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%%% Author:		Iver Ottosen
%%% Title:		Equivariant evaluation on free loop spaces
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%%% 1991 Mathematics Subject Classification: 55N91, 55P35. 			
%%% Keywords and phrases: Equivariant cohomology, free loop spaces.
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\title{Equivariant evaluation on free loop spaces}
\author{Iver Ottosen \footnote{
1991 {\em Mathematics Subject Classification}: 55N91, 55P35.
I would like to thank Universit\' e de Paris Nord and Universit\" at 
Bielefeld where this work was developed, supported by the European Union 
TMR network ERB FMRX CT-97-0107: Algebraic $K$-Theory, Linear Algebraic 
Groups and Related Structures.}}
%** end of header

\begin{document} 
\maketitle

\begin{abstract}
There is a natural evaluation map on the free loop space
$\Lambda X \to X^k$ which sends a loop to its values at the $k$th 
roots of unity. This map is equivariant with respect to the action 
of the cyclic group on $k$ elements $C_k$. We study the induced map in
$C_k$-equivariant cohomology with mod two coefficients 
in the cases where $k=2^m$ for $m\geq 1$.
\end{abstract}


\section{Introduction}
For each positive integer $k$ there is an evaluation map on the free
loop space $\la X \to X^k$ which is equivariant with respect to the 
action of the cyclic group $C_k$. The evaluation map sends a loop  
to its values at the $k$th roots of unity. 
As $k$ approaches infinity this evaluation 
distinguishes more loops. One hopes that the 
induced map in cohomology becomes more and more informative as
$k$ increases and hence provides new input to the ongoing problem of 
computing the cohomology of free loop spaces. Since evaluating at 
two different points of the circle gives homotopic maps, one should 
use $C_k$-equivariant cohomology in order to get interesting results. 

In this paper we use cohomology with $\Z /2$-coefficients. 
We consider the cases $k=2^m$ with $m\geq 1$. 
Under these assumptions we compute the map induced by evaluation in 
$C_k$-equivariant cohomology in terms of
the approximation functor $\ell$ introduced in [BO].

The functor $\ell$ comes equipped with a ring homomorphism 
as follows
$$\psi : \ell (H^*X) \to H^*(ES^1\times_{S^1} \la X).$$
It is an isomorphism for 1-connected spaces $X$ with 
polynomial cohomology. It is not hard to see that 
$H^*(BC_k)\otimes_{H^*(BS^1)} \ell (H^*X)$ approximates
$H^*(ES^1\times_{C_k}\la X)$ similarly. 

In \S 2 we introduce some notation and conventions which are fixed 
through the paper.
In \S 3 we review the definition and some basic properties of 
the functor $\ell$ and the related twisted de Rham functor $\dr$.

In \S 4 we show that $\ell$ and $\dr$ are functors from the category 
of connected unstable $\sa$-algebras to 
the category of unstable $\sa$-algebras. Here $\sa$ denotes the 
mod two Steenrod algebra. We give a useful pullback description
of $\ell (K)$ when $K$ is the cohomology of a product of Eilenberg-MacLane 
spaces of type $K(\Ftwo , n)$ with $n\geq 1$. The description
is similar to the one for the quadratic construction $C_2(K)$
as described in [HLS].

In \S 5 we generalize the exactness result for the approximation $\psi$
to 0-connected spaces with polynomial cohomology and a finite 2-group as 
fundamental group. The sections \S 4 and \S 5 give a convenient setup
for working with the functor $\ell$.  

In \S 7 we set up the computation. There is a commutative diagram
$$ \diagram
S^1 \times \la X \dto^{\eta} \drto^f & \\
ES^1\times_{C_k}\la X \rto^{ev} & ES^1\times_{C_k} X^k \\
BC_k \times X \uto_i \urto_\Delta 
\enddiagram $$
where $i$ comes from the constant loop inclusion $X \hookrightarrow \la X$
and $\eta$ comes from the $S^1$-action on $\la X$. The pair $(i,\eta)$ 
induces an injective map in cohomology by the pullback in 
\S 5. The map induced by the diagonal $\Delta$ was computed in [O]. 
The induced map of $f$ is computed in \S 7. From this data $ev^*$ is 
determined in \S 8. Finally in \S 9 we give an extension
of the evaluation map $ev^*$ to an algebraic setting. 


\section{Notation and conventions}
\label{notation}

We always use $\Ftwo=\Z /2$ coefficients unless otherwise is
specified. We say that a space $X$ is of finite type if $H_iX$ is
finitely generated for each $i$. 
The mod two Steenrod algebra is denoted $\sa$, the 
category of unstable $\sa$-modules is denoted $\U$ and
the category of unstable $\sa$-algebras is denoted $\K$. 
The category $\K_0$ is the full subcategory of $\K$ consisting of 
connected unstable $\sa$-algebras. 
For any $\sa$-module $M$ we define $\lambda: M \to M$ by
$\lambda(x)=Sq^{|x|-1}x$. 

The cyclic group of order $k$ is denoted $C_k$. We view it as a
subgroup of the circle group $S^1$ generated by 
$\zeta_k=\exp(2\pi i/k)$.
Assume that $k=2^m$ with $m\geq 0$.
For an $S^1$-space $Y$ and a $C_k$-space $Z$ we use the following
short hand notation for Borel constructions:
$$E_\infty Y:=ES^1\times_{S^1} Y \quad , 
\quad E_m Z:=ES^1\times_{C_k}Z$$
For integers $n$ and $m$ with $0\leq n<m\leq \infty$ there is a map 
$q_m^n$ induced by the quotient and a transfer map 
$\tau_n^m$ as follows
$$ q_m^n: H^*E_mY \to H^*E_nY \quad ; \quad
\tau_n^m: H^*E_nY \to H^*E_mY $$
Here $\tau_n^\infty$ is the $S^1$-transfer from 
Definition 4.4 in [BO].

The free loop space is the mapping space $\la X=X^{S^1}$. 
Precomposition with the inclusion
map $j_k: C_k \hookrightarrow S^1$ defines a map
$j_k^*:X^{S^1} \to X^{C_k}$ which is $C_k$-equivariant.
We get a map of Borel constructions.
\begin{equation*}
\label{generaleval}
1\times_{C_k}j_k^*:
ES^1\times_{C_k}X^{S^1}\to ES^1\times_{C_k}X^{C_k}
\end{equation*}
When $k=2^m$ we put $\cyc_mX=ES^1 \times_{C_k} X^k$ and we write
$ev_m:E_m \la X \to \cyc_m X$ for the evaluation map $1\times_{C_k}j_k^*$.

Finally, we fix the names on the algebra generators of the following 
cohomology groups.
$H^*(S^1)=\Lambda (v)$ where $|v|=1$, $H^*(BS^1)=\Ftwo [u]$
where $|u|=2$ and 
$$
H^*(BC_{2^m})= 
\begin{cases}
\Lambda (v) \otimes \Ftwo [u] & \text{ when } m\geq 2 \\
\Ftwo [t] & \text{ when } m=1
\end{cases}
$$
where $|t|=|v|=1$ and $|u|=1$. 

\section{The functors $\dr$ and $\ell$}

In this section we review the definitions of the approximation
functors $\dr$ and $\ell$ from [BO] and provide some 
extra material on them.

\begin{definition}
Let $R$ be a unital, connected $\Ftwo$-algebra. A linear map
$\lambda:R\to R$ is called a derivation over Frobenius if
$|\lambda x|=2|x|-1$ and $\lambda (xy)= x^2\lambda y + y^2 \lambda x$.
We refer to $(R,\lambda)$ as an algebra with derivation over
Frobenius, and use the abbreviation FGA for such gadgets. They form a
category $FGA$ where the morphisms are algebra morphisms respecting the
structure given by the derivations over Frobenius.
\end{definition}

\begin{definition}
Let $(R,\lambda )$ be a FGA.
The algebra $\dr (R)$ is the quotient of the free
commutative algebra on 
$$x, dx \text{ for } x \in R$$
where $|dx|=|x|-1$, by the ideal generated by the elements
\begin{align}
& d(x+y)+dx+dy \label{dr1} \\
& d(xy)+xdy+ydx \label{dr2} \\
& (dx)^2+d(\lambda x) \label{dr3}
\end{align}
There is a differential on $\Omega_{\lambda}(R)$ defined by
the formula $d(x)=dx$ as a derivation over $R$. 
\end{definition}

Clearly $\Omega_{-}$ is a functor from the category of FGA's to
the category of DGA's. The following result shows that the category of
FGA's is equipped with a tensor product. We omit the proof which is
an easy direct computation.

\begin{proposition}
\label{*}
Let $(A,\lambda )$ and $(B,\gamma )$ be FGA's and define an
$\Ftwo$-linear map $\lambda * \gamma : A\otimes B \to A\otimes B$
by 
$$\lambda * \gamma (a\otimes b)= \lambda (a) \otimes b^2 
+ a^2 \otimes \gamma (b)$$
Then $(A\otimes B, \lambda * \gamma)$ is an FGA.
\end{proposition}

The functor $\Omega_{-}$ preserves tensor products in the
following sense:

\begin{proposition} 
The canonical inclusion maps
$i: A \to A \otimes B$; $a \mapsto a\otimes 1$ and 
$j: B \to A\otimes B$; $b \mapsto 1\otimes b$
are morphisms of FGA's and the composite map 
$$ \begin{CD}
\kappa : \Omega_{\lambda} (A) \otimes \Omega_{\gamma} (B) 
@>i_*\otimes j_*>> 
\Omega_{\lambda * \gamma}(A\otimes B)\otimes 
\Omega_{\lambda * \gamma}(A\otimes B) @>\mu >> 
\Omega_{\lambda * \gamma}(A\otimes B) 
\end{CD} $$ 
is an isomorphism of DGA's.
\end{proposition}

\begin{proof}
$i$ and $j$ are morphisms of FGA's since an FGA is non negatively
graded. Thus $i_*$ and $j_*$ are morphisms of DGA's hence 
$\kappa$ is a morphism of DGA's. 
We check that there is a well defined map
$\Omega_{\lambda *\gamma}(A\otimes B) \to
\Omega_{\lambda} (A) \otimes \Omega_{\gamma} (B)$
such that $y\mapsto y$ and $dy\mapsto d_\otimes y$ for $y\in A\otimes B$
Such a map is clearly an inverse to $\kappa $. 
As for the usual de Rham complex one sees that elements of the form
(\ref{dr1}) and (\ref{dr2}) are mapped to zero. 
For $a\in A$ and $b\in B$ we have
\begin{align*}
& (d(a\otimes b))^2+d(\lambda * \gamma (a\otimes b)) = \\
& (d(a\otimes b))^2+d(\lambda (a) \otimes b^2
+a^2\otimes \gamma (b)) \mapsto \\
& (d(a) \otimes b + a\otimes d(b))^2+d(\lambda a) \otimes b^2
+a^2\otimes d(\gamma b) = 0
\end{align*}
and since (\ref{dr3}) is additive in $x$ we are done.
\end{proof}

The forgetful functor $U:\K \to FGA$ provide the main examples of
FGA's. For an object $K$ in $\K$ we let $\lambda x= Sq^{|x|-1}x$.
It is a derivation over Frobenius by the Cartan formula.

