ideals defining the secant varieties (cf.~\cite{Hauenstein-Ikenmeyer-Landsberg}).

Broadly speaking, our method starts by tensoring together several copies of the same tensor. This, depending on the interpretation of the codomain, may be viewed as a Kronecker power or Veronese embedding \cite{kaski2025universal}, and thus is the universal map for fixed degree polynomials. Next, we further tensor with several $2$-tensors corresponding to identity maps on various vector spaces. Finally, we group and contract some of the factors to exterior powers. This results in what we call a \emph{Kronecker-Koszul tensor}. A Kronecker-Koszul flattening is then a classical flattening of the Kronecker-Koszul tensor. Note that due to initially taking a tensor power, Kronecker-Koszul tensors are in general nonlinear as functions of the starting tensor.
See Section~\ref{sec:construction} for details of the construction. There we also comment on what we call {Kronecker-Young flattenings}, a further generalization which synthesizes the approach of Young flattenings into the method.

Our main result uses the \emph{tangency flattening}, a very particular Kronecker-Koszul flattening which depends quadratically on a tensor. See the paragraph preceding Corollary~\ref{cor:tan_flat} for the definition.

\begin{corollary}\label{cor:tan_flat}
    If $\dim V_1=\dim V_2=\dim V_3=n$, then $q(q-1)(n-2)+1$ minors of the tangency flattening provide equations for border rank $q$.
\end{corollary}

For these reasons, the existence of easily computable determinantal equations vanishing on secant varieties but not vanishing on cactus varieties, has long been considered unlikely or even impossible in the tensor-geometric community.
In this article, we illustrate that it is not the determinantal expression which creates the cactus barrier, but rather the linear embedding of the tensor product into a matrix space.
Thus the starting point of our approach in this article is focusing on \emph{nonlinear} maps from tensor spaces to matrix spaces.
To obtain bounds on matrix rank with linear embeddings, it is enough to investigate the embedded image of the Segre variety and appeal to linearity. With nonlinear maps on the other hand, we need to work more generally, but this does yield stronger results. We obtain a new method, which we dub \emph{Kronecker-Koszul flattenings}, or more generally \emph{Kronecker-Young flattenings}, that allows one to produce
explicit and efficiently computable determinantal equations of secant varieties that do not generally vanish on cactus varieties. In particular, Kronecker-Koszul flattenings require less computation time and memory than methods based on a more direct identification of ideals defining the secant varieties (cf.~\cite{Hauenstein-Ikenmeyer-Landsberg}).

Our main result uses the \emph{tangency flattening}, a very particular Kronecker-Koszul flattening which depends quadratically on a tensor. See the paragraph preceding Corollary~\ref{cor:tan_flat} for the definition.

While Kronecker-Koszul flattenings generalize Koszul flattenings (cf. \cite{langsberg-ottaviani-equationsforsecant, landsberg2008generalizations, landsberg-michalek-mamu, landsberg-michalek-haystack}), we also analyze in Section~\ref{sec:koszul} bounds that may be extracted via Koszul flattenings for the border rank (and border cactus rank) of structure tensors of certain algebras. It is known \cite{blaser2016degeneration} that a (finite) algebra is smoothable if and only if the associated structure tensor has minimal border rank. This is also one of the main challenges in application of the border apolarity method.
Further, two of the classical flattenings of structure tensors of algebras give rise to spaces of commuting matrices. Thus, many of the flattening methods cannot provide nontrivial bounds on border rank of such tensors. However, we show that it is possible to obtain lower bounds on border cactus rank of structure tensors of algebras strictly greater than the degree of the algebra, even using Koszul flattenings (see Proposition~\ref{prop:1de-koszul}).

\begin{corollary}\label{cor:flat bound}
 If $T$ has border rank at most $q$, then the rank of a Kronecker-Koszul flattening of $T$ is at most $F(q)\cdot \prod_{i=1}^r\prod_{j=1}^{s_i} \binom{d_i-|\lambda_{i,j}|}{d_{i,j}'}$. In particular, minors of the Kronecker-Koszul flattening provide equations for secant varieties of the Segre variety.
\end{corollary}
\begin{proof}
 Follows from Theorem \ref{thm:main}, as ranks of classical flattenings lower bound border rank of the tensor.
\end{proof}
The bounds in Theorem \ref{thm:main} and Corollary \ref{cor:flat bound} do not have to be tight. This can happen for various reasons, some more opaque than others:

In this section we present one explicit construction that we find particularly interesting. By \cite{kaski2025universal} the linear span of image of the $k$-th Kronecker power of tensors in $A\ot B \ot C$ is $S^k(A\ot B \ot C)$. As a $\GL(A)\times \GL(B)\times \GL(C)$ representation, the latter space has a decomposition into isotypic components
 \[S^k(A\ot B \ot C)=\bigoplus_{\lambda,\mu,\rho\vdash k}\left(S^\lambda(A)\ot S^\mu(B)\ot S^\rho(C)\right)^{\bigoplus K_{\lambda,\mu,\rho}},\]
 where $K_{\lambda,\mu,\rho}$ is the Kronecker coefficient. In particular, when $\lambda=\mu=1^k$ and $\rho=(k)$ we have $K_{\lambda,\mu,\rho}=1$. This gives a canonical projection
 \[\pi:S^k(A\ot B \ot C)\rightarrow \bigwedge^k A\ot \bigwedge^k B\ot S^k(C).\]
 An example of Kronecker-Young flattening is the classical flattening of $\pi(T^{\otimes k})$ given by
 \[\Phi_{\mathrm M,k}^C(T):\bigwedge^kA^*\ot\bigwedge^kB^*\rightarrow S^k(C).\]
 The advantage of these choices is that the image of the map $\Phi_{\mathrm M,k}^C(T)$ has a direct interpretation in terms of the flattening $T^C:C^*\rightarrow A\ot B$. Namely, the image of $\Phi_{\mathrm M,k}^C(T)$ is the linear span of all $k\times k$ minors of $T^C$ viewed as a matrix with entries from $C$. For this reason, we call $\Phi_{\mathrm M,k}$ the \emph{$k$-minor flattening}.
\begin{lemma}
 If $\dim A=\dim B=\dim C=n$ and a tensor $T\in A\ot B\ot C$ has border rank at most $n$, then the rank of the $k$-minor flattening $\Phi_{\mathrm M,k}^C(T)$ is at most $\binom{n}{k}$.
\end{lemma}
\begin{proof}
 The $n$-th secant variety of the Segre variety is the closure of the $\GL(A)\times \GL(B)\times \GL(C)$ orbit of the unit tensor $I_n:=\sum_{i=1}^n a_i\ot b_i\ot c_i$, where $a_i$, $b_i$, $c_i$ form respectively bases of $A$, $B$, $C$. As our constructions are continuous and equivariant, it is enough to prove that $\rk \Phi_{\mathrm M,k}^C(I_n)=\binom{n}{k} $. We note that $I_n^C(C^*)$ is the space of diagonal matrices. Clearly, the linear span of $k\times k$ minors of the diagonal matrices is the linear span of squarefree monomials of degree $k$ in $n$ variables and thus has dimension $\binom{n}{k}$.
\end{proof}

\begin{corollary}
 \label{cor:alg-tangency-sum}
 If $T_i\in R_i^*\otimes R_i^*\otimes R_i$, $i=1,2$ are structure tensors of $n_i$-dimensional commutative algebras $R_i$, then the structure tensor $T_1\oplus T_2$ of the $n:=(n_1+n_2)$-dimensional algebra $R_1\oplus R_2$ satisfies
 \[
 \rank\tang^A(T_1\oplus T_2) - n(n-1)(n-2) = \Bigl(\rank \tang^A(T_1) - n_1(n_1-1)(n_1-2)\Bigr) + \Bigl(\rank\tang^A(T_2) - n_2(n_2-1)(n_2-2)\Bigr).
 \]
\end{corollary}
\begin{proof}
 Along with the conclusions of Proposition~\ref{prop:koszul-equals-quad-on-algs}, observe that structure tensors of algebras are concise. Thus all but the first two terms of the right-hand side of \eqref{eq:tangency-direct-sum} simplify, yielding
 \[
 \rank\tang^A(T_1\oplus T_2) = \rank \tang^A(T_1) + \rank \tang^A(T_2) + 3n_1n_2(n_1+n_2-2).
 \]
 This rearranges to the Corollary.
\end{proof}
The following lemma is well-known to experts. It can be derived e.g.~from the results in \cite{jelisiejew2024concise}. We include the proof for the sake of completeness.
\begin{lemma}\label{lem:GorCactus}
 Let $R$ be a finite Gorenstein algebra and $T\in (R^*)^{\otimes k}\ot R$ its structure tensor corresponding to the map $R^k\rightarrow R$, $(r_1,\dots,r_k)\mapsto r_1\cdots r_k$. The cactus rank of $T$ equals $d=\dim_{\CC} R$.
\end{lemma}
\begin{proof}
 As $T$ is concise it is enough to prove that it has cactus rank at most $d$. Thus we have to exhibit a scheme of length $d$, which will be $\Spec R$, inside the Segre variety, so that $T$ is in its linear
span. As cactus rank is subadditive under direct sum we may assume $(R,\mathfrak m)$ is local.

\begin{theorem}
 \label{thrm:beat-cactus}
 The $(n(n-1)(n-2)+1)$-minors of the tangency flattenings for border rank $n$ in $\CC^n\otimes \CC^n\otimes \CC^n$ do not vanish on the $n$-th cactus variety for all $n\geq 14$.
\end{theorem}
\begin{proof}
By Lemma \ref{lem:GorCactus}, structure tensors of Gorenstein algebras have minimal cactus rank.
 Thus, it suffices to produce examples of structure tensors of $n$-dimensional Gorenstein algebras whose tangency flattenings have ranks strictly higher than $n(n-1)(n-2)$.

For example, for $n=14$ the minors in question are polynomials of degree $4370$ in $2744$ variables. Yet, they are explicit and we can determine their vanishing or nonvanishing very fast.
We will prove Theorem~\ref{thrm:beat-cactus-introversion} by lower-bounding border rank via rank of the tangency flattening in Corollary~\ref{cor:tan_flat} and exhibiting in Theorem~\ref{thrm:beat-cactus} a sequence of tensors of cactus rank $n$ on which the above-named minors of the tangency flattening do not vanish. The name ``tangency'' flattening is motivated by a conjectural connection to tangent spaces of Hilbert schemes of points -- see Conjecture~\ref{conj:tangency}. In Subsection~\ref{subsec:mamu}, we additionally show that tangency flattenings provide a new and completely elementary proof of the fact that $2\times2$ matrix multiplication tensor has border rank $7$.

