text:
\section*{Introduction}\label{sec:intro}
\addcontentsline{toc}{section}{\nameref{sec:intro}}

If $X$ is a spectrum then the smash power $X^{\wedge 2}$ carries
a $\Sigma_2$-action. Equipping $X$ with a map
	\[
	X^{\wedge 2}_{h\Sigma_2} \to X
	\]
is a way of encoding a coherently symmetric multiplication. In the
presence of this structure, there are natural operations
	\[
	Q^i: \mathrm{H}_*(X) \to \mathrm{H}_{*+i}X
	\]
on the mod 2 homology of $X$. These operations play a crucial
role in computations. For example, in the dual Steenrod
algebra $\mathcal{A}_*=\mathbb{F}_2[\zeta_1, \zeta_2, ...]$,
Steinberger \cite[\S III.2]{bmms} proved that, for $i\ge 1$:
	\[
	Q^{2^i}\overline{\zeta}_i = \overline{\zeta}_{i+1}.
	\]
This computation provides key non-formal input in many results, for example:
	\begin{itemize}
	\item B\"okstedt's \cite{bokstedt}
	computation of the topological Hochschild homology
	of $\mathbb{F}_2$ and $\mathbb{Z}$.
	\item The Hopkins-Mahowald theorem identifying $\mathrm{H}\mathbb{F}_p$
	as a Thom spectrum.
	\item Lawson's \cite{lawson} proof that the Brown-Peterson spectrum
	cannot be made $\mathbb{E}_{\infty}$.
	\end{itemize}

Since the success of Hill-Hopkins-Ravenel \cite{hhr}, there
has been much recent activity in calculations in $C_2$-equivariant
homotopy theory. To aid in this endeavor, the author set out in
\cite{wilson-old} to study $C_2$-equivariant power operations on
ordinary equivariant homology with mod 2 coefficients. As first
applications, the author gave an equivariant, cellular construction of $\mathrm{BP}$
with its action by complex conjugation, and proved an analogue
of the Hopkins-Mahowald theorem jointly with Mark Behrens \cite{behrens-wilson}.
Unfortunately, the treatment
in \cite{wilson-old} is a bit clunky, contains some errors in various
formulas, and did not include the Adem relations.

The purpose of this note is to give a corrected and expanded
account of these power operations. In order to more quickly provide
a reference
for those interested in calculating, we have decided to state the main
formulas governing the operations and sketch the proofs. We defer
the verification of certain categorical statements to a more detailed,
forthcoming paper. We believe our approach is interesting even
nonequivariantly and have given a separate and complete treatment
of that story in \cite{wilson-classical}.

\subsubsection*{Outline}

In \S\ref{sec:results} we briefly recall the set-up for equivariant homotopy
theory and then go on to state the properties of the equivariant power
operations. This includes a Cartan formula (Proposition \ref{prop:cartan}),
Adem relations (Theorem \ref{thm:adem}), and Nishida relations
(Proposition \ref{prop:nishida}). We end by computing
the action of power operations on the equivariant dual Steenrod algebra
(Theorem \ref{thm:action}).

In \S\ref{sec:sketch} we sketch the
categorical and computational input necessary to prove the theorems
from \S\ref{sec:results}. The main idea is to adapt the method in
\cite{wilson-classical} to the equivariant setting using several
basic calculations combined with tools from the theory of
$C_2$-$\infty$-categories and higher equivariant algebra.

Much of this paper is devoted to explaining how the theory
of power operations in $C_2$-equivariant homotopy theory
behaves like its nonequivariant counterpart (or, more accurately,
like the \emph{odd} primary nonequivariant counterpart).
However, there are several important instances where the theory is
different:
	\begin{itemize}
	\item The equivariant Cartan and Adem formulas contain terms which
	vanish after restriction to nonequivariant homotopy.
	\item The homology of equivariant extended powers of spheres
	is not so simple. For example, there is not, in general, a Thom
	isomorphism which reduces their computation to the homology
	of a classifying space.
	\item The power operations we build are necessarily stable.
	Nonequivariantly, all mod 2 power operations are stable. The analogous
	statement is false equivariantly, see \S\ref{ssec:counterexample}.
	\end{itemize}

\subsubsection*{Acknowledgements} We are grateful to Tom Bachmann,
Saul Glasman,
Jeremy Hahn, Mike Hill, Tyler Lawson, and Krishanu Sankar for
helpful conversations. We also thank Nick Kuhn, Cary Malkiewich, Sean Tilson,
Yan Zou, and an anonymous referee for comments on an earlier
incarnation of this material.

\subsubsection*{A historical remark} Our approach to power operations
relies heavily on their relationship to Tate constructions. This
has a long history. On the algebraic side, the relationship
between Tate-like constructions and operations goes back to Singer \cite{singer}.
Jones and Wegmann realized this algebraic construction topologically in
\cite{jones-wegmann}
(see especially their results 5.2(a) and 5.2(b)). An ad-hoc
version of the Tate-valued
Frobenius on $\mathrm{H}\mathbb{F}_p$ makes an appearance
in disguise during McClure's proof of the Nishida relations
in \cite[\S VIII.3]{bmms}, and there are hints of the same
in Steinberger's calculation of the action of power operations
on the dual Steenrod algebra \cite[\S III.2]{bmms}. More recent references,
where the relationship is more explicit,
include course notes of Lurie \cite{lurie-course}, a paper of Kuhn and McCarty
\cite[Cor. 2.13]{kuhn-mccarty}, and especially
a paper of Glasman-Lawson \cite{glasman-lawson}.
See also Theorem IV.1.15 in \cite{nikolaus-scholze} (attributed to Lurie).

\section{Results} \label{sec:results}

In \S\ref{ssec:preliminaries} we review enough equivariant
homotopy theory to be able to state our results. In
\S\ref{ssec:operations} we assert the existence of power operations
satisfying certain basic properties. In the remaining sections we
state the Cartan formulas for products and norms (Proposition
\ref{prop:cartan}), the Adem relations (Theorem \ref{thm:adem}),
the Nishida relations (Proposition \ref{prop:nishida}), and the action on the
equivariant dual Steenrod algebra (Theorem \ref{thm:action}).
In each case we give a clean
statement in terms of the total power operation and then extract
more explicit formulas from there. The formulas in terms of power
series identities are more than just a mnemonic; in \S\ref{sec:sketch}
we explain how the set-up
we have chosen produces these identities directly.

\begin{remark} The answers we get below closely align with
the properties of the motivic Steenrod operations \cite{voevodsky} upon making
the substitutions:
	\[
	Q^{i\rho} \leftrightarrow \mathrm{Sq}^{-2i},
	\,\, Q^{i\rho -1} \leftrightarrow \mathrm{Sq}^{-2i+1},\,\,
	a \leftrightarrow \rho, \,\, u \leftrightarrow \tau
	\]
Of course, we allow $i$ to be positive in our case, and the proofs
are necessarily different. It would be interesting to know of a more
precise relationship between the two stories.
\end{remark}

\subsection{Preliminaries} \label{ssec:preliminaries}
We establish just as many notions and notations as we need to state our results.
A nice treatment of most of the material we need is in \cite[\S 1-3, \S A, \S B]{hhr}.
\subsubsection{Spaces and spectra}
The homotopy theory of $C_2$-spaces arises from the category
$\mathbf{Top}^{C_2}$
of topological spaces equipped with a $C_2$-action by inverting
(in the $\infty$-categorical sense)
the collection of maps $W$ which become weak equivalences on underlying and
fixed point spaces. We write:
	\[
	\mathbf{Top}^{C_2}[W^{-1}] =: \mathsf{Spaces}^{C_2}.
	\]
Given a representation $V$ we may form the one-point compactification
$S^V$ and thus define functors $\Sigma^V$ and $\Omega^V$ on
$\mathsf{Spaces}^{C_2}_*$. If we let $\sigma$ denote the sign representation
and $\rho = 1+\sigma$ the regular representation, then the
$\infty$-category of \textbf{(genuine) $C_2$-spectra} is the homotopy limit:
	\[
	\mathsf{Sp}^{C_2} = \holim\left(\cdots \to \mathsf{Spaces}^{C_2}_*
	\stackrel{\Omega^\rho}{\longrightarrow} \mathsf{Spaces}^{C_2}_*\right).
	\]
This limit comes with a canonical infinite loop functor $\Omega^{\infty}:
\mathsf{Sp}^{C_2} \to \mathsf{Spaces}^{C_2}_*$ which admits a
left adjoint, the suspension spectrum functor $\Sigma^{\infty}$.
We abuse notation and write
$S^V$ for either the space or its suspension spectrum, and
similarly if $X$ is a $C_2$-space then we often write $X_+$ for
its suspension spectrum.

The
$\infty$-category $\mathsf{Sp}^{C_2}$ is stable and hence canonically
enriched in the $\infty$-category of spectra \cite[4.2.1.33,4.8.2.18]{HA}.
In particular, the objects
$S^0$ and $C_{2+}$ corepresent functors to spectra, denoted
$(-)^{C_2}$ and $(-)^u$, respectively. We refer to $X^{C_2}$ as the
\textbf{genuine fixed points} and $X^u$ as the \textbf{underlying spectrum}.

The $\infty$-category $\mathsf{Sp}^{C_2}$ is closed
symmetric monoidal under the smash product. We denote the internal
mapping object by $F(-,-)$.

\subsubsection{Homotopy groups}

If $V$ is a representation, then $S^V$ is an invertible object in
$\mathsf{Sp}^{C_2}$, so we can make sense of $S^{a+b\sigma}$
for any $a, b \in \mathbb{Z}$. We define groups indexed over
the representation ring $RO(C_2)$ by:
	\[
	\pi_{a+b\sigma}X := [S^{a+b\sigma}, X],
	\]
and we denote the direct sum of these groups by $\pi_{\star}X$.