Let $Y$ be a $S^1$-space with action map $\eta :S^1 \times Y \to Y$.
Then there is a map $d:H^*Y \to H^{*-1}Y$ defined by
$\eta^*(y)=1\otimes y+v\otimes dy$. Proposition 3.2 in [BO] shows
that $(H^*Y,d)$ is a DGA and $Sq^i(dy)=d(Sq^iy)$ such that
$(dy)^2=d(\lambda y)$. It is not necessary to assume that $Y$ is 
connected for these results even though we did so in [BO].
When $Y$ is a free loop space we get the following.

\begin{proposition}
For any connected space $X$ there is a morphism of DGA's
$$e: (\dr (H^*X),d) \to (H^*(\la X),d) \quad ; \quad e(x)=ev_0^*(x) 
\quad ; \quad e(dx)=dev_0^*(x)$$
which is natural in $X$.
For any pair of connected spaces $X$ and $Y$ there is a commutative diagram 
$$\begin{CD}
\dr (H^*X) \otimes \dr (H^*Y) @>\kappa>> \dr(H^*X \otimes H^*Y) \\
@Ve\otimes eVV @VeVV \\
H^*(\la X) \otimes H^*(\la Y) @>>> H^*(\la (X\times Y))
\end{CD}$$
where the lower horizontal map is the K\" unneth isomorphism.
\end{proposition} 

\begin{definition}
Let $(R,\lambda )$ be an FGA. The algebra $\ell (R)$ is the quotient
of the free commutative algebra on generators
$$\phi (x), q(x), \delta (x) \text{ for } x \in R \text{ and } u$$
of degree $|\phi (x)|=2|x|$, $|q(x)|=2|x|-1$, $|\delta (x)| = |x|-1$
and  $|u|=2$, by the ideal generated by the elements
\begin{align}
& \phi(a+b)+ \phi(a)+\phi(b) \label{r1} \\
& \delta(a+b)+\delta(a)+\delta(b) \label{r2} \\
& q(a+b)+q(a)+q(b)+\delta(ab) \label{r3} \\
& \delta(xy)\delta(z)+\delta(yz)\delta(x)+\delta(zx)\delta(y) \label{r4} \\
& \phi(xy) + \phi(x) \phi(y)+uq(x)q(y) \label{r5} \\
& q(xy) +q(x)\phi(y)+\phi(x)q(y) \label{r6} \\
& \delta(x)^2 +\delta(\lambda x) \label{r7} \\
& q(x)^2 +\phi(\lambda x)+\delta(x^2\lambda x) \label{r8} \\
& \delta(x)\phi(y)+\delta(xy^2) \label{r9} \\
& \delta(x)q(y) +\delta(x\lambda y)+\delta(xy)\delta(y) \label{r10}\\
& u\delta(x) \label{r11}
\end{align}
where $a,b,x,y,z$ are homogeneous elements in $R$ and $|a|=|b|$.
\end{definition}

\begin{theorem}
\label{DR}
Let $(R,\lambda)$ be an FGA. There is a natural algebra homomorphism
$DR: \ell (R) \to \dr (R)$ defined by
$$\phi (x) \mapsto x^2 \quad , \quad q(x) \mapsto xdx+\lambda x
\quad , \quad \delta (x) \mapsto dx \quad , \quad u\mapsto 0$$
If $R$ is a symmetric algebra $R=S(V)$ where $V$ is a non negatively graded 
$\Ftwo$-vector space with $V^n$ finitely generated for each $n$, there
is a short exact sequence
$$\begin{CD}
0 \to (u) @>>> \ell (R) @>DR>> \ker (d) \to 0
\end{CD}$$
\end{theorem}

\begin{proof}
This was shown in [BO] Proposition 8.1, Theorem 8.2 and Theorem 8.5.
\end{proof}

We can apply the results from chapter 6 of [BO] component vise and get
the following.

\begin{theorem}
\label{l-approx}
For any connected space $X$ there is an algebra homomorphism
$\psi :\ell(H^*X) \to H^*(ES^1\times_{S^1} \la X)$ defined by
$$\begin{array}{ccccccc}
\phi (x) & \mapsto &
\tau_1^\infty \circ ev_1^*(t\otimes x^{\otimes 2}) 
& \quad ; \quad &
q(x) & \mapsto & 
\tau_1^\infty \circ ev_1^*(1\otimes x^{\otimes 2}) \\ 
\delta (x) & \mapsto & \tau_0^\infty \circ ev_0^* (x) 
& \quad ; \quad &
u & \mapsto & \tau_1^\infty \circ ev_1^*(t^3\otimes 1^{\otimes 2})
\end{array}$$
It is natural in $X$ and 
the following diagram commutes
$$\begin{CD}
\ell (H^*X) @>\psi>> H^*(ES^1\times_{S^1} \la X) \\
@VDRVV @Vq_\infty ^0VV \\
\dr (H^*X) @>e>> H^*(\la X)
\end{CD}$$
\end{theorem}


\section{Unstable algebras.}

In this section we show that $\dr$ and $\ell$ restricts to
functors $\K_0 \to \K$.
We do this algebraically instead of using the result in the
next chapter, since it gives a better understanding of 
the functors. For the functor $\ell$ we follow the treatment
of the quadratic construction as close as possible. A certain 
pull back diagram appears, which will become useful later.

We use the notation $Sq_i(x):=Sq^{|x|-i}x$.
For a sequence of positive integers $n_1, \dots ,n_r$ we define
the following object in $\K_0$:
$$A(n_1, \dots , n_r)=
H^*(K(\Ftwo , n_1) \times \dots \times K(\Ftwo, n_r))$$

\begin{lemma}
\label{hcrplem}
The following relations are valid in any unstable $\sa$-module
\begin{align}
& Sq^{2i}(\lambda x)=\lambda (Sq^ix) \\
& Sq^{2i+1}(\lambda x)= (i+|x|)Sq^{i+|x|}Sq^ix 
\end{align}
\end{lemma}

\begin{proof}
Follows by the Adem relation. See the proof of Proposition 5.2 in [O].
\end{proof}

\begin{proposition}
\label{sadr}
For any $K\in ob(\K_0 )$ we can define an $\sa$-action on 
$\dr (K)$ by $Sq^i(dx)=d(Sq^ix)$ and the Cartan formula. With this 
action $\dr$ becomes a functor $\dr: \K_0 \to \K$.
The differential $d:\dr(K)\to \dr(K)$ is $\sa$-linear.
\end{proposition}

\begin{proof}
Let $dK$ be the $\sa$-module given by 
$(dK)^n=K^{n+1}$. The symmetric algebra $S(K\oplus dK)$ is an
$\sa$-module by the Cartan formula. $\dr (K)$ is by definition 
the quotient of this module by the ideal generated by
(\ref{dr1}),(\ref{dr2}) and (\ref{dr3}).
We verify that this ideal is closed under the $\sa$-action.
We have
$$
Sq^i(d(x+y)+dx+dy)=d(Sq^ix+Sq^iy)+d(Sq^ix)+d(Sq^iy)
$$
which is in the ideal (in fact (\ref{dr1}) is already zero 
in $dK$). Further
\begin{align*}
& Sq^i(d(xy)+xdy+ydx) = \\
& \sum_{j=0}^i \big( d(Sq^j(x)Sq^{i-j}(y))
+(Sq^jx)d(Sq^{i-j}y)+(Sq^jy)d(Sq^{i-j}x) \big) 
\end{align*}
which is in the ideal by (\ref{dr2}). We use Lemma \ref{hcrplem}
for the last type of elements:
\begin{align*}
& Sq^{2i}((dx)^2+d(\lambda x))=(Sq^i(dx))^2+d(Sq^{2i}(\lambda x)) =
(d(Sq^ix))^2+d(\lambda (Sq^ix)) \\
& Sq^{2i+1}((dx)^2+d(\lambda x))=d(Sq^{2i+1}(\lambda x))=
d((i+|x|)(Sq^ix)^2)=0
\end{align*}
We conclude that $\dr(K)$ has a natural $\sa$-module structure. 
It is a non negatively graded module since
$d1=0$ by (\ref{dr2}). 
We have $Sq^{|x|}(dx)=d(x^2)=0$ and clearly $Sq^i(dx)=0$
for $i>|x|$ so $\dr(K)$ is an unstable $\sa$-module. The Cartan 
formula holds by definition and 
$Sq^{|x|-1}(dx)=d(\lambda x)=(dx)^2$ so
$\dr (K)$ is an unstable $\sa$-algebra.
The differential on $\dr(K)$ is $\sa$-linear by definition. 
\end{proof}

\begin{definition}
For $K\in \K_0 $ we define the modified de Rham cohomology by
$H^*_{\lambda}(K)=H^*(\dr (K),d)$.
\end{definition}

Note that this cohomology theory has the
nice property of being a functor $H^*_{\lambda}: \K_0 \to \K$ as well
as it respects tensor products:
$H^*_{\lambda} (A\otimes B) \cong H^*_{\lambda}(A)\otimes H^*_{\lambda}(B)$.

\begin{definition}
Let $K$ be an unstable $\sa$-algebra. For $x\in K$ we define the elements
$St_0(x)=\sum u^i\otimes Sq_{2i}x$ and 
$St_1(x)=\sum u^i \otimes Sq_{2i+1}x$ in $\Ftwo [u] \otimes K$ 
where the sums are taken over all integers $i\geq 0$.
Further we let  $R(K)$ denote the sub $\Ftwo$-algebra of
$\Ftwo [u] \otimes K$ generated by $u$ and $St_0(x)$, $St_1(x)$ 
for $x\in K$.
\end{definition}

\begin{proposition}
For $i\geq 0$ one has $Sq^iR(K) \subseteq R(K)$ hence we have  
defined a functor $R: \K \to \K$. Explicitly, the following
formula holds for $\nu =0,1$:
\begin{equation}
\label{sar}
Sq^i St_{\nu}(x) =  
\sum_{j=0}^{\infty } \binom {|x|-j+1-\nu} {i-2j} 
u^{[\frac {i+1-\nu} 2]-j} St_{i+\nu} (Sq^jx) 
\end{equation}
where the lower index on the right should be taken modulo 2.
\end{proposition}

\begin{proof}
Let $St(x)=\sum t^i\otimes Sq_i(x)$ in $\Ftwo [t] \otimes K$ where
$|t|=1$. The following formula holds
\begin{equation}
\label{Stformula}
Sq^i(St(x))=\sum_{j=0}^\infty \binom {|x|-j} {i-2j} 
t^{i-2j}St(Sq^jx)
\end{equation}
It expresses that the diagonal map 
$\Delta: \cyc_1 (K) \to \Ftwo [t] \otimes K$ which is given by
$\Delta(1\otimes (x\otimes y+y\otimes x))=0$ and
$\Delta (1\otimes x^{\otimes 2})=St(x)$
is $\sa$-linear. Here $\cyc_1$ is the quadratic functor [HLS] page 58-59. 
The formulas for the $\sa$-module structure on 
$\cyc_1K$ can be found in [M].
There is an $\sa$-linear transfer map of degree -1 
$$\tau : \Ftwo [t] \to \Ftwo [u] \quad ; \quad \tau (t^{2i})=0 \quad ,
\quad \tau (t^{2i+1})=u^i$$
Note that the following formula holds for $\nu=0,1$:
\begin{equation}
\label{StTr}
St_\nu(x) = (\tau \otimes 1) (t^{1-\nu}St(x))
\end{equation}
It implies that $Sq^iSt_\nu (x)=(\tau \otimes 1)\circ Sq^i(t^{1-\nu}St(x))$.
By this and formula (\ref{Stformula}) we see that equation
(\ref{sar}) holds.
\end{proof}

\begin{proposition}
\label{defofst}
For $K\in ob(\K_0)$ there is an $\Ftwo$-algebra homomorphism 
\begin{align*}
& St:\ell (K) \to \Ftwo [u] \otimes K \quad ; \\
&\phi (x) \mapsto St_0(x) \quad , \quad q(x) \mapsto St_1(x)
\quad , \quad \delta (x) \mapsto 0 \quad , \quad u \mapsto u
\end{align*}
and $R(K)$ is the image of this map.
\end{proposition}

\begin{proof}
We must check that $St$ is well defined.
The elements (\ref{r2}), (\ref{r4}), (\ref{r7}),
(\ref{r9}), (\ref{r10}), (\ref{r11}) are clearly mapped to 
zero since $St\circ \delta=0$. The elements 
(\ref{r1}), (\ref{r3}) are mapped to zero since
$St\circ \phi$ and $St\circ q$ are additive and $St\circ \delta=0$.
For the elements (\ref{r5}) and (\ref{r6}) we proceed as follows. 
Let again $St(x)=\sum t^i \otimes Sq_i(x)$ instead of the above map.
By the Cartan formula we have $St(xy)=St(x)St(y)$ and applying 
(\ref{StTr}) to this we find
\begin{align*}
& St_1(ab)=St_1(a)St_0(b)+St_0(a)St_1(b) \\
& St_0(ab)=St_0(a)St_0(b)+uSt_1(a)St_1(b)
\end{align*}
Finally (\ref{r8}) is mapped to zero since Lemma \ref{hcrplem} implies
$St_0(\lambda x)=(St_1x)^2$.
By its definition $R(K)$ is the image of the map $St$.
\end{proof}

\begin{lemma}
\label{EMspace}
The Eilenberg-MacLane space $K(\Ftwo , n)$ 
with $n\geq 1$ has the following mod two cohomology algebra.
$$\Ftwo [\lambda ^k Sq^I\iota_n \, | \, I\text{ admissible }, \, 
e(I)\leq n-2 \, , \, k\geq 0]$$ 
\end{lemma}

\begin{proof}
The usual set of algebra generators is
$\{ Sq^J\iota_n \}$ where $J$ runs through the admissible 
sequences with excess $e(J)\leq n-1$.
We verify that this set is in fact the same as the set of generators
in the lemma. 