While Kronecker-Koszul flattenings generalize Koszul flattenings (cf. \cite{langsberg-ottaviani-equationsforsecant, landsberg2008generalizations, landsberg-michalek-mamu, landsberg-michalek-haystack}), we also analyze in Section~\ref{sec:koszul} bounds that may be extracted via Koszul flattenings for the border rank (and border cactus rank) of structure tensors of certain algebras. It is known \cite{blaser2016degeneration} that a (finite) algebra is smoothable if and only if the associated structure tensor has minimal border rank. This is also one of the main challenges in application of the border apolarity method.
Further, two of the classical flattenings of structure tensors of algebras give rise to spaces of commuting matrices. Thus, many of the flattening methods cannot provide nontrivial bounds on border rank of such tensors. However, we show that it is possible to obtain lower bounds on border cactus rank of structure tensors of algebras strictly greater than the degree of the algebra, even using Koszul flattenings (see Proposition~\ref{prop:1de-koszul}).

Finally, we show that under mild assumptions, Koszul flattenings may be replaced and improved upon by certain quadratic Kronecker-Koszul flattenings for the purposes of lower-bounding border rank of tensor (see Proposition~\ref{prop:quadratic-better-than-koszul}).

We will prove Theorem~\ref{thrm:beat-cactus-introversion} by (1) lower-bounding border rank via the rank of the tangency flattening (Corollary~\ref{cor:tan_flat}), and (2) exhibiting (in Theorem~\ref{thrm:beat-cactus}) a sequence of tensors of cactus rank $n$ on which the minors of size $n(n-1)(n-2)+1$ of the tangency flattening do not vanish.

Question: For each integer \(n\ge 14\), consider the Segre variety \(\operatorname{Seg}(\mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n))\subset \mathbb P(\mathbb C^n\otimes \mathbb C^n\otimes \mathbb C^n)\). Let its \(n\)-th secant variety be the Zariski closure of the union of linear spans of \(n\) points of the Segre variety, and let its \(n\)-th cactus variety be the Zariski closure of the union of linear spans of length-\(n\) subschemes of the Segre variety. The paper associates to tensors in \(\mathbb C^n\otimes \mathbb C^n\otimes \mathbb C^n\) certain matrix-valued maps called tangency flattenings; taking all minors of size \(n(n-1)(n-2)+1\) gives polynomial equations on the ambient projective space. Which statement holds for every such \(n\)?

All minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish on the \(n\)-th secant variety of \(\operatorname{Seg}(\mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n))\), but they do not vanish on its \(n\)-th cactus variety.

Choices:
B. All minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish on both the \(n\)-th secant variety and the \(n\)-th cactus variety of \(\operatorname{Seg}(\mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n))\).
C. All minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish on the \(n\)-th secant variety of \(\operatorname{Seg}(\mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n))\).
D. All minors of size \(n(n-1)(n-2)\) of the tangency flattenings vanish on the \(n\)-th secant variety of \(\operatorname{Seg}(\mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n)\times \mathbb P(\mathbb C^n))\), but they do not vanish on its \(n\)-th cactus variety.
E. For every tensor in \(\mathbb C^n\otimes \mathbb C^n\otimes \mathbb C^n\) outside the \(n\)-th cactus variety, some minor of size \(n(n-1)(n-2)+1\) of a tangency flattening is nonzero; equivalently, these minors cut out the \(n\)-th cactus variety set-theoretically.


Question:
Let
\[
X=\operatorname{Seg}(\mathbb{P}(\mathbb{C}^n)\times \mathbb{P}(\mathbb{C}^n)\times \mathbb{P}(\mathbb{C}^n))\subset \mathbb{P}(\mathbb{C}^n\otimes\mathbb{C}^n\otimes\mathbb{C}^n)
\]
be the Segre variety of rank-one tensors, and assume \(n\ge 14\). Denote by \(\sigma_n(X)\) the \(n\)-th secant variety, i.e. the Zariski closure of the union of linear spans of \(n\) points of \(X\), and by the \(n\)-th cactus variety the Zariski closure of the union of linear spans of all length-\(n\) zero-dimensional subschemes of \(X\). For the tangency flattenings associated to tensors in \(\mathbb{C}^n\otimes\mathbb{C}^n\otimes\mathbb{C}^n\), consider all minors of size \(n(n-1)(n-2)+1\). Which conclusion holds?

For every \(n\ge 14\), all minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish identically on the \(n\)-th secant variety \(\sigma_n(X)\), but they do not vanish identically on the \(n\)-th cactus variety of \(X\); equivalently, these minors give equations for \(\sigma_n(X)\) that fail on at least one point of the \(n\)-th cactus variety.

Choices:
B. For every \(n\ge 14\), all minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish identically both on the \(n\)-th secant variety \(\sigma_n(X)\) and on the \(n\)-th cactus variety of \(X\); equivalently, these minors cannot distinguish the secant variety from the cactus variety in this range.
C. For every \(n\ge 14\), all minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish identically on the \(n\)-th secant variety \(\sigma_n(X)\).
D. For every \(n\ge 14\), there exist tangency flattenings whose minors of size \(n(n-1)(n-2)+1\) vanish on the \(n\)-th secant variety \(\sigma_n(X)\) but such that, for each tensor in the \(n\)-th cactus variety, one can choose a tangency flattening for which all these minors also vanish.
E. For every \(n\ge 13\), all minors of size \(n(n-1)(n-2)+1\) of the tangency flattenings vanish identically on the \(n\)-th secant variety \(\sigma_n(X)\), but do not vanish identically on the \(n\)-th cactus variety of \(X\).

uniform nonvanishing-vs-existential failure on cactus


\label{mainth1}
Let $c,d,s\in\mathbb Z^+$ with $d\geq 2$, $1\leq c,s\leq d$ and $\gcd(cs,d)=1$, and let $p\geq5$ be a prime with $p\equiv s\pmod d$. Then, for any $r\in\mathbb Z^+$ with $(\f12+\alpha)^{\ast_r}(\f12+\alpha^{\ast_r})\not\eq0\pmod{p}$,  we have
\begin{equation}\label{mainth1eq}
\sum_{k=0}^{p^r-1}(2k+\alpha)\frac{(\alpha)_k^3(\f12)_k}{(1)_k^3(\f12+\alpha)_k}\equiv \alpha^{\ast_r}p^r -\frac{(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}p^{r+2}H_{\alpha^{\ast_r}p-\alpha^{\ast_{r-1}}}^{(2)}\pmod{p^{r+3}},
\end{equation}
where $\alpha=c/d$.

\begin{equation}\label{2ordharmonic}
H_{p-1}^{(2)}\eq H_{(p-1)/2}^{(2)}\eq 0\pmod{p},
\end{equation}

\begin{theorem}\label{mainth1}
Let $c,d,s\in\Z^+$ with $d\geq 2$, $1\leq c,s\leq d$ and $\gcd(cs,d)=1$, and let $p\geq5$ be a prime with $p\eq s\pmod d$. Then, for any $r\in\Z^+$ with $(\f12+\alpha)^{\ast_r}(\f12+\alpha^{\ast_r})\not\eq0\pmod{p}$,  we have
\begin{equation}\label{mainth1eq}
\sum_{k=0}^{p^r-1}(2k+\alpha)\f{(\alpha)_k^3(\f12)_k}{(1)_k^3(\f12+\alpha)_k}\eq \alpha^{\ast_r}p^r -\f{(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}p^{r+2}H_{\alpha^{\ast_r}p-\alpha^{\ast_{r-1}}}^{(2)}\pmod{p^{r+3}},
\end{equation}
where $\alpha=c/d$.
\end{theorem}

In the 1910s, Ramanujan announced some convergent hypergeometric series related to $1/\pi$ without proofs (cf. \cite{Berndt1994}), such as
\begin{equation}\label{RamaSer}
\sum_{k=0}^{\infty}(8k+1)\frac{(\f14)_k^4}{(1)_k^4}=\frac{2\sqrt2}{\sqrt{\pi}\Gamma(\f34)^2},
\end{equation}
where $(x)_n=x(x+1)\cdots(x+n-1)$ is the Pochhammer symbol and $\Gamma(x)$ is the classical Gamma function. \eqref{RamaSer} was finally confirmed by Hardy \cite{Hardy1923}.

In 1997, Van Hamme \cite{VanHamme1997} observed that the truncated forms of original Ramanujan-type series possess good congruence properties. For example, corresponding to \eqref{RamaSer}, Van Hamme conjectured the following supercongruence: for any prime $p\eq1\pmod4$,
\begin{equation}\label{VanHammeconj}
\sum_{k=0}^{(p-1)/4}(8k+1)\frac{(\f14)_k^4}{(1)_k^4}\equiv p\frac{\Gamma_p(\f12)\Gamma_p(\f14)}{\Gamma_p(\f34)}\pmod{p^3},
\end{equation}
where $\Gamma_p(x)$ denotes the $p$-adic Gamma function introduced by Morita \cite{Morita1975}. Nowadays, we usually refer to \eqref{VanHammeconj} as a $p$-adic analogue of \eqref{RamaSer}. All of Van Hamme's observations have now been confirmed by different authors using various techniques (see, e.g., \cite{Long2011,McCarthy-Osburn2008,Mortenson2008,Osburn-Zudilin2016,Swisher2015}). In particular, Swisher \cite{Swisher2015} proved that \eqref{VanHammeconj} holds modulo $p^4$ and established the following associated supercongruence: for any prime $p\eq3\pmod4$,
\begin{equation}\label{Swisherextension}
\sum_{k=0}^{(3p-1)/4}(8k+1)\frac{(\f14)_k^4}{(1)_k^4}\equiv -\frac{3}{2}p^2\frac{\Gamma_p(\f12)\Gamma_p(\f14)}{\Gamma_p(\f34)}\pmod{p^4}.
\end{equation}
Note that $(1/4)_k\eq0\pmod{p}$ for $(p+3)/4\leq k\leq p-1$ if $p\eq1\pmod4$ or $(3p+3)/4\leq k\leq p-1$ if $p\eq3\pmod4$. Therefore, both the upper limits of sums on the left-hand side of \eqref{VanHammeconj} and \eqref{Swisherextension} can be replaced with $p-1$. In 2022, Pan, Tauraso and Wang \cite{Pan-Tauraso-Wang2022} established some parametric extensions of \eqref{VanHammeconj} and \eqref{Swisherextension}. For instance, they showed that for any odd prime $p$ and $p$-adic integer $\alpha$ with $\<-\alpha\>_p\geq (p+1)/2$,
\begin{equation}\label{Pan-Tauraso-Wangres}
\sum_{k=0}^{p-1}\frac{2k+\alpha}{\alpha}\cdot\frac{(\alpha)_k^4}{(1)_k^4}\equiv p^2\alpha^{\ast}(2\alpha^{\ast}-1)\frac{\Gamma_p(1-2\alpha)}{\Gamma_p(1+\alpha)\Gamma_p(1-\alpha)^3}\pmod{p^4},
\end{equation}
where for any $p$-adic integer $x$, $\<x\>_{p^r}$ stands for the least nonnegative residue of $x$ modulo $p^r$ and $x^{\ast}=(x+\<-x\>_p)/p$ denotes Dwork's dash operation (cf. \cite{Dwork1969}). Clearly, \eqref{Swisherextension} is the special case $\alpha=1/4$ of \eqref{Pan-Tauraso-Wangres}.
For other parametric extensions of \eqref{VanHammeconj}, we refer the reader to \cite{Barman-Saikia2020, Guo2025, Guo-Schlosser2020, Liu-Wang2021, Liu-Wang2022, Pan-Tauraso-Wang2022}.