By dualizing the cofiber sequence $C_{2+} \to S^0 \to S^{\sigma}$
and using the shearing equivalence $C_{2+} \wedge S^{1-\sigma} \simeq
C_{2+}$
we learn that $C_{2+}$ is self-dual. In particular, we have a map
$S^0 \to C_{2+}$ dual to the collapse $C_{2+} \to S^0$. These give
rise to the transfer and restriction, respectively, after smashing
with $S^{a+b\sigma}$:
	\[
	\mathrm{tr}_{a+b\sigma}: \pi_{a+b}X^u \to \pi_{a+b\sigma}X
	\]
	\[
	\mathrm{res}: \pi_{a+b\sigma}X \to \pi_{a+b}X
	\]
If $b=0$ then we omit the subscript on the transfer.

If $X$ is equipped with a homotopy commutative product
$X \wedge X \to X$ then the homotopy groups carry the structure
of a graded ring with commutation rule given by
\cite[Lemma 2.12]{hu-kriz}:
	\[
	xy = (-1)^{km}(1-\mathrm{tr}(1))^{\ell n}yx, \quad x\in \pi_{k+\ell\sigma}X,
	y \in \pi_{m+n\sigma}X.
	\]
In particular, if $\mathrm{tr}(1)=2$, this is the usual rule
for a bigraded commutative ring.

\subsubsection{Eilenberg-MacLane spectra}

Associated to any abelian group $M$ there is a $C_2$-spectrum
which we will denote by the same symbol and which is uniquely characterized
by the requirements:
	\[
	\mathrm{res}: \pi_0M \stackrel{\simeq}{\to} \pi_0M^u = M
	\]
	\[
	\pi_iM=\pi_iM^u = 0, \quad i\ne0.
	\]
If the abelian group
$M$ is equipped with the structure of a commutative ring
then the corresponding $C_2$-spectrum is an $\mathbb{E}_{\infty}$-ring
in $\mathsf{Sp}^{C_2}$. We also have that $\mathrm{tr}(1) = 2$ so that
any $M$-module equipped with a homotopy commutative multiplication
satisfies the simpler bigraded commutativity rule on its homotopy groups.

\begin{convention} We will denote by $k$ both the field $\mathbb{F}_2$
and its corresponding $C_2$-spectrum.
\end{convention}

We will need two elements in the homotopy of $k$. The first comes from
the homotopy of the sphere:
	\[
	a: S^{-\sigma} \to S^0
	\]
is adjoint to the inclusion of fixed points. The second arises from the identity
$\mathrm{tr}(1) = 0 \in \pi_0k$, which produces a map
	\[
	u: S^{1-\sigma} =
	\mathrm{cofib}(S^0 \to C_{2+}) \to k
	\]
that is well-defined since $\pi_1k = 0$.

\subsubsection{Norms}

The symmetric monoidal structure on $\mathsf{Sp}^{C_2}$
can be enhanced to allow for tensor products indexed over finite $C_2$-sets.
This is one of the key innovations introduced in \cite[\S B]{hhr}.

To describe the homotopical features of this enhancement, we define,
for each $n$, an $\infty$-category $\left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[n]}}$
as follows. Begin with $\mathbf{Top}^{C_2 \times \Sigma_n}$ and invert
those maps which are $C_2$-equivalences to form an $\infty$-category
$\left(\mathsf{Spaces}^{C_2}\right)^{h\Sigma_{[n]}}$. Now form the limit
	\[
	\left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[n]}}:=
	\holim \left(\cdots \stackrel{\Omega^{\rho}}{\longrightarrow}
	\left(\mathsf{Spaces}^{C_2}\right)_*^{h\Sigma_{[n]}}
	\stackrel{\Omega^{\rho}}{\longrightarrow}
	\left(\mathsf{Spaces}^{C_2}\right)_*^{h\Sigma_{[n]}}\right).
	\]
The inclusion $\mathbf{Top}^{C_2} \to \mathbf{Top}^{C_2\times\Sigma_n}$
of $C_2$-spaces with trivial $\Sigma_n$-action induces functors
	\[
	\mathsf{Spaces}^{C_2} \longrightarrow
	\left(\mathsf{Spaces}^{C_2}\right)^{h\Sigma_{[n]}}
	\]
	\[
	\mathsf{Sp}^{C_2} \longrightarrow
	\left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[n]}}
	\]
which admit both left and right adjoints, denoted $(-)_{h\Sigma_{[n]}}$
and $(-)^{h\Sigma_{[n]}}$, respectively.

We will need to know that the assignment $X \mapsto X^{\wedge n}$ can be
refined to a functor
	\[
	(-)^{\wedge n}: \mathsf{Sp}^{C_2}\longrightarrow
	\left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[n]}},
	\]
which allows us to produce a homotopical symmetric power
	\[
	\mathrm{Sym}^{[n]}(X):= (X^{\wedge n})_{h\Sigma_{[n]}}.
	\]
The latter functor, at least, can be found in \cite[\S B.6.1]{hhr}.
In the case $n=2$ we may form the \textbf{norm}:
	\[
	N(X):= \left(X^{\wedge 2} \wedge
	\frac{C_2\times \Sigma_2}{\Delta}_+\right)_{h\Sigma_{[2]}} \in
	\mathsf{Sp}^{C_2}.
	\]
Note that we have natural maps of $C_2$-spectra:
	\[
	X^{\wedge 2} \to \mathrm{Sym}^{[2]}(X), \quad
	N(X) \to \mathrm{Sym}^{[2]}(X).
	\]

We will also need the generalization of the above discussion to
$k$-modules, where the only difference is that we use
$\otimes := \otimes_k$ instead of the smash product.

\subsection{The operations} \label{ssec:operations}

The most general setting in which our operations exist
is the following.

\begin{definition} A $k$-module $A$ is \textbf{equipped with
an equivariant symmetric multiplication} if we provide a map
$\mathrm{Sym}^{[2]}(A) \to A$, where we take the indexed
symmetric square in $\mathsf{Mod}_k$.
\end{definition}

Note that we do not require this product to be unital or associative.
An example of such a $k$-module is $F(X_+, k)$ where $X$ is a
$C_2$-space and the product arises from the diagonal.

\begin{definition} If $A_{\star}$ is an $RO(C_2)$-graded commutative
$k_{\star}$-algebra, we denote by $A_{\star}(\!(s,t)\!)$ the algebra:
	\[
	A_{\star}(\!(s,t)\!):= (A_{\star}[s]\llbracket t\rrbracket/(s^2=as+ut))[t^{-1}]
	\]
where $|s|=-\sigma$ and $|t|=\rho$.
We observe that homogeneous
elements in this algebra may be written uniquely in the form
	\[
	\sum_{i\in \mathbb{Z}} x_i st^{i-1} + \sum_{i\in \mathbb{Z}} y_i t^{i}
	\]
where $x_i, y_i \in A_{\star}$ vanish for $i$ sufficiently negative.
\end{definition}

In Construction \ref{cstr:operations} below, we will sketch the construction
of operations $Q^{i\rho -\varepsilon}: A_{\star} \to A_{\star+i\rho-\varepsilon}$
for $\varepsilon = 0,1$. These agree with the operations constructed
in a different manner in our previous treatment \cite{wilson-old}, and
the following theorem summarizes their first properties. We draw attention
to (iv)-(vi) which are corrected versions of the corresponding statements
in \cite{wilson-old}.

\begin{theorem}\label{thm:properties} Let $A$ be a $k$-module equipped with
an equivariant symmetric multiplication. Then there is
a natural operation:
	\[
	Q(s,t): A_{\star} \to A_{\star}(\!(s,t)\!)
	\]
with coefficients
	\[
	Q(s,t)x = \sum_i (Q^{i\rho-1}x) st^{i-1} + \sum_i (Q^{i\rho}x) t^{i}
	\]
satisfying the following properties:
	\begin{enumerate}[\upshape(i)]
	\item \textup{(Mackey)} For $\varepsilon=0,1$, the operation
	$Q^{i\rho-\varepsilon}$ restricts to $Q^{2i-\varepsilon}$ on
	underlying homotopy groups, and commutes with addition and transfer.
	\item \textup{(Loops)} For any representation $V$, we have
	$\Omega^V Q(s,t) = Q(s,t) \Omega^V$.
	\item \textup{(Squaring)} If $|x| = n\rho-\varepsilon$ for $\varepsilon=
	0,1$ then $Q^{n\rho-\varepsilon}(x) = x^2$.
	\item \label{vanishing} \textup{(Vanishing)} Suppose $|x| = a+b\sigma$
	and $\varepsilon = 0,1$. Then
	$Q^{i\rho-\varepsilon}x = 0$ if $i<a+\varepsilon$ and $i\le b$.
	\item \textup{(Cohomology)} If $A=F(X,k)$ for a pointed $C_2$-space
	$X$,
	then $Q^{i\rho-\varepsilon}$ acts by zero if $2i-\varepsilon>0$.
	\item \textup{(Bockstein)} Suppose that $A$ arises as the
	mod 2 reduction of a $\mathbb{Z}/4$-module. Then
	$Q^{i\rho -1}$ acts by $\beta Q^{i\rho}$ where $\beta$ is the Bockstein.
	\end{enumerate}
\end{theorem}

\subsection{Cartan formulas} \label{ssec:cartan}

In \cite{wilson-old} we stated an obviously incorrect version
of the Cartan formula (it did not even restrict to the usual
version on underlying homotopy). The correct version
of the Cartan formula, as well as a new formula for the
value of operations on a norm class, is as follows:

\begin{proposition}\label{prop:cartan}
	\begin{align*}
	Q(s,t)(x\otimes y) &= Q(s,t)x \otimes Q(s,t)y\\
	Q(s,t)(Nx) &= N(Q(w)x)
	\end{align*}
\end{proposition}

Here we have used $Nx$ to denote the class obtained from $x: \Sigma^ik^u
\to A^u$ as $\Sigma^{i\rho}k = N(\Sigma^ik^u) \to N(A^u)$.
We also use $w$ to denote a power series generator
for for the underlying total power operation $Q(w) = \sum Q^i(-) w^i$ and
define $N(w) = t$.

