When $R$ is one of the spectra $\ku$, $\ko$, $\tmf$, $\MTSpin^c$, $\MTSpin$, or $\MTString$, there is a standard approach to computing twisted $R$-homology groups of a space $X$ with the Adams spectral sequence, by using a change-of-rings isomorphism to simplify the $E_2$-page. This approach requires the assumption that the twist comes from a vector bundle, i.e. the twist map $X\to B\GL_1(R)$ factors through $B\O$. We show this assumption is unnecessary by working with Baker-Lazarev's Adams spectral sequence of $R$-modules and computing its $E_2$-page for a large class of twists of these spectra. We then work through two example computations motivated by anomaly cancellation for supergravity theories.

§ INTRODUCTION

There is a standard formula for computing Steenrod squares in the cohomology of the Thom space or spectrum of a vector bundle $V\to X$: if $U$ is the Thom class,

\begin{equation}
\label{vb_steen}
\Sq^n(Ux) = \sum_{i+j=n}Uw_i(V)\Sq^j(x).
\end{equation}

The ubiquity of the Steenrod algebra in computational questions in algebraic topology means this formula has been
applied to questions in topology and geometry, and recently even in physics, where it is used to run the Atiyah-Hirzebruch and Adams spectral sequences computing groups of invertible field theories.

It is possible to build Thom spectra using more general data than vector bundles, and recently these Thom spectra
have appeared in questions motivated by anomaly cancellation in supergravity theories [51, 40]. Motivated
by these applications (which we discuss more in <ref>), our goal in this paper is to
understand the analogue of (<ref>) for non-vector-bundle twists of commonly studied generalized cohomology
theories. We found that the most direct generalization of (<ref>) is true; in a sense, for the theories we study, these more general Thom spectra behave just like vector bundle Thom spectra for the purpose of computing their homotopy groups with the Adams spectral sequence.

§.§ Statement of results

Now for a little more detail: our main theorem and the language needed to define it. We use
Ando-Blumberg-Gepner-Hopkins-Rezk's approach to twisted generalized cohomology theories [2, 3],
which generalizes the notion of a local system. Twists of $\Z$-valued cohomology on a pointed, connected space $X$
are specified by local systems with fiber $\Z$, which are equivalent data to homomorphisms $\pi_1(X)\to
\mathrm{Aut}(\Z)$, or, since $\Z$ is discrete, to maps $X\to B\mathrm{Aut}(\Z)$.

Ando-Blumberg-Gepner-Hopkins-Rezk generalize this to $E_\infty$-ring spectra.[An $E_\infty$-ring
spectrum is the avatar in stable homotopy theory of a generalized cohomology theory with a commutative ring
structure.  Examples include ordinary cohomology, real and complex $K$-theory, and many cobordism
theories.],[In fact, much of Ando-Blumberg-Gepner-Hopkins-Rezk's theory works in greater
generality, but we only need $E_\infty$-ring spectra in this article.] If $R$ is an $E_\infty$-ring spectrum,
Ando-Blumberg-Gepner-Hopkins-Rezk define a notion of local system of free rank-$1$
$R$-module spectra that is classified by maps to an object called $B\GL_1(R)$, making $B\GL_1(R)$ the classifying
space for twists of $R$-homology. Given a twist $f\colon X\to B\GL_1(R)$, they then define a Thom spectrum
$Mf$, and the homotopy groups of $Mf$ are the $f$-twisted $R$-homology groups of $X$. This construction
simultaneously generalizes twisted ordinary homology, twisted $K$-theory, and the vector bundle twists mentioned

We are interested in twisted $R$-homology for several $E_\infty$-ring spectra, so our first step is to give
examples of twists. Most of these examples are known, but by using a theorem of Beardsley <cit.>, one can produce them in a unified way.

* (<ref>) There is a map $K(\Z/2, 1)\times K(\Z, 3)\to B\GL_1(\MTSpin^c)$, meaning
bordism can be twisted on a space $X$ by $H^1(X; \Z/2)\times H^3(X;\Z)$. The induced maps to
$B\GL_1(\ku)$ and $B\GL_1(\KU)$ recover the usual notion of $K$-theory twisted by $H^1(X; \Z/2)\times

*  (<ref>) There is a map $K(\Z/2, 1)\times K(\Z/2, 2)\to B\GL_1(\MTSpin)$, meaning
spin bordism can be twisted on a space $X$ by $H^1(X; \Z/2)\times H^2(X;\Z/2)$. The induced maps to
$B\GL_1(\ko)$ and $B\GL_1(\KO)$ recover the usual notion of $\KO$-theory twisted by $H^1(X; \Z/2)\times

* (<ref>) Let $\SK(4)$ be a classifying space for degree-$4$ supercohomology classes. Then
there is a map $K(\Z/2, 1)\times \SK(4)\to B\GL_1(\MTString)$, meaning
string bordism can be twisted on a space $X$ by $H^1(X; \Z/2)\times \SH^4(X)$. The induced map

\begin{equation}
K(\Z, 4)\longrightarrow \SK(4)\longrightarrow B\GL_1(\MTString)\longrightarrow B\GL_1(\tmf)
\end{equation}

recovers the Ando-Blumberg-Gepner twist of $\tmf$ (and $\Tmf$ and $\TMF$) by degree-$4$ cohomology classes.

Supercohomology refers to a generalized cohomology theory $\SH$ introduced by Freed <cit.> and Gu-Wen [64]: $\pi_{-2}\SH = \Z/2$ and $\pi_0\SH = \Z$, with the unique nontrivial $k$-invariant, and no other nonzero homotopy groups. We explicitly define $\SK(4)$ in (<ref>).

Though twists of $\tmf$ by degree-$4$ cohomology classes are relatively well-studied, this supercohomology
generalization appears to only be suggested at in the literature by various authors including [55, 72, 112],

and sees more of the homotopy type of $B\GL_1(\tmf)$. It would be interesting to study instances of this twist.

We call the twists in the above theorem fake vector bundle twists: when the twist is given by a vector bundle $V$, these cohomology classes appear as characteristic classes of $V$, but these twists exist whether or not there is a vector bundle with the prescribed characteristic classes.

If $R$ is one of the spectra mentioned in the above theorem, the Thom spectrum $Mf$ of a fake vector bundle twist $f\colon X\to B\GL_1(R)$ is an $R$-module spectrum. This grants us access to Baker-Lazarev's variant of the Adams spectral sequence [24].
[Baker-Lazarev [24]]
Let $p$ be a prime number and $R$ be an $E_\infty$-ring spectrum such that $\pi_0(R)$ surjects onto $\Z/p$, so that
$H \coloneqq H\Z/p$ acquires the structure of an $R$-algebra. For $R$-module spectra $M$ and $N$, let $N_R^*M\coloneqq \pi_{-*}\mathrm{Map}_R(M, N)$. Then there is an Adams-type spectral sequence with signature

\begin{equation}
E_2^{s,t} = \Ext_{H_R^*H}^{s,t}(H_R^*(M), \Z/p) \Longrightarrow \pi_{t-s}(M)_p^\wedge,
\end{equation}

which converges for all $M$ and all $E_\infty$-ring spectra $R$ we consider in this paper.

What Baker-Lazarev prove is more general than what we state here: we stated only the generality we need.

For $H\Z$, $\ko$, and $\ku$ ($p = 2$), and $\tmf$, ($p = 2$ and $p = 3$) $H_R^*H$ is known due to work of various
authors: let $\cA(n)$ be the subalgebra of the mod $2$ Steenrod algebra generated by $\Sq^1,\dotsc,\Sq^{2^n}$. Then, at $p = 2$,

* $H_{H\Z}^*H\cong\cA(0)$,

* $H_{\ku}^*H\cong\cE(1)\coloneqq\ang{\Sq^1, \Sq^2\Sq^1 + \Sq^1\Sq^2}$,

* $H_{\ko}^*H\cong\cA(1)$, and

* $H_{\tmf}^*H\cong\cA(2)$.

See (<ref>) and the surrounding text.
For $\tmf$ at $p = 3$, see <ref>. These algebras are small enough for computations to be tractable, so
if we can compute the $H_R^*H$-module structure on $H_R^*(Mf)$ for $f$ a fake vector bundle twist, we can run the
Adams spectral sequence and hope to compute $\pi_*(Mf)$. This is the content of our main theorem,

The first step is to understand $H_R^*(Mf)$ as a vector space. In <ref>, we establish a Thom isomorphism

\begin{equation}
H_R^*(Mf)\overset\cong\longrightarrow H^*(X;\Z/2)\cdot U,
\end{equation}

where $U\in H^0_R(Mf)$ is the Thom class. Using this, we can state our main theorem:
Let $X$ be a topological space.

* Given $a \in H^1(X;\Z/2)$ and $c \in H^3(X; \Z)$, let $f_{a,c}: X \rightarrow B\GL_1(\ku)$ be the corresponding fake vector bundle twist.   $H^*_{\ku}(M^{\ku}f_{a,c})$ is a $\mathcal{E}(1)$-module with $Q_0$-and $Q_1$-actions by

\begin{equation*}
\begin{aligned}
Q_0(Ux) &\coloneqq Uax + U Q_0(x)\\
Q_1(Ux) &\coloneqq U(c\bmod 2+a^3)x + UQ_1(x).
\end{aligned}
\end{equation*}

*  Given $a\in H^1(X; \Z/2)$ and $b \in H^2(X; \Z/2)$, let $f_{a,b}\colon X\to B\GL_1(\ko)$ be the
corresponding fake vector bundle twist. $H^*_{\ko}(M^{\ko}f_{a,b})$ is an $\mathcal{A}(1)$-module with $\Sq^1$-and $\Sq^2$-actions

\begin{equation*}
\begin{aligned}
\Sq^1(Ux) &\coloneqq U(ax + \Sq^1(x))\\
\Sq^2(Ux) &\coloneqq U(bx + a\Sq^1(x) + \Sq^2(x)).
\end{aligned}
\end{equation*}

* Given $a\in H^1(X;\Z/2)$, and $d\in SH^4(X)$, let $f_{a,d}\colon X\to
B\GL_1(\tmf)$ be the corresponding fake vector bundle twist. $H_\tmf^*(M^\tmf f_{a,d})$ is an $\cA(2)$-module with $\Sq^1$-and $\Sq^2$-action the same as (<ref>) above, and $\Sq^4$-action

\begin{equation*}
\Sq^4(Ux) = U(\delta x + (t(d) a+\Sq^1(t(d))) \Sq^1(x) + t(d)\Sq^2(x) + a\Sq^3(x) + \Sq^4(x)).
\end{equation*}

Furthermore, $H^*_{\tmf}(M^{\tmf}f_{a,d};\Z/3)$ is an $\cA^\tmf$-module with $\beta$ and $\mathcal{P}^1$ actions

\begin{equation*}
\begin{aligned}
\beta(Ux) &\coloneqq U\beta(x)\\
\cP^1(Ux) &\coloneqq U((d\bmod 3)x + \cP^1(x)).
\end{aligned}
\end{equation*}

With this, we have the input to the Baker-Lazarev spectral sequence in
<ref>, from which many computations open up. We give three examples of applications of our techniques.

* In <ref>, we use <ref> to compute low-dimensional $G$-bordism groups for $G = \Spin\times_{\set{\pm 1}}\SU_8$. These are the twisted spin bordism groups for a twist over $B(\SU_8/\set{\pm 1})$ which is not a vector bundle twist. In [51], we discussed an application of $\Omega_5^G$ to an anomaly cancellation question in $4$-dimensional $\cN = 8$ supergravity; using <ref>, we can give a much simpler calculation of $\Omega_5^G$ than appears in <cit.>.

* In <ref>, we study twisted string bordism groups for a non-vector bundle twist over $B((E_8\times E_8)\rtimes\Z/2)$, where $\Z/2$ acts on $E_8\times E_8$ by swapping the factors. These bordism groups have applications in the study of the $E_8\times E_8$ heterotic string; see [40] for more information. Here, we work through the $3$-primary calculation, simplifying a computation in [40].

* In <ref>, we reprove a result of Devalapurkar <cit.> describing $H\Z/2$ as a $\ku$-module Thom spectrum; Devalapurkar's proof uses different methods.

Our theorems proceed similarly for several different families of spectra. One naturally wonders if there are more families out there. Specifically, there is a spectrum for which many but not all of the ingredients of our proofs are present.
Let $\tmf_1(3)$ denote the connective spectrum of topological modular forms with a level structure for the congruence subgroup $\Gamma_1(3)\subset\SL_2(\Z)$ [68]. Is there a tangential structure $\xi\colon B\to B\O$ such that $\mathit{MT\xi}$ is an $E_\infty$-ring spectrum with an $E_\infty$-ring map $\mathit{MT\xi}\to \tmf_1(3)$ which is an isomorphism on low-degree homotopy groups?

If such a spectrum exists, then one could use our approach to run the Baker-Lazarev Adams spectral sequence to compute twisted $\tmf_1(3)$-homology; the needed change-of-rings formula for $\tmf_1(3)$ is due to Mathew <cit.>. We would be interested in learning if such a spectrum $\mathit{MT\xi}$ exists.

§.§ Outline

<ref> is about twists and Thom spectra. First, in <ref>, we review
Ando-Blumberg-Gepner-Hopkins-Rezk's theory of Thom spectra [2, 3] and discuss some constructions
and lemmas we need later in the paper. Then, in <ref>, we construct fake vector bundle twists for
the four families of ring spectra that we study in this paper: $\MTSO$ and $H\Z$ in <ref>; $\MTSpin^c$,
$\ku$, and $\KU$ in <ref>; $\MTSpin$, $\ko$, and $\KO$ in <ref>; and $\MTString$,
$\tmf$, $\Tmf$, and $\TMF$ in <ref>.

In <ref> we study the Adams spectral sequence for the Thom spectra of these twists.  We begin in
<ref> by reviewing how the change-of-rings story simplifies Adams computations for vector
bundle Thom spectra. Then, in <ref>, we introduce Baker-Lazarev's $R$-module Adams spectral
sequence [24]. In <ref> we prove <ref> computing the input to the
Baker-Lazarev Adams spectral sequence for the Thom spectra of our fake vector bundle twists.

We conclude in <ref> with some applications and examples of computations using the main
theorem: a twisted spin bordism example in <ref> and an application to U-duality anomaly
cancellation; a twisted string bordism example in <ref> motivated by anomaly cancellation in
heterotic string theory; and a twisted $\ku$-homology example in <ref> exhibiting $H\Z/2$ as the
$2$-completion of a $\ku$-module Thom spectrum.

§ THOM SPECTRA AND TWISTS À LA ANDO-BLUMBERG-GEPNER-HOPKINS-REZK

§.§ The Ando-Blumberg-Gepner-Hopkins-Rezk approach to Thom spectra

In this subsection we introduce Ando-Blumberg-Gepner-Hopkins-Rezk's theory of Thom spectra [2, 3] and recall the key facts we need for our theorems.[Here and throughout the paper, we work with
the symmetric monoidal $\infty$-category of spectra constructed by Lurie <cit.>, where by
“$\infty$-category” we always mean quasicategory. In <ref>, we use work of
Baker-Lazarev [24], who work with a different model of spectra, the $\Sph$-modules of
Elmendorf-Kriz-Mandell-May <cit.>. The equivalence between the $\infty$-category presented by the
model category of $\Sph$-modules and Lurie's $\infty$-category of spectra follows from work of
Mandell-May-Schwede-Shipley [90], Schwede [102], and Mandell-May [89]. Likewise, these
papers show that commutative algebras in the category of $\Sph$-modules correspond to $E_\infty$-rings in the
$\infty$-category of spectra.
] In this paper, we only need to work
with $E_\infty$-ring spectra, and we will state some theorems in only the generality we need, which is less general
than what Ando-Blumberg-Gepner-Hopkins-Rezk prove.

By an $\infty$-group we mean a grouplike $E_1$-space, which is a homotopically invariant version of topological group.

By an
abelian $\infty$-group $A$ we mean a grouplike $E_\infty$-space.