Recently, using the creative microscoping method (cf. \cite{Guo-Zudilin2019}), Guo and Zhao \cite{Guo-Zhao2026} studied some $q$-supercongruences from a very-well-poised ${}_6\phi_5$ basic hypergeometric identity. As consequences, they obtained the following results: for any prime $p\eq1\pmod4$ and integer $r\geq1$,
\begin{equation}\label{GuoZhaores1}
\sum_{k=0}^{(p^r-1)/2}(8k+1)\frac{(\f14)_k^3(\f12)_k}{(1)_k^3(\f34)_k}\equiv p^r\pmod{p^{r+3}},
\end{equation}
and for any prime $p\eq3\pmod4$ and odd integer $r\geq1$,
\begin{equation}\label{GuoZhaores2}
\sum_{k=0}^{p^r-1}(8k+1)\frac{(\f14)_k^3(\f12)_k}{(1)_k^3(\f34)_k}\equiv 3p^r\pmod{p^{r+2}}.
\end{equation}
Guo and Zhao \cite[Conjectures 7.1 and 7.2]{Guo-Zhao2026} also conjectured that \eqref{GuoZhaores1} still holds modulo $p^{r+5}$ for $p>5$ and \eqref{GuoZhaores2} can be extended to the modulus $p^{r+3}$ case as follows:
\begin{equation}\label{GuoZhaoconj7.2}
\sum_{k=0}^{p^r-1}(8k+1)\frac{(\f14)_k^3(\f12)_k}{(1)_k^3(\f34)_k}\equiv 3p^r+\frac{27}{4}p^{3r}\sum_{j=1}^{(p^r-3)/4}\frac{1}{j^2}\pmod{p^{r+3}}.
\end{equation}
This is our initial motivation.

Recall Whipple's ${}_5F_4$ formula (cf. \cite{Whipple1926})
\begin{align}\label{Whipple}
&\sum_{k=0}^{\infty}\frac{(a)_k(1+\frac{a}{2})_k(b)_k(c)_k(d)_k}{(1)_k(\frac a2)_k(1+a-b)_k(1+a-c)_k(1+a-d)_k}\notag\\
&\qquad=\frac{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+a-b-c-d)}{\Gamma(1+a)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)}.
\end{align}
Clearly, \eqref{RamaSer} is the special case $a=b=c=d=1/4$ of \eqref{Whipple}. Meanwhile, \eqref{Pan-Tauraso-Wangres} and \eqref{GuoZhaores2} are $p$-adic analogues of \eqref{Whipple} in the case $a=b=c=d=\alpha$ and the case $a=b=c=1/4,\ d=1/2$, respectively. Motivated by \eqref{Pan-Tauraso-Wangres}, it is natural to ask whether \eqref{GuoZhaores2} or \eqref{GuoZhaoconj7.2} has a parametric extension. This is the second motivation.

Before stating our main result, we first introduce some notations. For $n\in\mathbb Z^+=\{1,2,3,\ldots\}$ and $x\in\mathbb Z$, use $\<x\>_n$ to denote the least nonnegative residue of $x$ modulo $n$. Let $p$ be an odd prime. Similarly as in $\mathbb Z$, for any $p$-adic integer $x$ and $r\in\mathbb Z^+$, $\<x\>_{p^r}$ stands for the least nonnegative residue of $x$ modulo $p^r$ and $x^{\ast}=(x+\<-x\>_p)/p$ denotes Dwork's dash operation on $x$ (cf. \cite{Dwork1969}). For convenience, for $n\in\mathbb Z^+$, use $x^{\ast_n}$ to represent iterating the dash operation on $x$ $n$ times, that is,
$$
x^{\ast_1}=x^{\ast},\quad x^{\ast_n}=(x^{\ast_{n-1}})^{\ast}\quad (n=2,3,4,\ldots).
$$
In particular, set $x^{\ast_0}=x$. For $n\in\mathbb N=\{0,1,2,\ldots\},\ m\in\mathbb Z^+$, the $n$-th harmonic number of order $m$ is defined by
$$
H_n^{(m)}:=\sum_{k=1}^n\frac{1}{k^m}.
$$

Our main purpose is to establish a $p$-adic analogue of Whipple's formula \eqref{Whipple} with $a=b=c=\alpha$ and $d=\f12$.

Our main purpose is to establish a $p$-adic analogue of Whipple's formula \eqref{Whipple} with $a=b=c=\alpha$ and $d=\f12$.

Theorem \ref{mainth1} seems quite strange. When $\<-1/2-\alpha\>_p<\min\{\<-\alpha\>_p,(p-1)/2\}$, the summands on the left-hand side of \eqref{mainth1eq} are not always $p$-adic integers. However, the sum of these summands is a $p$-adic integer. This phenomenon renders it difficult for us to prove Theorem \ref{mainth1} directly using the formula \eqref{Whipple} and conventional congruence techniques. To overcome this obstacle, we find a new parametric WZ pair (cf. \cite{PWZ}) which allows us to transform the original sum to a computable form in the sense of congruence. This idea is crucial in our proof.

\begin{lemma}\label{dashres}
Let $c,d,s\in\Z^+$ with $d\geq2$, $1\leq c,s\leq d$ and $\gcd(cs,d)=1$. Then, for any prime $p\eq s\pmod{d}$, we have
$$
\alpha^{\ast}=\f{\<s^{-1}c\>_d}{d},
$$
where $s^{-1}$ stands for the inverse of $s$ modulo $d$.
\end{lemma}

\begin{proof}
Obviously,
\begin{align*}
\sum_{k=0}^{p^r-1}F(\alpha^{\ast_r}p^r,k)&=\sum_{k=0}^{p^r-1}(2k+\alpha^{\ast_r}p^r)\f{(\alpha^{\ast_r}p^r)_k^3(\f12)_k}{(1)_k^3(\f12+\alpha^{\ast_r}p^r)_k}\\
&=\alpha^{\ast_r}p^r+p^{3r}{\alpha^{\ast_r}}^3\sum_{k=1}^{p^r-1}\f{2k+\alpha^{\ast_r}p^r}{k^3}\f{(1+\alpha^{\ast_r}p^r)_{k-1}^3(\f12)_k}{(1)_{k-1}^3(\f12+\alpha^{\ast_r}p^r)_k}.
\end{align*}
Note that for $k\in\{1,2,\ldots,p^r-1\}$, $\ord_p(k)\leq r-1$, where $\ord_p$ stands for the $p$-adic order. Therefore,
$$
\ord_p\l(\f{p^{3r}}{k^2}\r)\geq r+2\quad \t{and}\quad \ord_p\l(\f{p^{4r}}{k^3}\r)\geq r+3.
$$
Moreover, it is easy to see that
$$
\f{(1+\alpha^{\ast_r}p^r)_{k-1}}{(1)_{k-1}^3}=\prod_{j=1}^{k-1}\f{j+\alpha^{\ast_r}p^r}{j}=\prod_{j=1}^{k-1}\l(1+\f{\alpha^{\ast_r}p^r}{j}\r)\eq1\pmod{p}
$$
and
\begin{align*}
\f{(\f12)_k}{(\f12+\alpha^{\ast_r}p^r)_k}&=\prod_{j=0}^{k-1}\f{\f12+j}{\f12+j+\alpha^{\ast_r}p^r}=\prod_{j=0}^{k-1}\f{1}{1+\f{\alpha^{\ast_r}p^r}{\f12+j}}\\
&\eq\begin{cases}1/(2\alpha^{\ast_r}+1)\pmod{p}\quad&\t{if}\ k\geq (p^r+1)/2,\\ 1\pmod{p}\quad&\t{if}\ k\leq (p^r-1)/2.\end{cases}
\end{align*}
Combining the above and in view of Lemmas \ref{u} and \ref{2ordharmonicdown}, we arrive at
$$
\sum_{k=0}^{p^r-1}F(\alpha^{\ast_r}p^r,k)\eq \alpha^{\ast_r}p^r+2(\alpha^{\ast_r})^3p^{3r}\sum_{k=1}^{(p^r-1)/2}\f{1}{k^2}+\f{2(\alpha^{\ast_r})^3p^{3r}}{2\alpha^{\ast_r}+1}\sum_{k=(p^r+1)/2}^{p^r-1}\f{1}{k^2}\eq \alpha^{\ast_r}p^r \pmod{p^{r+3}},
$$
as desired.
\end{proof}

\begin{lemma}\label{evalG}
Under the conditions of Theorem \ref{mainth1}, we have
\begin{equation}\label{evalGeq}
\sum_{l=0}^{a-1}G(\alpha+l,p^r)\eq \f{(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}p^{r+2}H_{\alpha^{\ast_r}p-\alpha^{\ast_{r-1}}}^{(2)}\pmod{p^{r+3}}.
\end{equation}
\end{lemma}