We may extract more explicit formulas by comparing coefficients
of $s^{\varepsilon}t^{n-\varepsilon}$ and using the distributive
law for the norm of a sum \cite[Prop. A.37]{hhr}:

\begin{corollary}[Explicit Cartan formulas]
	\begin{align*}
	Q^{n\rho}(x \otimes y) &=
	\sum_{i+j=n} Q^{i\rho}x \otimes Q^{j\rho}y
	+ u\sum_{i+j=n+1}Q^{i\rho-1}x \otimes Q^{j\rho-1}y\\
	Q^{n\rho-1}(x \otimes y) &=
	\sum_{i+j = n} \left(Q^{i\rho -1}x \otimes Q^{j\rho}y +
	Q^{i\rho}x \otimes Q^{j\rho -1}y\right) +
	a\sum_{i+j = s+1} Q^{i\rho-1}x \otimes Q^{j\rho-1}y\\
	Q^{n\rho}(Nx) &= N(Q^nx) + \sum_{i<j, i+j = 2n}
	\mathrm{tr}_{n\rho}(Q^ix \otimes Q^jx)\\
	Q^{n\rho -1}(Nx) &=\sum_{i<j, i+j=2n-1} \mathrm{tr}_{n\rho-1}
	(Q^ix\otimes Q^jy)
	\end{align*}
\end{corollary}

There are internal versions of these formulas if the maps
$N(A^u) \to \mathrm{Sym}^{[2]}(A) \to A$ and
$A^{\otimes 2} \to \mathrm{Sym}^{[2]}(A) \to A$ commute
with the equivariant symmetric multiplication.

\subsection{Adem relations} \label{ssec:adem}

The author's motivation for the change in our
treatment of power operations was to prove the Adem relations. We state
them first in a form reminscent of Bullett-Macdonald \cite{bullett-macdonald},
Steiner \cite{steiner}, and Bisson-Joyal \cite{bisson-joyal},
but we need some preliminaries.

\begin{definition} An \textbf{Adem object} is a $k$-module
$A$ equipped with an equivariant symmetric multiplication
such that there exists a dotted arrow
	\[
	\xymatrix{
	\mathrm{Sym}^{[2]}(\mathrm{Sym}^{[2]}(A)) \ar[r]\ar[d]&
	\mathrm{Sym}^{[2]}(A) \ar[d]\\
	\mathrm{Sym}^{[4]}(A) \ar@{-->}[r] & A
	}
	\]
making the diagram commute up to homotopy.
\end{definition}

\begin{definition} Given $Q(s,t): A_{\star} \to
A_{\star}(\!(s,t)\!)$ we extend to a ring homomorphism
	\[
	Q(s,t): A_{\star}(\!(c,d)\!) \to A_{\star}(\!(c,s,d,t)\!)
	\]
by the rule:
	\[
	Q(s,t)c = c+dst^{-1}, \quad Q(s,t)d = d+d^2t^{-1}.
	\]
Here the target ring is defined as:
	\[
	A_{\star}(\!(c,s,d,t)\!) :=
	\left(A_{\star}[c,d]\llbracket d,t\rrbracket/(c^2=ac+ud,
	s^2=as+ut)\right)[d^{-1},t^{-1}]
	\]
and the operation lands in this ring because of the vanishing
property in Theorem \ref{thm:properties}(\ref{vanishing}).
\end{definition}

\begin{theorem}[Adem relations] \label{thm:adem} Suppose $A$ is an Adem object.
Then, for any $x \in A$, the series
	\[
	Q(s,t) \circ Q(c,d) x \in A_{\star}(\!(c,s,d,t)\!)
	\]
is invariant under the transformation $c \leftrightarrow s$,
$d \leftrightarrow t$.
\end{theorem}

Comparing coefficients of $s^{\varepsilon}c^{\varepsilon'}$ we get
four different identities between series in $d$ and $t$. Using
a residue argument analogous to that in \cite{bullett-macdonald}
or \cite[\S 1]{bisson-joyal}, we can extract the following:

\begin{corollary}[Explicit Adem relations] With assumptions as above,
we have (with $x$ omitted):
	\begin{align*}
	Q^{i\rho -1}Q^{j\rho -1} &=
	\sum \binom{\ell-j-1}{2\ell-i}Q^{(i+j-\ell)\rho-1}Q^{\ell \rho -1}
	\\
	Q^{i\rho}Q^{j\rho} &=
	\sum \binom{\ell-j-1}{2\ell-i}Q^{(i+j-\ell)\rho} Q^{\ell\rho}
	+ u \sum
	\binom{\ell-j-1}{2\ell-1-i}Q^{(i+j-\ell+1)\rho -1}Q^{\ell\rho -1}
	\\
	Q^{i\rho -1}Q^{j\rho}&=
	\sum \binom{\ell-j-1}{2\ell-i}Q^{(i+j-\ell)\rho -1}Q^{\ell\rho} +
	a\sum \binom{\ell-j-1}{2\ell-i-1}Q^{(i+j-\ell+1)\rho-1}Q^{\ell\rho-1}\\
	Q^{i\rho}Q^{j\rho -1} &=
	\sum_{\ell} \binom{\ell-j}{2\ell-i} Q^{(i+j-\ell)\rho-1}Q^{\ell\rho}
	+\sum_{\ell}\binom{\ell-j}{2\ell+1-i}Q^{(i+j-\ell-1)\rho}Q^{(\ell+1)\rho-1}
	\\&+ a\sum \binom{\ell-j}{2\ell+1-i} Q^{(i+j-\ell)\rho-1}Q^{(\ell+1)\rho-1}
	\end{align*}
\end{corollary}

\subsection{Nishida relations} \label{ssec:nishida}

Recall that Hu and Kriz \cite{hu-kriz} computed the equivariant
dual Steenrod algebra to be
	\[
	\mathcal{A}^{C_2}_{\star} := \pi_{\star}(k \wedge k)
	= k_{\star}[\tau_i, \xi_{i+1}| i\ge 0]/(\tau_i^2 = (u+a\tau_0)\xi_{i+1} +
	a\tau_{i+1})
	\]
for certain generators with
	\[
	|\tau_i| = 2^i\rho -\sigma, \quad |\xi_i| = (2^i-1)\rho.
	\]
If $X$ is a $C_2$-spectrum then $k \wedge X$ is a equipped with a
right coaction
	\[
	k_{\star}X \to (k_{\star}X) \otimes_{k_{\star}} \mathcal{A}^{C_2}_{\star}
	\]
arising from the unit $k \wedge X \to k \wedge X \wedge k$. The Nishida
relations describe the relationship between power operations on
$k_{\star}X$, if they exist, and this coaction.

The Nishida relations from \cite{wilson-old} are correct as stated,
but we restate them here in a slightly different form. We will use formal
series:
	\[
	\tau(s,t) = s + \sum_{i\ge 0} \tau_i t^{2^i},
	\,\, \xi(t) = t+ \sum_{i\ge 1}t^{2^i}
	\]
as well as similar series for the conjugates
$\overline{\tau}(s,t)$ and $\overline{\xi}(t)$. Note
$\tau(s,t)+s$ is a power series in $t$.

\begin{proposition}\label{prop:nishida} Suppose $X$ is a $C_2$-spectrum
equipped with an equivariant symmetric multiplication
and $x \in \pi_{\star}(k \wedge X)$ is arbitrary. Then
	\[
	\sum \psi_R(Q^{i\rho-1}x) st^{i-1} +
	\sum \psi_R(Q^{i\rho}x) t^{i} =
	Q(\overline{\tau}(s,t), \overline{\xi}(t))\psi_R(x).
	\]
\end{proposition}

\begin{remark} The theorem applies more generally to
limits of spectra of the form $k \wedge X$ equipped with
the corresponding \emph{completed} right coaction,
e.g. to $k^{h\Sigma_{[2]}}$.
\end{remark}

Comparing coefficients yields the formulas:

\begin{corollary} With assumptions as above,
	\[
	\sum_i \psi_R(Q^{i\rho-1}x) t^{i-1} = \sum_r Q^{r\rho-1}(\psi_R(x))
	\overline{\xi}(t)^{r-1}
	\]
	\[
	\sum_i \psi_R(Q^{i\rho}x) t^i =
	\sum_r Q^{r\rho}(\psi_r(x))\overline{\xi}(t)^r +
	\sum_r Q^{r\rho-1}(\psi_R(x)) (\overline{\tau}(s,t) + s)\overline{\xi}(t)^{r-1}
	\]
\end{corollary}

\subsection{Action on dual Steenrod algebra} \label{ssec:action}

In \cite{wilson-old} we gave a partial computation of the
action of the power operations on the dual Steenrod algebra.
The proof given was incomplete, but the results are correct.
Now we are able to state a \emph{complete} calculation of the
action.

\begin{theorem}\label{thm:action} The total power operation on
$\mathcal{A}^{C_2}_{\star}$
is uniquely determined by the identities:
	\[
	\tau(c,d) + \xi(d)\tau(s,t)\xi(t)^{-1} = (c+dst^{-1})+\sum_{i\ge 0}
	(Q(s,t)\tau_i)(d^{2^i}+d^{2^{i+1}}t^{-2^i})
	\]
	\[
	\xi(d) + \xi(d)^2\xi(t)^{-1} =
	\sum_{i\ge 0}(Q(s,t)\xi_i)(d^{2^i} + d^{2^{i+1}}t^{-2^i})
	\]
\end{theorem}

Comparing coefficients leads to a recursive description of
$Q(s,t)\tau_i$ and $Q(s,t)\xi_i$ which can be solved explicitly.
We record this solution and a few other more explicit corollaries
of the above formula, including the main result from \cite{wilson-old}.

