\section{Reduction to codimension \texorpdfstring{$2$}{2} and a surface statement} \label{Sec:Reduction} The goal of this section is to explain a reduction for the Hodge conjecture for abelian varieties of CM type. 

\subsubsection{Recollections on the Hodge conjecture} For a smooth projective variety $X$ over $\mathbb{C}$ we let $\operatorname{CH}^{\ast}(X)$ denote the $\mathbb{Q}$-linear Chow ring of cycles on $X$ modulo rational equivalence, with $\operatorname{CH}^{i}(X)$ the codimension $i$ cycles. We similarly let $\rH^{\ast}(X, \mathbb{Q})$ denote the rational Betti (or singular) cohomology ring of the topological space $X(\mathbb{C})^{\mathrm{an}}$. This comes equipped with a Hodge decomposition
\begin{align}
    \rH^i(X, \mathbb{C}) = \bigoplus_{p+q=i} \rH^{p,q}(X, \mathbb{C}).
\end{align}
There is a multiplicative cycle class map $\operatorname{CH}^{\ast}(X) \to \rH^{2\ast}(X, \mathbb{Q})$ which takes $\operatorname{CH}^{p}(X)$ to $\operatorname{Hdg}^{p}(X,\mathbb{Q}):=\rH^{2p}(X,\mathbb{Q}) \cap \rH^{p,p}(X, \mathbb{C})$. We let $\operatorname{Alg}^{p}(X, \mathbb{Q}) \subset \operatorname{Hdg}^{p}(X,\mathbb{Q})$ denote the image of the cycle class map. The \emph{Hodge conjecture} asserts that $\operatorname{Alg}^{p}(X, \mathbb{Q})=\operatorname{Hdg}^{p}(X,\mathbb{Q})$, see e.g. \cite{DeligneHodge}, \cite{HodgeHodge}.


\subsection{Recollections on CM abelian varieties} We first recall some basics regarding CM abelian varieties following \cite[Section 1.2]{MilneHodgeClasses}.

\subsubsection{} For abelian varieties $A,B$ over a field $k$ we will write $\operatorname{Hom}^0(A,B) = \operatorname{Hom}(A,B) \otimes_{\mathbb{Z}} \mathbb{Q}$, where $\operatorname{Hom}$ denotes homomorphisms of abelian varieties over $k$. If $A=B$ we will denote this by $\operatorname{End}^0(A)$.

\subsubsection{} Let $L$ be a CM algebra and let $M$ be a field of characteristic zero. An \emph{abelian variety with CM by $L$} over $M$ is a pair $(A, \iota)$, where $A$ is an abelian variety over $M$ and $\iota$ is an injective ring homomorphism $\iota:L \to \operatorname{End}^0(A)$ such that $2 \operatorname{Dim} A = [L:\mathbb{Q}]$. 

\subsubsection{} Given a CM abelian variety $A$ over $\mathbb{C}$ with CM by a CM algebra $L$, we can write $L=L_1 \times \cdots \times L_n$ as a product of CM fields. We now choose a Galois CM field $E \subset \mathbb{C}$ containing $L_i$ for all $i$. Let $\Sigma=\operatorname{Hom}(L,E)$ be the set of ring homomorphisms from $L$ to $E$, equipped with its action of $G=\operatorname{Gal}(E/\mathbb{Q})$. As explained in \cite[Section (4.A)]{Hazama}, the CM abelian variety $A$ gives rise to a CM type $\Phi \subset \Sigma$ such that $\Sigma= \Phi \coprod \rho \Phi$ for a complex conjugation $\rho$. 