\begin{proof}
Clearly,
\begin{align*}
\sum_{l=0}^{a-1}G(\alpha+l,p^r)&=\sum_{l=0}^{a-1}\f{p^{3r}(p^r+2\alpha+2l)}{(\alpha+l)^3}\f{(\alpha+l)_{p^r}^3(\f12)_{p^r}}{(1)_{p^r}^3(\f12+\alpha+l)_{p^r}}\\
&=\f{p^{3r}(\alpha)_{p^r}^3(\f12)_{p^r}}{(1)_{p^r}^3(\f12+\alpha)_{p^r}}\sum_{l=0}^{a-1}\f{(p^r+2\alpha+2l)(\alpha+p^r)_l^3(\f12+\alpha)_l}{(\alpha+l)^3(\alpha)_l^3(\f12+\alpha+p^r)_l}.
\end{align*}
Then, by Lemma \ref{pochreduce}, we have
\begin{align*}
\sum_{l=0}^{a-1}G(\alpha+l,p^r)&=\f{p^{3r}(\alpha^{\ast_r})^3}{2(\f12+\alpha)^{\ast_r}}\prod_{j=1}^r\f{\Gamma_p(\alpha+p^j)^3\Gamma_p(\f12+p^j)\Gamma_p(1)^3\Gamma_p(\f12+\alpha)}{\Gamma_p(\alpha)^3\Gamma_p(\f12)\Gamma_p(1+p^j)^3\Gamma_p(\f12+\alpha+p^j)}\\
&\quad\times \sum_{l=0}^{a-1}\f{(p^r+2\alpha+2l)(\alpha+p^r)_l^3(\f12+\alpha)_l}{(\alpha+l)^3(\alpha)_l^3(\f12+\alpha+p^r)_l},
\end{align*}
where we have used the facts
$$
\l(\f12\r)^{\ast}=\f12\quad \t{and}\quad 1^{\ast}=1.
$$
Since $(\f12+\alpha)^{\ast_r}\not\eq0\pmod{p}$, we have
$$
\ord_p\l(\f{p^{3r}(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}\r)\geq 3r.
$$
It is easy to see that $\ord_p(\alpha+l)\leq r-1$ for $l\in\{0,1,2,\ldots,a-1\}$. It follows that
$$
\ord_p\l(\f{p^{4r}}{(\alpha+l)^3}\r)\geq r+3
$$
and
$$
\ord_p\l(\f{p^{3r}}{(\alpha+l)^2}\r)\geq r+2.
$$
For $k\in\{0,1,2,\ldots,a-1\}$ we have
$$
\f{(\alpha+p^r)_l}{(\alpha)_l}=\prod_{j=0}^{l-1}\f{\alpha+j+p^r}{\alpha+j}=\prod_{j=0}^{l-1}\l(1+\f{p^r}{\alpha+j}\r)\eq1\pmod{p}.
$$
In view of Lemma \ref{1/2+alpha} and the above, we obtain
\begin{equation}\label{evalGkey}
\sum_{l=0}^{a-1}G(\alpha+l,p^r)\eq\f{p^{3r}(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}\sum_{l=0}^{a-1}\f{(\f12+\alpha)_l}{(\alpha+l)^2(\f12+\alpha+p^r)_l}\pmod{p^{r+3}}.
\end{equation}

Suppose that $p=3$. By a similar argument as in the proof of Lemma \ref{evalF}, we have
\begin{align*}
\sum_{k=0}^{p^r-1}F(\alpha^{\ast_r}p^r,k)&\eq \alpha^{\ast_r}p^r+2(\alpha^{\ast_r})^3p^{3r}\sum_{k=1}^{(p^r-1)/2}\f{1}{k^2}+\f{2(\alpha^{\ast_r})^3p^{3r}}{2\alpha^{\ast_r}+1}\sum_{k=(p^r+1)/2}^{p^r-1}\f{1}{k^2}\pmod{p^{r+3}}.
\end{align*}
Note that
$$
\ord_p\l(\sum_{k=1}^{(p^r-1)/2}\f{1}{k^2}\r)\geq -2(r-1)\quad\t{and}\quad \ord_p\l(\sum_{k=(p^r+1)/2}^{p^r-1}\f{1}{k^2}\r)\geq -2(r-1).
$$
Since $\alpha^{\ast_r}=3/4\eq0\pmod{p}$ and $2\alpha^{\ast_r}+1=5/2\not\eq0\pmod{p}$, we still have
\begin{equation}\label{evalF'}
\sum_{k=0}^{p^r-1}F(\alpha^{\ast_r}p^r,k)\eq \alpha^{\ast_r}p^r\pmod{p^{r+3}}.
\end{equation}
Now, $a=(3p^r-1)/4\geq (p^r+1)/2$. From the proof of Lemma \ref{evalG}, we know
\begin{align*}
&\sum_{l=0}^{a-1}G(\alpha+l,p^r)\\
&\qquad\eq \f{p^{3r}(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}\sum_{l=(p^r+1)/2}^{a}\f{1}{(\alpha+a-l)^2}+\f{p^{3r}(\alpha^{\ast_r})^3(2\alpha^{\ast_r}-1)}{(\f12+\alpha)^{\ast_r}(2\alpha^{\ast_r}+1)}\sum_{l=1}^{(p^r-1)/2}\f{1}{(\alpha+a-l)^2}\pmod{p^{r+3}}.
\end{align*}
Also,
$$
\ord_p\l(\sum_{l=(p^r+1)/2}^{a}\f{1}{(\alpha+a-l)^2}\r)\geq -2(r-1)\quad\t{and}\quad \ord_p\l(\sum_{l=1}^{(p^r-1)/2}\f{1}{(\alpha+a-l)^2}\r)\geq -2(r-1).
$$
Then, in view of the facts $\alpha^{\ast_r}\eq0\pmod{p}$, $(1/2+\alpha)^{\ast_r}=1/4\not\eq0\pmod{p}$ and $2\alpha^{\ast_r}+1\not\eq0\pmod{p}$, we obtain
$$
\sum_{l=0}^{a-1}G(\alpha+l,p^r)\eq0\pmod{p^{r+3}}.
$$
This, together with \eqref{evalF'} gives
$$
\sum_{k=0}^{p^r-1}F(\alpha,k)=\sum_{k=0}^{p^r-1}F(\alpha^{\ast_r}p^r,k)-\sum_{l=0}^{a-1}G(\alpha+l,p^r)\eq\alpha^{\ast_r}p^r\pmod{p^{r+3}},
$$
which implies
\begin{equation*}
\sum_{k=0}^{3^r-1}(8k+1)\f{(\f14)_k^3(\f12)_k}{(1)_k^3(\f34)_k}\eq 3^{r+1}\pmod{3^{r+3}}.
\end{equation*}
Moreover, since
$$
\ord_3\l(\sum_{j=1}^{(3^r-3)/4}\f{1}{j^2}\r)\geq-2(r-1),
$$
we have
$$
\f{27}{4}3^{3r}\sum_{j=1}^{(3^r-3)/4}\f{1}{j^2}\eq 0\pmod{3^{r+3}}.
$$
Therefore, we arrive at
\begin{equation*}
\sum_{k=0}^{3^r-1}(8k+1)\f{(\f14)_k^3(\f12)_k}{(1)_k^3(\f34)_k}\eq 3^{r+1}+\f{27}{4}3^{3r}\sum_{j=1}^{(3^r-3)/4}\f{1}{j^2}\pmod{3^{r+3}},
\end{equation*}
as desired.

Theorem \ref{mainth1} seems quite strange. When $\<-1/2-\alpha\>_p<\min\{\<-\alpha\>_p,(p-1)/2\}$, the summands on the left-hand side of \eqref{mainth1eq} are not always $p$-adic integers. However, the sum of these summands is a $p$-adic integer. This phenomenon renders it difficult for us to prove Theorem \ref{mainth1} directly using the formula \eqref{Whipple} and conventional congruence techniques. To overcome this obstacle, we find a new parametric WZ pair (cf. \cite{PWZ}) which allows us to transform the original sum to a computable form in the sense of congruence. This idea is crucial in our proof.

In particular, putting $d=4,s=3,c=1$ and requiring $r$ to be odd in Theorem \ref{mainth1}, we have the following result.
\begin{corollary}\label{cor}
Guo and Zhao's conjectural supercongruence \eqref{GuoZhaoconj7.2} \cite[Conjecture 7.2]{Guo-Zhao2026} is true.
\end{corollary}
Moreover, Theorem \ref{mainth1} with $d=2,s=1,c=1$ gives that
\begin{equation}\label{cor1}
\sum_{k=0}^{p^r-1}(4k+1)\frac{(\f12)_k^4}{(1)_k^4}\equiv p^r \pmod{p^{r+3}},
\end{equation}
where we have used \eqref{2ordharmonic}. Note that the $r=1$ case of \eqref{cor1} is a stronger version of Van Hamme's (C.2) supercongruence \cite{VanHamme1997} and was first proved by Long \cite{Long2011}.

For the convenience of interested readers, in Table \ref{examples}, we provide some concrete examples of the parameters in Theorem \ref{mainth1} for future use.

\begin{table}[htbp]
    \centering
    \caption{Examples of the parameters in Theorem \ref{mainth1}}
    \label{examples}
    \renewcommand{\arraystretch}{1.5}
    \begin{tabular}{>{\centering\arraybackslash}p{2cm}
                      >{\centering\arraybackslash}p{2cm}
                      >{\centering\arraybackslash}p{2cm}
                      >{\centering\arraybackslash}p{3cm}
                      >{\centering\arraybackslash}p{3cm}}
        \toprule
        $d$ & $s$ & $\alpha$ & $\alpha^{\ast_r}$ & $(1/2+\alpha)^{\ast_r}$\\
        \midrule
        $2$ & $1$ & $1/2$  & $1/2$ & $1$\\
        $3$ & $1$ & $1/3$  & $1/3$ & $5/6$\\
        $3$ & $1$ & $2/3$  & $2/3$ & $1/6$\\
        $3$ & $1$ & $1/6$  & $1/6$ & $2/3$\\
        $3$ & $1$ & $5/6$  & $5/6$ & $1/3$\\
        $3$ & $2$ & $1/3$  & $(3-(-1)^r)/6$ & $(3+2(-1)^r)/6$\\
        $3$ & $2$ & $2/3$  & $(3+(-1)^r)/6$ & $(3-2(-1)^r)/6$\\
        $3$ & $2$ & $1/6$  & $(3-2(-1)^r)/6$ & $(3+(-1)^r)/6$\\
        $3$ & $2$ & $5/6$  & $(3+2(-1)^r)/6$ & $(3-(-1)^r)/6$\\
        $4$ & $1$ & $1/4$ & $1/4$ & $3/4$\\
        $4$ & $1$ & $3/4$ & $3/4$ & $1/4$\\
        $4$ & $3$ & $1/4$ & $(2-(-1)^r)/4$ & $(2+(-1)^r)/4$\\
        $4$ & $3$ & $3/4$ & $(2+(-1)^r)/4$ & $(2-(-1)^r)/4$\\
        \bottomrule
    \end{tabular}
\end{table}

We briefly outline this paper. In Section \ref{sec2}, we prove some properties of Dwork's dash operation and give some immediate applications which play essential roles in the subsequent proof. We shall prove Theorem \ref{mainth1} and Corollary \ref{cor} in Section \ref{sec3}.

The introduction explains why a direct proof of Theorem~\ref{mainth1} is hard and what method is used instead: when $\langle-1/2-\alpha\rangle_p<\min\{\langle-\alpha\rangle_p,(p-1)/2\}$, “the summands on the left-hand side of \eqref{mainth1eq} are not always $p$-adic integers,” although “the sum of these summands is a $p$-adic integer.” This “renders it difficult…to prove Theorem~\ref{mainth1} directly using the formula \eqref{Whipple} and conventional congruence techniques.”

To “overcome this obstacle,” the paper’s approach is to “find a new parametric WZ pair (cf. \cite{PWZ}) which allows us to transform the original sum to a computable form in the sense of congruence,” and “this idea is crucial in our proof.” Additionally, Section~\ref{sec2} establishes “properties of Dwork’s dash operation” with “immediate applications” that “play essential roles in the subsequent proof,” and then “We shall prove Theorem~\ref{mainth1} and Corollary~\ref{cor} in Section~\ref{sec3}.”