\begin{corollary} The action of power operations on
$\tau_n$ and $\xi_n$ is given by the identites:
	\begin{align*}
	t^{2^n}\sum_{r} Q^{r\rho}(\tau_n) t^r &=
	\left(\sum_{i\ge n+1}\tau_i t^{2^i}\right) +
	(\tau(s,t)+s)\xi(t)^{-1}\left(\sum_{i\ge n+1}\xi_i t^{2^i}\right)\\
	t^{2^n}\sum_r Q^{r\rho -1}(\tau_n) t^{r-1} &=
	\xi(t)^{-1}\left(\sum_{i\ge n+1}\xi_i t^{2^i}\right)\\
	t^{2^n}\sum_r Q^{r\rho}(\xi_n)t^r &= \sum_{i\ge n+1}\xi_i t^{2^i}
	+\xi(t)^{-1}\sum_{i\ge n} \xi_i^2t^{2^{i+1}}\\
	\sum_r Q^{r\rho-1}(\xi_n)t^{r-1} &= 0
	\end{align*}
\end{corollary}

\begin{corollary}
	The following formulas, for $k\ge 1$, hold for the action on
	$\tau_0$:
	\begin{align*}
	Q^{(2^k-1)\rho}\tau_0 &= \overline{\tau}_k, \\
	Q^{(2^k-1)\rho -1}\tau_0 &= \overline{\xi}_k.
	\end{align*}
\end{corollary}

\begin{corollary}
	The following formulas hold for the action on
	$\tau_i, \xi_i$, and their conjugates:
	\begin{align*}
	Q^{2^k\rho}\tau_k &= \tau_{k+1} + \tau_0\xi_{k+1}
	& Q^{2^k\rho}\overline{\tau}_k &=\overline{\tau}_{k+1} \\
	Q^{2^k\rho -1}\tau_k &= \xi_{k+1}
	& Q^{2^k\rho -1}\overline{\tau}_k &=\overline{\xi}_{k+1}\\
	Q^{2^k\rho}\xi_k &= \xi_{k+1} + \xi_1\xi_{k}^2
	& Q^{2^k\rho}\overline{\xi}_k &= \overline{\xi}_{k+1}
	\end{align*}
\end{corollary}

As an algebraic corollary, used in \cite{behrens-wilson}, we have:

\begin{corollary} The algebra $\mathcal{A}^{C_2}_{\star}$ is generated
by $\tau_0$ as a ring equipped with the operations $Q^{i\rho-\varepsilon}$.
More specifically, the elements $Q^{2^k\rho}Q^{2^{k-1}\rho}\cdots Q^{\rho}\tau_0$
and their Bocksteins generate $\mathcal{A}^{C_2}_{\star}$.
\end{corollary}

\section{Sketch of the proof}\label{sec:sketch}

\subsection{Review of the classical case}\label{ssec:review}
We begin by reviewing the classical setting, with some modifications.
It is possible \cite{wilson-classical}
to develop the entire theory of mod 2 power operations based
on the following categorical inputs:
	\begin{enumerate}[(A)]
	\item (Universal property of the Tate construction)
	\cite[\S I.3]{nikolaus-scholze}
	Let $\mathcal{F}$ be a collection of subgroups of
	a group $G$, closed under sub-conjugacy. Then the
	generalized Tate construction \cite[p.443]{greenlees1}
		\[
		(-)^{t\mathcal{F}}: \mathsf{Sp}^{hG} \to \mathsf{Sp}
		\]
	has the following universal property: the map
	$(-)^{hG} \to (-)^{t\mathcal{F}}$ is initial amongst
	all natural transformations of exact, lax symmetric monoidal
	functors whose target annihilates the stable subcategory
	of objects induced from subgroups
	$H \in \mathcal{F}$.
	\item (Universal property of the forgetful functor)
	\cite{nikolaus}
	The forgetful functor $U: \mathsf{Mod}_{\mathbb{F}_2}
	\to \mathsf{Sp}$ is initial amongst all exact, lax symmetric
	monoidal functors from $\mathsf{Mod}_{\mathbb{F}_2}$
	to $\mathsf{Sp}$.
	\item (Universal property of Tate powers) Let $G \subseteq \Sigma_n$
	be a subgroup and $\mathcal{T}$ the family of non-transitive subgroups
	of $G$. Define the Tate powers by
	$T_G(X):= (X^{\otimes n})^{t\mathcal{T}}$ and the divided power by
	$\Gamma^G(X):= (X^{\otimes n})^{hG}$.
	Then the map $\Gamma^G \to T_G$ is initial amongst natural transformations
	from $\Gamma^G$ to an exact functor, i.e. $T_G$ is the Goodwillie derivative
	of $\Gamma^G$ and is, in particular, exact. We do not
	know a reference that states the result in these words, but the
	actual calculation is well-known; we spell this out in
	\cite[2.2.1]{wilson-classical}.
	\end{enumerate}
We also need the following computational inputs:
	\begin{enumerate}[(I)]
	\item As a ring,
		\[
		\pi_*\mathbb{F}_2^{t\Sigma_2} = \mathbb{F}_2(\!(w)\!)
		\]
	where $w$ is the first Stiefel-Whitney class.
	\item There is a canonical
	equivalence $(\Sigma^{k}\mathbb{F}_2)^{\otimes 2}
	\simeq \Sigma^{2k}\mathbb{F}_2$ in $\mathsf{Mod}_{\mathbb{F}_2}^{h\Sigma_2}$.
	\item The map $\mathbb{F}_2 = \mathbb{F}_2 \wedge S^0 \to
	\mathbb{F}_2 \wedge \mathbb{F}_2$ induces, upon taking Tate
	fixed points for the trivial action of $\mathbb{F}_2$, the completed
	coaction:
		\[
		\psi_R: \mathbb{F}_2(\!(w)\!) \to \mathcal{A}_*(\!(w)\!)
		\]
	determined by
		\[
		\psi_R(w) = \sum \zeta_i w^{2^i} =: \zeta(w).
		\]
	\end{enumerate}
We now review how these facts imply the desired properties of
power operations. A more detailed treatment is given in \cite{wilson-classical}.

Suppose we are given an $\mathbb{F}_2$-module $A$ equipped
with a symmetric multiplication $\mathrm{Sym}^2(A) \to A$. By
(A), (B), and (C) there is an essentially unique transformation of
lax symmetric monoidal functors $U \to UT_2$, and so we may form
the \textbf{Tate-valued Frobenius}:
	\[
	A \to T_2(A) = (A^{\otimes 2})^{t\Sigma_2} \to A^{t\Sigma_2}.
	\]
We can then define operations $Q^i: A \to \Sigma^{-i}A$,
using (I), as the composite
	\[
	A \to A^{t\Sigma_2} \stackrel{w^{-i-1}}{\longrightarrow}
	\Sigma^{-i-1}A^{t\Sigma_2} \to \Sigma^{-i}A_{h\Sigma_2} \to \Sigma^{-i}A
	\]
The Frobenius then induces a total power operation
	\[
	Q(w): A_* \to A^{t\Sigma_2}_* \simeq A_*(\!(w)\!)
	\]
given as $Q(w)x = \sum_i (Q^ix) w^i$. The first properties follow
immediately from this definition, with the exception of the squaring
property, which eventually requires the computation (II) above.
The Cartan formula follows from the lax symmetric monoidal structures
on every functor in sight. The Adem relations follow from an important
lemma: every $\mathbb{F}_2$-module equipped with a symmetric multiplication
admits a canonical lift of the iterated Frobenius
	\[
	\xymatrix{
	&A^{\tau\Sigma_2\wr \Sigma_2}\ar[d]\\
	A \ar[r]\ar[ur] & (A^{t\Sigma_2})^{t\Sigma_2}
	}
	\]
where $(-)^{\tau G}$ denotes $(-)^{t\mathcal{T}}$ when $G \subseteq \Sigma_n$
is a subgroup. The proof of this uses the universality assertions,
naturality, and lax symmetric monoidal properties in each of
(A), (B), and (C) to verify that various diagrams commute.

If $A$ is, moreover, an Adem object, then there
is a further lift to $A^{\tau\Sigma_4}$ and this immediately implies
the Adem relations in the form stated by \cite{steiner} after Bullett-Macdonald
\cite{bullett-macdonald}.

The computation (III) immediately implies the Nishida relations
by examining the diagram
	\[
	\xymatrix{
	(\mathbb{F}_2 \wedge X) \ar[r] \ar[d] & (\mathbb{F}_2 \wedge X)
	\otimes (\mathbb{F}_2 \wedge \mathbb{F}_2) \ar[d]\\
	(\mathbb{F}_2 \wedge X)^{t\Sigma_2} \ar[r] &
	\left((\mathbb{F}_2 \wedge X)
	\otimes (\mathbb{F}_2 \wedge \mathbb{F}_2)\right)^{t\Sigma_2}
	}
	\]

Finally, the action on the dual Steenrod algebra is obtained by
applying the Nishida relations, in the form above, to the
element $w \in \pi_{-1}\mathbb{F}_2^{h\Sigma_2}$.