Assume that $I=(i_1,i_2, \dots , i_s)$ is admissible with 
excess $e(I)\leq n-1$. We have 
$\lambda Sq^I\iota_n=Sq^{(n-1+|I|,I)}\iota_n$.
Since $e(I)=i_1-i_2-\dots -i_s\leq n-1$ we have $n-1+|I|\geq 2i_1$
such that $(n-1+|I|,I)$ is admissible and clearly $e(n-1+|I|,I)=n-1$.
Thus the set of generators in the lemma is a subset of the usual set
of algebra generators. 

Conversely, if $J=(j_1,\dots ,j_r)$ is admissible 
and $e(J)\leq n-2$ then
$Sq^J\iota_n$ is one of the generators with $k=0$ in the lemma. 
Assume that
$e(J)=n-1$ and put $J(t)=(j_t, j_{t+1}, \dots ,j_r)$.
The map $t\mapsto e(J(t))$ is weakly decreasing since
$e(J(t))-e(J(t+1))=j_t-2j_{t+1}\geq 0$ and eventually it becomes
zero. Let $k=max \{t | e(J(t))=n-1 \}$. Then we have
$Sq^J\iota_n=\lambda^k Sq^{J(k+1)}\iota_n$ which is one of the 
generators in the lemma.
\end{proof} 

\begin{definition}
When $R$ is an FGA we let $J(R)$ denote the subring of $\ell(R)$ 
generated by the elements $\delta (x)$ for $x\in R$. By
(\ref{r9}), (\ref{r10}) and (\ref{r11}) we see that $J(R)$ is an
ideal in $\ell (R)$.
\end{definition}

\begin{proposition}
\label{kernel}
When $K=A(n_1, \dots , n_r)$ where
$n_i\geq 1$ for each $i$ we have $\ker (St)=J(K)$.
\end{proposition}

\begin{proof}
By definition $J(K)\subseteq \ker (St)$ so
it suffices to show that the induced map 
$St:\ell (K) / J(K) \to \Ftwo [u] \otimes K$
is injective. The algebra $\ell (K) / J(K)$ has generators 
$u$, $\phi(x)$, $q(x)$ for $x\in K$. The relations are that $\phi$ and
$q$ are additive and further
\begin{align*}
& \phi (xy) = \phi (x) \phi (y) +uq(x)q(y)\\
& q(xy)=q(x)\phi (y)+\phi (x)q(y)\\
& q(x)^2=\phi (\lambda x)
\end{align*}
Put $x_i=\iota_{n_i}$ for each $i$. We see that the following elements
are algebra generators: $\phi(Sq^{I_i}x_i)$, 
$q(\lambda ^{k_i} Sq^{I_i}x_i)$, $u$ where
$I_i$ is admissible, $e(I_i)\leq n_i-2$ and $0 \leq k_i$ for  
$1\leq i \leq r$.
It suffices to show that these are mapped to algebraic independent
elements. The coefficient to $u^0$ in 
$St_0(Sq^{I_i}x_i)$ is $(Sq^{I_i}x_i)^2$
and in $St_1(\lambda ^{k_i} Sq^{I_i}x_i)$ it is 
$\lambda^{k_i+1} Sq^{I_i}x_i$. Using the projection homomorphisms
$p_i: \Ftwo [u] \otimes K \to K$ given by $p_i(u^j\otimes x)=
\delta_{i,j}x$ and induction the algebraic independence 
follows by Lemma \ref{EMspace}.
\end{proof}

\begin{theorem}
\label{pullback}
For $K\in ob(\K_0)$ there is a commutative diagram in the category
of $\Ftwo$-algebras
$$
\begin{CD}
\ell (K) @>St>> R(K)\\
@VDRVV @VpVV \\
\ker (d) @>q>> K
\end{CD} 
$$
where $p:\Ftwo [u] \otimes K\to K$ and $q:\dr (K) \to K$ 
are the projections defined by $q(x)=x$, $q(dx)=0$ and
$p(x)=x$ and $p(u)=0$ for $x\in K$. 
If $K=A(n_1, \dots , n_r)$ where
$n_i\geq 1$ for each $i$ the diagram is a pull back square.
\end{theorem}

\begin{proof}
By evaluating on generators it is easy to check that the diagram
is commutative. Assume $K$ is the cohomology of a product of 
Eilenberg-MacLane spaces as above. The diagram is a pull back if and
only if the restriction 
$St: \ker (DR) \to \ker (p)$ is an isomorphism.
By Proposition \ref{kernel} and Theorem \ref{DR} we have
$$\ker(DR) \cap \ker (St) = (u) \cap J(K) = 0$$ so the 
restriction of $St$ is injective. The kernel of 
$p:\Ftwo [u] \otimes K \to K$ is $(u) \subseteq \Ftwo [u] \otimes K$
so $(u) \cap R(K)$ is the kernel of $p: R(K) \to K$.
Since $St(\phi (x)) =St_0(x)$, $St(q(x))=St_1(x)$ and $St(u)=u$ it is
obvious that the restriction of $St$ is surjective.
\end{proof}

\begin{theorem}
\label{saell}
The functor $\ell $ restricts to a functor $\ell: \K_0 \to \K$ 
For $K\in ob(\K_0 )$ the $\sa$-module structure on
$\ell (K)$ is given by the following formulas:
\begin{align}
Sq^i \delta (x) &= \delta (Sq^ix) \label{sa1} \\
Sq^i \phi (x) &= \sum_{j=0}^{\infty } \binom {|x|+1-j} {i-2j} 
u^{[\frac {i+1} 2]-j} \big( (i+1) \phi (Sq^jx) + i q(Sq^jx) \big) 
\label{sa2} \\
Sq^i q(x) &= \sum_{j=0}^{\infty } \binom {|x|-j} {i-2j} 
u^{[\frac i 2]-j} \big( i \phi (Sq^jx) + (i+1) q(Sq^jx) \big)
+\delta (Q^i(x)) \label{sa3} \\
Sq^iu &= \binom 2 i u^{[\frac i 2]+1} \label{sa4}
\end{align}
Here by convention a binomial coefficient is zero 
when its lower parameter is negative. The $Q^i$ operation in the 
last formula is defined by 
$$Q^i(x)=\sum_{r=0}^{[\frac i 2]} Sq^r(x)Sq^{i-r}(x)$$
There is a commutative diagram in the category $\K$
$$
\begin{CD}
\ell (K) @>St>> R(K)\\
@VDRVV @VpVV \\
\ker (d) @>q>> K
\end{CD} 
$$
It is natural in $K$ and when $K=A(n_1, \dots ,n_r)$ for a sequence
of positive integers $n_1, \dots ,n_r$ it is a pull back square. 
\end{theorem}

\begin{proof}
When $K=A(n_1, \dots , n_r)$ we
see that $\ell (K)$ is an unstable $\sa$-algebra by the pull back 
diagram in Theorem \ref{pullback}. 
We verify that (\ref{sa1})-(\ref{sa4}) are formulas for
this $\sa$-module structure by using the injectivity of the pair (St,DR). 
The verification of equation (\ref{sa1}) and 
(\ref{sa4}) is trivial. When we apply $DR$ to the right hand side of 
(\ref{sa2}) we only need to consider 
the $j=[\frac {i+1} 2]$ term since $DR(u)=0$.
This term equals 
$\phi (Sq^{i/2}x)$ for $i$ even and zero for $i$ odd.
hence we get $(Sq^{i/2}x)^2$ for $i$ even and zero for $i$ odd.
This equals $Sq^iDR(\phi(x))= Sq^i(x^2)$ as it should.
When we apply $DR$ to the right hand side of (\ref{sa3}) only the
$j=[\frac i 2]$ term contribute. This term equals
$q(Sq^rx)+\delta (Q^{2r}(x))$ for $i=2r$ and
$(|x|-r) \phi (Sq^rx) + \delta (Q^{2r+1}(x))$ for $i=2r+1$.
Thus we must check that
\begin{align*}
& (Sq^rx)d(Sq^rx)+\lambda (Sq^rx) + dQ^{2r}(x) = 
Sq^{2r}(xdx+\lambda x) \\
& (|x|-r)(Sq^rx)^2+dQ^{2r+1}(x) = Sq^{2r+1}(xdx+\lambda x)
\end{align*}
We have $dQ^i(x)=Sq^i(xdx)+(i+1)Sq^{i/2}(x)Sq^{i/2}(dx)$
by the definition of $Q^i(x)$ and this together with 
Lemma \ref{hcrplem} gives the desired result.
For the map $St$ the verification follows directly by formula
(\ref{sar}).

For a general connected unstable $\sa$-algebra $K$ we proceed as follows.
Let $T$ denote the free, graded, unital, non commutative algebra on 
$Sq^i$ for $i\geq 0$ where $Sq^0=1$. Let $F(K)$ be the free, unital,
graded, commutative algebra on $u$, $\phi (x)$, $q(x)$, 
$\delta (x)$ for $x\in K$. We let $T$ act on $F(K)$ by the formulas
(\ref{sa1})-(\ref{sa4}) and the Cartan formula.
Let $I(K) \subseteq F(K)$ denote the ideal generated by the
elements (\ref{r1})-(\ref{r11}). First we check that 
$T\cdot I(K) \subseteq I(K)$ such that we have a well defined
action $T\times \ell (K) \to \ell (K)$. 
Let $g(y_1, \dots , y_r)$ be one of the elements (\ref{r1})-(\ref{r11}).
Put $n_j=|y_j|$, $x_j=\iota_{n_j}$ and $A=A(n_1,\dots ,n_r)$.
Let $f:A(n_1, \dots , n_r)\to K$ be the morphism in $\K$ 
defined by $f(x_j)=y_j$.
By the first part of this proof we have $T\cdot F(A) \subseteq F(A)$
so naturality gives $Sq^i g(y_1, \dots ,y_r) \in I(K)$.