\subsubsection{} We may write (where we are using singular cohomology with $E$ coefficients)
\begin{align}
    \rH^1(A, E) = \bigoplus_{s \in \Sigma} \rH^1(A, E)_s
\end{align}
as a direct sum of $s$-isotypic components. Because $A$ has CM by $L$, the space $\rH^1(A, E)_s$ is a one-dimensional $E$-vector space. Identifying
\begin{align}
    \rH^r(A,E) = \bigwedge^r \rH^1(A, E),
\end{align}
we may thus identify
\begin{align}
      \rH^r(A,E) = \bigoplus_{\Delta} \rH^r(A,E)_{\Delta},
\end{align}
where $\Delta$ runs over size $r$ subsets of $\Sigma$ and $L^{\times}$ acts on $ \rH^r(A,E)_{\Delta}$ via $\prod_{s \in \Delta} s$. We may use this to describe the Hodge decomposition $p+q=r$
\begin{align}
    \rH^{p,q}(A,E):=\bigoplus_{\substack{\Delta \\ |(\Delta \cap \Phi)| = p, |(\Delta \cap \overline{\Phi})| = q}} \rH^r(A,E)_{\Delta}.
\end{align}
\subsubsection{} A result of Pohlmann, see \cite[Theorem 1]{Pohlmann} and \cite[Theorem 3.1]{gao2025hodgecyclesquadraticrelations} for the nonsimple case, describes the Hodge classes
\begin{align}
   \operatorname{Hdg}^p(A, \mathbb{Q}) \otimes E = \bigoplus_{\Delta} \rH^{2p}(A,E)_{\Delta},
\end{align}
where $\Delta$ runs over subsets of $\Sigma$ of size $2p$ with the property that for all $g \in G$ we have
\begin{align}
    | g\Delta \cap \Phi| = p =   | g\Delta \cap \overline{\Phi}|. 
\end{align}
So in degree $4$ we get
\begin{align} \label{eq:DegreeTwoHodgeClasses}
    \operatorname{Hdg}^2(A, \mathbb{Q}) \otimes E = \bigoplus_{\substack{\Delta \subset \Sigma, |\Delta|=4 \\ \forall g \in G |(g \Delta \cap \Phi)| =2 }} \rH^{4}(A,E)_{\Delta}.
\end{align}
For $\Delta$ satisfying the condition of \eqref{eq:DegreeTwoHodgeClasses} as above, we say that the \emph{Hodge conjecture holds for $(A, \Delta)$} if $\rH^{4}(A,E)_{\Delta}$ is in the image of the cycle class map tensored with $E$. Obviously the Hodge conjecture in codimension two for $A$ follows from the Hodge conjecture for $(A,\Delta)$ for all $\Delta$. 

\subsection{Results of Hazama} We first recall the following result of Hazama \cite{Hazama}.
\begin{Thm} \label{Thm:Hazama}
If the Hodge conjecture holds in codimension two for all CM abelian varieties over $\mathbb{C}$, then the Hodge conjecture holds for all CM abelian varieties.
\end{Thm}
\begin{proof}
This is a special case of \cite[Theorem 8.3]{Hazama}, see also \cite[Theorem 8.5]{Milne2008TateConjecture}.\footnote{Note that \cite[Lemmas 7.11, 7.12]{Hazama} are false as stated, but nonetheless \cite[Theorem 7.13]{Hazama} is still true.}
\end{proof}

\begin{Rem}
In fact, Hazama's result is much stronger. He defines for a Galois CM field $E$ of degree $2n$ with Galois group $G$, and a hyperplane arrangement $A_{\mathbb{A}(2^n)}$ in $\mathbb{R}^n$ a CM abelian variety $A_{\mathbb{A}(2^n)}(G;E)$. Then \cite[Theorem 8.3]{Hazama} states that the Hodge conjecture for $\rH^4(A_{\mathbb{A}(2^n)}(G;E))$ for all $G,E,A_{\mathbb{A}(2^n)}$ implies the generalized Hodge conjecture for all CM abelian varieties. 
\end{Rem}

\subsubsection{} We could now state a version of Theorem \ref{Thm:Hazama} in terms of the decomposition \eqref{eq:DegreeTwoHodgeClasses}, and this is exactly what \cite[Proposition 2.6]{OAI} is. 

\subsection{A further reduction} We will now explain how to state \cite[Proposition 3.1]{OAI} in our language.