\subsection{The equivariant setting: categorical input}

The $C_2$-equivariant setting proceeds in an identical fashion.
The idea is to first make the following adjustments to the
categorical input:
	\begin{itemize}
	\item Replace the use of $\infty$-categories with
	$C_2$-$\infty$-categories, in the sense of
	\cite{bdgns}.
	\item Replace the use of exact functors with
	$C_2$-exact functors in the sense of \cite[2.14]{nardin}.
	\item Replace the use of symmetric monoidal
	$\infty$-categories and lax symmetric monoidal functors
	with their indexed analogues. One can define
	a $C_2$-symmetric monoidal $\infty$-category
	as a product preserving functor
		\[
		\mathsf{Span}(\mathsf{Fin}^G) \to \mathsf{Cat}_{\infty}
		\]
	but the notion of lax $C_2$-symmetric monoidal functors
	takes a little longer to define.
	A suitable definition can be found in
	\cite[2.1.4.2]{wilson-thesis}.
	\item Replace the group $\Sigma_n$ with the
	$C_2$-groupoid $\Sigma_{[n]}$, i.e. the functor
	$\Sigma_{[n]}: \mathrm{Orbit}(C_2)^{\mathrm{op}} \to
	\mathsf{Gpd}$ which assigns to an orbit $T$ the
	groupoid of maps of $(C_2\times \Sigma_n)$-sets
	$U \to T$ which are $\Sigma_n$ torsors. This records
	the natural symmetries on indexed tensor powers
	in a $C_2$-symmetric monoidal $\infty$-category.
	\end{itemize}
With these changes, the claims (A), (B), and (C) remain true
in the $C_2$-equivariant setting, and the proofs are mostly
the same as their nonequivariant analogues. We note that,
without the symmetric monoidal structures, these statements
are readily provable using tools from \cite{nardin} and \cite{shah}.
However, the additional multiplicative structure requires more set-up.

\subsection{The equivariant setting: computational input}

The computation (I) is replaced
by
	\[
	\pi_{\star}k^{t\Sigma_{[2]}} = k_{\star}(\!(s,t)\!)
	\]
with notation as in \S\ref{ssec:operations}. Here some care is required.
First, recall that there is a $C_2$-space $\mathrm{B}_{C_2}\Sigma_2$
which classifies $\Sigma_2$-torsors on $C_2$-spaces,
and a corresponding universal torsor $\mathrm{E}_{C_2}\Sigma_2$.
The map $\mathrm{E}_{C_2}\Sigma_{2+} \to S^0$ is an equivalence
in $\left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[2]}}$, so we get
equivalences:
	\[
	F(\mathrm{E}_{C_2}\Sigma_{2+}, X)^{h\Sigma_{[2]}} \simeq
	X^{h\Sigma_{[2]}}
	\]
	\[
	(X \wedge \mathrm{E}_{C_2}\Sigma_{2+})_{h\Sigma_{[2]}}
	\simeq X_{h\Sigma_{[2]}}
	\]
for any $X \in \left(\mathsf{Sp}^{C_2}\right)^{h\Sigma_{[2]}}$. We will
use this to define a filtration on these functors.

If we let
$\tau$ denote the sign representation of $\Sigma_2$, then an explicit
model for $\mathrm{E}_{C_2}\Sigma_2$
is the unit sphere $S(\infty(\rho\otimes \tau))$
in the $C_2\times\Sigma_2$-representation $(\rho\otimes \tau)^{\oplus \infty}$.
The filtration
	\[
	\Sigma_{2+} = S(1\otimes \tau) \subseteq
	S(\rho \otimes \tau) \subseteq S((\rho+1)\otimes \tau)
	\subseteq \cdots
	\]
then produces a filtration on the functors $(-)^{h\Sigma_{[2]}}$ and
$(-)_{h\Sigma_{[2]}}$. Now, in general, if $\lambda$ is a one-dimensional
representation and $W$ is any representation, then the map
	\[
	S(\lambda) \times \mathrm{Hom}(\lambda, W) \to
	S(\lambda \oplus W) - S(W), \quad (x,\phi) \mapsto
	\frac{(x,\phi(x))}{|(x,\phi(x))|}
	\]
is an equivariant homeomorphism and so induces an equivalence
	\[
	S(\lambda)_+ \wedge S^{\lambda^{\vee}\otimes W} \simeq
	S(\lambda \oplus W)/S(W)
	\]
upon taking one-point compactifications. Applying this to $S(\tau)=\Sigma_2$
and $S(\sigma\otimes \tau) = \frac{C_2\times\Sigma_2}{\Delta}$, where
$\Delta$ is the diagonal subgroup, we learn that:
	\[
	\mathrm{gr}_{2j}\mathrm{Sym}^{[2]}(M) \simeq \Sigma^{j\rho}M^{\otimes 2},
	\]
	\[
	\mathrm{gr}_{2j+1}\mathrm{Sym}^{[2]}(M) \simeq \Sigma^{(j+1)\rho -1}
	N(M).
	\]
Unlike the classical setting, this filtration does not split in general.
For example, it does not split when $M$ is of the form $\Sigma^{j\rho +1}k$
or $\Sigma^{j\rho -2}k$ (see below \S\ref{ssec:counterexample}).

However, in the case $M=k$ or, more generally, $M=\Sigma^{j\rho-\varepsilon}k$
for $\varepsilon=0,1$, the filtration splits for degree reasons stemming
from a computation of the homology of a point, $k_{\star}$.
This gives an additive calculation
	\[
	k_{h\Sigma_{[2]}} \simeq \bigoplus_{j\ge 0}
	\Sigma^{j\rho}k \oplus \bigoplus_{j\ge 1}
	\Sigma^{j\rho-1}k
	\]
Dualizing produces an additive calculation for $k^{h\Sigma_{[2]}}$ and it
is not difficult to prove that the ring structure is given by
	\[
	k^{h\Sigma_2}_{\star}= k_{\star}[s]\llbracket t\rrbracket/(s^2=as+ut)
	\]
This calculation was obtained in a different way by Hu-Kriz \cite[6.27]{hu-kriz},
but the filtration above is useful in other contexts.

\begin{remark} Here there is a subtle point: there are two possible generators in degree
$-\sigma$ which satisfy the above relation.
The choice is determined by restriction to
$S(\rho\otimes \tau)/\Sigma_2 = S^{\sigma} \subseteq \mathrm{B}_{C_2}\Sigma_2$.
It is standard to choose the generator which restricts to
$1 \in \pi_0k =\pi_{-\sigma}\Sigma^{-\sigma}k$. We will \emph{not} make that choice.
Instead we choose $s$ to be the generator which restricts to $a$. The other
generator is given by
	\[
	\overline{s}:= s+a.
	\]
These two generators behave in an almost identical fashion, but with our
choice of generator the map
	\[
	k^{t\Sigma_{[2]}} \to \Sigma k_{h\Sigma_{[2]}} \to \Sigma k
	\]
will, on homotopy, record the coefficient of $st^{-1}$ (as opposed to $(s+a)t^{-1}$),
which is more convenient.
\end{remark}

To finish the analogue of (I), one checks that the Tate construction inverts the
class $t$. The filtration can also be used to show that, for $M$ equipped with
trivial $\Sigma_2$-action,
	\[
	M^{t\Sigma_{[2]}}_{\star} = M_{\star}(\!(s,t)\!)
	\]
even if $M$ is not free as a $k$-module. An equivariant symmetric multiplication
on a $k$-module $A$ is now enough to form the
\textbf{equivariant Tate-valued Frobenius}:
	\[
	A \to T_{[2]}(A):= (A^{\otimes 2})^{t\Sigma_{[2]}} \to A^{t\Sigma_{[2]}}.
	\]
\begin{construction}\label{cstr:operations}
The individual power operations are defined as the composites:
	\[
	Q^{i\rho -\varepsilon}:
	A \to A^{t\Sigma_{[2]}} \stackrel{\overline{s}^{\varepsilon}t^{-i-1+\varepsilon}}{\longrightarrow}
	\Sigma^{-i\rho -1+\varepsilon}A^{t\Sigma_{[2]}} \to \Sigma^{-i\rho+\varepsilon}
	A_{h\Sigma_{[2]}} \to \Sigma^{-i\rho+\varepsilon}A.
	\]
\end{construction}

The analogous computation to (II) is an equivalence
$(\Sigma^{j\rho}k)^{\otimes 2} \simeq \Sigma^{2j\rho}k$ in
$\mathsf{Mod}^{h\Sigma_{[2]}}_k$. One can show this either
by establishing the orientability of a bundle on $\mathrm{B}_{C_2}\Sigma_2$
or using the computation of $\mathrm{Sym}^2(\Sigma^{j\rho}k)$ referenced above.
The analogous fact does not hold for other representations, but this
turns out to suffice. Finally, the analogue of (III) is essentially the definition
of the Milnor generators from Hu-Kriz \cite[p.381]{hu-kriz}. The main theorems
in \S\ref{sec:results} now follow as in \S\ref{ssec:review}.

\subsection{A counterexample}\label{ssec:counterexample}

In the previous section we constructed a filtration on $\mathrm{Sym}^{[2]}(M)$
and remarked that it does not split in general. Let us explain what happens
in two examples to see how things differ from the classical case.

First, in the case $M=\Sigma^{n\rho +1}k$ it turns out that, while the
given filtration does not split, we can still describe the answer:
\begin{proposition}
	\[
	\mathrm{Sym}^{[2]}(\Sigma^{n\rho+1}k)
	\simeq
	\bigoplus_{j\ge n+1} \Sigma^{(n+j)\rho+1}k \oplus
	\bigoplus_{j\ge n+1}\Sigma^{(n+j)\rho}k
	\]
\end{proposition}

This situation is not too bad because the generators can actually
be described in terms of the power operations we have.
That is, we can choose generators so that, if $A$ is equipped with
an equivariant symmetric multiplication, and $x: \Sigma^{n\rho+1}k \to A$
is a class, then the image of the generator in degree $(n\rho+1)+(j\rho -\varepsilon)$
under the map
	\[
	\mathrm{Sym}^{[2]}(\Sigma^{n\rho+1}k) \to \mathrm{Sym}^{[2]}(A) \to A
	\]
is equal to $Q^{j\rho-\varepsilon}x$.