We have $\sa = T/J$ where $J\subseteq T$ is the two sided ideal  
generated by the Adem elements.
We must check that $J\cdot \ell (K)=0$ such that we have a well
defined action $\sa \times K \to K$. But this follows by a similar
naturality argument as above. Hence we have shown that
the formulas (\ref{sa1})-(\ref{sa4}) define a natural $\sa$-module
structure on $\ell (K)$, which satisfy the Cartan formula.
It is immediate from the formulas that the instability condition
holds for each generator of $\ell (K)$ so the action is unstable.
Using the relations (\ref{r1})-(\ref{r11}) one verifies that
$Sq^{|y|}y=y^2$ when $y$ is either of the generators
$\phi (x)$, $q(x)$, $\delta (x)$, u and so this relation holds
in general. Thus the action makes $\ell (K)$ an unstable $\sa$-algebra.
It is obvious that $p$ and $q$ are $\sa$-linear.
\end{proof}


\section{A more general condition for exactness of the 
approximation functors}

In this section we prove a stronger version of the main theorem
in [BO], Theorem 10.1, which states that $e$ and $\psi$ are
isomorphisms for 1-connected spaces with polynomial cohomology.
We start by an examination of the case $X=\bz =\rp$.

For a connected space $X$ one has $\pi_0 \la X \cong [S^1_+,X] \cong 
<\pi_1 X>$ where $<\cdot >$ denotes the set of conjugacy classes.
Especially $\pi_0 \la BG\cong <G>$ for any group $G$. By Lemma 7.11 of
[BHM] we have the following result.

\begin{lemma}
For a discrete group $G$
$$\Lambda BG = \coprod_{[g]\in <G>} \la_{[g]}BG \quad \text{ and } 
\quad \la_{[g]} BG \cong BC_G[g]$$
where $C_G[g]$ denotes the centralizer of $g$. 
\end{lemma}

As a special case $\la \bz \cong \la_0 \bz \sqcup \la_1 \bz$
where $\la_{\nu} \bz \simeq \bz$ for $\nu =0,1$.
Define a group action
$a_{\nu}:\Z \times \Z /2 \to \Z /2$ by $(r,s)\mapsto \nu r+s$.
Then $Ba_{\nu}: B\Z \times B \Z /2 \to B\Z /2$ is an $S^1$-action which is
the trivial one for $\nu = 0$.
Let $\bzi{\nu}$ denote the space $\bz$ with $S^1$-action $Ba_\nu$.

\begin{lemma}
\label{dt}
$(Ba_\nu )^*(t)=1\otimes t+\nu t\otimes 1$ for $\nu = 0,1$.
\end{lemma}

\begin{proof}
Since $\pi_1(Ba_{\nu})=a_{\nu}$ we have $H_1(Ba_{\nu};\Z )=a_{\nu}$.
The result follows by tensoring with $\Ftwo$ and dualizing.
\end{proof}

\begin{proposition}
\label{looprp}
For $\nu =0,1$ the following map is $S^1$-equivariant and also a homotopy 
equivalence
$$j_\nu: \bzi{\nu} \to \la_{\nu}\bz  \quad ; \quad 
j_\nu(x)(z)=Ba_{\nu}(z,x)$$
\end{proposition}

\begin{proof}
The conjugacy class of the neutral element corresponds to 
the component containing constant loops so 
the index of the target of $j_0$ is $0$ as stated. Let 
$i$ be the index of the target of $j_1$. 
Since $Ba_\nu$ is an action map we see that $j_\mu$ is 
equivariant and that $ev_0 \circ j_\nu=id$. 
Thus $\pi_q(j_\nu )$ is injective. But the homotopy groups
of its domain and target space are the same finite groups
so it is injective as well. By the Whitehead theorem 
$j_\nu$ is a homotopy equivalence. 
$H^*(\bzi{\nu})$ is a DGA and $dt=\nu$ by Lemma \ref{dt}. 
Since $j_{\nu}$ induces an isomorphism of DGA's we have $i=1$.
\end{proof}

\begin{proposition}
\label{rp}
The map $e:(\dr (H^*\bz),d) \to (H^*(\la \bz),d)$ is an isomorphism.
\end{proposition}

\begin{proof}
Let $R=H^*\bz=\Ftwo [t]$. 
It suffices to show that 
$(j_0\sqcup j_1)^* \circ e:\dr (R) \to R\oplus R$ 
is an isomorphism. Since $ev_0 \circ j_\nu=id$ it follows that 
$t\mapsto (t,t)$ and by Lemma \ref{dt} we see that $dt\mapsto (0,1)$.
In $\dr (R)$ we have $(dt)^2=d\lambda t=dt$ such that
$(1+dt)^2=1+dt$, $(dt)(1+dt)=0$ and clearly $dt+(1+dt)=1$.
This gives an algebra splitting:
$\dr (R) = (dt)\dr (R) \oplus (1+dt) \dr (R)$. 
The components are simply 
$(dt)\dr (R)= (dt)R$ and $(1+dt)\dr (R)= (1+dt)R$ 
thus the above map is an isomorphism.
\end{proof}

Recall that an action of a group $\pi$ on an Abelian group $M$ is said to be 
{\em nilpotent} if there is a finite $\pi$-filtration of $M$ such that
$\pi$ acts trivially on the filtration quotients. 

\begin{theorem}
\label{main1}
Let $X$ be a connected space of finite type. 
Then the two maps
$$e: \dr (H^*X) \to H^*(\la X) \quad \text{ and } \quad
\psi: \ell (H^*X) \to H^*(ES^1\times_{S^1} \la X)$$
are morphisms in $\K$ which are natural in $X$.
Assume further that the group $\pi_1 (X) \times \pi_1 (X)$
acts nilpotently on $H_i(\Omega X)$ for each $i$ and that the cohomology
of $X$ is a symmetric algebra
$H^*X=S(V)$ where $V$ is a graded $\Ftwo$-vector space with $V^n$
finitely generated for each $n$. 
Then both $e$ and $\psi $ are isomorphisms of unstable 
$\sa$-algebras.
\end{theorem}

\begin{remark}
Any action of a finite 2-group on a mod 2 vector space is nilpotent.
Hence the nilpotent action condition in the theorem always holds when
$\pi_1 X$ is a finite 2-group.
\end{remark}

\begin{proof}
We must check that the $\Ftwo$-algebra maps $e$ and $\psi$ are 
$\sa$-linear.
The $\sa$-action on $\dr (H^*X)$ was described in Proposition \ref{sadr}.
Since the differential $d$ on $H^*\la X$ is $\sa$-linear we see that
$e$ is $\sa$-linear.
The $\sa$-action on $\ell (H^*X)$ is given in Theorem \ref{saell}.
By Proposition 6.7 of [BO] we see that the $\Ftwo$-algebra
map $\psi$ is also $\sa$-linear. 

Making the further assumptions of nilpotent action and polynomial 
cohomology we first show that $e$ is an isomorphism. According to 
[S] there is a fiber square with common fiber $\Omega X$ as follows.
$$
\begin{CD}
\la X @>ev>> X \\
@V ev VV @V \Delta VV \\
X @> \Delta >> X \times X
\end{CD}
$$
The associated cohomology Eilenberg-Moore spectral sequence lies in the
second quadrant and has the following $E_2$-page 
$$E_2^{-p,q}= \Tor_p^{K\otimes K}(K,K)^q $$
where $K=H^*X=S(V)$. It converges strongly to $H^*(\la X)$ by [B] 4.1.
The $E_2$-page can be interpreted by Hochschild Homology of $K$ and
then via the Hochschild Konstant Rosenberg theorem by differential
forms
$$E_2^{-p,*}=HH_p(K) \cong \Omega^p(K)$$
Here the bidegree of $x_0 dx_1 \dots dx_p$ is 
$(-p,|x_0|+ \dots +|x_p|)$. Since all algebra generators for $E_2$
sits in $E_2^{0,*}$ or $E_2^{-1,*}$ we have $E_2=E_\infty$.
The assumption that $V^n$ is finitely generated for each $n$ 
combined with the formula for the bidegree shows that
for $m$ fixed $E_\infty^{-p,m+p}=0$ for $p$ sufficiently large. 
Hence the filtration of $H^m:=H^m(\la X)$ is finite for any $m$.
As in the proof of Theorem 10.1 in [BO] one easily sees that
$e(x)\in H^*(\la X)$ represents $x\in E_2^{0,*}$ for $x\in K$.
We now show that $e(dx)\in H^*(\la X)$ represents 
$dx \in E_2^{-1,*}$. By the naturality argument given
in the proof of Theorem 10.1 in [BO] it suffices to 
verify this when $X=K(\Ftwo, n)$ where $n\geq 1$ and $x=\iota_n$ is the 
fundamental class. For $n\geq 2$ this is done in [BO]
and for $n=1$ we proceed as follows. 
Since $E_\infty =E_2=\Omega (\Ftwo [\iota_1])$ we have
$E_\infty^{-p,*}=0$ for $p\geq 2$. Thus the filtration has length two:
$$H^n=F^{-1}H^n \supseteq F^0H^n \supseteq 0 \quad , \quad
E_\infty ^{-p,n+p} \cong F^{-p}H^n/ F^{-p+1}H^n$$
From the $E_2$ page we read that $F^{-1}H^0 \cong \Ftwo \oplus \Ftwo$
and $F^0H^0\cong \Ftwo$ so it suffices to show that $e(d\iota_1)\neq 0$
and $e(d\iota_1) \neq 1= (1,1)\in H^*(\la K(\Ftwo ,1))$. 
But this is a consequence of Proposition \ref{rp}.

Defining a compatible filtration of $\dr (K)$ one sees that $e$ is 
an isomorphism as in [BO]. The same kind of Serre spectral sequence
argument as given in [BO] shows that $\psi$ is an isomorphism.
\end{proof}


\section{Detection maps}

Let $X$ be a connected space of finite type. 
In this section we shall see how it is 
possible to describe the cohomology of the evaluation map
$ev_m:E_m\la X \to \cyc_m X$ via the functor $\ell$. 
The cohomology of its target 
space has a classical interpretation. For a positive integer $k$
and a positively graded $\Ftwo$-vector space $V$ we let $C_k(V)$ 
denote the group cohomology $C_k(V):=H^*(C_k;V^{\otimes k})$
where the action on the coefficient module is by cyclic permutation.
We put $\cyc_m (V):= C_{2^m}(V)$ for $m\geq 0$. 
Our reason for choosing the  same notation as for cyclic constructions 
is the following. By classical work of Dold and Steenrod 
there is an isomorphism of $\Ftwo$-algebras
\begin{equation}
\label{DoldSteenrodiso}
\Phi : \cyc_m (H^*X) \to H^*(\cyc_m X)
\end{equation}
In [O] we found formulas for an $\sa$-action on $\cyc_m (M)$ 
when $M$ is an unstable $\sa$-module. This gave a functor
$\cyc_m :\U \to \U$ such that (\ref{DoldSteenrodiso}) is an isomorphism of
unstable $\sa$-modules.

The cohomology of the domain space of $ev_m$ is of course unknown in
general but the $\ell$ functor gives an approximation for it. It is easy to see that Theorem 4.2 in [BO] also holds for non connected spaces $Y$.
Hence we have an isomorphism
$$\theta :H^*(BC_{2^m}) \otimes_{\Ftwo [u]} 
H^*(E_\infty \la X) \to H^*(E_m\la X)$$
for $m\geq 1$. Via this we get a morphism of unstable $\sa$-algebras
\begin{equation}
\label{Psi}
\Psi = \theta \circ (1\otimes \psi) : 
H^*(BC_{2^m}) \otimes_{\Ftwo [u]} \ell (H^*X) \to H^*(E_m\la X)
\end{equation}
We know that this is an isomorphism when $X=K(\Ftwo ,n)$ with $n\geq 1$
so we can attempt to compute the map $ev^*_m$ in these cases. By naturality
this would give information on the general case.