The introduction explains why a direct proof of the main theorem is hard and what method is used instead: when $\langle-1/2-\alpha\rangle_p<\min\{\langle-\alpha\rangle_p,(p-1)/2\}$, “the summands on the left-hand side of
\begin{equation}\label{mainth1eq}
\sum_{k=0}^{p^r-1}(2k+\alpha)\f{(\alpha)_k^3(\f12)_k}{(1)_k^3(\f12+\alpha)_k\eq \alpha^{\ast_r}p^r -\f{(\alpha^{\ast_r})^3}{(\f12+\alpha)^{\ast_r}}p^{r+2}H_{\alpha^{\ast_r}p-\alpha^{\ast_{r-1}}}^{(2)}\pmod{p^{r+3}},
\end{equation}
are not always $p$-adic integers,” although “the sum of these summands is a $p$-adic integer.” This “renders it difficult…to prove the main theorem directly using the formula
\begin{align}\label{Whipple}
&\sum_{k=0}^{\infty}\f{(a)_k(1+\f{a}{2})_k(b)_k(c)_k(d)_k}{(1)_k(\f a2)_k(1+a-b)_k(1+a-c)_k(1+a-d)_k}\notag\\
&\qquad=\f{\Gamma(1+a-b)\Gamma(1+a-c)\Gamma(1+a-d)\Gamma(1+a-b-c-d)}{\Gamma(1+a)\Gamma(1+a-b-c)\Gamma(1+a-b-d)\Gamma(1+a-c-d)}.
\end{align}
and conventional congruence techniques.”

To “overcome this obstacle,” the paper’s approach is to “find a new parametric WZ pair (cf. \cite{PWZ}) which allows us to transform the original sum to a computable form in the sense of congruence,” and “this idea is crucial in our proof.” Additionally, next the paper establishes “properties of Dwork’s dash operation” with “immediate applications” that “play essential roles in the subsequent proof,” and then later it proves the main theorem and also the following result.

\begin{corollary}\label{cor}
Guo and Zhao's conjectural supercongruence \eqref{GuoZhaoconj7.2} \cite[Conjecture 7.2]{Guo-Zhao2026} is true.
\end{corollary}

Question:
Let $\alpha=c/d$, where $c,d,s\in\mathbb Z_{>0}$ satisfy $d\ge 2$, $1\le c,s\le d$, and $\gcd(cs,d)=1$. Let $p\ge 5$ be a prime such that $p\equiv s\pmod d$, and let $r\in\mathbb Z_{>0}$. For a $p$-adic integer $x$, write $\langle -x\rangle_{p^r}$ for the least nonnegative residue of $-x$ modulo $p^r$, and define
$$
x^{\ast_r}:=\frac{x+\langle -x\rangle_{p^r}}{p^r}
$$
(so in particular $\alpha^{\ast_r}$ and $(\tfrac12+\alpha)^{\ast_r}$ are defined). Assume
$$
\bigl(\tfrac12+\alpha\bigr)^{\ast_r}\bigl(\tfrac12+\alpha^{\ast_r}\bigr)\not\equiv 0\pmod p.
$$
Here $(x)_k=x(x+1)\cdots(x+k-1)$ is the Pochhammer symbol, and $H_n^{(2)}=\sum_{j=1}^n \frac1{j^2}$ denotes the second-order harmonic number. Under these hypotheses, which explicit congruence for
$$
\sum_{k=0}^{p^r-1}(2k+\alpha)\frac{(\alpha)_k^3(\tfrac12)_k}{(1)_k^3(\tfrac12+\alpha)_k}
$$
holds modulo $p^{r+3}$?

Choices:
C. $$\sum_{k=0}^{p^r-1}(2k+\alpha)\frac{(\alpha)_k^3(\tfrac12)_k}{(1)_k^3(\tfrac12+\alpha)_k}\equiv \alpha^{\ast_r}p^r\pmod{p^{r+2}}.$$

denominator uses \tfrac12+\alpha^{\ast_r} instead of (\tfrac12+\alpha)^{\ast_r}

drops the explicit secondary correction term of order p^{r+2}

harmonic cutoff \alpha^{\ast_r}p-\alpha^{\ast_{r-1}} replaced by \alpha^{\ast_r}p-\alpha^{\ast_r}

raises correction from p^{r+2} to p^{r+3}, making the congruence too strong/trivial modulo p^{r+3}


\begin{corollary}\label{cor:integral_Hodge_S} Let $E$ be a smooth elliptic curve over $\bbC$ with $\End(E)=\Z$. Then there exists a bielliptic surface $S$ of type 2 whose Albanese variety $\Alb(S)$ is isomorphic to $E,$ and for which the integral Hodge conjecture for $1$-cycles on $E\times S$ fails. In particular, there exists a non-torsion integral Hodge class in $H^{4}(E\times S,\Z)$ that is not algebraic.
\end{corollary}

\begin{theorem}\label{thm:main_result} Let $E$ be a smooth complex elliptic curve with $\End(E)=\Z$. There exists a smooth projective complex surface $S$ with $\Alb(S)\cong E$ for which $\Ch(S)$ is representable, while $S$ admits no universal $0$-cycle.
\end{theorem}

Let $X$ be a smooth projective complex variety of dimension $d=\dim X$. Given a base-point $x_{0}\in X(\mathbf{C}),$ the universal morphism \begin{equation*}
    \al_{X}\colon X\longrightarrow\Alb(X)
\end{equation*}
to the Albanese variety of $X$ (see \cite[Theorem 1.4.4]{Murre}) induces the so-called Abel--Jacobi map for $0$-cycles on $X$ \begin{equation}
    \label{eq:abel-jacobi-map} \alpha_{X}\colon\Ch(X)_{\hom}\longrightarrow\Alb(X).
\end{equation}
The group homomorphism $\alpha_{X}$ no longer depends on the choice of the base-point and it is known to be surjective and regular. The latter asserts that, for any smooth projective complex variety $Z$ equipped with a base-point $z_{0}\in Z(\mathbf{C})$, and for any codimension-$d$ cycle $[\Gamma]\in\CH^{d}(Z\times X)$, the assignment
\begin{equation}\label{eq:psi}
    \psi_{(Z,[\Gamma])}\colon Z\longrightarrow\Alb(X),\quad z\longmapsto \alpha_{X}\circ[\Gamma]_{\ast}(z-z_{0})
\end{equation}
defines a morphism of algebraic varieties; see \cite[Definition 1.6.1]{Murre}. In fact, the Abel--Jacobi map $\alpha_{X}$ also enjoys the universal property of being initial among all regular homomorphisms from $\Ch(X)_{\hom}$ to abelian varieties; see \cite[Example 1.8(b)]{Murre}.
\par Since the homomorphism $\alpha_{X}$ in \eqref{eq:abel-jacobi-map} is regular and surjective, there exists a codimension-$d$ cycle $[\Gamma]\in\CH^{d}(\Alb(X)\times X)$ and a positive integer $n$ such that the morphism $\psi_{(\Alb(X),[\Gamma])}$ in \eqref{eq:psi}, with respect to the chosen base-point $0_{\Alb(X)}\in\Alb(X)(\mathbf{C})$, is equal to $n\cdot\Id_{\Alb(X)}$; see \cite[Corollary 1.6.3]{Murre}.\par

The present work is concerned with a strengthening of this statement, in the sense of the following property of the variety $X$, first introduced and studied by Voisin; see \cite{Voisin2024CycleCO,Voisin2025}.

\begin{definition}[\cite{voisin_a}, Definition 1.1]\label{def:universal_0-cycle} We say that $X$ admits a \emph{universal $0$-cycle} if there exists a codimension-$d$ cycle $[\Gamma]\in\CH^{d}(\Alb(X)\times X)$ such that the morphism $\psi_{(\Alb(X),[\Gamma])}$ in \eqref{eq:psi}, with respect to the base-point $0_{\Alb(X)}\in\Alb(X)(\mathbf{C})$, is the identity map $\Id_{\Alb(X)}$ on the Albanese variety of $X$.
\end{definition}

If $X$ is a curve, then the Poincaré divisor on $\Jac(X)\times X$ induces a universal $0$-cycle on $X$. In contrast, Voisin showed in \cite{Voisin2025} that for each $d\geq2$, there exists a smooth projective complex $d$-fold $X$ admitting no universal $0$-cycle; see \cite[Corollary  0.14]{Voisin2025}.\par
In \cite{colliotthélène2025notessurlapplicationdalbanese}, Colliot--Th\'{e}l\`ene considered whether the existence of a universal $0$-cycle persists under additional geometric assumptions on a smooth projective complex variety $X$. In particular, he asked whether this property holds for varieties with representable $\Ch$-group, that is, for which the homomorphism $\alpha_{X}$ in \eqref{eq:abel-jacobi-map} is an isomorphism; see \cite{Mumford1969,Roitman_1972}. Building on earlier ideas of Benoist--Ottem \cite{benoist-ottem}, Voisin subsequently constructed in \cite{voisin_a} a smooth projective complex threefold $X$ with representable $\Ch$-group but admitting no universal $0$-cycle.\par

This leads to the following question, raised by Totaro after Voisin’s talk at the workshop \emph{Hodge Theory and Algebraic Cycles} (Clay Mathematics Institute, September 2025).

\begin{ques}\label{ques:2-dim_example}
Does there exist a smooth projective complex surface $S$ whose Chow group of $0$-cycles $\CH_0(S)$ is representable, but which admits no universal $0$-cycle?
\end{ques}

The main result of the present paper answers Question \ref{ques:2-dim_example} in the affirmative.