We also have a computation of $\mathrm{Sym}^{[2]}(\Sigma^{n\rho -2}k)$:
\begin{proposition}
	\[
	\mathrm{Sym}^{[2]}(\Sigma^{n\rho-2}k)
	\simeq \Sigma^{2n\rho-4}k \oplus
	\bigoplus_{j\ge n} \Sigma^{(n+j)\rho -2}k \oplus
	\bigoplus_{j\ge n}\Sigma^{(n+j)\rho-3}k
	\]
\end{proposition}

In this case, most classes will correspond to named power operations,
but the bottom class cannot, for degree reasons. In other words:
there is no stable operation which refines the square $x \mapsto x^2$
for classes in degree $n\rho -2$. We imagine the situation becomes
worse the further one strays from regular representations.

redpajama_set_name: RedPajamaArXiv


text:
X-Men: Dark Phoenix: Jessica Chastain will be playing…
Warning: Mild spoilers ahead.
Diana Othman
The talented and beautiful Oscar-nominated thespian Jessica Chastain and us! Photo credit: Diana Othman for Geek Crusade.
When it was first revealed earlier this year that Golden Globe nominee Jessica Chastain was joining the cast of X-Men: Dark Phoenix, it was widely assumed that she would be playing the Shi'ar Queen Lilandra Neramani
For one, she and pal James McAvoy, who is reprising his role as Charles Xavier, exchanged cheeky social media messages to one another as a fun way of announcing her involvement. And every X-fan knows that Lilandra is the professor's sometime lover and ally.
And secondly, it's an adaptation of the Dark Phoenix saga, man! So the Shi'ar Empire, an alien race ruled by Lilandra that crosses path with the Phoenix force, is bound to be featured in the movie, right? But Chastain threw us off by surreptitiously commenting on her Instagram that she is, in fact, not playing the Shi'ar empress.
But it looks like Entertainment Weekly may have just let slip the character she will be playing. In a massive exclusive, EW released several first-look images of the wildly anticipated X-Men sequel including one of Chastain in character below:
Jessica Chastain's mystery role in X-Men: Dark Phoenix is said to be a manipulative alien shape-shifter. Photo: Entertainment Weekly.
While much of the article touches on new plot details and Sophie Turner's role as Jean Grey and subsequently, Phoenix, it also states that the movie "introduces an alien shape-shifter, played by Jessica Chastain, who tries to manipulate Phoenix for her own secret agenda."
An alien shape-shifter from Marvel's cosmic universe? Is Chastain playing Veranke, the Skrull Queen?! Even though EW didn't officially mention the character by name, fans are willing to bet their honour that this is indeed Chastain's role. But I have to say, at this point, there could be a lot more to this than meets the eye.
If Chastain really is playing the Skrull Queen, there are two issues that are keeping us X-fans up at night. Firstly, aren't Skrulls going to be featured in the upcoming Captain Marvel movie which is being developed at Disney? When have the two separate studios – Disney and 20th Century Fox, which owns the movie rights to the X-Men – ever shared their Marvel assets? Unless, Disney really is close to buying over 20th Century Fox and the studio's entertainment assets including the X-Men series of films, as we reported earlier.
Secondly, what exactly are Skrulls doing in X-Men: Dark Phoenix? In the original comics series, the only alien race that plays a significant part in the story is the Shi'ar Empire. It has to be said though that the Skrulls' empire is situated quite close to the Shi'ar Empire. I just hope Bryan Singer and team didn't just throw in a war between the two races to try and make the movie more "epic" because that would just make it too messy.
What do you think, geeks? Is Chastain going Skrull in X-Men: Dark Phoenix? Or what other role do you think she might be playing?
Previous articlePIX: A glimpse of Fantastic Beasts 2!
Next articleFIRST LOOK: Sophie Turner smoulders as Dark Phoenix
Curly haired and Crazy. About Movies.
Molly's Game review: Jessica Chastain is fierce as real life poker queen
FIRST LOOK: Sophie Turner smoulders as Dark Phoenix
James Franco's Multiple Man may well be a hard R

redpajama_set_name: RedPajamaCommonCrawl


Where can i look for a native speaking paper author?
Essays will be definitely composed by scholars later or sooner in their curriculum.
Best write my essay service which guarantees timely shipping. Writing services are so popular these days there's a superb chance they could direct you in the proper direction. The companies we 've reviewed and have included on our list provide a warranty on all.
will finish your assignment based on the academic criteria that are best.
In locating the appropriate service to buy essay on the internet the point is that students wish to choose which the best one is. Whenever there's a need to dictate for an argumentative essay instance term and research paper, and you would love to locate a expert provider of these services, you will need to think about a few nuances. Has existed for years, helping learners.
Our custom made paper authors are prepared to assist you with essay writing of any kind and difficulty. You therefore are well on your way to writing a paper that was wonderful in the forseeable future and learned how to improve in composing essays. Simply supply the writer with also the date by which you need your paper and the details.
solution for you. Kind of service is utilized to spot the.
The severe point in locating the service that is proper to buy essay on the net is that on which the complete best one is students wish to decide. When there's a need to dictate for instance an argumentative essay, research and term paper, and you'd like to locate a expert provider of such services, you will have to think about several nuances. To learn more about how to locate a essay writing service's purchase price browse the inputs to the webpage.
As you aren't able to compose your own assignments the thought of growing specialist writer appears possible. Together with our affordable essay writing service, you will not need to be concerned about not being in a position since you're likely to be working together with our specialist writers to complete your newspaper professionally or punctually. If you're looking for a professional writer to compose my own essays', you've located a website to put an order!
Our payment method is secure and you may pick any payment alternative suitable to you. When the purpose has been identified, now you can begin exploring on topics which you find intriguing. Therefore, according to our policies, you're entirely eligible to acquire your money back.
Provides you a exceptional chance of getting your essay. It's likely to order a services then put an order for a article of work if you're fighting to compose an essay. Proceed to the purchase page and select which type of paper you expect from us.
us, you are going to find the best article ASAP. Essays play a major role to comprehend the subject. If you don't know how to compose a superb analytical essay, then adhere to the tips you can.
Education is vital, and every mission is the major concern of student. In the event of unsatisfactory documents modifications can be requested by the client through guarantee and our revision coverage. Whether your office's policy isn't unenforceable under regulation depends upon the state.
You're surely not meant to neglect on your studies, which is MeowEssay professional writing service is the solution for you. People ask why our service is a popular with this sort of a huge number of students. It has an excellent reputation and a large number of clients.
College papers don't always provide you with the quality. The work that you are able to do onscreen, the paper you will need to print. Composing a superb paper may include its share of challenges together with moment like research.
You are able to get a inexpensive essay from our group of expert writers. Our crew of writers is always available to provide the assistance you need irrespective of whether you will require a person proofread to you, or edit it to discover the best outcomes. Possessing the knowledge that is extensive in the disciplines that are several, our writers at the several fields are prepared to create essays that are fine and distinctive.
This entry was posted on Saturday, April 13th, 2019 at 12:40 am and is filed under Uncategorized.	You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.


text:
Eurovision NewsOddsCalendarSongs & VideosRotterdam 2020TicketsResults & Points 2019FactsOGAEQuizselect year/country
Photo: KAN Israeli Public Broadcasting Corporation
Vote for friendship comedy: Israel offers Cyprus voting alliance
As part of a new comedy video, Israel is reaching out to Cyprus with a special Eurovision offer
By Sagi Klein 3 January 2019
Tags:IsraelCyprus
The Israeli Public Broadcaster KAN published a comedy video, with the ambition to win again Eurovision Song Contest. As part of the comic video, the presenter Elia Grinfeld, suggesting his plan to KAN for a "Double" Eurovision victory, by offering Cyprus a Voting Alliance. Grinfeld explaining that many countries vote for their neighboring countries, while Israel doesn't have any Eurovision-partcipating neighbors (by land).
Geographically, Cyprus is the closest country to Israel among all the countries participating in the Eurovision Song Contest and it's considered by Israel as a close friend. KAN posted the clip in it's official YouTube channel, with the option of Greek subtitles.
Israel already won Eurovision two years in a row, when it won the first time in 1978 with the song "A-Ba-Ni-Bi" and again in 1979 with the song "Hallelujah". Back then, Israel hosted the contest in 1979, in Jerusalem, but when Israel won again in 1979, it was decided to give up thier right to host the contest in 1980.
KAN offering Cyprus voting alliance (subtitles offered also in Greek)
About the author: Sagi Klein (Israel)
Sagi Klein is our correspondent from Israel. Sagi sees Eurovision as a unique platform that brings people together and bridging differences between them.
IsraelCyprus
Found an error in the text? Please notify us
Italy boycotted the 1981 Eurovision Song Contest, saying that it was too old fashioned
Eurovision Song Contest is one of the longest running recurring television broadcasts in the world
select year or country
Eurovision Results, Voting & Points
Complete database with all the votings, points, songs and lyrics from Eurovision Song Contest history:
In the first ever Eurovision Song Contest (1956), Luxembourg asked Switzerland to vote on its behalf. And the winner was: Switzerland!


EvilValentine CharmingBunny jasmin. LiNaLi69. Demigalvan. MelenaURlove.
STARKJHOSEPmaagda98ErnnestoANDRENALYNExxx .SEXTANKAMarryJolieTattoMoonSEXTANKA .LiNaLi69AuriziysNatusika12Darejan .tesztbtsStrawberry1KissANDRENALYNExxxPerfectQueenX .danielarodriguezSweetDanielleVKatyKeatSweetDanielleV .KatyKeatDamianBlackDickMiahNitaTheSchoolGirlx .VirginiaaCassieTSSEXTANKABIGSTRONGMEN .VanessaPrincessSashaSophieSofiaBBWKatyKeat .1LonelyMoonNatusika12eradaxxWildJustyne .PerfectQueenXTheSchoolGirlxNatusika12maagda98 .MELINAMULIEREAdamMonsterDickMANMANMANStefanye .
CassieTSEstebanHandsomealessiamitch0BigSquirt4U .UrFavXDollTheSchoolGirlxWildJustyneCassieTS .ALPHAxMALEStefanyeBIGSTRONGMENKarolino .HottyMartaSweetDanielleVLiNaLi69VanessaPrincess .OhPamelaFruiteradaxxLiNaLi69NATHANX7 .kayleesmokeAnastasiaAnthemANDRENALYNExxxRubyStyles .EstebanHandsome1LonelyMoonTattoMoonTheSchoolGirlx .SEXTANKAGabyOsbornexLoveMeHarderxPhoebiena .RubyStylesMellanieHalleBlackHotx4uErnnesto .LaVieEstBelleSashaSophieBlueGypsyNinaLeiX .UrLifeLUANNAHOT23xLoveMeHarderxJaneBurgess .