We shall use two detection maps, $i_m$ and $\eta_m$, which we now describe.
In the following a lower index 0 on a space means the space equipped
with the trivial $S^1$-action. Firstly, the constant loop inclusion
$i:X_0 \to \la X$ gives maps
$$i_m := E_m i : BC_{2^m} \times X \to E_m \la X$$
Secondly, the action map $\eta: S^1\times (\la X)_0 \to \la X$ gives maps
$$E_m\eta : E_m(S^1\times (\la X)_0) \to E_m \la X$$
The domain space can be rewritten
$E_m(S^1\times (\la X)_0) \cong E_m(S^1)\times \la X$ and the projection
on the second factor $pr_2: E_m(S^1) \to S^1/C_{2^m}$ is a homotopy 
equivalence. For our purpose it is better with a map in the opposite direction
of $pr_2$. We choose a point $e \in ES^1$ and define
$s_0: S^1 \to ES^1\times S^1$ by $s_0(z)=(ze,z)$. It is an $S^1$-map so we 
can define
\begin{align*}
& s_m = s_0/C_{2^m} : S^1/C_2^m \to E_m(S^1) \\
& s_0 = s_0/S^1 : * \to E_\infty (S^1)
\end{align*}
Since $pr_2 \circ s_m=id$ we see that $s_m$ induces an isomorphism in 
cohomology. Our second detection map is the composite
\begin{equation}
\label{etadef}
\begin{CD}
\eta_m : S^1 /C_{2^m} \times \la X @>s_m\times 1>> E_m(S^1)\times \la X
@>\cong>> E_m (S^1\times (\la X)_0) \\
@>E_m\eta >> E_m \la X 
\end{CD}
\end{equation}
The following result explains why we call $i_m$ and $\eta_m$ detection maps.

\begin{theorem}
\label{detection}
Let $X$ be a connected space of finite type and let 
$m\geq 2$ be an integer. There is a commutative digram
\begin{equation}
\label{upperdiagram}
\begin{CD}
\Lambda (v) \otimes \ell (H^*X) @>1\otimes St>> 
\Lambda (v) \otimes \Ftwo [u] \otimes H^*X \\
@V1\otimes \psi VV @V\cong VV \\
\Lambda (v) \otimes H^*(E_\infty \la X) @>1\otimes i_\infty^*>>
\Lambda (v) \otimes H^*(BS^1 \times X) \\
@V\cong VV @V\cong VV \\
H^*(E_m\la X) @>i_m^* >> H^*(BC_{2^m} \times X)
\end{CD}
\end{equation}
and a commutative diagram
\begin{equation}
\label{lowerdiagram}
\begin{CD}
\Lambda (v) \otimes \ell (H^*X) @>1\otimes DR>> 
\Lambda (v) \otimes \dr (H^*X) \\
@V1\otimes \psi VV @V1\otimes e VV \\
\Lambda (v) \otimes H^*(E_\infty \la X) @>1\otimes q_\infty^0>>
\Lambda (v) \otimes H^*(\la X) \\
@V\cong VV @V\cong VV \\
H^*(E_m\la X) @>\eta_m^* >> H^*(S^1/C_{2^m} \times \la X)
\end{CD}
\end{equation}
Thus the map 
$$(i_m^*,\eta_m^*):H^*(E_m \la X) \to H^*(BC_{2^m} \times X)
\oplus H^*(S^1/C_{2^m}\times \la X)$$
is injective when $X=K(\Ftwo ,n)$ with $n\geq 1$.
\end{theorem}

\begin{proof}
By the argument given in the proof of Proposition 7.1 of [BO] we see that
the lower square in (\ref{upperdiagram}) commutes. By Proposition 7.2 of 
[BO] the upper square in this diagram also commutes (in [BO] we assumed
that $\la X$ was connected, but it is easy to see that this assumption
is not necessary). 

The composite (\ref{etadef}) defining $\eta_m$ sits over the following
composite via quotient maps:
$$
\begin{CD}
* \times \la X @>s_\infty \times 1>> E_\infty (S^1) \times \la X
@>\cong >> E_\infty (S^1\times (\la X)_0)  
@>E_\infty \eta >> E_\infty \la X
\end{CD}
$$
which is given by $(*,\omega ) \mapsto [e,\omega ]$. So the top square
in the following diagram commutes. Mapping down by
projections $pr_1$ we get the whole commutative cube:
$$ \diagram
S^1/C_{2^m}\times \la X \rrto^{\eta_m } \drto \ddto   
& & E_m \la X  \drto^{Q_m^\infty} \xto'[d][dd] \\
& *\times \la X \rrto^(.3){Q_0^\infty} \ddto
& & E_\infty \la X \ddto \\
S^1/C_{2^m} \xto'[r]^j[rr] \drto
& & BC_{2^m} \drto \\
& * \rrto
& & BS^1
\enddiagram $$
Here $Q_m^\infty$ and $Q_0^\infty$ are quotient maps and $j$ is the
fiber inclusion of the fibration following it in the cube. 
From this cube and the definition of $\theta$ in Theorem 4.2 of [BO]
we see that the lower square of (\ref{lowerdiagram})
commutes. The top square commutes by Theorem \ref{l-approx}.

When $X=K(\Ftwo ,n)$ for $n\geq 1$ the vertical maps in (\ref{upperdiagram})
and (\ref{lowerdiagram}) are all isomorphisms
so $(i_m^*,\eta_m^* )$ is injective by Theorem \ref{saell}.
\end{proof}


\section{The twisted diagonal map}

Let $Y$ be an $S^1$-space with action map $\eta: S^1\times Y \to Y$ and
associated differential $d$ on $H^*Y$. 

\begin{definition}
For $m \geq 0$ we define the map
\begin{eqnarray*}
f_{Y,m}^\prime : S^1 \times Y &\to & ES^1\times Y^{2^m} \\
(z,y) & \mapsto & (ze,zy,z\xi_m y, z\xi_m^2 y, \dots , z\xi_m^{2^m-1}y)
\end{eqnarray*}
where $\xi_m= exp(2\pi i /2^m)$ and $e$ is a point in $ES^1$. 
We let $C_{2^m}$ act on the domain space by 
$\xi_m \cdot (z,y)= (\xi_mz,y)$ and on the target space by
$$\xi_m \cdot (e,y_1, \dots , y_{2^m}) = (\xi_me,y_2, \dots , y_{2^m},y_1)$$
Then the above map is $C_{2^m}$-equivariant. Passing to the quotients
we get a map
$$f_{Y,m}: S^1/ C_{2^m} \times Y \to \cyc_m Y$$
which we call the twisted diagonal of order $2^m$.
\end{definition}

\begin{remark}
$f_{Y,m}$ is natural in $Y$ with respect to $C_{2^m}$-equivariant maps. 
\end{remark}

\begin{remark}
\label{etaremark}
Our reason for introducing the twisted diagonal map is that it can be used
to compute $\eta_m^* \circ ev_m^*$ since there is a commutative diagram
$$
\begin{CD}
S^1/C_{2^m} \times \la X @>\eta_m >> E_m\la X \\
@Vf_{\la X,m}VV @Vev_mVV \\
E_m\la X @>E_m(ev_0)>> \cyc_m X
\end{CD}
$$
\end{remark}

\begin{lemma}
\label{twisttrivial}
Let $X$ be a space with trivial $S^1$-action and $m\geq 1$ an integer.
The map $f_{X,m}^*:H^*(\cyc_m X) \to H^*(S^1\times X)$ 
satisfies
$$f_{X,m}(1\otimes x^{\otimes 2^m})= 
1\otimes x^{2^m}+v\otimes x^{2^m-2}\lambda x$$
\end{lemma}

\begin{proof}
There is a factorization
$$ \begin{CD}
f_{X,m}:S^1/C_{2^m} \times X @>i\times 1>> ES^1/C_{2^m}\times X
@>1\times \Delta_m>> \cyc_mX 
\end{CD} $$
where $i:S^1 \to ES^1$ is the inclusion associated to $e$ and 
$\Delta_m$ is the diagonal $\Delta_m(y)=(y,\dots ,y)$.
When $m=1$ the result follows by Steenrod's formula for the diagonal.
For $m\geq 2$ we use Theorem 6.7 of [O] where the map 
$(1\times \Delta_m)^*$ was determined. We find
$$f_{X,m}^*(1\otimes x^{\otimes 2^m})=
1\otimes x^{2^m} +v \otimes Q_{m-1}^{2^{m-1}|x|-1}(x)$$
The operation $Q_n^i$ is defined in Definition 5.5 of [O].
We have $Q_1^{2|x|-1}(x)=x^2\lambda x$ giving
$Q_n^{2^n|x|-1}(x)=x^{2^{n+1}-2}\lambda x$ in general. 
\end{proof}

\begin{proposition}
\label{twisttr}
There is a commutative diagram
$$ \begin{CD}
H^*(\cyc_m Y) @>f_{Y,m}^*>> H^*(S^1/C_{2^m} \times Y) \\
@A\tau_{m-1}^m AA @A\tau_{m-1}^m \otimes 1AA \\
H^*(\cyc_{m-1} (Y^2)) @>(f_{Y,m}^\prime /C_{2^{m-1}})^* >> 
H^*(S^1/C_{2^{m-1}} \times Y) \\
@V\cyc_{m-1}(\Delta_1)^* VV @| \\
H^*(\cyc_{m-1} Y) @>f_{Y,m-1}^* >> H^*(S^1/C_{2^{m-1}}\times Y)
\end{CD} $$
where $\Delta_1 : Y \to Y\times Y$ denotes the diagonal $\Delta_1(y)=(y,y)$. 
Especially we have $f_{Y,m}^*(1\otimes Na_1 \otimes \dots \otimes a_{2^m})=
v\otimes d(a_1 \dots a_{2^m})$.
\end{proposition}

\begin{proof}
It is obvious that the upper square commutes. The lower square is
induced by a homotopy commutative diagram of spaces.
The norm class equals
$\tau_0^m (1\otimes a_1 \otimes \dots \otimes a_{2^m})$ so we can
find its image under $f_{Y,m}^*$ by use of the diagram.
From the factorization 
$$ \begin{CD} 
f_{Y,0}: S^1\times Y @>\Delta_1 \otimes 1 >> S^1\times S^1 \times Y
@>i\times \eta>> ES^1 \times Y
\end{CD} $$
we see that $f_{Y,0}^*(1\otimes y)=1\otimes y +v\otimes dy$ and the
result follows.
\end{proof}

\begin{lemma}
\label{twists1}
For $m\geq 1$ the map $f_{S^1,m}^*:H^*(\cyc_m S^1) \to 
H^*(S^1/C_{2^m} \times S^1)$ is zero on all classes of positive degrees
except for $1\otimes N v\otimes 1\otimes \dots \otimes 1$ which is
mapped to $v\otimes 1$.
\end{lemma}

\begin{proof}
When $m=1$ this is a special case of Lemma 4.11 and Theorem 4.12 of
[BO]. The differential on $H^*(S^1)$ is given by $dv=1$ since
$\eta ^*(v)=1\otimes v+v\otimes 1$. 