Let $R$ be an $E_\infty$-ring spectrum. The group of units of $R$ is the abelian $\infty$-group $\GL_1(R)$
defined to be the following pullback:

\begin{equation}
\label{GL1_defn}
\begin{tikzcd}
{\GL_1(R)} & {\Omega^\infty R} \\
{\pi_0(R)^\times} & {\pi_0(R)}\,.
\arrow[from=1-1, to=1-2]
\arrow[from=1-1, to=2-1]
\arrow[from=2-1, to=2-2]
\arrow[from=1-2, to=2-2]
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
\end{tikzcd}
\end{equation}

The pullback (<ref>) takes place in the $\infty$-category of abelian $\infty$-groups.

As the three legs of the
pullback diagram (<ref>) are functorial in $R$, $\GL_1(R)$ is also functorial in $R$.

Since $\GL_1(R)$ is an $\infty$-group, it has a classifying space $B\GL_1(R)$; we refer to a map $X\to B\GL_1(R)$
as a twist of $R$ over $X$. There is a sense in which $B\GL_1(R)$ carries the universal local system of
$R$-lines, or free $R$-module spectra of rank $1$: see <cit.>.

If $A$ is a commutative ring and $R = HA$, then the equivalence of abelian $\infty$-groups $\pi_0\colon
\Omega^\infty HA\overset\simeq\to A$ induces an equivalence of abelian $\infty$-groups $\GL_1(R)\simeq

Let $\cat{Mod}_R$ denote the $\infty$-category of $R$-module spectra and $\cat{Line}_R$ denote the
$\infty$-category of $R$-lines, and let $\pi_{\le\infty}(X)$ denote the
fundamental $\infty$-groupoid of a space $X$. The
identification $\abs{\cat{Line}_R}\overset\simeq\to B\GL_1(R)$ <cit.> allows us to
reformulate the inclusion $\cat{Line}_R\hookrightarrow\cat{Mod}_R$ as a functor $M\colon
\pi_{\le\infty}(B\GL_1(R))\to \cat{Mod}_R$, which one can think of as sending a point in $B\GL_1(R)$ to the
$R$-line which is the fiber of the universal local system of $R$-lines on $B\GL_1(R)$. In the rest of this paper,
we will simply write $X$ for $\pi_{\le\infty}(X)$, as we will never be in a situation where this causes ambiguity.

Let $R$ be an $E_\infty$-ring spectrum and $f\colon X\to B\GL_1(R)$ be a twist of $R$. The Thom spectrum
$M^Rf$ of the map $f$ is the colimit of the $X$-shaped diagram

\begin{equation}
\label{Thom_defn}
X\overset{f}{\longrightarrow} B\GL_1(R)\longrightarrow \cat{Mod}_R\,.
\end{equation}

When $R$ is clear from context, we will write $Mf$ for $M^Rf$.

By construction, $Mf$ is an $R$-module spectrum. If the reader is familiar with the definition of a Thom spectrum
associated to a virtual vector bundle, this definition is related but more general.

[Thom spectra from vector bundles]

Let $V\to X$ be a virtual stable vector bundle of rank zero; $V$ is classified by a map $f_V\colon X\to B\O$. There
is a map of abelian $\infty$-groups $J\colon B\O\to B\GL_1(\Sph)$ called the $J$-homomorphism, where $B\O$
has the abelian $\infty$-group structure induced by direct sum of (rank-zero virtual) vector bundles [116]. Theorems of Lewis <cit.> and
Ando-Blumberg-Gepner-Hopkins-Rezk <cit.> together imply that the Thom spectrum $X^V$ in
the usual sense is naturally equivalent to the Thom spectrum $M(J\circ f_V)$ in the
Ando-Blumberg-Gepner-Hopkins-Rezk sense.

[Trivial twists]

Suppose that the map $f\colon X\to B\GL_1(R)$ is null-homotopic. Then by definition, the colimit
of (<ref>) is $R\wedge X_+$; more precisely, a null-homotopy of $f$ induces an equivalence of $R$-module
spectra $Mf\simeq R\wedge X_+$.

We will need the following fact a few times.

Let $g\colon R_1\to R_2$ be a map of $E_\infty$-ring spectra and $f\colon X\to B\GL_1(R_1)$ be a twist. Then there
is an equivalence of $R_2$-module spectra

\begin{equation}
M^{R_2}(g\circ f)\overset\simeq\longrightarrow M^{R_1}f\wedge_{R_1} R_2.
\end{equation}

When $R_1 = \Sph$, Ando-Blumberg-Gepner-Hopkins-Rezk <cit.> mention that this lemma is a
straightforward consequence of a different, equivalent definition of the Thom spectrum <cit.>.

We will show that the diagram

\begin{equation}
\label{Thom_mod}
\begin{tikzcd}
{B\GL_1(R_1)} & {B\GL_1(R_2)} \\
{\cat{Mod}_{R_1}} & {\cat{Mod}_{R_2}}
\arrow["g", from=1-1, to=1-2]
\arrow["M^{R_1}"', from=1-1, to=2-1]
\arrow["{\text{--}\wedge_{R_1} R_2}"', from=2-1, to=2-2]
\arrow["M^{R_2}", from=1-2, to=2-2]
\end{tikzcd}
\end{equation}

is (homotopy) commutative, where just as above we identify the spaces $B\GL_1(R_i)$ with their
fundamental $\infty$-groupoids. Once we know this, the lemma is immediate from the colimit definition of $M^{R_2}(g\circ
f)$ in <ref>: replace $M^{R_2}\circ g\circ f$ with $(\bl\wedge_{R_1}R_2)\circ M^{R_1}\circ f$.

The key obstacle in establishing commutativity of (<ref>) is that $g\colon B\GL_1(R_1)\to B\GL_1(R_2)$
comes from maps of spectra via (<ref>), but $\bl\wedge_{R_1}R_2$ has a more module-theoretic flavor. The
resolution, which is the same as in the proof of <cit.>, is that the three other pieces
of the pullback (<ref>) defining $\GL_1$, namely $\Omega^\infty$, $\pi_0$, and $\pi_0(\bl)^\times$, have
module-theoretic interpretations: there are homotopy equivalences of abelian $\infty$-groups $\Omega^\infty
R\overset\simeq\to \End_R(R)$, and likewise $\pi_0(R)\overset\simeq\to\pi_0(\End_R(R))$ and
$\pi_0(R)^\times\overset\simeq\to \pi_0(\End_R(R))^\times$. And all of these identifications are compatible with
the tensor product functor $\cat{Mod}_{R_1}\to\cat{Mod}_{R_2}$, thus also likewise for their classifying spaces,
establishing commutativity of (<ref>).

The usual Thom diagonal for a Thom space $X^V$ gives $H^*(X^V;\Z/2)$ the structure of a module over
$H^*(X;\Z/2)$. One can generalize this for $R$-module Thom spectra as follows.

Let $R$ be an $E_\infty$-ring spectrum and $f\colon X\to B\GL_1(R)$ be a twist. The Thom diagonal for $Mf$
is an $R$-module map

\begin{equation}
\label{TD_equation}
Mf\overset{\Delta^t}{\longrightarrow} Mf\wedge R\wedge X_+
\end{equation}

defined by applying the Thom spectrum functor to the maps $f\colon X\to B\GL_1(R)$ and $(f, 0)\colon X\times X\to
B\GL_1(R)$: if $\Delta\colon X\to X\times X$ is the diagonal map, then $f = \Delta^*(f, 0)$, so $\Delta$ induces
the desired map $\Delta^t$ of $R$-module Thom spectra in (<ref>).

See Beardsley <cit.> for a nice coalgebraic interpretation of the Thom diagonal.

§.§ Constructing non-vector-bundle twists

Let $X$ and $Y$ be $E_\infty$-spaces and $f_1\colon X\to Y$ and $f_2\colon Y\to B\GL_1(\Sph)$ be $E_\infty$-maps. Ando-Blumberg-Gepner <cit.> show that the $E_\infty$-structure on $f_2\circ f_1$ induces an $E_\infty$-ring structure on $M(f_2\circ f_1)$.

Let $R$ be an $E_\infty$-ring spectrum. The data of an $E_\infty$-ring map $M(f_2\circ f_1)\to R$ induces a map
$T_{f_1,f_2}\colon Y/X\to B\GL_1(R)$.

An $E_\infty$-ring map of this kind is often called an $M(f_2\circ f_1)$-orientation of $R$.

Ando-Blumberg-Gepner-Hopkins-Rezk <cit.> show that the $M(f_2\circ f_1)$-orientation of $R$ is equivalent to a null-homotopy of the map

\begin{equation}
X \overset{f_1}{\longrightarrow} Y\overset{f_2}{\longrightarrow}
B\GL_1(\mathbb S)\overset{1_R}{\longrightarrow} B\GL_1(R),
\end{equation}

where $1_R$ denotes the unit map for $R$. That is, we have a map $g\colon Y\to B\GL_1(R)$ and a nullhomotopy of
$f_1\circ g\colon X\to Y\to B\GL_1(R)$, which is precisely the data needed to descend $g$ to the cofiber of $f_1$,
giving us a map $Y/X\to B\GL_1(R)$ as desired.

A theorem of Beardsley [18] uses a special case of <ref> to obtain many commonly-studied twists of various cohomology theories. We will usually apply it for maps to $B\O$ and implicitly compose with the $E_\infty$-map $J\colon B\O\to B\GL_1(\Sph)$, like in <ref>.

For $R = M(f_2\circ f_1)$, there is a natural equivalence
$M^RT_{f_1,f_2}\overset\simeq\to M^{\Sph}f_2$.

In this paper we consider twisted $R$-(co)homology for several different ring spectra $R$. These spectra are organized into several families: in each family there is a Thom spectrum $Mf$, another ring spectrum $R$, and a map of ring spectra $Mf\to R$ which is an isomorphism on homotopy groups in low degrees. In the context of a specific family, we will refer to $Mf$ as the big sibling and $R$ as the little sibling. The four families we consider in this paper are $(\MTSO, H\Z)$, $(\MTSpin, \ko)$, $(\MTSpin^c, \ku)$, and
$(\MTString, \tmf)$:[In the homotopy theory literature, it is common to refer to bordism spectra $\mathit{MSO}$, $\mathit{MSpin}$, etc., corresponding to the bordism groups of manifolds with orientations, resp. spin structures, on the stable normal bundle. In the mathematical physics literature, one sees $\MTSO$, $\MTSpin$, etc., corresponding to the same structures on the stable tangent bundle. If $\xi\colon B\to B\O$ is a tangential structure such that the map $\xi$ is a map of abelian $\infty$-groups, as is the case for $\O$, $\SO$, $\Spin^c$, $\Spin$, and $\String$, there is a canonical equivalence $\mathit{M\xi}\overset{\simeq}{\to}\mathit{MT\xi}$. For other tangential structures, this is not necessarily true: in particular, $\mathit{MPin}^\pm\simeq\MTPin^\mp$.]

* The map $\Omega_0^\SO\overset\cong\to\Z$ counting the number of points refines to a map of $E_\infty$-ring spectra $\MTSO\to H\Z$. Work of Thom <cit.> shows this map is an isomorphism on homotopy groups in degrees $3$ and below.

* The Atiyah-Bott-Shapiro map $\MTSpin^c\to\ku$ [7] was shown to be a map of $E_\infty$-ring spectra by Joachim [74], and Anderson-Brown-Peterson [5] showed this map is an isomorphism on homotopy groups in degrees $3$ and below.

* Joachim [74] also showed the real Atiyah-Bott-Shapiro map $\MTSpin\to\ko$ [7] is a map of $E_\infty$-ring spectra, and Milnor [87] showed this map is an isomorphism on homotopy groups in degrees $7$ and below.

* Ando-Hopkins-Rezk [10] produced a map of $E_\infty$-ring spectra $\sigma\colon \MTString\to\tmf$, which Hill <cit.> shows is an isomorphism on homotopy groups in degrees $15$ and below.

For all of these cases but $\MTString$, one can $2$-locally decompose the big sibling into a sum of cyclic modules over the little sibling: Wall [113] produced a $2$-local equivalence

\begin{equation}
\MTSO_{(2)}\overset\simeq\longrightarrow H\Z_{(2)} \vee \Sigma^4 H\Z_{(2)} \vee \Sigma^5 H\Z/2\vee\dotsb,
\end{equation}

and Anderson-Brown-Peterson [5] produced $2$-local equivalences

\begin{align}
\MTSpin_{(2)} &\overset\simeq\longrightarrow \ko_{(2)}\vee \Sigma^8 \ko_{(2)}\vee \Sigma^{10}(\ko\wedge J)_{(2)}\vee \dots\\
\MTSpin^c_{(2)} &\overset\simeq\longrightarrow \ku_{(2)}\vee \Sigma^4 \ku_{(2)} \vee\Sigma^8 \ku_{(2)} \vee \Sigma^8 \ku_{(2)}\vee\dotsb,
\end{align}

where $J$ is a certain spectrum such that $\Sigma^2\ko\wedge J$ is the Postnikov $2$-connected cover of $\ko$.[The decomposition is implicit in [5]; see Bahri-Gilkey [22] for an explicit reference.]

It is not known whether $\tmf$ is a summand of
$\MTString$ (see, e.g., [78, 41, 81, 97, 42]) so we do not know if there is a splitting like in
the three other cases.

Associated to congruence subgroups $\Gamma\subset\SL_2(\Z)$ there are “topological modular forms with level
structure:” Hill-Lawson [68] construct
$E_\infty$-ring spectra $\TMF(\Gamma)$, $\Tmf(\Gamma)$, and $\tmf(\Gamma)$ with maps between them like
for vanilla $\tmf$. The case $\Gamma = \Gamma_1(3)$ is especially interesting,
as all but one of the several ingredients we need for the proof of our main theorem are known to be true for
$\tmf(\Gamma_1(3))$ (usually written $\tmf_1(3)$): by work of Mathew <cit.>, there is a change-of-rings theorem allowing one to
simplify $2$-primary Adams spectral sequence computations to an easier subalgebra (see <ref>),
but it is not yet known how to construct an $E_\infty$-ring Thom spectrum $M$ with an orientation $M\to\tmf_1(3)$
that is an isomorphism on low-degree homotopy groups.[See Wilson [117] for results on closely
related questions.]
The existence of such a spectrum $M$ would lead to
generalizations of our main theorems to twists of $\tmf_1(3)$-homology,[A theorem of Meier <cit.> suggests this may also apply to twists of $\tmf_1(n)$-homology for other values of $n$.] and we would
be interested in learning whether this is possible.

§.§.§ Twists of $\MTSO$ and $H\Z$

We walk through the implications of <ref> in a relatively simple setting, addressing

* what cohomology classes define twists of $\MTSO$ and $H\Z$ by way of <ref>,

* what the corresponding twisted bordism and cohomology groups are, and

* what <ref> implies the Thom spectrum of the universal twist is.

Letting $f_2\colon X\to B\O$ be the identity and $f_1\colon X\to Y$ be $B\SO\to B\O$, we obtain twists of
$\MTSO$-oriented ring spectra, notably $\MTSO$ and $H\Z$, by maps to $B\SO/B\O\simeq K(\Z/2, 1)$. The map $B\O\to
B\O/B\SO$ admits a section defined by regarding a map to $K(\Z/2, 1)$ as a real line bundle, so these twists are
given by real line bundles in the sense of <ref>. Specifically, a class $a\in H^1(X;\Z/2)$ defines
a twist $f_a\colon X\to B\GL_1(\MTSO)$ by interpreting $a$ as a map $X\to B\O/B\SO$ and invoking
<ref>, and $a$ defines a second twist $g_a$ by choosing a real line bundle $L_a$ with $w_1(L_a) =
a$ (a contractible choice) and making the vector bundle twist as in <ref>, but $f_a\simeq g_a$ and so
$M^\MTSO f_a\simeq \MTSO\wedge X^{L_a-1}$. Thus in a sense this example is redundant, as the main theorems of this
paper are long known for vector bundle twists, but we include this example because we found it a useful parallel to
to other families we study.

Let $\Omega_*^\SO(X, a)\coloneqq\pi_*(M^\MTSO f_a)$. Using the vector bundle interpretation of this twist,
$\Omega_*^\SO(X, a)$ has an interpretation as twisted oriented bordism groups, specifically the bordism groups
of manifolds $M$ with a map $h\colon M\to X$ and an orientation on $TM\oplus h^*L_a$. Alternatively, one could
think of this as the bordism groups of manifolds $M$ with a map $h\colon M\to X$ and a trivialization of the class
$w_1(M) - h^*a$; this perspective will be useful in later examples of non-vector-bundle twists.