Let $X$ be a smooth projective complex variety of dimension $d=\dim X$. Given a base-point $x_{0}\in X(\mathbf{C}),$ the universal morphism \begin{equation*}
 \al_{X}\colon X\longrightarrow\Alb(X)
\end{equation*}
to the Albanese variety of $X$ (see \cite[Theorem 1.4.4]{Murre}) induces the so-called Abel--Jacobi map for $0$-cycles on $X$ \begin{equation}
 \label{eq:abel-jacobi-map} \alpha_{X}\colon\Ch(X)_{\hom}\longrightarrow\Alb(X).
\end{equation}
The group homomorphism $\alpha_{X}$ no longer depends on the choice of the base-point and it is known to be surjective and regular. The latter asserts that, for any smooth projective complex variety $Z$ equipped with a base-point $z_{0}\in Z(\mathbf{C})$, and for any codimension-$d$ cycle $[\Gamma]\in\CH^{d}(Z\times X)$, the assignment
\begin{equation}\label{eq:psi}
 \psi_{(Z,[\Gamma])}\colon Z\longrightarrow\Alb(X),\quad z\longmapsto \alpha_{X}\circ[\Gamma]_{\ast}(z-z_{0})
\end{equation}
defines a morphism of algebraic varieties; see \cite[Definition 1.6.1]{Murre}. In fact, the Abel--Jacobi map $\alpha_{X}$ also enjoys the universal property of being initial among all regular homomorphisms from $\Ch(X)_{\hom}$ to abelian varieties; see \cite[Example 1.8(b)]{Murre}.
\par Since the homomorphism $\alpha_{X}$ in \eqref{eq:abel-jacobi-map} is regular and surjective, there exists a codimension-$d$ cycle $[\Gamma]\in\CH^{d}(\Alb(X)\times X)$ and a positive integer $n$ such that the morphism $\psi_{(\Alb(X),[\Gamma])}$ in \eqref{eq:psi}, with respect to the chosen base-point $0_{\Alb(X)}\in\Alb(X)(\mathbf{C})$, is equal to $n\cdot\Id_{\Alb(X)}$; see \cite[Corollary 1.6.3]{Murre}.\par

The present work is concerned with a strengthening of this statement, in the sense of the following property of the variety $X$, first introduced and studied by Voisin; see \cite{Voisin2024CycleCO,Voisin2025}.

\begin{definition}[\cite{voisin_a}, Definition 1.1]\label{def:universal_0-cycle} We say that $X$ admits a \emph{universal $0$-cycle} if there exists a codimension-$d$ cycle $[\Gamma]\in\CH^{d}(\Alb(X)\times X)$ such that the morphism $\psi_{(\Alb(X),[\Gamma])}$ in \eqref{eq:psi}, with respect to the base-point $0_{\Alb(X)}\in\Alb(X)(\mathbf{C})$, is the identity map $\Id_{\Alb(X)}$ on the Albanese variety of $X$.
\end{definition}

If $X$ is a curve, then the Poincaré divisor on $\Jac(X)\times X$ induces a universal $0$-cycle on $X$. In contrast, Voisin showed in \cite{Voisin2025} that for each $d\geq2$, there exists a smooth projective complex $d$-fold $X$ admitting no universal $0$-cycle; see \cite[Corollary 0.14]{Voisin2025}.\par
In \cite{colliotthélène2025notessurlapplicationdalbanese}, Colliot--Th\'{e}l\`ene considered whether the existence of a universal $0$-cycle persists under additional geometric assumptions on a smooth projective complex variety $X$. In particular, he asked whether this property holds for varieties with representable $\Ch$-group, that is, for which the homomorphism $\alpha_{X}$ in \eqref{eq:abel-jacobi-map} is an isomorphism; see \cite{Mumford1969,Roitman_1972}. Building on earlier ideas of Benoist--Ottem \cite{benoist-ottem}, Voisin subsequently constructed in \cite{voisin_a} a smooth projective complex threefold $X$ with representable $\Ch$-group but admitting no universal $0$-cycle.\par

\begin{ques}\label{ques:2-dim_example}
Does there exist a smooth projective complex surface $S$ whose Chow group of $0$-cycles $\CH_0(S)$ is representable, but which admits no universal $0$-cycle?
\end{ques}

The main result of the present paper answers Question \ref{ques:2-dim_example} in the affirmative.

\begin{equation}
    \label{eq:abel-jacobi-map} \alpha_{X}\colon\Ch(X)_{\hom}\longrightarrow\Alb(X).
\end{equation}

\begin{equation}\label{eq:psi}
    \psi_{(Z,[\Gamma])}\colon Z\longrightarrow\Alb(X),\quad z\longmapsto \alpha_{X}\circ[\Gamma]_{\ast}(z-z_{0})
\end{equation}

The main result of the present paper answers Question \ref{ques:2-dim_example} in the affirmative.

\subsection{Bielliptic surfaces}\label{subsec:bielliptic} Throughout this subsection, the base field $k$ is assumed to be algebraically closed.\par
Bielliptic surfaces constitute one of the classes of smooth projective surfaces of Kodaira dimension zero in the Enriques--Kodaira classification; see \cite{beauville_surfaces,BombieriMumford1977}. Every bielliptic surface $S$ admits a presentation \begin{equation*}\label{eq:presentation_S}
 S\cong\bigl(E\times_k F\bigr)/\Gi,\end{equation*}
where $E$ and $F$ are elliptic curves, and $\Gi$ is a finite group acting freely on $E\times_k F.$ The action is by translations on $E$ and by automorphisms on $F$. Moreover, one has an isomorphism $$F/\Gi\cong\bbP^{1}_k.$$\par
Bielliptic surfaces were classified by Bagnera--De Franchis into seven types; see \cite{bielliptic,bielliptic_suwa}. The classification is summarized in Table \ref{tab:bielliptic}.
\renewcommand{\arraystretch}{1.5} \begin{table}[h]
\centering
\caption{Classification of Bielliptic Surfaces (Bagnera--de Franchis)}
\label{tab:bielliptic}
\begin{tabular}{c c l c}
\hline
\textbf{Type} & \textbf{Group $\Gi$} & \textbf{$\NS(S)_{\tor}$} & \textbf{Order of $K_S$ in $\Pic(S)$} \\
\hline
1 & $\mathbb{Z}/2$ & $\Z/2\times\Z/2$ & 2 \\
2 & $\mathbb{Z}/2 \times \mathbb{Z}/2$ & $\Z/2$ & 2 \\
3 & $\mathbb{Z}/4$ & $\Z/2$ & 4 \\
4 & $\mathbb{Z}/4\times \mathbb{Z}/2$ & $0$ & 4 \\
5 & $\mathbb{Z}/3$ & $\Z/3$ & 3 \\
6 & $\mathbb{Z}/3 \times \mathbb{Z}/3$ & $0$ & 3 \\
7 & $\mathbb{Z}/6$ & $0$ & 6 \\
\hline
\end{tabular}
\end{table}
\par The Albanese morphism of $S$ coincides with the projection $$\phi_{S}\colon S\longrightarrow E/\Gi,$$ whose fibres are elliptic curves isomorphic to $F$. The induced Abel--Jacobi map on $0$-cycles,
$$(\phi_{S})_{\ast}\colon\Ch(S)_{\hom}\longrightarrow E/\Gi$$
is an isomorphism; see \cite[Example (2)]{bloch-srinivas}. Equivalently, the Chow group of $0$-cycles on a bielliptic surface is representable.\par
There is also a natural elliptic fibration $$\psi_{S}\colon S\longrightarrow (F/\Gi)\cong\bbP^{1}_{k}.$$
This fibration admits multiple fibres corresponding to the branch points of the quotient morphism $F \to F/\Gi$, with multiplicities equal to the corresponding ramification indices.\par
Let $f$ and $e$ denote the classes in $\NS(S)$ of the smooth fibres of $\phi_S$ and $\psi_S$, respectively. A direct computation shows \begin{equation}\label{eq:intersection_e_f}e^2=0,\quad,f^2=0,\quad e\cdot f=|\Gi|.\end{equation}

\section{Main results}\label{sec:final}
\subsection{A bielliptic surface of type 2 admitting no universal $0$-cycle}
Theorem \ref{thm:main_result} follows immediately from the result below.
\begin{theorem}\label{thm:main_result_body}
Let $k$ be an algebraically closed field of characteristic $\neq 2$, and let $E$ be an elliptic curve over $k$ with $\End(E)=\mathbb{Z}$. Set $\Delta := \Spec k[[t]]$, and fix an algebraic closure $K$ of its function field. Let $\mathcal S \to \Delta$ be the degeneration constructed in Theorem \ref{thm:intro_degeneration_S}, together with the induced morphism
$$
\phi_{\mathcal S} \colon \mathcal S \longrightarrow E \times_k \Delta .
$$
Then for any algebraically closed field extension $F/K$, there exists no correspondence
$$
[\Gamma] \in \CH^{2}(E \times_k S_F)
$$
that induces a splitting of the pushforward map
\begin{equation*}
(\phi_{S_F})_{\ast} \colon \CH_0(S_F)_{\hom} \longrightarrow E_{F} .
\end{equation*}
\end{theorem}
\begin{proof} Suppose, for contradiction, that there exists an algebraically closed field extension $F/K$ and a correspondence
$$
[\Gamma] \in \CH^{2}(E \times_k S_F)
$$
such that the induced endomorphism $$(\phi_{S_F})_{\ast}\circ[\Gamma]_{\ast}\in\End\bigl(E_{F}\bigr)$$ is the identity.\par
Let $$S_0=\sum_{i=1}^nR_i$$ denote the special fibre of the family $\ca S\to\Delta,$ as in \eqref{it:special_fibre} of Theorem \ref{thm:intro_degeneration_S}. For each $i,$ let $\al_i\colon\Alb(R_i)\to E$ be the morphism of Albanese varieties induced by the restriction of $\phi_{\mathcal S}$ to $R_i$. By Corollary \ref{cor:obstruction}, the assumption above implies that the sum morphism
$$\sum_{i=1}^n\al_i\colon\bigoplus_{i=1}^n\Alb(R_i)\longrightarrow E$$
admits a section. \par
This contradicts condition \eqref{it:albanese_S_0} of Theorem \ref{thm:intro_degeneration_S}. The resulting contradiction completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:main_result}] Let $E$ be an elliptic curve over $\bbC$ with $\End(E)=\Z$. Choose a countable algebraically closed subfield $k_0\subset\bbC$ over which $E$ is defined; thus $$E\cong E_0\times_{k_{0}}\bbC$$ for some elliptic curve $E_{0}$ over $k_{0}$.\par
Let $$\ca S\longrightarrow\Delta:=\Spec k_{0}[[t]]$$ be the strictly semistable degeneration constructed in Theorem \ref{thm:intro_degeneration_S}, together with the induced morphism $$\phi_{\mathcal S} \colon \mathcal S \longrightarrow E_0 \times_{k_{0}} \Delta .$$\par
Since $k_0$ is countable, the inclusion $k_0\subset\bbC$ factors through the Laurent series field $k_0((t))$. Set $$S:=S_{k_{0}((t))}\times_{k_{0}((t))}\bbC.$$
By \eqref{it:generic_fibre} of Theorem \ref{thm:intro_degeneration_S}, the surface $S$ is a bielliptic surface of type 2, and the morphism $$\phi_{S}\colon S\longrightarrow E$$ induced by $\phi_{\ca S}$ coincides with the Albanese fibration of $S$.\par
In particular, the pushforward homomorphism $$(\phi_{S})_{\ast} \colon \CH_0(S)_{\hom} \longrightarrow E $$ agrees with the Abel--Jacobi map for $0$-cycles on $S$. It is well known that bielliptic surfaces have representable $\Ch$-group; see \cite[Example (2)]{bloch-srinivas}. Applying Theorem \ref{thm:main_result_body}, we conclude that there exists no correspondence $$
[\Gamma] \in \CH^{2}(E \times S)
$$
that induces a splitting of the Abel--Jacobi map
\begin{equation*}
(\phi_{S})_{\ast} \colon \CH_0(S)_{\hom} \longrightarrow E.
\end{equation*}
Equivalently, the surface $S$ admits no universal $0$-cycle; cf. Definition \ref{def:universal_0-cycle}. This completes the proof of Theorem \ref{thm:main_result}.
\end{proof}
\subsection{A counterexample to the integral Hodge conjecture}
In this final section, we apply Theorem \ref{thm:main_result} to prove Corollary \ref{cor:integral_Hodge_S}. In particular, we construct a threefold $X$ of Kodaira dimension zero for which the integral Hodge conjecture fails for $1$-cycles. In contrast to the counterexamples of Benoist--Ottem \cite{benoist-ottem}, the non-algebraic Hodge class in degree $4$ arising from our construction is non-torsion.\par
Let $X$ be a smooth projective variety over $\bbC$ of dimension $d= \dim X$. We begin by recalling how the non-existence of a universal $0$-cycle on $X$ is related to the failure of the integral Hodge conjecture. Throughout, all (co)homology groups are taken to be Betti (co)homology, and the subscript "tf" denotes the quotient by torsion.\par