Assume that $m\geq 2$. Since the target space is $\Lambda (v) \otimes
\Lambda (v)$ all classes of higher degree than two is mapped to zero.
The degree two class
$$1\otimes (1+T+\dots +T^{2^{m-1}-1}) 
(v\otimes 1^{\otimes 2^{m-1}-1})^{\otimes 2} =
\tau_1^m(1\otimes (v\otimes 1^{\otimes 2^{m-1}-1})^{\otimes 2})$$
is mapped to zero by Proposition \ref{twisttr} and the $m=1$ case.
All other elements of degree at most two are norm elements and
Proposition \ref{twisttr} takes care of these.
\end{proof}

We now introduce an expansion formula which will be used in the 
proof of the next theorem and again later on. Let $M$ be an
$\Ftwo$-vector space and $k\geq 1$ an integer. The cyclic group
$C_k$ act on $M^{\otimes k}$ by cyclic permutation of the factors.
For $x,y\in M$ we let $T_k(x,y)$ denote the subset of $M^{\otimes k}$ 
consisting of all strings 
$a_1\otimes \dots \otimes a_k$ with 
$a_i=x$ or $a_i=y$ for each $i$. Note that $T_k(x,y)$ is stable 
under the $C_k$ action and that
$(x+y)^{\otimes k}$ is the sum of the elements in $T_k(x,y)$. 
For an orbit $\beta \in T_k(x,y)/C_k$ we let 
$|\beta |$ denote the length of $\beta$ and $s\beta$ the sum 
of the elements in $\beta$. Thus we have the following expansion formula
among $C_k$-invariants in $M^{\otimes k}$:
\begin{equation}
\label{invexp}
(x+y)^{\otimes k} = \sum s\beta
\end{equation}
where the summation is taken over $\beta \in T_k(x,y) / C_k$.

\begin{theorem}
\label{fcomp}
For $m\geq 1$ we have
$$f_{Y,m}^*(1\otimes y^{\otimes 2^m})=1\otimes y^{2^m} + 
v\otimes y^{2^m-2}(ydy+\lambda y)$$
\end{theorem}

\begin{proof}
Let $Y_0$ denote the space $Y$ with trivial $S^1$-action. The action map
$\eta :S^1\times Y_0 \to Y$ is $S^1$-equivariant so by naturality there
is a commutative diagram
$$
\begin{CD}
S^1/C_{2^m} \times S^1\times Y_0 @>f_{S^1\times Y_0,m}>>
\cyc_m (S^1 \times Y_0) \\
@V1\times \eta VV @V \cyc_m (\eta ) VV \\
S^1/C_{2^m} \times Y @>f_{Y,m}>> \cyc_m (Y)
\end{CD}
$$
The map $\gamma: Y\to S^1 \times Y$ defined by $\gamma (y) =(1,y)$ 
has the property $\eta \circ \gamma =id$ thus
$$f_{Y,m} = \cyc_m (\eta ) \circ f_{S^1\times Y_0,m} \circ (1\times \gamma )$$
We pull back step by step according to this factorization.
$$\cyc_m (\eta)^*(1\otimes y^{\otimes 2^m})=
1\otimes (1\otimes y+v\otimes dy)^{\otimes 2^m}=
\sum 1\otimes s\beta$$  
where the summation is taken over 
$\beta \in T_{2^m}(1\otimes y, v\otimes dy)$.
Consider an orbit $\beta $ with $|\beta |=2^i$ where $1\leq i \leq m$.
A representative for this has the form
$\overline \beta = (a_1 \otimes \dots \otimes a_{2^i})^{\otimes 2^{m-i}}$.
By Proposition \ref{twisttr} we have
\begin{equation}
\label{fbeta}
f_{S^1\times Y_0,m}^*(1\otimes s\beta)=
(\tau_{m-i}^m \otimes 1)\circ f_{S^1\times Y_0,m-i}^*
(1\otimes(a_1 \dots a_{2^i})^{\otimes 2^{m-i}})
\end{equation}
Since $v$ is an exterior class we see that (\ref{fbeta}) is zero
when two or more of the elements $a_1, \dots , a_{2^i}$ equals
$v\otimes dy$. If there is exactly one of these we have
$a_1 \dots a_{2^i}=v\otimes y^{2^i-1}dy$ and (\ref{fbeta}) equals 
zero unless $i=m$ by Lemma \ref{twists1}.
When $i=m$ we have a norm element so Proposition \ref{twisttr} gives 
\begin{align*}
f_{S^1\times Y_0,m}^*(1\otimes s \beta) &=v\otimes d_{S^1\times Y_0}
(v\otimes y^{2^m-1}dy) \\
&=v\otimes d_{S^1}(v)\otimes y^{2^m-1}dy + 
v\otimes v\otimes d_{Y_0}(y^{2^m-1}dy) \\
&=v\otimes 1 \otimes y^{2^m-1}dy
\end{align*}
We have now taken care of all the terms except of
$1\otimes (1\otimes y)^{\otimes 2^m}$ and 
$1\otimes (v\otimes dy)^{\otimes 2^m}$ which corresponds to orbits
of length one. By Lemma \ref{twisttrivial} the first is mapped as follows
$$f_{S^1\times Y_0,m}^*(1\otimes (1\otimes y)^{\otimes 2^m})=
1\otimes 1\otimes y^{\otimes 2^m} +v\otimes 1\otimes y^{2^m-2}\lambda y$$
and Lemma \ref{twists1} gives that the second is mapped to zero.
We conclude that
$$f_{S^1\times Y_0,m}^*\circ \cyc_m(\eta )^*(1\otimes y^{\otimes 2^m})=
1\otimes 1\otimes y^{\otimes 2^m} + 
v\otimes 1\otimes y^{2^m-2}(ydy+\lambda y)$$
The result follows by applying $1\otimes \gamma ^*$.
\end{proof}

\begin{corollary}
\label{etaev}
For $m\geq 2$ one has
$$\eta_m^* \circ ev_m^* (1\otimes x^{\otimes 2^m})=
1\otimes e(x^{2^m})+v\otimes e(x^{2^m-2}(xdx+\lambda x))$$
\end{corollary}

\begin{proof}
Follows directly from Theorem \ref{fcomp} and Remark \ref{etaremark}.
\end{proof}


\section{Formulas for the evaluation maps}

In this section we find a formula for $ev_m^*(1\otimes x^{\otimes 2^m})$ 
in terms of the map (\ref{Psi}). This is done by the two detection
maps $\eta_m$ and $i_m$. Note that $i_m\circ ev_m=\Delta_m$
where $\Delta_m = E_m(D_m)$ with $D_m:X \to X^{2^m}$ the diagonal map.
We computed $\Delta_m^*$ in [O]. The result involved some notation
which we now introduce. In this section $K$ always denotes 
an unstable $\sa$-algebra.

\begin{definition}
For integers $k\geq 1$ and $n\geq 0$ we define $\lambda_k : K \to K$ and
$Q_k^n:K\to K$ by
\begin{eqnarray}
\lambda_k (x) &=& x^{2^k-2}\lambda x \\
Q_1^n(x) &=& \sum_{i=0}^{[\frac n 2]} Sq^i(x)Sq^{n-i}(x) \\
Q_k^n(x) &=& \sum_{i=0}^n Sq^i(x^{2^k-2})Q_1^{n-i}(x) \quad , \quad k\geq 2
\end{eqnarray}
We put $Q_k^n=0$ when $n\leq -1$.
\end{definition}

\begin{definition}
For integers $k\geq 1$ we define three maps $K \to \Ftwo [u] \otimes K$
as follows
\begin{eqnarray}
F_{k+1}(x) &=& \sum_{i=0}^\infty u^{2^{k-1} (|x|-i)} \otimes 
(Sq^ix)^{2^k} \\
G_{k+1}^\prime (x) &=& \sum_{j=0}^\infty u^{2^{k-1}|x|-j} \otimes
Sq^{2j}(\lambda_k x) \\
G_{k+1}^{\prime \prime} (x) &=& \sum_{j=0}^\infty u^{2^{k-1}|x|-j} \otimes
Q_k^{2j-1}(x) 
\end{eqnarray}
\end{definition}

Note that the sums are finite. $F_m$ from Definition 6.5 of [O] 
is the same as above and $G_m$ from Definition 6.5 of [O] equals the sum 
$G_m^\prime +G_m^{\prime \prime}$. 
Thus by Theorem 6.7 of [O] we have
$$\Delta_m^*(1\otimes x^{\otimes 2^m})=F_m(x)+v(G_m^{\prime}(x)+
G_m^{\prime \prime}(x))$$
for $m\geq 2$.

\begin{proposition}
\label{stongen}
For $x\in K$ and $k\geq 1$ the following equations hold.
\begin{align}
& St(\phi(x^{2^k}))=F_{k+1}(x) \label{steq1} \\
& St(uq(\lambda_kx))=G_{k+1}^{\prime}(x) \label{steq2} \\
& St(\phi(x^{2^k-1})q(x)) = G_{k+1}^{\prime \prime}(x) \label{steq3}
\end{align}
\end{proposition}

\begin{proof}
Equation (\ref{steq1}) and (\ref{steq2}) follows easily from the
definition of $St$ in Proposition \ref{defofst}. We first prove
(\ref{steq3}) for $k=1$. Since $St$ is a ring homomorphism we have that
$$St(\phi (x) q(x)) = \sum_{j=0}^{\infty} u^j\otimes
\sum_{r+s=j} Sq_{2r}(x)Sq_{2s+1}(x) $$
So we must show that
\begin{equation}
\label{qeqn1}
Q_1^{2|x|-2j-1}(x)= \sum_{r+s=j} Sq_{2r}(x) Sq_{2s+1}(x)
\end{equation}
We can rewrite the left hand side using the definition followed by
the substitution $t=|x|-i$.
\begin{align}
Q_1^{2|x|-2j-1}(x) &= \sum_{t=0}^{|x|-j-1} 
Sq^t(x)Sq^{2|x|-2j-1-t}(x) \nonumber \\
&=\sum_{i=j+1}^{|x|} Sq_i(x)Sq_{2j+1-i}(x)  \nonumber \\
&=\sum_{i=j+1}^{2j+1} Sq_i(x)Sq_{2j+1-i}(x) \label{qeqn2}
\end{align}
The map, $f_j: \{ 0,1, \dots ,j \} \to \{ j+1, j+2, \dots , 2j+1 \}$
defined by $f_j(r)=2(j-r)+1$ for $r\leq [j/2]$ and $f_j(r)=2r$ otherwise,
is a bijection from the set of summation indexes of (\ref{qeqn1})
to the set of summation indexes of (\ref{qeqn2}). Since the $r$'th term
of the right hand side of (\ref{qeqn1}) equals the $i=f_j(r)$'th term
of (\ref{qeqn2}) the result follows.

Finally we prove (\ref{steq3}) for $k\geq 2$. By (\ref{r6}) we have
$q(x^{2^k-2})=0$ such that 
$\phi (x^{2^k-1})q(x)=\phi (x^{2^k-2})\phi(x) q(x)$ by (\ref{r5}).
Thus we can apply the $k=1$ case as follows:
\begin{align*}
St(\phi (x^{2^k-1})q(x)) &=St(\phi (x^{2^k-2})) St(\phi(x) q(x)) \\
&= \sum_{j=0}^{\infty} u^j \otimes 
\sum_{r+s=j} Sq_{2r}(x^{2^k-2})Q_1^{2|x|-2s-1}(x)
\end{align*}
and we must prove the following equation for each $j$:
\begin{equation}
\label{qeqn3}
Q_k^{2^k|x|-2j-1}(x)=\sum_{r+s=j} Sq_{2r} (x^{2^k-2}) Q_1^{2|x|-2s-1}(x)
\end{equation}
By its definition and the fact that $Sq^i(y^2)=0$ we can rewrite
the left hand side as
$$Q_k^{2^k|x|-2j-1}(x)=\sum_{t=0}^\infty Sq^{2t}(x^{2^k-2})
Q_1^{2^k|x|-2(j+t)-1}(x)$$
We substitute $t=(2^{k-1}-1)|x|-r$ and get that
$$Q_k^{2^k|x|-2j-1}(x)=\sum_{r=-\infty}^{(2^{k-1}-1)|x|} 
Sq_{2r}(x^{2^k-2})Q_1^{2|x|-2(j-r)-1}(x)$$
If $r<0$ the first factor in each summand is zero so we can start the
summation at $r=0$. If $r>j$ the second factor in each summand is zero and 
if $r>(2^{k-1}-1)|x|$ the first factor in each summand is zero. So we
can end the summation at $j$ obtaining the formula (\ref{qeqn3}).
\end{proof}