<Ref> then implies the Thom spectrum of

\begin{equation}
\label{MSO_O1}
\rightarrow}B\O_1\overset\sigma\longrightarrow B\O\longrightarrow B\GL_1(\mathbb S)\longrightarrow B\GL_1(\MTSO),
\end{equation}

is equivalent to $\MTO$. <Ref> implies the Thom spectrum of (<ref>) is $\MTSO\wedge
(B\O_1)^{\sigma-1}$, so we have reproved a theorem of Atiyah: $\MTSO\wedge
(B\O_1)^{\sigma-1}\simeq\MTO$ <cit.>.

The twist of $H\Z$ defined by $a$ recovers the usual notion of integral cohomology twisted by a class in $H^1(X; \Z/2)$.

§.§.§ Twists of $\MTSpin^c$, $\ku$, and $\KU$

Our next family of examples includes bordism and complex $K$-theory. In <ref> we use <ref> to construct a map $K(\Z/2, 1)\times K(\Z, 3)\to B\GL_1(\MTSpin^c)$, defining twists of $\MTSpin^c$, $\ku$, and $\KU$ by classes in $H^1(\bl;\Z/2)$ and $H^3(\bl;\Z)$. These recover the usual twists of $K$-theory by these cohomology classes studied by [46, 99, 11, 1] (<ref>), and in <ref> we use work of Hebestreit-Joachim <cit.> to describe the homotopy groups of the corresponding $\MTSpin^c$-module Thom spectra as bordism groups of manifolds with certain kinds of twisted structures.

The Atiyah-Bott-Shapiro orientation [7, 74] defines ring homomorphisms $\mathit{Td} \colon
\MTSpin^c\to\ku\to\KU$, so by <ref> there are maps

\begin{equation}
\label{ku_cofib_twist}
B\O/B\Spin^c\longrightarrow B\GL_1(\MTSpin^c)\overset{\mathit{Td}}{\longrightarrow} B\GL_1(\ku)\longrightarrow
\end{equation}

i.e. twists of $\MTSpin^c$, $\ku$, and $\KU$ by maps to $B\O/B\Spin^c$.[The map (<ref>) is
nowhere near a homotopy equivalence; for example, it misses the “higher twists” of $\KU$ studied in, e.g., [50, 95].]

The map $K(\Z/2, 1)\to B\O$ defined by the tautological line bundle induces a homotopy equivalence of spaces

\begin{equation}
\label{ospinc_equiv}
B\O/B\Spin^c\overset\simeq\longrightarrow K(\Z/2, 1)\times K(\Z, 3),
\end{equation}

implying that $\MTSpin^c$, $\ku$, and $\KU$ can be twisted over a space $X$ by classes $a\in H^1(X;\Z/2)$ and $c\in

We want to apply the third isomorphism theorem to the sequence of maps of abelian
$\infty$-groups $B\Spin^c\to B\SO\to B\O$ to obtain a short exact sequence

\begin{equation}
\label{OSpinc_SES}
\shortexact*{B\SO/B\Spin^c}{B\O/B\Spin^c}{B\O/B\SO}.
\end{equation}

It is not immediate how to do this in the $\infty$-categorical setting, but we can do it. Instead of a short exact
sequence, we obtain a cofiber sequence, and in a stable $\infty$-category, the third isomorphism theorem for
cofiber sequences is a consequence of the octahedral axiom.

The $\infty$-category of abelian $\infty$-groups is not stable, as it is equivalent to the $\infty$-category of
connective spectra, but this $\infty$-category embeds in the stable $\infty$-category $\cat{Sp}$ of all spectra,
allowing us to make use of stability in certain settings: specifically, cofiber sequences $A\to B\to C$ of abelian
$\infty$-groups for which the induced map $\pi_0(B)\to\pi_0(C)$ is surjective; these cofiber diagrams map to
cofiber diagrams in $\cat{Sp}$, so we may invoke the octahedral axiom in $\cat{Sp}$. All cofiber sequences of
abelian $\infty$-groups we
discuss in this paper satisfy this $\pi_0$-surjectivity property, so we will not discuss it further. In particular,
we obtain the cofiber sequence (<ref>).
Throughout this paper, whenever we write a short exact sequence of abelian $\infty$-groups, we mean a cofiber

A similar argument allows one to deduce that fiber and cofiber sequences coincide for abelian $\infty$-groups from
the analogous fact for stable $\infty$-categories, assuming the same $\pi_0$-surjectivity hypothesis.

Since $B\Spin^c$ is the
fiber of $\beta w_2\colon B\SO\to K(\Z, 3)$, which is a map of abelian $\infty$-groups since $\beta w_2$ satisfies
the Whitney sum formula for oriented vector bundles, the cofiber $B\SO/B\Spin^c$ is equivalent, as abelian
$\infty$-groups, to $K(\Z, 3)$. Here, $\beta\colon H^k(\bl;\Z/2) \rightarrow H^{k+1}(\bl; \Z) $ is the Bockstein.
Likewise, $B\SO$ is the fiber of $w_1\colon
B\O\to K(\Z/2, 1)$, which is a map of abelian $\infty$-groups, so $B\O/B\SO\simeq K(\Z/2, 1)$.

The quotient $B\O\to B\O/B\SO\simeq K(\Z/2, 1)$ admits a section given by the tautological real line bundle
$K(\Z/2, 1)\simeq B\O_1\to B\O$; composing $K(\Z/2, 1)\to B\O$ with the quotient $B\O\to B\O/B\Spin^c$ we obtain a
section of (<ref>). That section splits (<ref>), which implies the proposition statement.

Given classes $a\in H^1(X;\Z/2)$ and $c\in H^3(X;\Z)$, we call the twist $f_{a,c}\colon X\to B\GL_1(\MTSpin^c)$ that <ref> associates to $a$ and $c$ the fake
vector bundle twist for $a$ and $c$, and likewise for the induced twists of $\ku$ and $\KU$.

The twist $f_{a,c}$ arises from a vector bundle twist if there is a vector bundle $V\to X$ such that $w_1(V) = a$ and
$\beta(w_2(V)) = c$, but there are choices of $X$, $a$, and $c$ for which no such vector bundle exists, e.g. if $c$ is not $2$-torsion.

Now that we have defined these twists, we get to the business of interpreting them.

Given $X$, $a$, and $c$ as above, let $\Omega_*^{\Spin^c}(X, a, c)$ denote the groups of bordism classes of manifolds $M$ with a map $f\colon M\to X$ and trivializations of
$w_1(M) - f^*(a)$ and $\beta(w_2(M)) - f^*(c)$.

This notion of twisted bordism, in the special case $a=0$, was first studied by Douglas <cit.>, and implicitly appears in Freed-Witten's work [58] on anomaly cancellation.

There is a natural isomorphism $\pi_*(M^{\MTSpin^c}f_{a,c})\overset\cong\to \Omega_*^{\Spin^c}(X, a, c)$.

Hebestreit-Joachim [67] use a different framework for twists based on May-Sigurdsson's parametrized homotopy
theory [92]; Ando-Blumberg-Gepner <cit.> prove a comparison theorem that allows us to pass between May-Sigurdsson's framework and Ando-Blumberg-Gepner-Hopkins-Rezk's. Additionally, Hebestreit-Joachim work with twisted spin bordism and $\KO$-theory, but for the complex case the arguments are essentially the same.

With $X$, $a$, and $c$ as above, the homotopy groups of $M^\KU f_{a,c}$ are naturally isomorphic to the twisted $K$-theory groups of [46, 99, 11, 1].

This is because the methods in [1] are nearly the same as ours, allowing for a direct comparison.[Alternatively, one could use a uniqueness result of Antieau-Gepner-Gómez <cit.> that $[K(\Z, 3), B\GL_1(\KU)]\cong\Z$ to reduce to checking whether these notions of twisted $K$-theory agree on a single example.]

<Ref> computes a few example of $\MTSpin^c$-module Thom spectra for us.

* Letting $X = Y = B\O/B\Spin^c$ and $f_1 = \id$, <ref> implies that the Thom spectrum of
the universal twist $B\O/B\Spin^c\to B\GL_1(\MTSpin^c)$ is $\MTO$. From a bordism point of view, this is the
fact that since $a$ and $c$ pull back from $K(\Z/2, 1)\times K(\Z, 3)$, they can be arbitrary classes, so the
required trivializations of $w_1(M) - f^*(a)$ and $\beta(w_2(M)) - f^*(c)$ are uniquely specified by $a =
w_1(M)$ and $c = \beta(w_2(M))$, so this notion of twisted structure is no structure at all.

* Let $Y$ be as in the previous example and let $f_1\colon X\to Y$ be the map $K(\Z, 3)\simeq
B\SO/B\Spin^c\to B\O/B\Spin^c$. <Ref> says the Thom spectrum of

\begin{equation}
K(\Z, 3)\longrightarrow B\O/B\Spin^c\longrightarrow B\GL_1(\MTSpin^c)
\end{equation}

is equivalent to $\MTSO$. We stress that this  twist by $K(\Z,3)$ does not come from a vector bundle because
all vector bundle twists of $\MTSpin^c$ are torsion and of the form $\beta(w_2(M))$, but the universal twist over $K(\Z,3)$ is not.

The equivalence of spaces $B\O/B\Spin^c\simeq K(\Z/2, 1)\times K(\Z, 3)$ from (<ref>) is not an equivalence of $\infty$-groups.

Suppose that this is an equivalence of $\infty$-groups. Then the inclusion $K(\Z/2, 1)\to K(\Z/2, 1)\times K(\Z, 3)\to B\O/B\Spin^c$ is a map of $\infty$-groups, so the composition

\begin{equation}
\label{pullback_to_BZ2}
\varphi\colon K(\Z/2, 1)\longrightarrow K(\Z/2, 1)\times K(\Z, 3)\longrightarrow B\O/B\Spin^c\longrightarrow B\GL_1(\MTSpin^c)
\end{equation}

is a map of $\infty$-groups. By work of Ando-Blumberg-Gepner <cit.>, this implies the Thom spectrum $M\varphi$ is an $E_1$-ring spectrum. We will explicitly identify $M\varphi$ and show this is not the case.

We saw above that the map $K(\Z/2, 1)\to B\O/B\Spin^c$ factors through the map $K(\Z/2, 1)\to B\O$ defined by the tautological line bundle $\sigma\to B\O_1\simeq K(\Z/ 2, 1)$, meaning that the twist (<ref>) is the vector bundle twist of $\MTSpin^c$ for the tautological line bundle $\sigma\to B\O_1$. Applying <ref> with $R_1 = \Sph$ and $R_2 = \MTSpin^c$, we conclude $M\varphi\simeq \MTSpin^c\wedge (B\O_1)^{\sigma-1}$. Bahri-Gilkey [21, 22] identify this spectrum with $\MTPin^c$, which is known to not be an $E_1$-ring spectrum: for example, a $E_1$-ring structure induces a graded ring structure on homotopy groups, making $\pi_k(\MTPin^c)$ into a $\pi_0(\MTPin^c)$-module for all $k$, but $\pi_0\MTPin^c\cong\Z/2$ and $\pi_2(\MTPin^c)\cong\Z/4$ <cit.>.

§.§.§ Twists of $\MTSpin$, $\ko$, and $\KO$

The real analogue of <ref> is very similar; we summarize the story here, highlighting the
differences. Again there is an Atiyah-Bott-Shapiro ring spectrum map $\MTSpin\overset{\widehat A}{\to}
\ko\to\KO$ [7, 74, 10], allowing us to use <ref> to produce a sequence of maps

\begin{equation}
B\O/B\Spin\longrightarrow B\GL_1(\MTSpin)\overset{\widehat A}{\longrightarrow} B\GL_1(\ko)\longrightarrow
\end{equation}

Freed-Hopkins <cit.> use the $\infty$-group $B\O/B\Spin$ to study vector bundle twists of spin
bordism; they call it $\mathbf P$.

The map $K(\Z/2, 1)\to B\O$ defined by the tautological line bundle induces a homotopy equivalence of spaces

\begin{equation}
\label{OSpin}
B\O/B\Spin\overset\simeq\longrightarrow K(\Z/2, 1)\times K(\Z/2, 2),
\end{equation}

implying $\MTSpin$, $\ko$, and $\KO$ can be twisted over a space $X$ by classes $a\in H^1(X;\Z/2)$ and $b\in

The proof is nearly the same as the proof of <ref>: fit $B\O/B\Spin$ into a split cofiber sequence with $B\SO/B\Spin\simeq K(\Z/2, 2)$ (because $B\Spin\to B\O$ is the fiber of $w_2\colon B\SO\to
K(\Z/2, 2)$) and $B\O/B\SO\simeq K(\Z/2, 1)$.

We call the twist $f_{a,b}\colon X\to B\GL_1(\MTSpin)$ associated to $a$ and $b$ the fake
vector bundle twist for $a$ and $b$, and likewise for the induced twists of $\ko$ and $\KO$.

The space of homotopy self-equivalences of $K(\Z/2, 1)\times K(\Z/2, 2)$ is not connected: for example, if $a$
denotes the tautological class in $H^1(K(\Z/2, 1);\Z/2)$ and $b$ is the tautological class in $H^2(K(\Z/2,
2);\Z/2)$, the homotopy class of maps $\Phi\colon K(\Z/2, 1)\times K(\Z/2, 2) \to K(\Z/2, 1)\times K(\Z/2, 2)$
defined by the classes $(a, a^2+b)$ is not the identity and is invertible. The choice of identification we made
in (<ref>) matters: if one uses a different identification, one obtains a different notion of fake vector
bundle twist and a different formula in <ref> to make <ref> true.

(<ref>) is not an equivalence of $\infty$-groups.

One can prove this lemma in the same way as <ref>, by pulling back along the section $K(\Z/2, 1)\to
B\O/B\Spin$ and observing that the Thom spectrum $\MTSpin\wedge (B\O_1)^{\sigma-1}$ is not a ring spectrum in much
the same way:[This is the first place where the choice of identification (<ref>) has explicit
consequences, as promised in <ref>: if we compose with the identification of $K(\Z/2, 1)\times K(\Z/2,
2)$ given by the classes$(a, a^2+b)$ described in that remark, we would instead obtain $\MTSpin\wedge
(B\O_1)^{3\sigma-3}$. This is not a ring spectrum either, as it can be identified with $\MTPin^+$ <cit.>, and $\pi_0(\MTPin^+)\cong\Z/2$ and $\pi_4(\MTPin^+)\cong\Z/16$ [61].] using the equivalence
$\MTSpin\wedge (B\O_1)^{\sigma-1}\simeq\MTPin^-$ <cit.> and the groups $\pi_0(\MTPin^-)\cong\Z/2$ and
$\pi_2(\MTPin^-)\cong\Z/8$ [6, 76] to show $\MTPin^-$ is not a ring spectrum. There is also another nice
proof, which we give below.

If $X$ is a space and $Y$ is an $\infty$-group, the set $[X, Y]$ has a natural group structure. Therefore it
suffices to find a space such that $[X, B\O/B\Spin]$ and $[X, K(\Z/2, 1)\times K(\Z/2, 2)]$ are non-isomorphic

To calculate the addition in $[\bl, B\O/B\Spin]$, we use the fact that if two maps $f,g\colon X\to \O/B\Spin$ factor through $B\O$, meaning they are represented by rank-zero virtual vector bundles $V_f, V_g\to X$, then $f+g$ is the image of $V_f\oplus V_g$ under $B\O\to B\O/B\Spin$. This implies that for classes in the image of that quotient map, if we use (<ref>) to identify two classes $\phi_1,\phi_2\in[X,
B\O/B\Spin]$ with pairs $\phi_i = (a_i\in H^1(X;\Z/2), b_i\in H^2(X;\Z/2))$, then addition follows the Whitney sum

\begin{equation}
\label{whitney_sum_OSpin}
(a_1, b_1) \oplus (a_2, b_2) = (a_1 + a_2, b_1 + b_2 + a_1a_2).
\end{equation}

This is different from the componentwise addition on $K(\Z/2, 1)\times K(\Z/2, 2)$: for example, $[B\Z/2, K(\Z/2,
1)\times K(\Z/2, 2)]\cong\Z/2\oplus\Z/2$, but the map $[B\Z/2, B\O]\to [B\Z/2, B\O/B\Spin]$ is surjective, so using (<ref>), one can show that $[B\Z/2, B\O/B\Spin]\cong\Z/4$.