\begin{proof}[Proof of Corollary \ref{cor:integral_Hodge_S}]Let $E$ be an elliptic curve over $\bbC$ with $\End(E)=\Z$.
By Theorem \ref{thm:main_result}, there exists a bielliptic surface $S$ of type 2 with $\Alb(S)\cong E$ such that $S$ admits no universal $0$-cycle (cf. Definition \ref{def:universal_0-cycle}).
By Lemma \ref{lem:int_Hodge_conj_vs_uni_0_cycl}, it follows that the cohomology class $$\delta_{S}\in H^{4}\bigl(E\times S,\Z\bigr)_{\mathrm{tf}}$$ constructed in \eqref{eq:Hodge_class} is a Hodge class which is not algebraic.\par
This provides a counterexample to the integral Hodge conjecture in degree $4$
for $E\times S$.\end{proof}

Recall that ruled surfaces have representable Chow groups of $0$-cycles and do admit a universal $0$-cycle; this follows directly from the corresponding result for curves.\par
More generally, let $X$ be a non-uniruled smooth projective surface with geometric genus $p_{g}(X)=0$ and irregularity $q(X)\neq0$. By work of Beauville, the structure of such surfaces is well understood; see \cite[Chapter VI]{beauville_surfaces}. Specifically, $X$ is birational to a quotient \begin{equation*}\label{eq:structure_X}\bigl(C\times D\bigr)/\Gi,\end{equation*}
where $C$ and $D$ are smooth projective curves and $\Gi$ is a finite group acting faithfully on both curves, without fixed points on $C \times D$. Moreover, at least one of the curves is elliptic. Up to exchanging $C$ and $D$, one has $$C/\Gi\cong E,\qquad D/\Gi\cong\mathbf{P}^{1}_{\mathbf{C}},$$ where $E$ is an elliptic curve. In this situation, the induced fibration \begin{equation}\label{eq:fibration_X}\phi_{X}\colon X\longrightarrow E\end{equation} coincides with the Albanese morphism of $X$.\par
It was shown in \cite{voisin_a} that if the index of the fibration $\phi_X$ in \eqref{eq:fibration_X}, defined as the greatest common divisor of the degrees $\deg(C/E)$, where $C \subset X$ ranges over irreducible curves dominating $E$, is equal to one, then $X$ admits a universal $0$-cycle. In particular, this condition is satisfied when the group $\Gi$ is cyclic; see \cite[Proposition 4.4]{voisin_a}.\par
Our example in Theorem \ref{thm:main_result} arises from a very general bielliptic surface $S$ of type 2 (see \cite{bielliptic, bielliptic_suwa}) whose Albanese variety $\Alb(S)$ is isomorphic to the prescribed elliptic curve $E$. In this situation, the Albanese fibration $\phi_{S}\colon S\to \Alb(S)\cong E$ has index 2. As observed in \cite[$\S 2.2$]{voisin_a}, this condition alone does not suffice to imply that $S$ admits no universal $0$-cycle. The proof of Theorem \ref{thm:main_result} therefore rests on the following geometric result.

\begin{theorem}\label{thm:intro_degeneration_S} Let $k$ be an algebraically closed field of characteristic $\neq2,$ and let $E$ be an elliptic curve over $k$ with $\End(E)\cong\mathbb{Z}$. Set $\Delta:=\Spec k[[t]],$ and fix an algebraic closure $K$ of its function field. Then there exists a regular, flat, projective scheme $\mathcal{S}\to\Delta,$ together with a morphism $$\phi_{\mathcal{S}}\colon\mathcal{S}\longrightarrow E\times\Delta$$ such that the following properties hold:
\begin{enumerate}
    \item\label{it:generic_fibre} The geometric generic fibre $S_{K}$ is a bielliptic surface of type $2$, and the induced morphism $\phi_{S_{K}}\colon S_{K}\to E\times_{k} K$ coincides with the Albanese fibration of $S_{K}$.
    \item\label{it:special_fibre} The special fibre $S_{0}=\sum_{i=1}^{n}R_{i}$ is a reduced simple normal crossings divisor on $\mathcal{S}$ whose dual graph is a chain; that is, each intersection $R_{i}\cap R_{i+1}$ is a smooth irreducible curve, and $R_{i}\cap R_{j}=\varnothing$ whenever $j\notin\{i-1,i,i+1\}$.
     \item\label{it:albanese_S_0} For each $i,$ let $\al_{i}:\Alb(R_{i})\to E$ denote the homomorphism of Albanese varieties induced by the restriction of $\phi_{\mathcal{S}}$ to $R_{i}$. Then the sum morphism $$\sum_{i=1}^{n}\al_{i}\colon\bigoplus_{i=1}^{n}\Alb(R_{i})\longrightarrow E$$ does not admit a section.
\end{enumerate}
\end{theorem}
As we shall explain, the non-existence of a section in \eqref{it:albanese_S_0} provides the key obstruction to the existence of a universal $0$-cycle on the geometric generic fibre of the family $\mathcal{S}\to\Delta$; see Theorem \ref{thm:obstruction} and Corollary \ref{cor:obstruction}.\par
We conclude the introduction with the following application of Theorem \ref{thm:main_result} to the integral Hodge conjecture.
\begin{corollary}\label{cor:integral_Hodge_S} Let $E$ be a smooth elliptic curve over $\mathbf{C}$ with $\End(E)=\mathbb{Z}$. Then there exists a bielliptic surface $S$ of type 2 whose Albanese variety $\Alb(S)$ is isomorphic to $E,$ and for which the integral Hodge conjecture for $1$-cycles on $E\times S$ fails. In particular, there exists a non-torsion integral Hodge class in $H^{4}(E\times S,\mathbb{Z})$ that is not algebraic.
\end{corollary}
Corollary \ref{cor:integral_Hodge_S} adds a further example to the list of threefolds with torsion canonical bundle for which the integral Hodge conjecture fails for $1$-cycles. This should be compared with the counterexamples of Benoist--Ottem \cite{benoist-ottem}, where the obstruction arises from a non-algebraic torsion cohomology class in degree 4. In contrast, our construction relies on a different degeneration strategy: rather than degenerating the elliptic curve $E,$ we degenerate the surface $S$. This idea is inspired by recent work of the author \cite{Alexandrou,alexandrou2025torsionhigherchowcycles} and by work of Schreieder \cite{Schreieder2025}. \par
The paper is organized as follows. In $\S \ref{sec:preliminaries}$ we review the necessary background material. Section \ref{sec:obstruction} introduces an obstruction to the existence of a universal $0$-cycle on a smooth projective variety $X$ with representable Chow group of $0$-cycles. In $\S \ref{sec:degen_bielliptic}$ we prove Theorem \ref{thm:intro_degeneration_S}. Finally, $\S \ref{sec:final}$ is devoted to the proofs of our main results, including Theorem \ref{thm:main_result} and Corollary \ref{cor:integral_Hodge_S}.

For surfaces of the form \eqref{eq:structure_X} with Albanese fibration \eqref{eq:fibration_X}, it is recalled that Voisin shows: if the index of \(\phi_X\) is \(1\), then \(X\) admits a universal \(0\)-cycle (and this holds in particular when \(\Gi\) is cyclic).

For Theorem~\ref{thm:main_result}, the surface \(S\) is taken to be “a very general bielliptic surface of type 2” with \(\Alb(S)\cong E\), for which the Albanese fibration \(\phi_S\) has index \(2\); and it is noted (following \cite[\S 2.2]{voisin_a}) that “this condition alone does not suffice to imply that \(S\) admits no universal \(0\)-cycle.”

The stated strategy is instead to use a degeneration: “The proof of Theorem~\ref{thm:main_result} therefore rests on” Theorem~\ref{thm:intro_degeneration_S}, constructing a regular flat projective family \(\mathcal S\to\Delta\) whose geometric generic fibre \(S_K\) is such a bielliptic surface (with its Albanese fibration), and whose special fibre \(S_0=\sum R_i\) is an SNC divisor with chain dual graph. The key geometric input is that for the induced maps \(\al_i:\Alb(R_i)\to E\), “the sum morphism \(\sum_i \al_i:\bigoplus_i \Alb(R_i)\to E\) does not admit a section” \eqref{it:albanese_S_0}.

As announced, “the non-existence of a section in \eqref{it:albanese_S_0} provides the key obstruction to the existence of a universal \(0\)-cycle on the geometric generic fibre of the family \(\mathcal S\to\Delta\)” (via Theorem~\ref{thm:obstruction} and Corollary~\ref{cor:obstruction}), yielding the nonexistence of a universal \(0\)-cycle for \(S\) in Theorem~\ref{thm:main_result}.

For surfaces of the form \eqref{eq:structure_X} with Albanese fibration \begin{equation}\label{eq:fibration_X}\phi_{X}\colon X\longrightarrow E\end{equation}, it is recalled that Voisin shows: if the index of \(\phi_X\) is \(1\), then \(X\) admits a universal \(0\)-cycle (and this holds in particular when \(\Gi\) is cyclic).

To prove the main theorem, the surface \(S\) is taken to be “a very general bielliptic surface of type 2” with \(\Alb(S)\cong E\), for which the Albanese fibration \(\phi_S\) has index \(2\); and it is noted (following Voisin, \S 2.2) that “this condition alone does not suffice to imply that \(S\) admits no universal \(0\)-cycle.”