\begin{theorem}
\label{spaceev}
Let $X$ be a connected space  of finite type and let $x,y \in H^*X$. For  
integers $m\geq 2$ the following equation holds
$$ev_m^*(1\otimes x^{\otimes 2^m})=
\Psi \big( \phi (x^{2^{m-1}}) + v\phi(x^{2^{m-1}-1})q(x) 
+vu q(x^{2^{m-1}-2}\lambda x) \big)$$
For $m=1$ we have 
$$ev_1^*(1\otimes x^{\otimes 2})=\Psi (\phi (x)+tq(x))\quad \text{ and } \quad
ev_1^*(1\otimes (1+T)x\otimes y)=\Psi (t\delta (xy))$$
Further the square with transfer maps, the square with quotient maps
and the triangle in the following diagram commutes.
$$ \diagram
& H^*(\cyc_m X) \rto^-{ev_m^*} \dto<1ex>^{q_m^{m-1}}
& H^*(E_m\la X) \dto<1ex>^{q_m^{m-1}}\\
& H^*(\cyc_{m-1}(X^2)) \uto<1ex>^{\tau_{m-1}^m} \dto_{\Delta_1^*} \rto 
& H^*(E_{m-1}\la X) 
\uto<1ex>^{\tau_{m-1}^m} \\
& H^*(\cyc_{m-1}X) \urto_{ev_{m-1}^*}
\enddiagram $$
where $\Delta_1: X \to X\times X$ denotes the diagonal. 
\end{theorem}

\begin{proof}
By naturality it is enough to prove the formula for $ev_m^*$ when
$X=K(\Ftwo ,n)$ with $n\geq 1$ and $x=\iota_n$ is the 
fundamental class.
By Theorem \ref{detection} it suffices to check that 
the right hand side maps correctly by $\eta_m^*$ and
$i_m^*$. By Theorem 6.7 of [O] and Proposition \ref{stongen}
it does by $i_m^*$. 
For $\eta_m^*$ we find
\begin{align*}
& 1\otimes DR(\phi (x^{2^{m-1}}) +v \phi (x^{2^{m-1}-1})q(x)
+vuq(x^{2^{m-1}-2}\lambda x))= \\
& x^{2^m} +vx^{2^m-2}(xdx+\lambda x)
\end{align*}
which is the correct result by Corollary \ref{etaev}.
The formulas for $m=1$ follows from Lemma 6.4 and Proposition 6.5
of [BO].

The squares in the diagram commute since there is a corresponding 
commutative diagram of spaces with evaluation and quotient maps. 
The triangle comes from a homotopy commutative diagram of spaces.
\end{proof}

\begin{remark}
The map $ev_m^*$ is $H^*(BC_{2^m})$-linear and elements in its domain
which are not of highest symmetry are hit by the transfer.
So we have given a complete description of $ev_m^*$ in terms of 
the approximation $\Psi$.
\end{remark}

\section{Algebraic evaluation maps}

In this section we show that the formulas for the map $ev_m^*$ also
define a ring map when we replace $H^*X$ by a general FGA. 
This allows us to state our main theorem in a nice form.

%\begin{proposition}
%Let $M$ be a non negatively graded $\Ftwo$-vector space with $M^q$ finitely
%generated for each $q$. Let $m\geq 1$ be an integer.
%Choose a basis for each $M^q$ and let $B_m(M)$ denote the corresponding
%product basis for $M^{\otimes 2^m}$. It possesses a $C_{2^m}$-action
%by cyclic permutation. For an orbit $\beta \in B_m(M)/C_{2^m}$ we
%write $|\beta |$ for the length of the orbit, $s\beta$ for the sum 
%of the elements in $\beta$ and $\bar \beta$
%for an representative of $\beta$.
%For $m\geq 2$ the group cohomology 
%$H^*(C_{2^m};M^{\otimes 2^m})$ has the following basis
%$$u^i \otimes s\beta \quad , \quad 1\otimes s\gamma \quad , \quad
%vu^i\otimes \bar \beta$$
%where $\beta$ and $\gamma$ are orbits with
%$|\beta |\leq 2^{m-1}$ and $|\gamma |=2^m$ and $i\geq 0$.
%For $m=1$ the group cohomology $H^*(C_2;M^{\otimes 2})$
%has basis
%$$t^i\otimes \beta \quad ,\quad 1\otimes s\gamma$$
%where $|\beta|=1$, $|\gamma |=2$ 
%\end{proposition}

Recall that the deviation from linearity of a set map $F:A\to B$ 
between $\Ftwo$-algebras is defined as
$\Delta F(x,y)=F(x+y)+F(x)+F(y)$.
If $L:A\to B$ is linear and we let $G(x)=L(x)F(x)$ we have
\begin{equation}
\label{deveqn}
\Delta G(x,y)=L(x)F(y)+L(y)F(x)+(L(x)+L(y))\Delta F(x,y)
\end{equation}

\begin{definition}
Let $K$ be an FGA.
The maps $f_i,g_i,h_i:K \to \ell (K)$ for integer $i\geq 0$ are defined by
$f_0(x)=f_1(x)=0$, $g_0(x)=\delta (x)$ and
\begin{eqnarray*}
f_i(x) =& uq(x^{2^{i-1}-2}\lambda x) & \text{ for } i\geq 2 \\
g_i(x) =& \phi (x^{2^{i-1}-1})q(x) & \text{ for } i\geq 1 \\
h_i(x) =& f_i(x)+g_i(x) & \text{ for } i\geq 0
\end{eqnarray*}
\end{definition}

\begin{lemma}
\label{productanddev}
For $w=f,g,h$ and $m\geq 1$ the following formulas are valid.
\begin{align}
& w_m(ab)=\phi (a^{2^{m-1}})w_m(b)+ \phi (b^{2^{m-1}})w_m(a) 
\label {wrel1} \\
& \Delta w_m(x,y)= \sum_{i=0}^{m-1} \sum_{r+s=2^{m-i}}
w_i(x^ry^s) \label{wrel2}
\end{align}
\end{lemma}

\begin{proof}
It is of course enough to prove this for $w=f$ and $w=g$.
We first prove (\ref{wrel1}). For $m=1$ and $w=f$ it is trivial.
For $m=1$ and $w=g$ it holds by (\ref{r6}). The $m=2$ case
requires a little computation. For $w=f$ we have
\begin{align*}
f_2(ab)&=uq(\lambda (ab))=uq(a^2\lambda b + b^2\lambda a) \\
&=u(\phi (a^2)q(\lambda b)+\phi (b^2)q(\lambda a))=
\phi (a^2) f_2(b)+ \phi (b^2) f_2(a)
\end{align*}
and for $w=g$ we find
\begin{align*}
g_2(xy) &= (\phi (x) \phi (y) +uq(x)q(y))(\phi (x) q(y) + q(x)\phi (y)) \\
&= \phi (x)^2 g_2(y) +\phi (y)^2 g_2(x)+uq(y)^2g_2(x)+uq(x)^2g_2(y) \\
&= (\phi (x)^2+uq(x)^2)g_2(y)+(\phi (y)^2+uq(y)^2)g_2(x) \\
&= \phi (x^2)g_2(y)+\phi (y^2)g_2(x)
\end{align*}
Note that when $m\geq 3$ the following formula holds for both
$w=f$ and $w=g$.
\begin{equation}
\label{wmw2}
w_m(x) = \phi (x^{2^{m-1}-2})w_2(x)
\end{equation}
So for these $m$ we have
\begin{align*}
w_m(ab) &= \phi ((ab)^{2^{m-1}-2})w_2(ab) \\
&= \phi (a^{2^{m-1}-2})\phi (b^{2^{m-1}-2})
(\phi (a^2)w_2(b)+\phi(b^2)w_2(a)) \\
&= \phi (a^{2^{m-1}}) w_m(b) +\phi (b^{2^{m-1}}) w_m(a)
\end{align*}

Next we prove (\ref{wrel2}).
From (\ref{wrel1}) and $\delta (a^2)=0$ we see that 
$w_i(a^2)=0$ for any $i\geq 0$. Thus the terms with both $r$ and $s$ even
are zero. For $m=1$ we have $f_1=0$ and $g_1=q$ such that
$\Delta g_1(x,y)=\delta (xy)=g_0(xy)$. So here (\ref{wrel2}) holds.
For $m=2$ we have $f_2(x)=uq(\lambda x)$ which is linear. 
$g_2(x)=\phi (x) q(x)$ so by (\ref{deveqn}) and (\ref{wrel1}) we have
$$\Delta g_2(x,y)= g_1(xy) + (\phi (x)+ \phi (y))\delta(xy)= 
g_1(xy)+\delta (xy^3)+\delta (x^3y)$$
Thus (\ref{wrel2}) also holds for $m=2$. 

For $m\geq 3$ we have $w_m(x)=\phi (x^{2^{m-2}})w_{m-1}(x)$ by
(\ref{wmw2}) so (\ref{deveqn}) and (\ref{wrel1}) gives
$$\Delta w_m(x,y)= w_{m-1}(xy)+(\phi (x^{2^{m-2}}) + \phi (y^{2^{m-2}}))
\Delta w_{m-1}(x,y)$$
Assume that $\Delta w_{m-1}(x,y)$ is given by (\ref{wrel2}). By (\ref{wrel1})
we have $\phi (a^{2^{m-2}})w_i(b)=w_i(a^{2^{m-1-i}}b)$ for $1\leq i \leq m-2$.
This formula also holds for $i=0$ trivially for $w=f$ and by (\ref{r9})
for $w=g$. So we have
$$\Delta w_m(x,y) = w_{m-1}(xy)+ \sum_{i=0}^{m-2} \sum_{r+s=2^{m-1-i}}
\big( w_i(x^{2^{m-1-i}+r}y^s)+w_i(x^ry^{2^{m-1-i}+s})\big)$$
Since $w_i(a^2)=0$ for any $i\geq 0$ this sum equals the right hand side
of (\ref{wrel2}).
\end{proof}

The cyclic functor $C_{2^m}: \U \to \U$ was introduced in Theorem 4.8 of [O]. 
We denote it $\cyc_m=C_{2^m}$. 
Recall that there are transfer and restriction
maps as follows
$$\tau_{m-1}^m: \cyc_{m-1}(M\otimes M) \to \cyc_m(M) \quad \text{ and } \quad
q_m^{m-1}: \cyc_m(M) \to \cyc_{m-1}(M\otimes M)$$
where $M\in ob(\U )$. Formulas for these were given in section 3 of [O].

\begin{proposition}
\label{crestricts}
For each $m\geq 1$ the functor $\cyc_m: \U \to \U$ restricts to a functor
$\cyc_m: \K \to \K$.
\end{proposition}

\begin{proof}
Let $K$ be an unstable $\sa$-algebra. We must verify that the 
Cartan formula and the relation $Sq^{|x|}x=x^2$ hold in 
$\cyc_m(K)$. This is done by induction on $m$. 
For $m=1$ the result is well known. Assume that it holds for $m-1$.
Let $\tau=\tau_{m-1}^m$ and 
$q=q_m^{m-1}$. We know that these are $\sa$-linear. 
Using Frobenius reciprocity we can then check
the Cartan formula when one of the factors in the product is
hit by the transfer:
\begin{align*}
Sq^n(a\tau (b)) &= Sq^n (\tau (q(a)b)) = \tau (Sq^n(q(a) b)) \\
& = \tau ( \sum_{i+j=n} Sq^i(q(a))Sq^j (b))
= \sum_{i+j=n} \tau (q(Sq^i(a)) Sq^j(b)) \\
& = \sum_{i+j=n} Sq^i(a) \tau (Sq^j(b)) = 
\sum_{i+j=n} Sq^i(a) Sq^j(\tau (b))
\end{align*}
Hence it suffices to verify the Cartan formula for a product of two
highest symmetry classes. For $x,y \in K$ with $|x|=r$ and $|y|=s$
there is a morphism of unstable $\sa$-algebras
$f:H^*(K(\Ftwo ,r)\times K(\Ftwo ,s)) \to K$ 
given by $f(\iota_r \otimes 1)=x$ and $f(1\otimes \iota_s)=y$.
If we apply $C_{2^m}$ to the cohomology of a space we know that
this is isomorphic to the cohomology of the cyclic construction of
order $2^m$ on the space. So here the Cartan formula holds.
So by the map $f$ we see that it also holds for the product
of $v^iu^j \otimes x^{\otimes 2^m}$ and $v^k u^n \otimes y^{\otimes 2^m}$.
The relation $Sq^{|x|}x=x^2$ is easy to verify directly from the
formulas for the $\sa$-action.
\end{proof}