Given $X$, $a$, and $b$ as above, let $\Omega_*^{\Spin}(X, a, b)$ denote the groups of bordism classes of manifolds $M$ with a map $f\colon M\to X$ and trivializations of
$w_1(M) - f^*(a)$ and $w_2(M) - f^*(b)$.

B.L. Wang <cit.> first studied these twists of spin bordism in the case $a = 0$.

There is a natural isomorphism $\pi_*(M^{\MTSpin}f_{a,b})\overset\cong\to \Omega_*^{\Spin}(X, a, b)$.

With $X$, $a$, and $b$ as above, the homotopy groups of $M^\KO f_{a,b}$ are naturally isomorphic to the twisted $\KO$-theory groups of [46, 67].

One can show this by appealing to Antieau-Gepner-Gómez' calculation <cit.> that there is a unique nontrivial twist of $\KO$ over $K(\Z/2, 2)$.

<Ref> implies the Thom spectrum of the universal twist of $\MTSpin$ over $B\O/B\Spin$ is $\MTO$, and of the universal twist over $K(\Z/2, 2)\simeq B\SO/B\Spin$ is $\MTSO$. The latter equivalence was first shown by Beardsley <cit.>.

§.§.§ Twists of $\MTString$, $\tmf$, $\Tmf$, and $\TMF$

The final family we consider in this paper is string bordism and topological modular forms. The story has a similar shape: we obtain twists by $B\O/B\String$, and we simplify $B\O/B\String$ to define fake vector bundle twists. However, in <ref> we learn that $B\O/B\String$ is not homotopy equivalent to a product of Eilenberg-Mac Lane spaces. For this reason, the fake vector bundle twist uses a generalized cohomology theory called supercohomology and denoted $\SH$ (<ref>); we finish this subsubsection by studying cohomology classes associated to a degree-$4$ supercohomology class, which we will need in the proof of <ref>.

If $V\to X$ is a spin
vector bundle, it has a characteristic class $\lambda(V)\in H^4(X;\Z)$ such that $2\lambda(V) = p_1(V)$; a
string structure on $V$ is a trivialization of $\lambda$. It is not hard to check that $\lambda$ is additive
in direct sums,

so defines a map of abelian
$\infty$-groups $\lambda\colon B\Spin\to K(\Z, 4)$. The fiber of this map is an $\infty$-group
$B\String$, which is the classifying space for string structures.

Unlike for $K$-theory, there are three different kinds of topological modular forms: a connective spectrum $\tmf$,
a periodic spectrum $\TMF$, and a third spectrum $\Tmf$ which is neither connective nor periodic. All three are
$E_\infty$-ring spectra, and there are ring spectrum maps $\tmf\to\Tmf\to\TMF$.
Ando-Hopkins-Rezk [10] constructed a ring spectrum map $\sigma\colon\MTString\to\tmf$, so
<ref> gives us twists of $\tmf$, $\Tmf$, and $\TMF$ from $B\O/B\String$:

\begin{equation}
B\O/B\String\to B\GL_1(\MTString)\overset\sigma\to B\GL_1(\tmf)\to B\GL_1(\Tmf)\to B\GL_1(\TMF).
\end{equation}

Like in <ref> and <ref>, the section $B\O/B\SO\to B\O$ defines a homotopy equivalence of spaces

\begin{equation}
\label{ostring_SES}
B\O/B\String\overset\simeq\longrightarrow K(\Z/2, 1)\times B\SO/B\String,
\end{equation}

and there is a central extension of abelian $\infty$-groups

\begin{equation}
\label{spin_string_SES}
\shortexact*[\iota][]{\underbracket{B\Spin/B\String}_{K(\Z, 4)}}{B\SO /
B\String}{\underbracket{B\SO/B\Spin}_{K(\Z/2, 2)}},
\end{equation}

but now something new happens.

(<ref>) is not split.

A splitting of (<ref>) defines a section $s\colon B\SO/B\String\to B\Spin/B\String$, meaning $s\circ\iota = \id$. Therefore the map $\lambda\colon B\Spin\to B\Spin/B\String\overset\simeq\to K(\Z, 4)$ factors through $B\SO$:

\begin{equation}\begin{tikzcd}
B\Spin & {B\Spin/B\String} & {K(\Z, 4)}. \\
B\SO & {B\SO/B\String}
\arrow["\simeq"', from=1-2, to=1-3]
\arrow[from=1-1, to=1-2]
\arrow[from=1-1, to=2-1]
\arrow[from=2-1, to=2-2]
\arrow["\iota"', shift right, from=1-2, to=2-2]
\arrow["s"', shift right, from=2-2, to=1-2]
\arrow["\lambda", curve={height=-18pt}, from=1-1, to=1-3]
\end{tikzcd}\end{equation}

We let $\mu$ denote the extension of $\lambda$ to $B\SO$. Brown <cit.> shows that $H^4(B\SO;\Z)\cong\Z$ with generator $p_1$, so for any class $x\in H^4(B\SO;\Z)$, the pullback of $x$ to $B\Spin$ is some integer multiple of $p_1$. But the pullback of $\mu$ is $\lambda$, which is not an integer multiple of $p_1$, so we have found a contradiction.

We want an analogue of the fake vector bundle twists from <ref> and <ref> for $\MTString$, $\tmf$,
$\Tmf$, and $\TMF$, but since we just saw that $B\SO/B\String$ is not a product of Eilenberg-Mac Lane spaces, we
have to figure out what exactly it is. The answer turns out to be the analogue of an Eilenberg-Mac Lane space for a
relatively simple generalized cohomology theory.

Postnikov theory implies that if $E$ is a spectrum with only two nonzero homotopy groups
$\pi_m(E) = A$ and $\pi_n(E) = B$ (assume $m < n$ without loss of generality), then $E$ is classified by the data
of $m$, $n$, $A$,
$B$, and the $k$-invariant $k_E\in [\Sigma^m HA, \Sigma^{n+1}HB]$, a stable cohomology operation.

Let $\SH$ be the spectrum with $\pi_{-2}(\SH) = \Z/2$, $\pi_0(\SH) = \Z$, and the $k$-invariant $k_\SH =
\beta\circ\Sq^2\colon H^*(\bl;\Z/2)\to H^{*+3}(\bl;\Z)$. The generalized cohomology theory defined by $\SH$ is
called (restricted) supercohomology.[The adjective “restricted” is to contrast this theory with
“extended” supercohomology of Kapustin-Thorngren [77] and Wang-Gu [115]. See <cit.>.]

Just as the Eilenberg-Mac Lane spectrum $H\Z$ is assembled from Eilenberg-Mac Lane spaces $K(\Z, n)$ and there is a natural isomorphism $H^n(X, \Z)\overset\cong\to [X, K(\Z, n)]$, if one defines $\SK(n)$ to be the abelian $\infty$-group which is the extension

\begin{equation}
\label{sk_defn}
\shortexact{K(\Z, n)}{\SK(n)}{K(\Z/2, n-2)}{}
\end{equation}

classified by $\beta(\Sq^2(T))\in H^{n+1}(K(\Z/2, n-2); \Z)$, where $T\in H^{n-2}(K(\Z/2, n-2);\Z/2)$ is the tautological class and $\beta$ is the integral Bockstein, then the spaces $\SK(n)$ assemble into a model for the spectrum $\SH$ and there is a natural isomorphism $\SH^n(X)\overset\cong\to [X, \SK(n)]$.

There is an equivalence of abelian $\infty$-groups $B\SO/B\String\overset\simeq\to \SK(4)$. Moreover, the space of such equivalences is connected. Therefore there is a natural isomorphism of abelian groups $[X, B\SO/B\String]\cong\SH^4(X)$.

The point of the last sentence in <ref> is that in our proof, we do not specify an
isomorphism, so a priori there could be ambiguity like in <ref>.  But since the space of such
identifications is connected, there is a unique identification in the homotopy category, which suffices for the
calculations we make in this paper.

We are trying to identify the extension (<ref>) of abelian $\infty$-groups to relate it to $\SH$.
Because $B\SO/B\String$ is an abelian $\infty$-group, this extension, a priori classified by $H^5(K(\Z/2, 2), \Z)$,
actually is classified by the stabilization $[\Sigma^2 H\Z/2, \Sigma^5 H\Z]$: this extension is equivalent data to a
fiber sequence of connective spectra, so we get to use stable Postnikov theory. Our first step is to understand
$[\Sigma^2 H\Z/2, \Sigma^5 H\Z]$.

For all $k\in\Z$, $[H\Z/2, \Sigma^k H\Z]\cong [H\Z, \Sigma^{k-1}H\Z/2]$.

This follows by using the universal coefficient theorem to relate both groups to homology groups: the short exact sequences in the universal coefficient theorem simplify to identify the two groups in the lemma statement with $H_{k-1}(H\Z; \Z/2)$, resp. $H_{k-1}(H\Z/2;\Z)$ (the latter because the homology of $H\Z/2$ is torsion). Both of these groups are isomorphic to $\pi_{k-1}(H\Z\wedge H\Z/2)$, so the lemma follows.

$[\Sigma^2 H\Z/2, \Sigma^4 H\Z] = 0$ and $[\Sigma^2 H\Z/2, \Sigma^5 H\Z]\cong\Z/2$.

By <ref>, we need to compute $[H\Z, \Sigma^i H\Z/2] = H^i(H\Z; \Z/2)$ for $i = 1, 2$.
Let $\cA$ denote the mod $2$ Steenrod algebra; then $H^*(H\Z; \Z/2)\cong\cA\otimes_{\cA(0)}\Z/2$ <cit.>. This vanishes in
degree $1$ and is isomorphic to $\Z/2$ in degree $2$.

<Ref> implies (<ref>) is classified by a nonzero element of $[\Sigma^2 H\Z/2,
\Sigma^5 H\Z]$. And by definition, $\SK(4)$ is an extension of $K(\Z/2, 2)$ by $K(\Z, 4)$ classified by $\beta\circ\Sq^2$, which is a
nonzero element of $[\Sigma^2 H\Z/2, \Sigma^5 H\Z]$. Since this group is isomorphic to $\Z/2$ by <ref>,
these two nonzero elements must coincide, so there is an equivalence of abelian $\infty$-groups $B\SO/B\String\simeq \SK(4)$. There is a homotopy
type of such equivalences, and $\pi_0$ of that homotopy type is a torsor over $[\Sigma^2 H\Z/2, \Sigma^4H\Z]$,
which vanishes by <ref>, so the space of identifications is connected.

The map $K(\Z/2, 1)\to B\O$ defined by the tautological line bundle induces a homotopy equivalence of spaces

\begin{equation}
\label{ostring_equiv}
B\O/B\String\overset\simeq\longrightarrow K(\Z/2, 1)\times SK(4),
\end{equation}

implying that $\MTString$, $\tmf$, and $\TMF$ can be twisted over a space $X$ by classes $a\in H^1(X;\Z/2)$ and $d\in
\SH^4(X)$.

We call the twists associated to $a$ and $d$ in <ref> the
fake vector bundle twists for $\MTString$, $\tmf$, $\Tmf$, and $\TMF$.

Another consequence of <ref>, applied to the proof strategy of <ref>, is
that, even though $\lambda\in H^4(B\Spin;\Z)$ does not pull back from $B\SO$, its image in $\SH^4(B\Spin)$
does pull back from a class $\lambda\in \SH^4(B\SO)$. This is a theorem of Freed <cit.>, with additional proofs given by Jenquin <cit.> and Johnson-Freyd and
Treumann <cit.>.

The map $K(\Z, 4) \simeq B\Spin/B\String\to B\SO/B\String$ means degree-$4$ ordinary cohomology classes also
define degree-$4$ twists of string bordism and topological modular forms. Twists of this sort have already been studied, so we compare our twists to the literature.

Given $X$, $a$, and $d$ as in <ref>, let $\Omega_*^\String(X, a, d)$ denote the groups of bordism classes of manifolds $M$ equipped with maps $f\colon M\to X$ and trivializations of $w_1(M) - f^*(a)\in H^1(M;\Z/2)$ and $\lambda(M) - f^*(d)\in\SH^4(M)$.

A priori we only defined $\lambda$ as a characteristic class of oriented vector bundles; for an unoriented vector bundle $V$, $\lambda(V)$ is be defined to be $\lambda(V\oplus\Det(V))$, as the latter bundle is oriented. <Ref> first appears in work of B.L. Wang <cit.> in the special case when $a = 0$ and $d$ comes from ordinary cohomology.

There is a natural isomorphism $\pi_*(M^{\MTString}f_{a,d})\overset\cong\to \Omega_*^{\String}(X, a, d)$.

This follows from work of Hebestreit-Joachim [67], much like <ref>. Though they do not discuss the $\MTString$ case explicitly, their proof can be adapted to our setting. See <cit.>.

We can also compare with preexisting twists of $\tmf$.

The fake vector bundle twist defined by $K(\Z, 4)\to \SK(4)\to B\GL_1(\tmf)$ is homotopy equivalent to the twist $K(\Z, 4)\to B\GL_1(\tmf)$ constructed by Ando-Blumberg-Gepner <cit.>.

This equivalence is not obvious, because Ando-Blumberg-Gepner construct their twist in a different way: beginning with a map
$\phi\colon \Sigma_+^\infty K(\Z, 3)\to \tmf$ and using the adjunction <cit.> between
$\Sigma_+^\infty$ and $\GL_1$. However, their argument builds $\phi$ out of the map $\lambda\colon B\Spin\to
B\Spin/B\String\simeq K(\Z, 4)$, allowing one to pass our construction through their argument and conclude that
our twist, as a class in $[K(\Z, 4), B\GL_1(\tmf)]$, coincides with Ando-Blumberg-Gepner's.

Though these twists by degree-$4$ cohomology are relatively well-studied, there are not so many examples of
lower-degree twists of string bordism or topological modular forms in the literature. See
Freed-Hopkins-Teleman <cit.>, Johnson-Freyd <cit.>, and Tachikawa-Yamashita <cit.> for some examples.

Just as in <ref>, <ref> calculates some
$\MTString$-module Thom spectra for us: over $B\O/B\String$ we get $\MTO$; over $B\SO/B\String$ we get $\MTSO$, and
over $K(\Z, 4)$ we get $\MTSpin$. The last example is due to Beardsley <cit.>.

Like in <ref>, (<ref>) is not an equivalence of $\infty$-groups. The same two proofs are available to us: pulling back to $K(\Z/2, 1)$ and showing we do not obtain an $E_1$-ring spectrum, and comparing the group structures on $[\RP^\infty, B\O/B\String]$ and $[\RP^\infty, K(\Z/2, 1)\times B\SO/B\String]$. For the second proof, one observes that $[\RP^\infty, B\O/B\String]\cong\Z/8$ but $[\RP^\infty, K(\Z/2, 1)\times B\SO/B\String]$ has at least four elements of order $4$, then concludes.

For the first proof, we obtain $\MTString\wedge (B\O_1)^{\sigma-1}$ like before; to our knowledge, this notion of bordism has not been studied.[By analogy with $\SO$ and $\O$ and $\Spin$ and $\Pin^-$, one could call this tring$-$ bordism. We hope there is a better name for this spectrum.] However, since this is a vector bundle Thom spectrum, the change-of-rings trick shows that in topological degrees $15$ and below, the $E_2$-page of the Adams spectral sequence computing $\Omega_*^\String((B\O_1)^{\sigma-1})_2^\wedge$ is isomorphic to $\Ext_{\cA(2)}^{s,t}(H^*((B\Z/2)^{\sigma-1};\Z/2), \Z/2)$ (see <ref> for notation and an explanation). Davis-Mahowald <cit.> have computed these Ext groups, and from their computation it directly follows using the Adams spectral sequence that $\pi_0\cong\Z/2$ and $\pi_3\cong\Z/8$, so just like for $\MTPin^c$ and $\MTPin^-$, $\MTString\wedge (B\O_1)^{\sigma-1}$ does not admit an $E_1$-ring spectrum structure.