\subsubsection{} \label{subsub:DefiningB} Continuing with the notation from above, in particular the fixed abelian varieties $A$ with $CM$ by $L$ and CM type $\Phi$, we choose $\Delta \subset \Sigma$ of size $4$ satisfying the condition in \eqref{eq:DegreeTwoHodgeClasses}. Write $\Delta = (s_1, s_2, c \circ s_3, c \circ s_4)$ with $s_1, s_2,s_3,s_4 \in \Phi$. First, we associate to $i=1, \cdots, 4$ an auxiliary abelian variety $B_i$ which has CM by $E$ (not by $L$!) and whose CM type is given by 
\begin{align}
    \Phi_i:= \ \{ g \in G=\operatorname{Hom}(E,E) : g \circ s_i \in \Phi\}.
\end{align}
Let $B=B_1 \times \cdots \times B_4$. Note that $\operatorname{Id} \in \Phi_i$ for all $i$, while $c \in \overline{\Phi}_i$ for all $i$. Moreover, condition \eqref{eq:DegreeTwoHodgeClasses} for $\Delta$ translates into the following identity of indicator functions (this is the analogue of the relation $v_1 + v_2 = v_3 + v_4$ occuring in \cite{OAI})
\begin{align} \label{eq:DegreeTwoHodgeClassesII}
\mathbf{1}_{\Phi_1} - \mathbf{1}_{\overline{\Phi}_1} + \mathbf{1}_{\Phi_2} - \mathbf{1}_{\overline{\Phi}_2} = \mathbf{1}_{\Phi_3} - \mathbf{1}_{\overline{\Phi}_3} + \mathbf{1}_{\Phi_4} - \mathbf{1}_{\overline{\Phi}_4}.
\end{align}
Define $U(i)=H^1(B_i, \mathbb{Q})$ for $i=1, \cdots, 4$ and let $\lambda=(\lambda_1, \lambda_2, \lambda_3, \lambda_4)=(\operatorname{Id}, \operatorname{Id}, c , c)$. 
\begin{Lem} \label{Lem:AbelianVarietyTransfer}
There is $t \in \operatorname{Hom}^0(A,B)$ such that the pullback $\otimes_{i} U(i)_{E,\lambda_i} \subset \rH^4(B,E)$ is given by $\rH^4(A,E)_{\Delta} \subset \rH^4(A,E)$. In particular if the $E$-vector space $\otimes_{i} U(i)_{E,\lambda_i}$ is contained in the image of the $E$-linear cycle class map, then the Hodge conjecture holds for $(A, \Delta)$.
\end{Lem}
\begin{proof}
For $i=1, \cdots, 4$, the morphism $s_i:L \to E$ factors through a simple factor of $L$; let us call these factors $L_{j_i}$ (they do not have to be distinct) and let us write $\Sigma_{j_i}=\operatorname{Hom}(L_{j_{i}}, E)$. Each $i$ determines an isogeny factor $A_{j_i}$ of $A$, and we will construct $t$ by constructing $t_i \in \operatorname{Hom}(A_{j_i},B_i)$ for $i=1, \cdots, 4$. \smallskip 

The morphism $t_i$ is constructed as follows. After choosing an $L_{j_i}$-linear isomorphism $\rH^1(A_{j_i}, \mathbb{Q}) \simeq L_{j_i}$ and an $E$-linear isomorphism $\rH^1(B_i, \mathbb{Q}) \simeq E$, we consider the map (induced by the trace map)
\begin{align}
    \rH^1(B_i, \mathbb{Q}) = E \xrightarrow{\operatorname{Tr}_{E/s_i(L_{j_i})}} s_i(L_{j_i}) \xrightarrow{s_i^{-1}} L_{j_i} = \rH^1(A_{j_i}, \mathbb{Q}).
\end{align}
The lemma follows if we can show that a) this is a map of rational Hodge structures and b) that this takes $\rH^1(B_i, E)_{g}$ to $\rH^1(A_{j_i}, E)_{g \circ s_i}$ after extending scalars to $E$. After extending scalars to $E$, we can identify the map with the $E$-linear map
\begin{align}
    \bigoplus_{G} E &\to \bigoplus_{\Sigma_{j_i}} E
\end{align}
which is the direct sum of the maps for $s \in \Sigma_{j_i}$
\begin{align}
\bigoplus_{\substack{ g \in G \\ g \circ s_i = s}} E \to E
\end{align}
given by $(e_g) \mapsto \sum e_g$. Claim b) follows directly from this description, and Claim a) follows from the definition of the CM type $\Phi_i$.
\end{proof}