We can now give the final description of the maps induced by the
equivariant evaluations in mod two cohomology. 

\begin{theorem}
Let $K$ be an FGA with multiplication $\mu : K\otimes K \to K$. 
There is a natural $H^*(C_{2^m})$-linear ring homomorphism
$$Ev_m:\cyc_m(K) \to H^*(C_{2^m}) \otimes _{\Ftwo [u]} \ell (K)$$
for each $m\geq 1$. When $m=1$ it is defined by  
$1\otimes x^{\otimes 2} \mapsto \phi(x) +t q(x)$ and
$1\otimes (1+T)a\otimes b \mapsto  t\delta (ab)$
where $t^2=u$.
For $m\geq 2$ it is defined by the following two conditions
\begin{align}
& Ev_m(1\otimes x^{\otimes 2^m})= 
\phi (x^{2^{m-1}})+v\phi (x^{2^{m-1}-1})q(x) +vuq(x^{2^{m-1}-2}\lambda x) 
\label{evcond1} \\
& Ev_m\circ \tau_{m-1}^m= (\tau_{m-1}^m \otimes 1) \circ Ev_{m-1}\circ \mu_*
\label{evcond2}
\end{align}
where the second condition relates maps in the following diagram.
$$
\diagram
& \cyc_m (K) \rto^-{Ev_m} \dto<1ex>^{q_m^{m-1}}
& H^*(C_{2^m}) \otimes_{\Ftwo [u]} \ell (K) \dto<1ex>^{q_m^{m-1}\otimes 1}\\
& \cyc_{m-1}(K^{\otimes 2}) \uto<1ex>^{\tau_{m-1}^m} \dto_{\mu_*} \rto 
& H^*(C_{2^{m-1}})\otimes_{\Ftwo [u]} \ell (K) 
\uto<1ex>^{\tau_{m-1}^m \otimes 1} \\
& \cyc_{m-1}(K) \urto_{Ev_{m-1}} 
\enddiagram
$$
The following equation holds for $m\geq 2$:
\begin{equation}
\label{evprop}
(q_m^{m-1}\otimes 1)\circ Ev_m=Ev_{m-1}\circ \mu_* \circ q_m^{m-1}
\end{equation}
If $K$ is a connected unstable $\sa$-algebra then $Ev_m$ is a morphisms in 
$\K$ for each $m\geq 1$. Finally if $K=H^*X$ for a connected space
$X$ of finite type then we have a commutative diagram in $\K$ as follows.
$$\begin{CD}
H^*(\cyc_m X) @>ev_m^*>> H^*(E_m \Lambda X) \\
@A\cong AA @A\Psi AA \\
\cyc_m (H^*X) @>Ev_m>> H^*(C_{2^m})\otimes _{\Ftwo [u]} \ell (H^*X)
\end{CD}$$
\end{theorem}

\begin{proof}
We first check that $Ev_1$ is well defined. $\phi (x)$ and $uq(x)$ are
linear in $x$ and $\delta (ab)$ is bilinear in $(a,b)$. From this
it follows directly that most relations in $H^*(C_2;K^{\otimes 2})$
are respected. In fact it suffices to verify that the following
type is respected.
$$1\otimes (x+y)^{\otimes 2}=
1\otimes x^{\otimes 2}+1\otimes y^{\otimes 2} + 1\otimes (1+T) x\otimes y$$
But since $\phi$ is linear this means   
that $tq(x+y)=tq(x)+tq(y)+t\delta(xy)$ which is true by (\ref{r3}).

Next we check that $Ev_1$ is a ring homomorphism. 
The following relations must be respected:
\begin{align*}
(1\otimes x^{\otimes 2})(1\otimes y^{\otimes 2}) &=
1\otimes (xy)^{\otimes 2} \\
(1\otimes x^{\otimes 2})(1\otimes (1+T) a\otimes b) &= 
1\otimes (1+T) xa \otimes xb \\
(1\otimes (1+T) a\otimes b)(1\otimes (1+T) c\otimes d) &=
1\otimes (1+T) ac\otimes bd \\
& + 1\otimes (1+T) ad\otimes bc
\end{align*}
The third is respected since both sides maps to zero.
For the first and the second we must check that
\begin{align*}
&(\phi (x) +tq(x))(\phi (y) +tq(y))=\phi (xy)+tq(xy) \\
&(\phi (x) +tq(x))t\delta (ab)= t\delta (x^2ab)
\end{align*}
which follows by (\ref{r5}), (\ref{r6}) and (\ref{r9}), (\ref{r11}).

For the higher evaluation maps we use induction on $m$. 
Assume that the maps
$Ev_1, \dots ,Ev_{m-1}$ have the stated properties. 
We first verify that $Ev_m$ is well defined. All classes in its domain
are hit by the transfer except for those of highest symmetry. So by
induction it suffices to check that the defining conditions for
$Ev_m$ respect the relation coming from the expansion (\ref{invexp}) 
$$1\otimes (x+y)^{\otimes 2^m}= \sum 1\otimes s\beta$$
where the summation is taken over $\beta \in T_{2^m}(x,y) /C_{2^m}$.
We rewrite this formula as
\begin{equation}
\label{welldefined}
1\otimes (x+y)^{\otimes 2^m} +1\otimes x^{\otimes 2^m} 
+1\otimes y^{\otimes 2^m} = \sum_{k=1}^m \sigma_k \text{ where }
\sigma_k=\sum_{|\beta |=2^k} 1\otimes s\beta
\end{equation}
When we apply $Ev_m$ to the left hand side we get $v\Delta h_m(x,y)$ by
(\ref{evcond1}).
We must check that this equals $Ev_m$ applied to the right hand 
side where $Ev_m$ was defined by (\ref{evcond2}). Thus we must find
$Ev_m(\sigma_k)$ where $1\leq k \leq m$.
Let $\beta$ be an orbit of length $2^k$. It has a representative
of the form $(a_1 \otimes \dots \otimes a_{2^k})^{\otimes 2^{m-k}}$.
Let $t(\beta )$ denote the order of the set
$\{ a_i \, | \, 1\leq i \leq 2^k \text{ and } a_i=x \}$.
By (\ref{evcond2}) we have
\begin{equation}
\label{type}
Ev_m(1\otimes s\beta ) =vh_{m-k} (a_1 \dots a_{2^k})=
vh_{m-k}(x^{t(\beta )}y^{2^k-t(\beta )})
\end{equation}
If $t(\beta )$ is even $Ev_m(1\otimes s\beta)=0$ since $h_i(a^2)=0$ for
all $i\geq 0$. Let $N(2^k,n)$ denote order of the set
$\{ \beta \, | \, |\beta | =2^k \text{ and } t(\beta )= n \}$.
A string $a_1 \otimes \dots \otimes a_{2^k}$ with an odd
number of $a_i$'s equal to $x$ cannot be broken into two equal halfs
so we have 
$$N(2^k,2j+1)= \frac 1 {2^k} \binom {2^k} {2j+1}=
\frac {(2^k-1)(2^k-2)\dots (2^k-2j)} {(2j+1)!}$$
The 2-adic valuation of this number is
\begin{align*}
v_2(N(2^k,2j+1))= & v_2(2^k-2) + v_2(2^k-4) + \dots +v_2(2^k-2j) \\
& -v_2(2) -v_2(4) -\dots - v_2(2j) =0
\end{align*}
so $N(2^k,2j+1)$ is odd and we conclude that
$$Ev_m(\sigma_k ) =\sum_{j=0}^{2^{k-1}-1}vh_{m-k}(x^{2j+1}y^{2^k-2j-1})
=\sum_{r+s=2^k} vh_{m-k}(x^ry^s)$$
Taking the sum over $1\leq k \leq m$ and substituting $i=m-k$ we obtain
$$Ev_m(\sum_{k=1}^m \sigma_k )=\sum_{i=0}^{m-1} \sum_{r+s=2^{m-i}}
vh_i(x^ry^s)$$
By (\ref{wrel2}) this equals $v\Delta h_m(x,y)$ so the relation
(\ref{welldefined}) is respected and $Ev_m$ is well defined.

Next we verify (\ref{evprop}). For highest symmetry element we have
\begin{align*}
Ev_{m-1}\circ \mu_* \circ q_m^{m-1}(1\otimes x^{\otimes 2^m}) &=
Ev_{m-1}\circ \mu_*(1\otimes (x\otimes x)^{\otimes 2^{m-1}}) \\
&= Ev_{m-1}(1\otimes (x^2)^{\otimes 2^{m-1}}) = \phi (x^{2^{m-1}}) \\
&=(q_m^{m-1}\otimes 1) \circ Ev_m (1\otimes x^{\otimes 2^m})
\end{align*}
For elements hit by the transfer the result follows since
$\mu_* \circ q_m^{m-1} \circ \tau_{m-1}^m=0$ and
$(q_m^{m-1}\otimes 1)\circ (\tau_m^{m-1}\otimes 1)=0$.

We can now use Frobenius reciprocity and induction to show that $Ev_m$ is a 
ring map. Assume that $Ev_{m-1}$ is a ring map. 
We have
\begin{align*}
Ev_m(a\cdot \tau_{m-1}^m(b)) &= Ev_m(\tau_{m-1}^m (q_m^{m-1}(a) \cdot b))\\
&= (\tau_{m-1}^m \otimes 1) \circ Ev_{m-1} \circ 
\mu_* (q_m^{m-1}(a) \cdot b)\\
&=(\tau_{m-1}^m \otimes 1)
(Ev_{m-1}\circ \mu_* \circ q_m^{m-1}(a) \cdot Ev_{m-1}\circ \mu_* (b)) \\
&= (\tau_{m-1}^m \otimes 1)
((q_m^{m-1}\otimes 1)\circ Ev_m(a) \cdot Ev_{m-1}\circ \mu_* (b)) \\
&=Ev_m(a)\cdot (\tau_{m-1}^m \otimes 1) \circ Ev_{m-1} \circ \mu_* (b) \\
&=Ev_m(a) \cdot Ev_m (\tau_{m-1}^m(b))
\end{align*}
Hence it suffices to check the following relation
$$Ev_m(1\otimes x^{\otimes 2^m}) \cdot 
Ev_m(1\otimes y^{\otimes 2^m}) = Ev_m(1\otimes (xy)^{\otimes 2^m})$$
which is valid by (\ref{wrel2}).

The statement for $K=H^*X$ follows from Theorem \ref{spaceev} and Proposition
2.3 of [O].
\end{proof}





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\end{thebibliography}

\begin{flushleft}
Iver Ottosen\\ 
Fakult\" at f\" ur Mathematik \\ 
Universit\" at Bielefeld\\
Postfach 100131\\ 
33501 Bielefeld, Germany \\
Email: ottosen@Mathematik.Uni-Bielefeld.DE
\end{flushleft}







\end{document}



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