In the proof of <ref> we will need to understand the mod $2$ cohomology classes naturally associated to a degree-$4$ supercohomology class $d$. The quotient $t\colon \SH\to \Sigma^{-2}H\Z/2$ gives us a degree-$2$ class $t(d)$, sometimes called the Gu-Wen layer of $d$.

To proceed further, we study the Serre spectral sequence associated to the fibration $K(\Z, 4)\to \SK(4)\to K(\Z/2, 2)$. Let $\overline \delta\in H^4(K(\Z, 4);\Z/2)$ be the mod $2$ reduction of the tautological class; this defines a class in $E_2^{0,4}$ of our Serre spectral sequence, which we also call $\overline\delta$.

The class $\overline\delta\in E_2^{0,4}$ survives to the $E_\infty$-page.

The only possible differential that could be nonzero on $\overline\delta$ is the transgressing $d_5$, which pulls back from the transgressing $d_5$ on $\overline\delta$ in the Serre spectral sequence for the universal fibration with fiber $K(\Z, 4)$, namely $K(\Z, 4)\to E(K(\Z, 4))\to B(K(\Z, 4)) \simeq K(\Z, 5)$. In the universal fibration, $d_5(\overline\delta)$ is the mod $2$ tautological class $\epsilon\in H^5(K(\Z, 5);\Z/2)$, so in the fibration with total space $\SK(4)$, $d_5(\overline\delta)$ is the pullback of $\epsilon$ by the classifying map $\beta\circ\Sq^2\colon K(\Z/2, 2)\to K(\Z, 5)$. Thus $\epsilon\mapsto (\beta\Sq^2(B))\bmod 2 = \Sq^1\Sq^2(B)$, where $B\in H^2(K(\Z/2, 2);\Z/2)$ is the tautological class, but $\Sq^1\Sq^2(B) = \Sq^3(B) = 0$, as $B$ has degree $2$. Thus $d_5(\overline\delta)= 0$.

This is an unstable phenomenon: for $n > 2$, a similar argument shows the transgressing differential on the mod $2$ tautological class of $K(\Z, n)$ is nonzero, so no analogue of $\overline\delta$ exists in the cohomology of $\SK(n)$.

We want to lift $\overline\delta\in E_\infty^{0,4}$ to an element $\delta$ of $H^4(\SK(4);\Z/2)$. If $B$ is the tautological class of $K(\Z/2, 2)$, then there is an ambiguity between $\delta$ and $\delta + B^2$. To resolve this ambiguity, pull back across the map $\lambda\colon B\SO\to \SK(4)$. By comparing the Serre spectral sequences for the fibrations $K(\Z, 4)\to \SK(4)\to K(\Z/2, 2)$ and $B\Spin\to B\SO\to K(\Z/2, 2)$, one learns that $\lambda^*(\delta)$ is either $w_4$ or $w_4 + w_2^2$. Choosing the former allows us to uniquely define $\delta$.

There is a unique class $\delta\in H^4(\SK(4);\Z/2)$ such that $\lambda^*(\delta) = w_4$.

Phrased differently, associated to every $d\in\SH^4(X)$ is a class $\delta\in H^4(X;\Z/2)$, such that if there is an oriented vector bundle $V\to X$ with $d = \lambda(V)$, then $\delta = w_4(V)$. The same line of reasoning also shows that $\lambda^*(t(d)) = w_2$.

§ COMPUTING THE INPUT TO BAKER-LAZAREV'S ADAMS SPECTRAL SEQUENCE

§.§ Review: the change-of-rings theorem for vector bundle Thom spectra

We begin by reviewing how the story goes for vector bundle Thom spectra, where we can take advantage of a general
change-of-rings theorem. This is a standard technique dating back to work of Anderson-Brown-Peterson [6]
and Giambalvo [60, 61, 62]; see Beaudry-Campbell <cit.> for a nice introduction.

Let $\mathcal{B}$ be a graded Hopf algebra and $\mathcal{C} \subset \mathcal{B}$ be a graded Hopf subalgebra. If
$M$ is a graded $\mathcal C$-module and $N$ is a graded $\mathcal{B}$-module, then there is a natural isomorphism

\begin{equation}\label{eq:changeofrings}
\Ext^{s,t}_{\mathcal{B}}(\mathcal{B} \otimes_{\mathcal C} M, N) \overset{\cong}{\longrightarrow}\Ext^{s,t}_{\mathcal{C}}(M, N)
\end{equation}

For the little siblings we consider, we have the following isomorphisms of $\cA$-modules:

\begin{align}
H^*(H\Z;\Z/2) &\cong \cA \otimes_{\cA(0)}\Z/2\\
H^*(\ku;\Z/2) &\cong \cA \otimes_{\cE(1)} \Z/2\\
H^*(\ko; \Z/2) &\cong \cA \otimes_{\cA(1)} \Z/2\\
H^*(\tmf; \Z/2) &\cong \cA \otimes_{\cA(2)} \Z/2.
\end{align}

Here $\cA(n)$ is the subalgebra of $\cA$ generated by $\set{\Sq^1,\Sq^2,\Sq^4,\dotsc,\Sq^{2^n}}$ and $\cE(1) =
\ang{Q_0, Q_1}$, where $Q_0 = \Sq^1$ and $Q_1 = \Sq^1\Sq^2 + \Sq^2\Sq^1$. The isomorphisms in (<ref>)
were proven by Wall <cit.> ($H\Z$), Adams [8] ($\ku$), Stong [107] ($\ko$), and
Hopkins-Mahowald [69] ($\tmf$).

To use <ref>, we need to make $\cA(0)$, $\cA(1)$, $\cA(2)$, and $\cE(1)$
into Hopf subalgebras of $\cA$. This is equivalent to specifying how these algebras interplay with the cup product,
which the Cartan formula answers. For the Steenrod squares, this is standard; we also have $Q_i(ab) = aQ_i(b) +
Q_i(a)b$ for $i = 0,1$.

<Ref>, paired with (<ref>), greatly simplifies many computations: for any spectrum which splits as $ X= R \wedge Y$ where $R$ is
one of $H\Z$, $\ku$, $\ko$, or $\tmf$, the $E_2$-page of the Adams spectral sequence computing the $2$-completed homotopy groups of $X$ (or the
$R$-homology of $Y$) is identified with Ext groups over $\cA(0)$, $\cE(1)$, $\cA(1)$, or $\cA(2)$. These are much smaller than the entire
$2$-primary Steenrod algebra, so the Ext groups are easier to calculate; thus one often hears the slogan that
$\ko$-, $\ku$-, and $\tmf$-homology groups are relatively easy to compute with the Adams spectral
sequence,[The Adams spectral sequence computing $H\Z$-homology is essentially a repackaging of the
Bockstein spectral sequence; see May-Milgram [88].] and by (<ref>) and (<ref>),
those computations also compute , spin, and string bordism (the latter in dimensions $15$ and below). See
Douglas-Henriques-Hill [45] for a nice related computation of vector bundle twists of string bordism.

Another way to phrase this is that, though (<ref>) is about the little siblings only, combining it
with (<ref>) allows us to write down change-of-rings results for the Adams spectral sequences of
the big siblings. Specifically, there is an $\cA(0)$-module $W_1$, an $\cE(1)$-module $W_2$, and an $\cA(1)$-module
$W_3$ such that

\begin{align}
H^*(\MTSO;\Z/2) &\cong \cA\otimes_{\cA(0)} W_1\\
H^*(\MTSpin^c;\Z/2) &\cong\cA\otimes_{\cE(1)} W_2\\
H^*(\MTSpin; \Z/2) &\cong \cA\otimes_{\cA(1)} W_3,
\end{align}

so that the $E_2$-pages of the Adams spectral sequences computing the $2$-completions of $\Omega_*^\SO$,
$\Omega_*^{\Spin^c}$, and $\Omega_*^\Spin$ are the Ext groups of $W_1$, $W_2$, and $W_3$, respectively, over
$\cE(1)$, $\cA(1)$, and $\cA(2)$ respectively.
Explicitly, these modules begin in low degrees with (compare (<ref>))

\begin{align}
W_1 &\cong \Z/2 \oplus \Sigma^4\Z/2 \oplus \Sigma^5 \cA(0) \oplus \Sigma^8\Z/2 \oplus \Sigma^8 \Z/2
\oplus\dotsb\\
W_2 &\cong \Z/2 \oplus \Sigma^4 \Z/2 \oplus\Sigma^8\Z/2 \oplus \Sigma^8\Z/2\oplus
\Sigma^{10}\cE(1)\oplus\dotsb\\
W_3 &\cong \Z/2\oplus \Sigma^8\Z/2 \oplus \Sigma^{10}\cA(1)/\Sq^3 \oplus\dots
\end{align}

Often, though, what one wants is twisted. For vector bundle twists in the sense of <ref>, this is not a
problem: if $f\colon X\to B\GL_1(R)$ is a vector bundle twist specified by a rank-$r$ virtual vector bundle $V\to
X$, or strictly speaking by the rank-$0$ virtual vector bundle $V - r\coloneqq V - \underline\R^r$, then $f$
factors through $B\GL_1(\Sph)$, so <ref> provides a natural homotopy equivalence[For
a different, less abstract proof of this splitting, see <cit.> or <cit.>.]

\begin{equation}
Mf\overset{\simeq}{\longrightarrow} R\wedge X^{V - r}.
\end{equation}

Thus, for the ring spectra $R$ we discussed above, one can also use the change-of-rings isomorphism to simplify the computation of twisted $R$-homology for vector bundle twists: for $\ko$, the $E_2$-page is

\begin{equation}
E^{s,t}_2 = \Ext^{s,t}_{\cA(1)}(H^*(X^{V-r}; \Z/2), \Z/2) \Rightarrow ko_{t-s}(X)^{\wedge}_2,
\end{equation}

and the other choices of $R$ are analogous. The $\cA$-action (and hence also the $\cA(n)$ and $\cE(1)$-actions) on $H^*(X^{V-r};\Z/2)$ is easy to compute: the Thom isomorphism tells us the cohomology as a vector space, and the Stiefel-Whitney classes of $V$ twist the Steenrod squares as described in <cit.>.

This is a powerful generalization: many bordism spectra of interest arise as twists in this way, including pin$\pm$ bordism and all of the bordism spectra studied in [23, 32, 119, 121, 39, 54].

§.§ Baker-Lazarev's $R$-module Adams spectral sequence

For $R$ an $E_\infty$-ring spectrum,[Baker-Lazarev work with commutative algebras in
Elmendorf-Kriz-Mandell-May's $\Sph$-modules; as we discussed in Footnote <ref>, we may
equivalently work with $E_\infty$-ring spectra.] Baker-Lazarev [24] develop an $R$-module spectrum
generalization of the Adams spectral sequence which reduces to the usual Adams spectral sequence when $R = \Sph$.

For $R$-modules $H$ and $M$, the $R$-module $H$-homology of $M$ is

\begin{equation}
H^R_*(M) \coloneqq \pi_*(H \wedge_R M),
\end{equation}

and the $R$-module $H$-cohomology of $M$ is

\begin{equation}
H_R^*(M) \coloneqq \pi_{-*}\mathrm{Map}_R(M, H).
\end{equation}

For the purposes of this paper, $R$ will be one of the little siblings. For each such $R$, there is a canonical
isomorphism $\pi_0(R)\overset\cong\to\Z$, which lifts to identify the Postnikov quotient $\tau_{\le
0}R\overset\simeq\to H\Z$; as $\tau_{\le 0}R$ is an $R$-module spectrum via the quotient map $R\to\tau_{\le 0}R$,
this data provides a canonical $R$-algebra structure on $H\Z$. Composing with the mod $n$ reduction map $H\Z\to
H\Z/n$, we also obtain canonical $E_\infty$ $R$-algebra structures on $H\Z/n$ for all $n$. For $n = 2$ we have the
following isomorphisms of “algebras of $R$-module cohomology operations:”

Let $R$ be one of the little siblings and $H = H\Z/2$ with the $R$-algebra structure defined above. Then there are

\begin{align}
R &= H\Z\,, \quad H_R^{*}H \cong \cA(0)\\
R &= \ko\,, \quad H_R^{*}H \cong \cA(1)\\
R &= \ku\,, \quad H_R^{*}H \cong \cE(1)\\
R &= \tmf\,, \quad H_R^{*}H \cong \cA(2),
\end{align}

and dualizing gives the corresponding algebras of homology operations, e.g. $H^{H\Z}_*H\cong \cA(0)_*$.

This theorem was proven in pieces: the part for $H\Z$ is standard;
for $\ko$ and $\ku$ this is due to Baker <cit.>; and for $\tmf$ it is due to
Henriques [44].

We now present the spectral sequence; let $M$ and $N$ be $R$-modules, and let $H$ be a commutative $R$-ring spectrum.

Let $M$ and $N$ be $R$-modules and $H$ be an $E_\infty$ $R$-algebra, and suppose that $H^R_*H$ is a flat
$\pi_*(H)$-module. Then there is a spectral sequence of Adams type, natural in $M$, $N$, $H$, and $R$, with

\begin{equation}
E^{s,t}_2 = \Ext^{s,t}_{H_R^* H}(H^*_R M,H^*_R N),% \Rightarrow M^R_*N\,.
\end{equation}

and if $N$ is connective and $M$ is a cellular $R$-module spectrum with finitely many cells in each
degree,[This condition on $M$ is the analogue in $\cat{Mod}_R$ of the notion of a CW spectrum with
finitely many cells in each degree. If $M$ is the $R$-module Thom spectrum associated to a map $f\colon X\to
B\GL_1(R)$, which is the only case we consider in this paper, then this condition on $M$ is met if $X$ is a CW
complex with finitely many cells in each dimension.] then this spectral sequence converges to the homotopy groups
of the ($R$-module) $H$-nilpotent completion of $N_*^RM$.

Without the flatness assumption, one only has a description of the $E_1$-page, and it is more
complicated.[In applications, this may be less bad than it seems: for example, McNamara-Reece <cit.> interpret the $E_1$-page of the classical Adams spectral sequence in the context of quantum gravity.]
For example, this issue occurs when $R = \Sph$ and $H = \ku$, $\ko$, or $\tmf$; see [83, 37, 79, 26, 14, 15]. However, if $p$ is a prime number, $\pi_*(H\Z/p)\cong\Z/p$ is a field, so the flatness
assumption is satisfied for all $R$; as this is the only case we consider in this paper, we say no more about the
flatness assumption in <ref>.

The notion of the $H$-nilpotent completion of a spectrum is due to Bousfield <cit.>. When $H =
H\Z/p$, $p$ prime, this is the usual $p$-completion <cit.>. Thus if the homotopy groups of $N
\wedge_R M$ are finitely generated abelian groups, this as usual detects free and $p^k$-torsion summands, but not
torsion for other primes.

When $R = \Sph$, <ref> reduces to the classical $H$-based Adams spectral sequence, with its standard
convergence results. We will apply <ref> when $R$ is one of the little siblings, $H = H\Z/p$ for $p$
prime, and $N = R$: there is a canonical homotopy equivalence $R\wedge_R M\simeq M$, so in this setting
Baker-Lazarev's spectral sequence takes as input $\Ext_{H_R^*H}(H_R^*(M), \Z/2)$, and converges to the
$p$-completed homotopy groups of $M$.

For Thom spectra $H_R^*$ is easy.

For any $E_\infty$-ring spectrum $R$ such that $H\coloneqq H\Z/2$ is an $R$-algebra and $H_R^*H$ is a Poincaré
duality algebra, then for any map $f\colon X\to B\GL_1(R)$, there is an isomorphism $H^*(X;\Z/2)\to H_R^*(Mf)$.

This means that $H_R^*(Mf)$ is a free $H^*(X;\Z/2)$-module on a class $U\in H_R^0(Mf)$, which is the Thom class in
this setting. The Poincaré duality algebra condition is met for all of the little siblings $R$ we consider in this
paper, thanks to the explicit identifications above.

Apply <ref> with $R_1 = R$ and $R_2 = H\Z/2$ to learn that $Mf\wedge_R H\Z/2$, the object
whose homotopy groups are $H^R_*(Mf)$, is the Thom spectrum of a twist $f'\colon X\to B\GL_1(H\Z/2)$. By
<ref>, $B\GL_1(H\Z/2)$ is contractible, so $f'$ is null-homotopic, so by <ref>,
$Mf\wedge_R H\Z/2\simeq X_+\wedge H\Z/2$. Taking homotopy groups, we learn the analogue of the theorem statement
for $H^R_*(Mf)$.