Rathcline takes its name from the townland and parish of Rathcline. It is located in southwest County Longford, to the east of Lough Ree and north of the River Inny. Rathcline barony was formed from the territories of The Callow, Lanesboro to Ballymahon) and the territory of Clanconnor .Rathcline Castle was built around the 9th Century by the Ó Cuinn (O'Quinn) clan. Later it was fought for and taken by the Uí Fhearghail (O'Farrell) clan and subsequently taken over by the Normans around the end of the 12th Century.
The Corlea Iron Age bog road which was built in 148 BC across the boglands in proximity to the River Shannon. The oak road is the largest of its kind to have been uncovered in Europe and can be seen at the Visitor Centre (Keenagh).


text:
Archive for the 'Film' Category
October 17, 2012 @ 6:28 pm | Filed under: Film
A couple of months back, I wrote the following note to my friend Ron, at whose urging Scott and I had watched The Tree of Life. "I could write a thousand words about the film," I'd warned him the day after we saw it, and a day or two later I did go on for nearly that long—an unpolished meditation on the movie, which I'm pasting below.
Tree of Life…my thousand words aren't going to happen in any kind of articulate order, not with all the other writing sitting on top of me. But I'm still thinking about it, in that close-your-eyes-and-replay-a-moment way that doesn't happen often. I owe you thanks—if not for you I might not have hung in long enough, I'd have been like 'this is lovely but…'—I'm aware, and mostly unapologetic, that my film tastes have degraded a bit toward a preference for easy entertainment that doesn't make me work. (With TV I'm more discerning, because it's shorter. I enjoy smart, difficult TV—but a show is usually under an hour. The fact is that at the end of the day, I'm completely done in and films are long. Hard, under those circumstances. I'd feel guilty about it except that I know I'm still pushing myself to read challenging things and to think and assimilate. By 9pm, when we have a chance to sit down alone in front of the television, I'm just…done.)
But I wanted to see what moved you so deeply about it, so I hung in there, like physically sat up so I wouldn't drift off…and oh! Both of us were sort of breathless about it. I felt my brain make a shift in the middle. At first I felt somewhat resentful or impatient with the long silences, even as I acknowledged that they were visually quite wonderful, not just the lovely images but the range of thoughts & emotions passing over an actor's face while the camera lingered long, long, long in a particular shot—resentful because so little was being said overtly, and I knew I was being asked to think, to write much of the internal dialogue myself, to read the symbols. It irked me a little; I don't come to films to do more writing. But gradually, I gave myself over to what the film was asking of me, and it was one of those rare transformative experiences where you know you're walking to bed a slightly different person than you woke up.
Am intrigued by the conflict presented in the opening, nature vs. grace, and the way the images belied the words. She's talking about 'the way of nature' (human nature) and 'the way of grace' (presented as opposing or contradictory paths) but on the word 'grace' we get that glorious shot of sunflowers (flowers that turn their faces toward the light)…and all through the rest of the film, it's like that; grace is always shown via a sublime natural image. Looking up at the sky through trees means reaching for God. Touching the grass is the moment the grown son begins to heal & reconnect with the spiritual. Water seemed to mean the current of life carrying us forward (toward the churning waterfall, death), always with overtones of death, a hint of something inescapable to come. Bathing the baby, the baptism, the incident at the pool. Water, too, was almost always in the best, happiest family moments—the spraying of the hose, usually the mother, once the father—though he carried the hose with a kink in it to stop the water flowing out, exerting control (fruitless in the end)—such a strong image there, of his attempting to control his life, his family, and being ultimately powerless to.
In many ways the father seemed to represent a kind of Old Testament God —stern, demanding, controlling, punitive, never satisfied, requiring obeisance and craving love, almost demanding love. Through the mother and the imagery surrounding her, we see nature and harmony and beauty, a constant yearning upward towards the sublime, a bit of a mother earth/mother goddess thing going on there, except powerless, easily overpowered or dominated by man, so more like simply earth.
The white linens blowing in the windows, nearly always associated with the mother, gentleness, peace, happiness—sometimes the mother and children play in the curtains, draping them over their faces. (Suggesting a shroud? Peaceful death?) Very often, the curtains are moved by wind—the breath of God, inspiration, the 'moving of the spirit.'
I spent a long time wondering what Terrence Malick's views are, and what he was trying to suggest….a belief in an active Spirit or Force? Certainly his characters expressed a yearning upward, a longing for there to be some meaning or purpose behind the seemingly random chain of creation (and oh how I loved those parts of the film, the millennia of life surging forward)—and the delicate and lovely beach reunions at the end would suggest a belief in soul, redemption, reconciliation. But always I felt a tension between what was being said and what was being shown. The moments of purest joy were always connected to nature. The 'way of grace' was to walk in the natural world, to delight like children in sensual experience. In the boys' happiest moments they are tumbling around like puppies, like wolf cubs, or running, exulting in the sheer grace of the human body. (Especially in that one powerful scene when the father was away on business and the boys tore through the house, celebrating his absence, bursting outside into the yard, letting the door bang behind them—pure joy in their freedom.) The mother, too, running with them, being playful with them.
There's lots more to say, I'm not doing more than pouring out notes here, but then again it's barely six a.m. and it's a wonder I'm thinking at all.
Tags: film, Terrence Malick, The Tree of Life
A Million Lights Are Dancing & There You Are, A Shooting Star
February 14, 2011 @ 8:36 am | Filed under: Film
Here is what love is:
He watched this movie with me because it jumped back into my mind, full force, every single lyric intact, my entire being once more suffused with a yearning to wear ribbon barrettes and flowing skirts while dancing on roller skates, and I couldn't live another night without having all that synthesizery goodbadness etched into his brain too.
He endured every terrible, terrible line of dialogue and crater-sized plot hole, and he even found the "Whenever You're Away from Me" number as charming as I do.
This clip is NOT that number. "Charming" is not the word for it. A Giant Glittering Mess of Audio-Visual Cheesepuffery is what this is.
Worth noting: we realized that the same summer I watched Xanadu on an endless loop, Scott watched Kramer vs. Kramer 37 times. This may explain a lot about the way each of us turned out.
December 1, 2010 @ 7:23 am | Filed under: Film
An almost certainly incomplete list of the films Jane and Scott have watched in their Film Club this year. Scott's pretty sure there were more John Wayne movies, among others.
(alphabetical, not in order viewed)
Benny and Joon
The Buddy Holly Story
The Cowboys
The Horse Soldiers
Max Dugan Returns
McLintock
Mr. Smith Goes to Washington
Mr. and Mrs. Smith
My Darling Clementine
Ocean's Eleven
Romancing the Stone
Supercop
Trouble Along the Way
You Had Me at "Studio Ghibli Does The Borrowers"
September 29, 2010 @ 6:29 am | Filed under: Books, Film
How did I not know about this? Studio Ghibli's latest release (this summer, in Japan) was The Borrowers?
Official title: Karigurashi no Arrietty (The Borrower Arrietty)
I see (via the SDSU Children's Literature blog) that Miyazaki wrote the screenplay but did not direct the film. Still, my hopes are high.
The post links to an interesting article at the Daily Yomiuri about the slight differences between Arrietty's father in the book and the film:
…while Pod is not particularly loquacious in the novel, neither is he reserved. In a scene after Pod has discovered that Arrietty has spoken to "the boy," Pod speaks quite a bit.
When Arrietty defends herself, saying the boy has agreed to deliver a letter she has written to Borrowers living elsewhere, he appears to take some grim satisfaction in his scornfully elaborate explication of the uselessness of Arrietty's act: "…do you see your mother walking across two fields and a garden…two fields full of crows and cows and horses and what-not, to take a cup of tea with your Aunt Lupy, whom she never much liked anyway?" There is a lavishness in Pod's amplification that can only be achieved through the protracted use of language. Not for him, the effect of understated brevity. Ghibli's Pod, on the other hand, barely speaks unless strictly necessary—and even then he sometimes remains reticent in situations when a tad more back and forth might be deemed obligatory. His lack of words might seem unduly taciturn in a Western context, but, as in the original, he is portrayed as a sympathetic character. Just not a very talkative one.
Tags: I, Studio Ghibli, The Borrowers
The Voyage of the Dawn Treader, the Trailer
June 17, 2010 @ 6:35 am | Filed under: Film
I just watched the trailer for the new Narnia movie. Dawn Treader is is one of my favorite Narnia books and I've been anxious about the movie; so much potential for getting it wrong; so many things I desperately want them to get right.
I don't know…some worrisome glimpses there. Looks like they've added a conflict subplot for Edmund—back in England, the war is on, and they won't let a mere "squirt" join up. "But I'm a king!" he huffs to Lucy. Argh. Even worse, later in the trailer the White Witch appears in some sort of vision to tempt him. Really? Really? Edmund is so beyond that. After his fall and redemption in LWW, he's one of the staunchest, most honorable young men in either world.
Equally puzzling: Eustace is barely present in the trailer. All the focus is on Edmund and Lucy, and Ian McKellan's*,** voice uttering vague yet grand pronouncements about their adventure just beginning. No dragon. Scarcely any indication that Eustace is along for the journey at all. Perhaps in this early trailer, they're targeting fans who know the films better than the books?
The Dufflepuds look good, though.
*I wrote "Patrick Stewart" before. I knew it was Ian; nearly made a Gandalf reference; I think I must have had Patrick's name lodged in my mind because of Scott's dramatic recitation yesterday.
**Except!! I am totally wrong. It's Liam Neeson. LOL! Thanks, Robin, for the heads-up! Oh, these actors with their sonorous voices!
Tags: Narnia, Narnia movies, The Chronicles of Narnia, Voyage of the Dawn Treader
IMAX Beavers: Thanks for the Tip
July 17, 2007 @ 7:38 am | Filed under: Film, Nature Study
I've studded my Netflix queue with your suggestions, and the Beavers * movie a couple of you raved about sounded so intriguing we bumped it to the top of the list.
Whoa.
Stunning movie. The cinematography will knock your socks off. I kept thinking, this can't be real, it's like really good CGI…But it's real. Up close, delightfully personal, and captivating. Rose and Beanie watched the movie three times, I think. (It isn't very long, alas.) And I snuck in an encore viewing myself—I couldn't resist showing it to Scott, you see.
There were beavers on the undeveloped land surrounding our old neighborhood; they had a big lodge downstream of the place the kids called the Rock Store, named after a splashy afternoon spent collecting and displaying stony wares on a big flat boulder where the creek ran wide. To get to the Rock Store, you had to hike through a field that had once been young woods. The beavers had turned it to meadow, and we marveled at the stubby, pointy tree trunks left behind, so exactly like pictures we'd seen in books.
As the creek wended its way toward the marshy basin where a bald eagle was rumored to fish, it gurgled past trees whose lower trunks had been wrapped with chicken wire: the attempt of a concerned neighbor to save our woods from the enterprising beaver clan. Undaunted, the beavers turned their attention to a location half a mile away, where slender trees shaded the small pond that welcomes people into the neighborhood. One by one, those trees began to topple.
The film's narrator remarked that after humans, beavers wreak more change upon the landscape than any other animal.
They dance, too; did you know that?
Excellent tip, ladies. I hadn't realized you could get IMAX films on DVD—you can bet we'll be checking out a good many more. The whales movie, for starters…
*Link added per Brigid's request…always happy to oblige our Brigid!
Your Movie Suggestions
July 5, 2007 @ 7:10 am | Filed under: Film
Look at this list! I'm going to put in bold the films we've already seen ("we" meaning my kids, really, since I've seen many more titles on the list than they have).
Heidi (Sarah wants to know which version?)
Beatrix Potter Collection
Peter Pan, Cathy Rigby version (look on eBay)
Little House on the Prairie, Disney version
Frontier House
Bridge to Terebithia
Secret of Roan Inish
Anne of Avonlea (1975 BBC version)
IMAX movies like the one on beavers (This one is going at the top of my queue!)
Night at the Museum
Bringing Up Baby
Darby O'Gill and the Little People
You Can't Take it With You
Building Big with David Macaulay (a few scary scenes for more sensitive types)
The Pacifier
Cheaper by the Dozen (1950s version)
Akeelah and the Bee (slight language issues)
I Remember Mama
Pride of the Yankees
Spirit of Saint Louis
Emma, both the Kate Beckinsale & Gwyneth Paltrow versions were recommended (I'm partial to the Kate one, myself)
Pride & Prejudice, BBC version (but of course!!)
Nanny McPhee
National Velvet (old one with E. Taylor)
Fairy Tale: A True Story
Iron Will
The Black Stallion
Yankee Doodle Dandy
Captains Courageous
Miss Potter
A Little Princess (not Shirley Temple version)
BBC Planet Earth series
Yours, Mine, & Ours (Lucille Ball/Henry Fonda version)
Pollyanna
Lassie (2005 version)
Natl Geographic Inside the Vatican
Life with Father
Martha Stewart dvds
Mr Blanding…
Railway Children
Connections w/ James Burke (This is available online, too, did you know? I saw a link at Sandra Dodd's site.)
Hayley Mills movies (we've seen Pollyanna, Summer Magic, The Parent Trap)
and this list by Colleen, with this caveat: "…some
have language and/or mild sexual content that others might be
uncomfortable with. Some would definitely be more appropriate for
older, mature children. We tend to worry much more about violence at
our house…"
The Snow Walker
Mad Hot Ballroom
The Ron Clark Story (unique chance for my homeschooled kids to get a look at inner city New York schooling)
Mrs. Palfrey at the Claremont
Rabbit Proof Fence
Follow Me, Boys!
The Apple Dumpling Gang
Song Catcher
Children of Heaven
Series wise, we've enjoyed The Waltons, All Creatures Great and Small,
and the early years of Monarch of the Glen. My kids also liked As Time
Goes By, although mainly this was a treat for their parents!!
Thanks, and keep 'em coming!
What Should I Put in My Netflix Queue?
In the comments of yesterday's post, Faith mentioned Netflix and Ana mentioned the movie version of The Railway Children. That reminded me that I'd been meaning to mention that video myself—we were given a copy as a gift a few years ago, and we watched it again last weekend, and ohhhh I do love that movie. It's one of those rare cases where I like the movie version just as well as I like the book—and I like the book a lot. It's my favorite Nesbit novel.
Anyway, Faith and Ana reminded me that I keep meaning to stack our Netflix queue with some good movies for the girls and me to watch this summer. What are your family's favorites? Hit me!