\subsubsection{} Finally, we can state a version of \cite[Proposition 3.1]{OAI}. 
\begin{Thm} \label{Thm:ReductionToSurface}
If there is a smooth projective surface $S$ over $\mathbb{C}$ together with maps of $\mathbb{Q}$-Hodge structures 
\begin{align}
    h_i:U(i) \to \rH^1(S, \mathbb{Q})
\end{align}
for $i=1,2,3,4$, and nonzero vectors $\alpha_i \in  U(i)_{E,\lambda_i}$ such that 
\begin{align}
    h_1(\alpha_1) \cup h_2(\alpha_2) \cup h_3(\alpha_3) \cup h_4(\alpha_4) \not = 0
\end{align}
then the Hodge conjecture holds for $(A, \Delta)$. 
\end{Thm}
\begin{proof}
Assume without loss of generality that $S$ is connected. By Lemma \ref{Lem:AbelianVarietyTransfer}, it suffices to prove that the one-dimensional $E$ vector space
\begin{align}
    \bigotimes_{i} U(i)_{E,\lambda_i} \subset \rH^4(B,E)
\end{align}
is in the image of the cycle class map tensored with $E$. By the Hodge conjecture for homomorphisms of abelian varieties and the theory of the Albanese variety, there are morphisms $f_i:S \to B_i$ such that $h_i \cdot m_i = f_i^{\ast} $ for some nonzero integer $m_i$ (this is explained in \cite[Lemma 2.2]{OAI}). Let $f=(f_1, \cdots, f_4)$ and for $n=\operatorname{Dim} B$, we let $Z=f_{\ast}[S] \in \operatorname{CH}^{n-2}(B)$. If $x=\alpha_1 \otimes \alpha_2 \otimes \alpha_3 \otimes \alpha_4$, then the projection formula tells us that 
\begin{align}
    \int_{B} Z \cup x & = m_1 h_1(\alpha_1) \cup \cdots \cup m_4 h_4(\alpha_4) \not=0, 
\end{align}
showing that $Z \not=0$. Next, let $e_i$ be the idempotent element of $E \otimes_{\mathbb{Q}} E \subset \operatorname{End}^0(B_i) \otimes_{\mathbb{Q}} E$ whose induced map on $U(i)_E=\rH^1(B_i,E)$ is projection onto $U(i)_{E,\lambda_i}$. Let $p_{i}=p_{i,1}$ be the Chow--K\"unneth idempotent inducing the projection $\rH^{\ast}(B_i, \mathbb{Q}) \to \rH^1(B_i, \mathbb{Q})$; this is algebraic by \cite[Main Theorem]{DeningerMurre}. Then the composition $e_i p_i$ acts on cohomology by projection $\rH^{\ast}(B_i, E) \to \rH^{1}(B_i, E) \to U(i)_{E,\lambda_i}$. We can now consider $e = \otimes_i e_i p_i$, which induces the projection $\rH^4(B,E) \to \otimes_i U(i)_{E, \lambda_i}$. \smallskip 

Consider $x=\alpha_1 \otimes \alpha_2 \otimes \alpha_3 \otimes \alpha_4$ and note that $ex=x$. Under the $E$-linear intersection product 
\begin{align}
    \rH^{2n-4}(B,E) \times \rH^{4}(B,E) \to E
\end{align}
we have seen that $Z$ pairs nontrivially with $x=ex$. This implies that $e^tZ$ also pairs nontrivially with $x$, and thus $e^t\rH^{2n-4}(B,E)$ is nonzero. It is moreover one-dimensional since $e \rH^4(B,E)$ is. Now, because numerical and homological equivalence agree for cycles on CM abelian varieties, see \cite{MR230336}, the pairing 
\begin{align}
    \operatorname{Alg}^{n-2}(B,E) \times \operatorname{Alg}^2(B,E) \to E
\end{align}
is perfect. After applying the projector $e$ and its transpose $e^t$, the pairing 
\begin{align}
    e^t\operatorname{Alg}^{n-2}(B,E) \times e\operatorname{Alg}^2(B,E) \to E
\end{align}
is also perfect. Because $e^t \operatorname{Alg}^{n-2}(B,E)$ is nonzero and spanned by $e^t Z$, it follows that $e \operatorname{Alg}^{2}(B,E)$ is nonzero and one-dimensional. Since $e \rH^4(B,E)$ is one-dimensional, it follows that $e \operatorname{Alg}^{2}(B,E)=e \rH^4(B,E)$. Thus $x=ex$ is in the image of the $E$-linear cycle class map. 
\end{proof}