To pass from homology to cohomology, we need some version of the universal coefficient theorem. This would follow
if $H\Z/2$ were shifted Spanier-Whitehead self-dual in $\cat{Mod}_R$, i.e. if we were given data of a
homotopy equivalence of $R$-module spectra $\mathrm{Map}_R(H\Z/2, R)\overset\simeq\to \Sigma^k H\Z/2$ for some $k$:
this correspondence would allow one to identify $R$-module $H\Z/2$-homology and $R$-module $H\Z/2$-cohomology after
a shift by $k$. Baker <cit.> shows that whenever $H^*_RH$ is a Poincaré self-duality algebra, then $H\Z/2$ is
shifted Spanier-Whitehead self-dual, and Poincaré self-duality holds for $\cA(0)$, $\cA(1)$, $\cE(1)$, and
$\cA(2)$, so we can conclude.

For most of our applications we will take $H = H \Z/2$.
[$\tmf$ at the prime $3$]

We will also work with an interesting odd-primary example, where $H = H\Z/3$ and $R = \tmf$. Let $\cA_3 \coloneqq
H^*H$, which is the mod $3$ Steenrod algebra, and let $\cA^{\tmf}\coloneqq H^*_\tmf H$; Henriques and Hill, using
the work of Behrens [20] and unpublished work of Hopkins-Mahowald, showed that

\begin{equation}\label{eq:Atmf_relations}
\cA^{\tmf}\cong \Z/3\ang{\beta, \cP^1}/(\beta^2, \beta(\cP^1)^2\beta - (\beta\cP^1)^2 - (\cP^1\beta)^2, (\cP^1)^3).
\end{equation}

Curiously, Rezk showed that $H^*(\tmf;\Z/3)$ is not isomorphic to $\cA_3\otimes_{\cA^{\tmf}}\Z/3$: see <cit.>.

The map $\phi\colon H_\tmf^*H\to H^*H$ sends $\beta$ to the Bockstein of $0\to\Z/3\to\Z/9\to\Z/3\to 0$ and $\cP^1$ to the first Steenrod power. However, unlike in the previous examples we studied, $\phi$ is not injective! The relation $\beta(\cP^1)^2 + \cP^1\beta\cP^1 + (\cP^1)^2\beta = 0$ is present in $\cA_3$ but not in $\cA^{\tmf}$ (see, e.g., <cit.>).

Baker-Lazarev's <ref> implies that for any $\tmf$-module spectrum $M$, $H_\tmf^*(M)$ carries a natural $\cA^{\tmf}$-module action, and there is an Adams spectral sequence

\begin{equation}
E_2^{s,t} = \Ext_{\cA^{\tmf}}^{s,t}(H_\tmf^*(M), \Z/3) \Longrightarrow \pi_{t-s}(M)_3^\wedge.
\end{equation}

In general, we will let $H_\tmf^*(M)$ refer to the mod $2$ $\tmf$-module cohomology and denote the mod $3$
$\tmf$-module cohomology by $H_\tmf^*(M; \Z/3)$. Because $(\Z/3)^\times$ is nontrivial, $B\GL_1(H\Z/3)$ is not
contractible, so the proof of <ref> does not directly generalize to this setting; however, as
$B\GL_1(H\Z/3)\cong B(\Z/3)^\times$ (see <ref>), for any twist $f\colon X\to B\GL_1(\tmf)$
factoring through a simply connected space, the induced twist of $H\Z/3$ is trivial and the argument goes through
to show $H_{\tmf}^*(M^\tmf f;\Z/3)\cong H^*(X;\Z/3)$. As $\SK(4)$ is simply connected, this includes the fake
vector bundle twists of $\tmf$ whose components in $H^1(\bl;\Z/2)$ vanish.

Like for the mod $2$ subalgebras of the Steenrod algebra that we discussed, we will want to know how $\cA^{\tmf}$ acts on products. The map $\cA^{\tmf}\to\cA_3$ is a map of Hopf algebras <cit.>, allowing us to use the Cartan formula and multiplicativity of the Bockstein in $\cA_3$ to conclude that in $\cA^{\tmf}$,

\begin{align}
\cP^1(ab) &= \cP^1(a)b + a\cP^1(b)\,,\\
\beta(ab) &= \beta(a)b + (-1)^{\abs a}a\beta(b).
\end{align}

When $R$ is one of the little siblings, <ref> implies that for any $R$-module spectrum $M$,
Baker-Lazarev's spectral sequence calculates $pi_*(M)_2^\wedge$ as the Ext of something over an algebra much
smaller than $\cA$ — one of $\cA(0)$, $\cE(1)$, $\cA(1)$, or $\cA(2)$. Thus the change-of-rings approach to
computing $\pi_*(R\wedge Y)_2^\wedge$ that we described in <ref> generalizes to other
$R$-modules $M$, in particular when $M$ is an $R$-module Thom spectrum — we just have to figure out $H_R^*(M)$.
This will be the main result of the next section.

§.§ Proof of the main theorem

At this point, we know from the previous section that even for non-vector-bundle Thom spectra $M^Rf$ over $R =
H\Z$, $\ku$, $\ko$ and $\tmf$, we can work over $\cE(1)$, $\cA(1)$, and $\cA(2)$ to compute the $E_2$-page of
Baker-Lazarev's Adams spectral sequence, implying that a change-of-rings formula for these Thom spectra exists.
Our next step is to
determine the $\cE(1)$-, $\cA(1)$-, and $\cA(2)$-modules $H_R^*(M^Rf)$. We describe the actions of the generators of
$\cE(1)$, $\cA(1)$, and $\cA(2)$ below in <ref>; however, it is not yet clear that they
satisfy the Adem relations, so we describe these modules over freer algebras, then later in the proof of
<ref> we show they are compatible with the Adem relations, hence are in fact $H_R^*H$-modules.

Let $X$ be a space.

* Given $a\in H^1(X;\Z/2)$, let $M_{H\Z}(a, X)$ be the $\Z/2[s_1]$-module which is a free
$H^*(X;\Z/2)$-module on a single generator $U$, and with $s_1$-action

\begin{equation}
s_1(Ux) \coloneqq U(ax + \Sq^1(x)).
\end{equation}

* Given $a\in H^1(X;\Z/2)$ and $c\in H^3(X;\Z)$, let $M_\ku(a, c, X)$ be the $\Z/2\ang{q_0, q_1}$-module
which is a free $H^*(X;\Z/2)$-module on a single generator $U$, and with $q_0$- and $q_1$-actions given by

\begin{equation}\label{eq:kuActions}
\begin{aligned}
q_0(Ux) &\coloneqq U(ax + Q_0(x))\\
q_1(Ux) &\coloneqq U((c\bmod 2+a^3)x + Q_1(x)).
\end{aligned}
\end{equation}

* Given $a\in H^1(X;\Z/2)$ and $b\in H^2(X;\Z/2)$, let $M_\ko(a, b, X)$ be the $\Z/2\ang{s_1, s_2}$-module
which is a free $H^*(X;\Z/2)$-module on a single generator $U$, and with $s_1$- and $s_2$-actions

\begin{equation}
\label{ko_s1s2}
\begin{aligned}
s_1(Ux) &\coloneqq U(ax + \Sq^1(x))\\
s_2(Ux) &\coloneqq U(bx + a\Sq^1(x) + \Sq^2(x)).
\end{aligned}
\end{equation}

* Given $a\in H^1(X;\Z/2)$, and $d\in SH^4(X)$, let $M_\tmf(a,  d, X)$ be the
$\Z/2\ang{s_1, s_2, s_4}$-module
which is a free $H^*(X;\Z/2)$-module on a single generator $U$, with $s_1$- and $s_2$-actions given
by (<ref>) with $b = t(d)$, and $s_4$-action given by

\begin{equation}
s_4(Ux) = U(\delta x + t(d) a+\Sq^1(t(d))) \Sq^1(x) + t(d)\Sq^2(x) + a\Sq^3(x) + \Sq^4(x)).
\end{equation}

* Given $d\in \SH^4(X)$, let $M_\tmf'(d, X)$ be the $\Z/3\ang{\beta, p^1}/(\beta^2)$-module which is a
free $H^*(X;\Z/2)$-module on a single generator $U$ and $\beta$- and $p_1$-actions specified by

\begin{equation}
\begin{aligned}
\beta(Ux) &\coloneqq U\beta(x)\\
p^1(Ux) &\coloneqq U((d\bmod 3)x + \cP^1(x)).
\end{aligned}
\end{equation}

The mod $3$ reduction of the supercohomology class $d$ is defined as usual as the image of $d$ after passing to the
mod $3$ Moore spectrum $\Sph/3$:

\begin{equation}
[X, \Sigma^4\SH] \longrightarrow [X, \Sigma^4\SH\wedge \Sph/3]\overset\cong\longrightarrow [X, \Sigma^4 H\Z/3],
\end{equation}

because $H\Z/2\wedge \Sph/3\simeq 0$. Thus $d\bmod 3$ is well-defined as a class in $H^4(X;\Z/3)$.

Keep the notation from <ref>.

* The action of $s_1$ on $M_{H\Z}(a, X)$ squares to $0$, so the $\Z/2[s_1]$-module structure on $M_{H\Z}(a,
X)$ refines to an $\cA(0)$-module structure with $\Sq^1(x)\coloneqq s_1(x)$.

* The actions of $q_0$ and $q_1$ on $M_\ku(a, c, X)$ both square to $0$, so the $\Z/2\ang{q_0,
q_1}$-module structure on $M_\ku(a, c, X)$ refines to an $\cE(1)$-module structure, where for $i = 0,1$,
$Q_i(x)\coloneqq q_i(x)$.

* The actions of $s_1$ and $s_2$ on $M_\ko(a, b, X)$, and of $s_1$, $s_2$, and $s_4$ on $M_\tmf(a, b, c,
X)$, satisfy the Adem relations with $s_i$ in place of $\Sq^i$, hence refine to an $\cA(1)$-module structure on
$M_\ko(a, b, X)$ and an $\cA(2)$-module structure on $M_\tmf(a, c, d, X)$.

* The actions of $\beta$ and $p^1$ on $M_\tmf'(c, X)$ satisfy the relations in (<ref>), hence refine the $\Z/3\ang{\beta, p^1}/(\beta^2)$-module structure on $M_\tmf'(c, X)$ to
an $\cA^\tmf$-module structure, where the Bockstein acts as $\beta$ and $\cP^1$ acts as $p^1$.

Rather than prove this directly, we will obtain it as a corollary of <ref>. This theorem
says that the modules defined in <ref> are $H_R^*$ of the Thom spectra for the corresponding

Let $X$ be a topological space.

* Given $a\in H^1(X;\Z/2)$, let $f_a\colon X\to B\GL_1(H\Z)$ be the corresponding fake vector bundle twist.
Then there is an isomorphism of $\cA(0)$-modules

\begin{equation}
H_{H\Z}^*(M^{H\Z} f_{a})\overset\cong\longrightarrow M_{H\Z}(a, X).
\end{equation}

* Given $a\in H^1(X;\Z/2)$ and $c\in H^3(X;\Z)$, let $f_{a,c}\colon X\to B\GL_1(\ku)$ be the corresponding
fake vector bundle twist. Then there is an isomorphism of $\cE(1)$-modules

\begin{equation}
H_\ku^*(M^\ku f_{a,c})\overset\cong\longrightarrow M_{\ku}(a, c, X).
\end{equation}

* Given $a\in H^1(X;\Z/2)$ and $b\in H^2(X;\Z/2)$, let $f_{a,b}\colon X\to B\GL_1(\ko)$ be the
corresponding fake vector bundle twist. Then there is an isomorphism of $\cA(1)$-modules

\begin{equation}
H_\ko^*(M^\ko f_{a,b})\overset\cong\longrightarrow M_{\ko}(a, b, X).
\end{equation}

*  Given $a\in H^1(X;\Z/2)$, and $d\in SH^4(X)$, let $f_{a,d}\colon X\to
B\GL_1(\tmf)$ be the corresponding fake vector bundle twist. Then there is an isomorphism of $\cA(2)$-modules

\begin{equation}
H_\tmf^*(M^\tmf f_{a,d})\overset\cong\longrightarrow M_{\tmf}(a, d, X),
\end{equation}

and an isomorphism of $\cA^\tmf$-modules

\begin{equation}
H_\tmf^*(M^\tmf f_{0, d}; \Z/3)\overset\cong\longrightarrow M_{\tmf}'(d, X).
\end{equation}

In the last isomorphism, we turn off degree-$1$ twists so that we have a Thom isomorphism for mod $3$ cohomology.

Keep the notation from <ref>.

Twisted $\Z$-homology The $E_2$-page of Baker-Lazarev's Adams spectral sequence computing $\pi_*(M^{H\Z}
f_{a})_2^\wedge$ is isomorphic as $\Ext_{\cA(0)}(\Z/2)$-modules to $\Ext_{\cA(0)}^{s,t}(M_{H\Z}(a, X), \Z/2)$.

Twisted $\ku$-homology The $E_2$-page of Baker-Lazarev's Adams spectral sequence computing $\pi_*(M^\ku
f_{a,c})_2^\wedge$ is isomorphic as $\Ext_{\cE(1)}(\Z/2)$-modules to $\Ext_{\cE(1)}^{s,t}(M_\ku(a, c, X),
\Z/2)$.

Twisted $\ko$-homology The $E_2$-page of Baker-Lazarev's Adams spectral sequence computing $\pi_*(M^\ko
f_{a,b})_2^\wedge$ is isomorphic as $\Ext_{\cA(1)}(\Z/2)$-modules to $\Ext_{\cA(1)}^{s,t}(M_\ko(a, b, X),
\Z/2)$.

Twisted $\tmf$-homology

* The $E_2$-page of Baker-Lazarev's Adams spectral sequence computing $\pi_*(M^\tmf f_{a, d})_2^\wedge$
is isomorphic as $\Ext_{\cA(2)}(\Z/2)$-modules to $\Ext_{\cA(2)}^{s,t}(M_\tmf(a ,d, X), \Z/2)$.

* The $E_2$-page of Baker-Lazarev's Adams spectral sequence computing $\pi_*(M^\tmf f_{0, d})_3^\wedge$
is isomorphic as $\Ext_{\cA^\tmf}(\Z/3)$-modules to $\Ext_{\cA^\tmf}^{s,t}(M_\tmf'(d, X), \Z/3)$.

In <ref> we saw that the twists of $H\Z$ discussed above are all vector bundle twists, so that the
$H\Z$ part of <ref> follows from the standard change-of-rings argument; the same is
true for the twists of $\MTSO$ appearing below in <ref>. In both cases, the other
calculations are new.

The analogue of <ref> is true for a few standard variants of the Adams spectral
sequence. For example, one could switch the order of $H_R^*(Mf)$ and $\Z/2$ in $\Ext_{H_R^*H}$ and obtain the
$E_2$-page of Baker-Lazarev's Adams spectral sequence computing twisted $R$-cohomology. One could also work out a
version of <ref> in terms of $R$-module $H$-homology with its $H^R_*H$-comodule

Recall the modules $W_1$, $W_2$, and $W_3$ from <ref>.

Keep the notation from <ref>.

Twisted oriented bordism The $E_2$-page of Baker-Lazarev's Adams spectral sequence
computing $(\Omega_*^{\SO}(X, a))_2^\wedge$ is isomorphic as $\Ext_{\cA(0)}(\Z/2)$-modules to
$\Ext_{\cA(0)}^{s,t}(M_{H\Z}(a, X)\otimes W_1, \Z/2)$.

Twisted bordism The $E_2$-page of Baker-Lazarev's Adams spectral sequence
computing $(\Omega_*^{\Spin^c}(X, a, c))_2^\wedge$ is isomorphic as $\Ext_{\cE(1)}(\Z/2)$-modules to
$\Ext_{\cE(1)}^{s,t}(M_\ku(a, c, X)\otimes W_2, \Z/2)$.