The stated strategy is instead to use a degeneration: “The proof of the main theorem therefore rests on” the following theorem.

\begin{theorem}\label{thm:intro_degeneration_S} Let $k$ be an algebraically closed field of characteristic $\neq2,$ and let $E$ be an elliptic curve over $k$ with $\End(E)\cong\Z$. Set $\Delta:=\Spec k[[t]],$ and fix an algebraic closure $K$ of its function field. Then there exists a regular, flat, projective scheme $\mathcal{S}\to\Delta,$ together with a morphism $$\phi_{\mathcal{S}}\colon\mathcal{S}\longrightarrow E\times\Delta$$ such that the following properties hold:
\begin{enumerate}
    \item\label{it:generic_fibre} The geometric generic fibre $S_{K}$ is a bielliptic surface of type $2$, and the induced morphism $\phi_{S_{K}}\colon S_{K}\to E\times_{k} K$ coincides with the Albanese fibration of $S_{K}$.
    \item\label{it:special_fibre} The special fibre $S_{0}=\sum_{i=1}^{n}R_{i}$ is a reduced simple normal crossings divisor on $\mathcal{S}$ whose dual graph is a chain; that is, each intersection $R_{i}\cap R_{i+1}$ is a smooth irreducible curve, and $R_{i}\cap R_{j}=\varnothing$ whenever $j\notin\{i-1,i,i+1\}$.
     \item\label{it:albanese_S_0} For each $i,$ let $\al_{i}:\Alb(R_{i})\to E$ denote the homomorphism of Albanese varieties induced by the restriction of $\phi_{\mathcal{S}}$ to $R_{i}$. Then the sum morphism $$\sum_{i=1}^{n}\al_{i}\colon\bigoplus_{i=1}^{n}\Alb(R_{i})\longrightarrow E$$ does not admit a section.
\end{enumerate}
\end{theorem}

Thus one constructs a regular flat projective family \(\mathcal S\to\Delta\) whose geometric generic fibre \(S_K\) is such a bielliptic surface (with its Albanese fibration), and whose special fibre \(S_0=\sum R_i\) is an SNC divisor with chain dual graph. The key geometric input is the preceding theorem’s assertion that for the induced maps \(\al_i:\Alb(R_i)\to E\), the sum morphism \(\sum_i \al_i:\bigoplus_i \Alb(R_i)\to E\) does not admit a section.

As announced, “the non-existence of a section in” this property provides the key obstruction to the existence of a universal \(0\)-cycle on the geometric generic fibre of the family \(\mathcal S\to\Delta\) (via the following theorem and corollary), yielding the nonexistence of a universal \(0\)-cycle for \(S\) in the main theorem.

\begin{theorem}
\label{thm:obstruction}
Let $k$ be an algebraically closed field and let $C$ be a smooth projective curve over $k$. Set $\Delta := \Spec k[[t]]$, and fix an algebraic closure $K$ of its function field.

Let $p \colon \mathcal{X} \to \Delta$ be a flat, projective morphism of relative dimension $d$ with regular total space, and let $\phi_{\mathcal{X}} \colon \mathcal{X} \to C \times_k \Delta$ be a surjective morphism.
Denote by $X_0$ the special fibre of $p$, by
$\phi_{X_0} \colon X_0 \to C$ the induced morphism, and by $X_{K
}$ the geometric generic fibre of $p$. Write $\phi_{X_K} \colon X_K \to C_K$
for the base change of $\phi_{\mathcal{X}}$ to $K$.\par
We further assume the following:
\begin{enumerate}
    \item \label{it:X_0} $X_{0}=\sum_{i=1}^{n}X_{0i}$ is a reduced simple normal crossings divisor on $\mathcal{X}$ whose dual graph is a chain. Precisely, each intersection $X_{0i}\cap X_{0i+1}$ is smooth and irreducible of dimension $d-1,$ and $X_{0i}\cap X_{0j}=\varnothing$ whenever $j\notin\{i-1,i,i+1\}$.
    \item \label{it:correspondence}
    There exists a correspondence
$$
[\Gamma] \in \CH^{d}(C \times_k X_{K})
$$
inducing a splitting of the pushforward map
$$
(\phi_{X_K})_\ast\colon \Ch(X_{K})_{\hom}\longrightarrow\jac(C_{K}).
$$

\end{enumerate}
\par
For each $i,$ let $\al_{i}\colon\Alb(X_{0i})\to \jac(C)$ be the homomorphism of Albanese varieties induced by the restriction of $\phi_{X_0}$ to $X_{0i}$. Then the induced sum morphism $$\sum_{i=1}^{n}\al_{i}\colon\bigoplus_{i=1}^{n}\Alb(X_{0i})\longrightarrow\jac(C)$$
admits a section.
\end{theorem}

\begin{corollary}\label{cor:obstruction}
We use the notation and assumptions of Theorem \ref{thm:obstruction}, with condition \eqref{it:correspondence} replaced by the following:
\begin{enumerate}
    \item[\((2')\)]\label{it:correspondence-prime} There exists an algebraically closed field extension $F/K$ and a correspondence
$$
[\Gamma] \in \CH^{d}(C \times_k X_{F})
$$
inducing a splitting of the pushforward map
$$
(\phi_{X_F})_\ast\colon \Ch(X_{F})_{\hom}\longrightarrow\jac(C_{F}).
$$
\end{enumerate}
Assume moreover that $\End(\jac(C))=\Z$. Then the sum morphism $$\sum_{i=1}^{n}\al_{i}\colon\bigoplus_{i=1}^{n}\Alb(X_{0i})\longrightarrow\jac(C)$$
admits a section.
\end{corollary}

\label{thm:main_result} Let $E$ be a smooth complex elliptic curve with $\End(E)=\mathbb{Z}$. There exists a smooth projective complex surface $S$ with $\Alb(S)\cong E$ for which $\Ch(S)$ is representable, while $S$ admits no universal $0$-cycle.,

Question:
Let $E$ be a smooth complex elliptic curve with $\operatorname{End}(E)=\mathbb{Z}$. For a smooth projective complex surface $S$, let
\[
\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\longrightarrow \operatorname{Alb}(S)
\]
be the Abel--Jacobi map for $0$-cycles. Say that $\CH_0(S)$ is representable if $\alpha_S$ is an isomorphism. Say that $S$ admits a universal $0$-cycle if there exists a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ such that the induced morphism
\[
\psi_{(\operatorname{Alb}(S),[\Gamma])}\colon \operatorname{Alb}(S)\to \operatorname{Alb}(S),\qquad a\mapsto \alpha_S\bigl([\Gamma]_*(a-0_{\operatorname{Alb}(S)})\bigr),
\]
is the identity map on $\operatorname{Alb}(S)$. Under these assumptions on $E$, which conclusion holds?

There exists a smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$ such that $\CH_0(S)$ is representable, i.e. $\alpha_S$ is an isomorphism, but $S$ does not admit a universal $0$-cycle; equivalently, there is no codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ for which $\psi_{(\operatorname{Alb}(S),[\Gamma])}=\operatorname{Id}_{\operatorname{Alb}(S)}$.

Choices:
B. For every smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$, if $\CH_0(S)$ is representable, then $S$ admits a universal $0$-cycle; equivalently, whenever $\alpha_S$ is an isomorphism there exists a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ such that $\psi_{(\operatorname{Alb}(S),[\Gamma])}=\operatorname{Id}_{\operatorname{Alb}(S)}$.
C. There exists a smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$ such that $\CH_0(S)$ is representable, i.e. $\alpha_S$ is an isomorphism.
D. There exists a smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$ such that $\CH_0(S)$ is representable and $S$ admits a universal $0$-cycle after base change to some algebraically closed field extension; equivalently, there is an algebraically closed extension $F/\mathbb{C}$ and a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)_F\times S_F)$ for which $\psi_{(\operatorname{Alb}(S)_F,[\Gamma])}=\operatorname{Id}_{\operatorname{Alb}(S)_F}$.
E. There exists a smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$ such that $\CH_0(S)$ is representable, but $S$ does not admit a universal $0$-cycle only in the weaker sense that no codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ makes $\psi_{(\operatorname{Alb}(S),[\Gamma])}$ equal to $n\,\operatorname{Id}_{\operatorname{Alb}(S)}$ for any positive integer $n$.

index-1/cyclic sufficient criterion promoted to universal statement for all representable surfaces

dropped the failure of a universal $0$-cycle

existence over an algebraically closed extension $F/K$ treated as if it could occur over a base change from $\mathbb{C}$ despite the obstruction using $\operatorname{End}(E)=\mathbb{Z}$

confusion between absence of identity splitting and absence of any Murre multiple $n\operatorname{Id}$


Question:
Let $E$ be a smooth complex elliptic curve with $\operatorname{End}(E)=\mathbb Z$. For a smooth projective complex surface $S$, write
\[
\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)
\]
for the Abel--Jacobi map. Say that $\CH_0(S)$ is representable if $\alpha_S$ is an isomorphism. Say that $S$ admits a universal $0$-cycle if there exists a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ such that, with base point $0_{\operatorname{Alb}(S)}\in \operatorname{Alb}(S)(\mathbb C)$, the morphism
\[
\psi_{(\operatorname{Alb}(S),[\Gamma])}\colon \operatorname{Alb}(S)\to \operatorname{Alb}(S),\qquad a\mapsto \alpha_S\bigl([\Gamma]_{*}(a-0_{\operatorname{Alb}(S)})\bigr)
\]
is the identity map on $\operatorname{Alb}(S)$. Under these assumptions, which existence statement holds?

There exists a smooth projective complex surface $S$ such that $\operatorname{Alb}(S)\cong E$, the Abel--Jacobi map $\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)$ is an isomorphism, and $S$ does not admit a universal $0$-cycle.

Choices:
B. There exists a smooth projective complex surface $S$ such that $\operatorname{Alb}(S)\cong E$, the Abel--Jacobi map $\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)$ is an isomorphism, and $S$ admits a universal $0$-cycle.
C. There exists a smooth projective complex surface $S$ such that $\operatorname{Alb}(S)\cong E$ and the Abel--Jacobi map $\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)$ is an isomorphism.
D. For every smooth projective complex surface $S$ with $\operatorname{Alb}(S)\cong E$, if the Abel--Jacobi map $\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)$ is an isomorphism, then there exists a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ such that $\psi_{(\operatorname{Alb}(S),[\Gamma])}=\operatorname{Id}_{\operatorname{Alb}(S)}$.
E. There exists a smooth projective complex surface $S$ such that $\operatorname{Alb}(S)\cong E$, the Abel--Jacobi map $\alpha_S\colon \CH_0(S)_{\mathrm{hom}}\to \operatorname{Alb}(S)$ is an isomorphism, and there is a codimension-$2$ cycle $[\Gamma]\in \CH^2(\operatorname{Alb}(S)\times S)$ for which $\psi_{(\operatorname{Alb}(S),[\Gamma])}=n\cdot \operatorname{Id}_{\operatorname{Alb}(S)}$ for some integer $n=1$.

degeneration obstruction from nonexistence of a section

dropped the conclusion that no universal 0-cycle exists

replaced an existential theorem by a universal implication from representability to existence of a universal 0-cycle

confuses Murre's existence of $n\cdot\operatorname{Id}$ with the stronger requirement of $\operatorname{Id}$ itself