text:
CRITICALLY ACCLAIMED UNDERWATER EXPLORATION GAME ABZÛ DESCENDS ON SWITCH THIS MONTH
November 13, 2018 by Justin Germino
CALABASAS, CA, Calif. – Nov. 13, 2018 – Independent publisher 505 Games and developer Giant Squid have announced their award-winning inspirational aquatic journey ABZÛ is coming to Nintendo Switch on Thursday, Nov. 29 priced at £14.99 . Available for pre-order beginning today, players who purchase ABZÛ ahead of its launch will receive a limited-time 10% discount on the meditative oceanic adventure game.
ABZÛ is a beautiful underwater journey that immerses players in a vibrant ocean world full of mystery and bursting with color and life. Playing as the Diver, you'll discover hundreds of unique species based on real creatures and form a powerful connection with the abundant sea life. But this underwater world can be dangerous, broken and toxic – and your mission is to unlock the mysteries of the deep and change the world around you for the better.
Developed by Giant Squid Studios – led by Matt Nava, the artistic mind behind Journey® and Flower® – and featuring a score by Grammy-nominated and two-time BAFTA-winning composer Austin Wintory, ABZÛ is told in the form of mythology and folklore, while capturing the dream-like feeling of underwater exploration. The name is derived from ancient Sumerian language; AB, meaning water, and ZÛ, meaning to know: ABZÛ is the ocean of wisdom.
First launched in 2016 for PlayStation 4, Xbox One and PC, ABZÛ was lauded by critics and garnered numerous awards and nominations including Best Original Score for a Video Game or Interactive Media from the International Film Music Critics Association (winner), Artistic Achievement from the 2017 BAFTA British Academy Games Awards, Music of the Year from the 2017 Game Audio Network Awards (winner), Best Original Game from the 2016 Joystick Awards and Best Art Direction from the 2016 Game Awards.
ABZÛ is rated PEGI 7 and available now for pre-order here: https://www.nintendo.co.uk/Games/Nintendo-Switch-download-software/ABZ–1467719.html
For more information on ABZÛ, please visit: www.abzugame.com or follow the ABZÛ community at: Twitter: https://twitter.com/abzugame | Facebook: https://www.facebook.com/AbzuGame/
ABOUT 505 GAMES
505 Games is a global video game publisher focused on offering a broad selection of titles for players of all ages and levels. The company publishes and distributes premium and free-to-play games on leading console, PC and handheld platforms as well as for mobile devices and social networks.
Publishing highlights in premium games include No Man's Sky, Last Day of June, PAYDAY 2, Terraria, Portal Knights, Assetto Corsa, ABZÛ, Virginia, Brothers – A Tale of Two Sonsand How to Survive. Free-to-play publishing highlights include Battle Islands, Gems of Warand Hawken. Distribution highlights include Stardew Valley, Dead by Daylight and Inside/Limbo. Upcoming titles include Remedy's CONTROL, OVERKILL'S The Walking Dead, Indivisible (from the development team behind Skullgirls), Underworld Ascendant and Koji Igarashi's next project, Bloodstained: Ritual of the Night.
505 Games has offices in California, United Kingdom, France, Germany, Italy, Spain and China. It works with a network of distribution and sub-licensing partners in all other markets. For more information on 505 Games and its products please visit www.505games.com.
ABOUT GIANT SQUID
Giant Squid is a video game development studio based in Santa Monica, California. The team is led by the creative vision of the award-winning artist Matt Nava. Previously the Art Director at thatgamecompany, Matt was a core contributor to the success of the critically acclaimed games Flower and Journey. Dedicated to creating deeply immersive and meaningful experiences, Giant Squid is hard at work on their first project, ABZÛ. Learn more about the company at www.giantsquidstudios.com.
Justin Germino
Owner at DragonBlogger.com
I have been obsessed with computers, tech, gadgets and games since the early 1980's having grown up on the Commodore 64 and Amiga computers. By day I work in the IT Security Industry and have been in IT for over 20 years. On my spare time I am a Vlogger, Blogger, Streamer, Gadget Reviewer, affiliate marketer, influencer and entertainer. I am also an avid movie fan, TV Show fan, Anime fan, video game fan and fan of trying anything and everything new.
@dragonblogger
Content Creator/Owner of https://t.co/iWP6afuOc9 for Tech, Game, Entertainment and streamer at https://t.co/594fXMdMIA
ENCODYA Dystopian Game Featurette Trailer https://t.co/EMlySkAfip - 12 hours ago
Latest posts by Justin Germino (see all)
TerraMaster F5-422 NAS with 10GbE Networking and up to 80TB Max Storage Capacity - January 21, 2021
BIOSTAR UNVEILS THE ALL NEW Z590 VALKYRIE MOTHERBOARDS - January 20, 2021
Tohu Adventure Game Has a Release Date - January 18, 2021
Categories Press Releases Post navigation
iBUYPOWER Launches Same Day Gaming PC Shipping Program to Beat DIY Prices
5 Android Apps that Make College Life a Lot Simpler
Share Feedback We Want to Hear From You