Twisted spin bordism The $E_2$-page of Baker-Lazarev's Adams spectral sequence
computing $(\Omega_*^{\Spin}(X, a, b))_2^\wedge$ is isomorphic as $\Ext_{\cA(1)}(\Z/2)$-modules to
$\Ext_{\cA(1)}^{s,t}(M_\ko(a, b, X)\otimes W_3, \Z/2)$. For $t-s < 8$, this is isomorphic to
$\Ext_{\cA(1)}^{s,t}(M_\ko(a, b, X), \Z/2)$.

Twisted string bordism For $t-s < 16$, the $E_2$-pages of Baker-Lazarev's Adams spectral sequences
computing $(\Omega_*^\String(X, a, d))_2^\wedge$, resp. $(\Omega_*^\String(X, 0, d))_3^\wedge$, are isomorphic
to $\Ext_{\cA(2)}^{s,t}(M_\tmf(a, d, X), \Z/2)$, resp. $\Ext_{\cA^{\tmf}}(M_\tmf'(d, X), \Z/3)$, as modules
over $\Ext_{\cA(2)}(\Z/2)$, resp. $\Ext_{\cA^{\tmf}}(\Z/3)$.

All five parts of the theorem have similar proofs, so we walk through
the full proof in two cases — $R = \ku$, whose proof carries through for $H\Z$, $\ko$, and $\tmf$ at $p = 3$ with
minor changes; and $R = \tmf$ at $p = 2$, where the presence of supercohomology means the proof is slightly

Now we specialize to $R= \ku$ and a fake vector bundle twist $f_{a,c}\colon X\to B\GL_1(\ku)$. To begin, use
<ref> to learn that $H_\ku^*(M^\ku f_{a,c})\cong H^*(X;\Z/2)$ as $\Z/2$-vector spaces.

(In the more familiar case where the twist is given by a vector bundle, this is the Thom isomorphism.) Next, the
Thom diagonal (<ref>) and the Cartan formula provide a formula for $Q_i(Ux)$, $i = 0,1$, in terms of
$Q_i(U)$ and $Q_i(x)$. In particular, this formula implies that if we can show $Q_0(U) = Ua$ and $Q_1(U) =
U(a^3+c)$, then the $\cE(1)$-module action defined on $M_\ku(X, a, c)$ in <ref> is identified
with $H_\ku^*(Mf_{a,c})$.

By the naturality of cohomology operations, it suffices to compute $Q_0(U)$ and $Q_1(U)$ for the

the universal twist over $B\O/B\Spin^c$. <Ref> then allows us to infer what the cohomology
operations on the Thom class have to be in order to recover the correct $\cA$-module structure on the Thom spectrum
after applying the universal twist.

Let $f\colon B\O/B\Spin^c\to B\GL_1(\MTSpin^c)$ be the universal fake vector bundle twist,
$M^{\MTSpin^c}f$ be its associated Thom spectrum, and $M^{\ku}f$ be the $\ku$-module Thom spectrum obtained by
composing $f$ with the map $B\GL_1(\MTSpin^c)\to B\GL_1(\ku)$ induced by the Atiyah-Bott-Shapiro map. The
Atiyah-Bott-Shapiro map is $3$-connected, so the map $M^{\MTSpin^c}f\to M^{\ku}f$ is also $3$-connected. Thus, for
example, $\pi_0(M^{\ku}f)\cong\pi_0(M^{\MTSpin^c} f)$; by <ref> $M^{\MTSpin^c}f\simeq\MTO$,
so $\pi_0(M^{\ku} f)\cong\Omega_0^\O\cong\Z/2$. This and similar ideas will determine $Q_0(U)$ and $Q_1(U)$ for us:

in particular we will find $\Sq^1(U) = Ua$ and $Q_1(U) = U(c \bmod 2+a^3)$ because this is the unique
choice that is compatible with the known homotopy groups of the Thom spectra of the universal twists from <ref>: $\MTO$ over $K(\Z/2, 1)\times K(\Z, 3)$, $\MTSO$ over $K(\Z, 3)$, and $\MTPin^c$ over $K(\Z/2, 1)$.

We first consider $Q_0$: $Q_0(U)$ is either $0$ or $Ua$.

For either of the two options for $Q_0(U)$, one can explicitly
write the $\cE(1)$-module structure on $H_\ku^*(M^\ku f)$ in low degrees. Then, using Baker-Lazarev's Adams
spectral sequence, one finds that if $Q_0(U) = 0$, $\pi_0(M^\ku f)_2^\wedge\cong \pi_0(M^{\MTSpin^c}f)$ has at
least $4$ elements, but since $Mf \simeq\MTO$, we know this group is $\Omega_0^\O\cong\Z/2$. Thus $Q_0(U) = Ua$.

There are three options for $Q_1(U)$: $0$, $U c\bmod 2$, and $U(c\bmod 2+a^3)$. In order to verify the $Q_1$
action, we pull back to $K(\Z, 3)$ and $K(\Z/2, 1)$ separately, and then argue in a similar way.

* For $f\colon K(\Z, 3)\to B\GL_1(\MTSpin^c)$, $M^{\MTSpin^c} f\simeq\MTSO$, which is incompatible with
$Q_1(U) = 0$; the argument is similar to that for $Q_0$.

* For $f\colon K(\Z/2, 1)\to B\GL_1(\MTSpin^c)$, $M^{\MTSpin^c} f\simeq \MTPin^c$.
In $H_\ku^*(M^\ku f)$, $Q_1(U)\ne 0$, which one can show by pulling back further along

\begin{equation}
M^\ku f\wedge H\Z/2\longrightarrow M^\ku\wedge_{\ku} H\Z/2.
\end{equation}

Thus $Q_1(U) = U(c\bmod 2+a^3)$.

Using the fact that $\cE(1) = \langle Q_0, Q_1\rangle$ and applying the Cartan formula recovers the actions in (<ref>).

Because the fake vector bundle twist for $\tmf$ uses supercohomology, its part of the proof is different enough
that we go into the details. The reduction to the computation of $\Sq^1(U)$, $\Sq^2(U)$, and $\Sq^4(U)$ in the case
of the universal twist proceeds in the same way as for $\ku$. In <ref> we computed
$H^*(B\SO/B\String;\Z/2)$ in low degrees; this and the Künneth formula imply that in the mod $2$ cohomology of
$K(\Z/2, 1)\times B\SO/B\String$, $H^1$ is spanned by $a$, $H^2$ is spanned by $\set{a^2, t(d)}$, and $H^4$ is
spanned by $\set{a^4, a^2 t(d), a\Sq^1t(d), \delta, t(d)^2}$. Therefore there are
$\lambda_1,\dotsc,\lambda_8\in\Z/2$ such that

\begin{align}
\Sq^1(U) &= U\lambda_1 a\\
\label{eq:Sq2tmf}
\Sq^2(U) &= U(\lambda_2 a^2 + \lambda_3 t(d))\\
\Sq^4(U) &= U(\lambda_4 a^4 + \lambda_5 a^2t(d) + \lambda_6 a\Sq^1t(d) + \lambda_7\delta + \lambda_8 t(d)^2).\label{eq:Sq4tmf}
\end{align}

We finish the proof by indicating how to find $\lambda_1$ through $\lambda_8$.
To find $\lambda_7$, consider the twist pulled back to $f\colon K(\Z, 4)\simeq B\Spin/B\String\to B\O/B\String$.
Like in the proof for twists of $\ku$, the action of $\Sq^4$ on the Thom class can be detected on either
$M^{\MTString}f$ or $M^{\tmf}f$; as we discussed in <ref>, $M^{\MTString}f\simeq\MTSpin$,
so $\pi_3(M^{\MTString}f)\cong\Omega_3^\Spin = 0$, and since the map $M^{\MTString}f\to M^{\tmf}f$ is
sufficiently connected, $\pi_3(M^{\tmf}f) = 0$ as well. In $H_{\tmf}^*(M^{\tmf}f)$, the only options for $\Sq^4(U)$
are $0$ or $U$ times the tautological class. One can run the Baker-Lazarev Adams spectral sequence for these two
options and see that only the latter choice is compatible with $\pi_3(M^{\tmf}f) = 0$.[To do so, it will
be helpful to know $\Ext_{\cA(2)}(C\nu, \Z/2)$, where $C\nu$ is the $\cA(2)$-module with two $\Z/2$ summands in
degrees $0$ and $4$, joined by a $\Sq^4$. These Ext groups have been computed by Bruner-Rognes <cit.>.] Thus $\lambda_7 = 1$.

For the other coefficients, we pull back to vector bundle twists for various vector bundles $V\to X$, where we know
$\Sq^k(U) = Uw_k(V)$, $a\mapsto w_1(V)$, $t(d)\mapsto w_2(V)$, and $\delta\mapsto w_4(V)$. Choosing vector bundles with auspicious values of $w_1$, $w_2$, and $w_4$ quickly determines the remaining coefficients.

* Pulling back the twist to $K(\Z/2, 1)\simeq B\O_1$ gives the Thom spectrum  $\tmf\wedge
(B\O_1)^{\sigma-1}$, where $\sigma\to B\O_1$ is the tautological line bundle. As $w_1(\sigma)\ne 0$ but $w_2(\sigma) = 0$ and $w_4(\sigma) = 0$, we can plug these
Stiefel-Whitney classes into  (<ref>) (with $w_1(V)$ in place of $a$, $w_2(V)$ in place of $t(d)$,
and $w_4(V)$ in place of $\delta$ as usual) to conclude
$\lambda_1 = 1$, $\lambda_2 = 0$, and $\lambda_4 = 0$.

* Let $V \coloneqq \mathcal{O}(1)\oplus\mathcal{O}(2)\to\mathbb{CP}^2$. If $\alpha\in H^2(\mathbb{CP}^2;\Z/2)\cong\Z/2$ is the unique nonzero element, then $w_1(V) = 0$, $w_2(V) = \alpha$, and $w_4(V) = 0$. Plugging this into (<ref>), we find $\Sq^2(U) = U\alpha = U\lambda_3\alpha$, so $\lambda_3 = 1$. And plugging $w_1(V)$, $w_2(V)$, and $w_4(V)$ into (<ref>), we obtain $\Sq^4(U) = 0 = U\lambda_8\alpha^2$, so $\lambda_8 = 0$.

* Let $x$, resp. $y$ be the nonzero classes in $H^1(\RP^2\times\RP^2;\Z/2)$ pulled back from the first, resp. second copy of $\RP^2$, and let $\sigma_x,\sigma_y\to\RP^2\times\RP^2$ be the real line bundles satisfying $w_1(\sigma_x) = x$ and $w_1(\sigma_y) = y$. Now let $V \coloneqq \sigma_x\oplus \sigma_y^{\oplus 3}$; then $w_1(V) = x+y$, $w_2(V) = xy+y^2$, and $w_4(V) = 0$. Plugging into (<ref>), we have $\Sq^4(U) = 0 = U\lambda_5x^2y^2$, so $\lambda_5 = 0$.

* Repeat the preceding example, but with $\RP^1\times\RP^3$ in place of $\RP^2\times\RP^2$; this time, $w_1(V) = x+y$, $w_2(V) = xy+y^2$, and $w_4(V) = xy^3$. Plugging into (<ref>), we have $\Sq^4(U) = Uxy^3 = U(1+\lambda_6)xy^3$, so $\lambda_6 = 0$.

§ APPLICATIONS

In this section, we give examples in which we use <ref>
to make computations of twisted (co)homology groups.

§.§ U-duality and related twists of spin bordism

Let $G$ be a topological group and

\begin{equation}
\shortexact*{\set{\pm 1}}{\wG}{G}{}
\end{equation}

be a central extension classified by $\beta\in H^2(BG;\set{\pm 1})$. Then the central extension

\begin{equation}
\label{spin_G_xtn}
\shortexact*[][p]{\set{\pm 1}}{\Spin\times_{\set{\pm 1}}\wG}{\SO\times G}{}
\end{equation}

is classified by $w_2 + \beta\in H^2(B(\SO\times G);\Z/2)$. One can
prove this is the extension by pulling back along $\SO \rightarrow \SO \times G$ and $G \rightarrow \SO \times G$ and observing that both
pulled-back extensions are non-split.
Therefore given an oriented vector bundle
$E\to X$ and a principal $G$-bundle $P\to X$, i.e. the data of an $\SO\times G$ structure on $E$, a lift of this
data to a $\Spin\times_{\set{\pm 1}}\wG$-structure is a trivialization of $w_2(E) + f_P^*(\beta)$, where
$f_P\colon X\to BG$ is the classifying map of $P\to X$. That is, if $\xi$ denotes the composition

\begin{equation}
\xi\colon B(\Spin\times_{\set{\pm 1}}\wG)\overset{Bp}{\longrightarrow} B\SO\times BG\to B\SO\to B\O,
\end{equation}

then a $\xi$-structure on $E$ is equivalent to a $(BG, \beta)$-twisted spin structure, meaning that by <ref> the Thom spectrum $\mathit{MT\xi}$ is canonically equivalent to the $\MTSpin$-module Thom spectrum
$Mf_{0,\beta}$ associated to the fake vector bundle twist $f_{0,\beta} \colon BG\to B\GL_1(\MTSpin)$. This Thom spectrum
may or may not split as $\MTSpin\wedge X$ for a spectrum $X$: a sufficient condition is the existence of a
vector bundle $V\to BG$ such that $w_2(V) = \beta$, as we discussed in <ref>, but as we will see soon,
there are choices of $(G, \beta)$, even when $G$ is a compact, connected Lie group, for which no such $V$ exists.
For these $G$ and $\beta$, <ref> significantly simplifies the calculation of $\xi$-bordism.

As an example, consider $G = \SU_8/\set{\pm 1}$ and $\beta$ the nonzero element of $H^2(BG;\Z/2)
\cong\Hom(\pi_1(G), \Z/2) \cong\Z/2$, corresponding to the central extension

\begin{equation}
\shortexact*{\set{\pm 1}}{\SU_8}{\SU_8/\set{\pm 1}}.
\end{equation}

In [51], we studied $\Omega_*^{\Spin\times_{\set{\pm 1}}\SU_8}$ as part of an argument that the
$E_{7(7)}(\R)$ U-duality symmetry of four-dimensional $\cN = 8$ supergravity is anomaly-free. Speyer [104]
shows that all representations of $G$ are spin, so $\beta\ne w_2(V)$ for any vector bundle $V\to BG$ induced from a
representation of $G$, and this can be upgraded to show $Mf_{0,\beta}\not\simeq\MTSpin\wedge X$ for
any spectrum $X$ (see <cit.>). This precludes the standard shearing/change-of-rings argument for
computing $\Spin\times_{\set{\pm 1}}\SU_8$ bordism, and indeed in <cit.> we had to give a more
complicated workaround. However, thanks to <ref>, we can now argue over $\cA(1)$. We need as input
the low-degree cohomology of $B(\SU_8/\set{\pm 1})$.

$H^*(B(\SU_8/\set{\pm 1});\Z/2)\cong\Z/2[\beta, b, c, d, e, \dots]/(\dots)$ with $\abs \beta = 2$, $\abs b = 3$,
$\abs c = 4$, $\abs d = 5$, and $\abs e = 6$; there are no other generators below degree $7$ and no relations below
degree $7$. The Steenrod squares are

\begin{equation}\label{eq:SteenrodSU8/Z2}
\begin{aligned}
\Sq(\beta) &= \beta + b + \beta^2\\
\Sq(b) &= b+d + b^2\\
\Sq(c) &= c+e + \Sq^3(c) + c^2\\
\Sq(d) &= d + b^2 + \Sq^3(d) + \Sq^4(d) + d^2.
\end{aligned}
\end{equation}

Now we can calculate $H_\ko^*(M^\ko f_{0,\beta})$; by <ref>, the map
$M^{\MTSpin} f_{0,\beta}\to M^\ko f_{0,\beta}$ is an isomorphism on homotopy groups in degrees we care about, so
$M^{\ko}f_{0,\beta}$ suffices. <Ref> tells us $\Sq^1(U) = 0$ and $\Sq^2(U) =
U\beta$; to make more computations, use the Cartan formula and the Steenrod squares in <ref>:
Then proceeding with the relation in (<ref>) yields

\begin{gather}
\begin{aligned}
\Sq^1(U\beta) &= U\Sq^1(\beta) + \Sq^1(U)\beta = Ub\\
\Sq^2(U\beta) &= U\Sq^2(\beta) + \Sq^1(U)\Sq^1(\beta) + \Sq^2(U)\beta = U(2\beta^2) = 0\\
%    \Sq^2(Ub) &= \Sq^2(b)U + Ub\beta \Sq^1(b)Sq^1(U) =  (d+b\beta)U\,, \notag
\end{aligned}\\
\begin{aligned}
\Sq^1(Ub) &= U\Sq^1(b) + \Sq^1(U)b = 0\\
\Sq^2(Ub) &= U\Sq^2(b) + \Sq^1(U)\Sq^1(b) + \Sq^2(U)b = U(d+b\beta)
\end{aligned}\\
\begin{aligned}
\Sq^1(U(d+b\beta)) &= U\Sq^1(d+b\beta) + \Sq^1(U)(d+b\beta) = U(2b^2) = 0\\
\Sq^2(U(d+b\beta)) &= U\Sq^2(d+b\beta) + \Sq^1(U)\Sq^1(d+b\beta) + \Sq^2(U)(d+b\beta)= 0.
\end{aligned}
\end{gather}

See the red piece of <ref>, left, for a picture of this data.
This calculation implies the vector space generated by $\set{U, U\beta, Ub, U(d+b\beta)}$ is an $\cA(1)$-submodule
of $H_\ko^*(M^\ko f_{0,\beta})$; specifically, it is isomorphic to the “seagull” $\cA(1)$-module $M_0\coloneqq
\cA(1)\otimes_{\cA(0)}\Z/2$. This is an $\cA(1)$-module whose $\cA(1)$-action does not compatibly extend to an
$\cA$-action. Continuing to compute $\Sq^1$- and $\Sq^2$-actions as in (<ref>), we learn that there
is an isomorphism of $\cA(1)$-modules

\begin{equation}
\label{spin-SU8_Hko}
H_{\ko}^*(M^\ko f_{0,\beta})\cong \textcolor{BrickRed}{M_0} \oplus
\textcolor{Green}{\Sigma^4 M_0} \oplus
\textcolor{MidnightBlue}{\Sigma^4 M_1} \oplus
\textcolor{Fuchsia}{\cA(1)} \oplus P,
\end{equation}

where $P$ is concentrated in degrees $6$ and above (so we can and will ignore it), and
$\textcolor{MidnightBlue}{M_1}$ is an $\cA(1)$-module which is isomorphic to either $M_0$ or $C\eta\coloneqq
\cA(1)\otimes_{\cE(1)}\Z/2$. We draw the decomposition (<ref>) in <ref>, left.

[scale=0.6, every node/.style = font = ]
in 0, ..., 11
at (-2, ) $\y$;

(0.4,1.7) node $U\beta$;
(0.4,3) node $Ub$;
(0,5.3) node $U(d+b\beta)$;

Left: the $\cA(1)$-module structure on $H_{\ko}^*(M^\ko f_{0,\beta})$ in low degrees, where
$\beta\in H^2(B(\SU_8/\set{\pm 1});\Z/2)$ is the generator. The pictured submodule contains all elements in degrees
$5$ and below. We have not determined $\Sq^3(Uc)$ – it may be $0$, in which case the blue summand would vanish in
degrees $7$ and above. In either case, the pictured $\cA(1)$-module cannot arise as the restriction of an
$\cA$-action to $\cA(1)$, indicating that the fake vector bundle twist $f_{0,\beta}$ of $\ko$ cannot arise from a
vector bundle.  Right: the $E_2$-page of the corresponding $\ko$-module Adams spectral sequence, which as discussed
in <ref> also computes the $2$-completion of $\Omega_*^{\Spin\times_{\set{\pm 1}}\SU_8}$.

The change-of-rings isomorphism (<ref>) and Koszul duality <cit.> allow us to compute
$\Ext_{\cA(1)}(M_0)\cong\Z/2[h_0]$ and $\Ext_{\cA(1)}(C\eta)\cong\Z/2[h_0, v_1]$ with $h_0$ in bidegree $(t-s, s)
= (0, 1)$ and $v_1$ in bidegree $(t-s, s) = (2, 1)$ <cit.>. Therefore we can draw
the $E_2$-page of the Adams spectral sequence computing the twisted $\ko$-homology associated to the fake vector
bundle twist $f_{0,\beta}\colon B(\SU_8/\set{\pm 1})\to B\GL_1(\ko)$ in <ref>, right, and by
<ref> this also computes the corresponding twisted spin bordism groups, which we
saw above are $\Omega_*^{\Spin\times_{\set{\pm 1}}\SU_8}$. This spectral sequence collapses on the $E_2$-page in
degrees $5$ and below, using $h_0$-linearity of differentials, so we have made the following computation.

\begin{equation}
\begin{aligned}
\Omega_0^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong \Z\\
\Omega_1^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong 0\\
\Omega_2^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong 0\\
\Omega_3^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong 0\\
\Omega_4^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong \Z^2\\
\Omega_5^{\Spin\times_{\set{\pm 1}}\SU_8} &\cong \Z/2.
\end{aligned}
\end{equation}

There are a few other choices of compact Lie groups $G$ and classes $\beta \in H^2(BG;\Z/2)$ such that $\beta$ is
not equal to $w_2$ of any representation, including

* $\SU_{4n}/\set{\pm 1}$ for $n > 1$, where $\beta$ corresponds to the double cover
$\SU_{4n}\to\SU_{4n}/\set{\pm 1}$ [104],

* $\PSO_{8n}$, where $\beta$ corresponds to the double cover $\SO_{8n}\to\PSO_{8n}$ [71],

* $\mathrm{PSp}_n$ and the double cover $\mathrm{Sp}_n\to \mathrm{PSp}_n$ for $n > 1$, and

* $E_7/\set{\pm 1}$ and the double cover $E_7\to E_7/\set{\pm 1}$.

For the last two items, the proof is analogous to <cit.> for $\SU_8/\set{\pm 1}$: compute the low-degree mod $2$ cohomology of $BG$ and use this to show that if $\beta$ is $w_2$ of a representation $V$, the $\cA$-action on the cohomology of the corresponding Thom spectrum violates the Adem relations.

For all of these choices of $G$ and $\beta$, one can define (at a physics level of rigor) unitary quantum field theories with fermions
and a background $\wG$ symmetry, such that $-1\in\wG$ acts by $-1$ on fermions and by $1$ on bosons. Then, as
described in [120, 109], these theories can be defined on manifolds with differential
$\Spin_n\times_{\set{\pm 1}}\wG$ structures, so by work of Freed-Hopkins [54], the anomaly field theories of
these QFTs are classified using the bordism groups $\Omega_*^{\Spin\times_{\set{\pm 1}}\wG}$, and computations such
as <ref> are greatly simplified using <ref>.

Though we focused on invertible field theories in this section, there are other applications of twisted spin
bordism groups. For example, Kreck's modified surgery [75] uses twisted spin bordism to classify closed,
smooth $4$-manifolds whose universal covers are spin up to stable diffeomorphism: given such a manifold $M$, one
shows that $w_1(M)$ and $w_2(M)$ pull back from $B\pi_1(M)$, then considers twisted spin bordism for the fake
vector bundle twist over $B\pi_1(M)$ given by $w_1(M)$ and $w_2(M)$.

Often one computes these bordism groups with Teichner's James spectral sequence <cit.>, a
version of the Atiyah-Hirzebruch spectral sequence for spin bordism that can handle non-vector-bundle twists.
However, extension questions in this spectral sequence can be difficult, and it is helpful to have the Adams
spectral sequence to resolve them (see [94] for an example for a vector bundle twist). Therefore
<ref> could be a useful tool for studying stable diffeomorphism classes of
$4$-manifolds, since not all of the relevant twists come from vector bundles.

§.§ Twists of string bordism

A story very similar to that of <ref> takes place one level up in the Whitehead tower for $B\O$. Many
supergravity theories require spacetime manifolds $M$ to satisfy a Green-Schwarz condition specified by a
Lie group $G$ and a class $c\in H^4(BG;\Z)$, which Sati-Schreiber-Stasheff [106] characterize as data of a
spin structure on $M$, a principal $G$-bundle $P\to M$ and a trivialization of $\lambda(M) - c(M)$, i.e. the data
of a $(BG, c)$-twisted string structure on $M$ (see also [100, 101, 105]). In many example theories of
interest, this twist does not come from a vector bundle, including

the $E_8\times E_8$ heterotic string and the CHL string <cit.>.

The corresponding twisted string bordism groups are used to study anomalies and defects for these theories;
anomalies were touched on in <ref>, and the use of bordism groups to learn about defects is through
the McNamara-Vafa cobordism conjecture [93].

<Ref> allow us to use the Adams spectral sequence at $p = 2$ and $p = 3$ to calculate these
twisted string bordism groups in dimensions $15$ and below, which suffices for applications to superstring theory.
(Calculations at primes greater than $3$ are easier and can be taken care of with other methods.) We will show an
example computation, relevant for the $E_8\times E_8$ heterotic string at $p = 3$; for applications of <ref> part (<ref>) to twisted string bordism at $p = 2$, see <cit.> and [17], and for more $p
= 3$ calculations, see [17].

Because $E_8$ is a connected, simply connected, simple Lie group, there is an isomorphism $c\colon
H^4(BE_8;\Z)\overset\cong\to\Z$ uniquely specified by making the Chern-Weil class of the Killing form positive; let
$c$ be the preimage of $1$ under this isomorphism. Bott-Samelson <cit.> showed that, interpreted as a map
$BE_8\to K(\Z, 4)$, $c$ is $15$-connected.

For $i = 1,2$, let $c_i\in H^4(BE_8\times BE_8;\Z)$ be the copy of $c$ coming from the $i^{\mathrm{th}}$ copy of
$E_8$. Let $\Z/2$ act on $E_8\times E_8$ by switching the two factors; then in the Serre spectral sequence for the
fibration of classifying spaces induced by the short exact sequence

\begin{equation}
\label{E82_Serre}
\shortexact*{E_8\times E_8}{(E_8\times E_8)\rtimes\Z/2}{\Z/2},
\end{equation}

the class $c_1 + c_2\in E_2^{0,4} = H^4(BE_8\times BE_8;\Z)$ survives to the $E_\infty$-page and lifts uniquely to
define a class $c_1 + c_2\in H^4(B((E_8\times E_8)\rtimes\Z/2);\Z)$. The Green-Schwarz condition for the $E_8 \times
E_8$ heterotic string asks for an $(E_8\times E_8)\rtimes\Z/2$-bundle $P\to M$ and a trivialization of $\lambda(M) -
(c_1+c_2)(P)$, so we want to compute $\Omega_*^\String(B((E_8\times E_8)\rtimes\Z/2), c_1+c_2)$.
<Ref> allows us to use the change-of-rings theorem to simplify the Adams spectral sequence at $p = 2,3$ for this
computation; we will give the $3$-primary computation here and point the interested reader to <cit.>
for the longer $2$-primary computation.

The $(B((E_8\times E_8)\rtimes\Z/2), c_1+c_2)$-twisted string bordism groups lack $3$-primary torsion in degrees
$11$ and below.

Just like for $\Spin\times_{\set{\pm 1}}\SU_8$ bordism and [51] in <ref>, the computation
in [40] does not take advantage of the change-of-rings theorem, works over the entire Steenrod algebra, and
is significantly harder than our proof here.

Recall the notation $\cA^{\tmf}$, $\beta$, and $\cP^1$ from <ref>.
By <ref>, the Thom spectrum for $(B((E_8\times E_8)\rtimes\Z/2), c_1+c_2)$-twisted string bordism
is identified with the $\MTString$-module Thom spectrum $M^{\MTString}f_{0,c_1+c_2}$, where $f_{0,c_1+c_2}$ is the
fake vector bundle twist defined by the image of the class $c_1+c_2\in H^4(B(E_8\times E_8)\rtimes\Z/2)$ in
$M^{\tmf}f_{0,c_1+c_2}$ be the $\tmf$-module Thom spectrum induced by the
Ando-Hopkins-Rezk map $\sigma\colon \MTString\to\tmf$. As a consequence of <ref>, the $(B((E_8\times
E_8)\rtimes\Z/2), c_1+c_2)$-twisted string bordism groups are isomorphic to $\pi_*(M^\tmf f_{0,c_1+c_2})$ in degrees $15$ and below, and
<ref> describes the $\cA^{\tmf}$-module structure on $H_\tmf^*(M^\tmf f_{0,c_1+c_2};\Z/3)$ (and
hence the input to the Adams spectral sequence) in terms of the $\cA^{\tmf}$-module structure on $H^*(B(E_8\times

Let $x\coloneqq (c_1 + c_2)\bmod 3$ and $y\coloneqq c_1c_2\bmod 3$. Then $H^*(B(E_8\times
E_8)\rtimes\Z/2;\Z/3)\cong \Z/3[x, \cP^1(x), \beta\cP^1(x), y, \dots]/(\dots)$; there are no other generators below
degree $12$, nor any relations below degree $12$.

The actions of $\cP^1$ and $\beta$ are as specified via the names of the generators.

Because $H^*(B\Z/2;\Z/3)$ vanishes in positive degrees, the Serre spectral sequence for (<ref>) collapses
at $E_2$ to yield an isomorphism to the ring of invariants

\begin{equation}
H^*(B(E_8\times E_8)\rtimes\Z/2;\Z/3) \overset\cong\longrightarrow (H^*(BE_8\times BE_8;\Z/3))^{\Z/2}.
\end{equation}

The lemma thus follows once we know $H^*(BE_8;\Z/3)\cong\Z/3[c\bmod 3, \cP^1(c\bmod 3), \beta\cP^1(c\bmod 3),
\dotsc]/(\dotsc)$, where we have given all generators and relations in degrees $11$ and below. Because $c\colon BE_8\to K(\Z, 4)$ is $15$-connected <cit.>, we may replace $BE_8$ with $K(\Z,
4)$, and the mod $3$ cohomology of $K(\Z, 4)$ was computed by Cartan [33] and Serre [103]; see
Hill <cit.> for an explicit description.

To compute $H_\tmf^*(M^\tmf f_{0,c_1+c_2})$, we also need to know $\cP^1(U)$ and $\beta(U)$. The latter vanishes
for degree reasons; the former is $\cP^1(U)=  Ux$ by <ref>. Then as usual we compute on all
classes in degrees $11$ and below using the Cartan formula.

Let $N_1\coloneqq \cA^{\tmf}/(\beta, (\cP^1)^2, \beta\cP^1\beta)$ and $N_2\coloneqq \cA^\tmf/(\beta, \beta\cP^1,
\cP^1\beta(\cP^1)^2)$. Then there is a map of $\cA^{\tmf}$-modules

\begin{equation}
\label{tmf_decomp}
H_\tmf^*(M^\tmf f_{0,c_1+c_2})\longrightarrow \textcolor{BrickRed}{N_2} \oplus
\textcolor{Green}{\Sigma^8 N_1} \oplus \textcolor{MidnightBlue}{\Sigma^8 N_1}
\end{equation}

which is an isomorphism in degrees $11$ and below.

We draw the decomposition (<ref>) in <ref>, left. The next step is to compute the Ext
groups of $N_1$ and $N_2$ over $\cA^{\tmf}$. To do so, we will repeatedly use the fact that a short exact sequence
of $\cA^{\tmf}$-modules induces a long exact sequence in Ext; see <cit.> for more information on this
technique, including how to depict the long exact sequence in an Adams chart along with some examples. Let $C\nu$
denote the $\cA^{\tmf}$-module consisting of two $\Z/3$ summands in degrees $0$ and $4$ linked by a nontrivial
$\cP^1$-action. Then there are short exact sequences

\begin{gather}
\label{Cnu_SES}
\shortexact{\textcolor{RubineRed}{\Sigma^4\Z/3}}{C\nu}{\textcolor{Periwinkle}{\Z/3}},\\
\label{qn_SES}
\shortexact{\textcolor{Cerulean}{\Sigma^5\Z/3}}{N_1}{\textcolor{PineGreen}{C\nu}},\\
\label{2qn_SES}
\shortexact{\textcolor{RawSienna}{\Sigma^4 N_1}}{N_2}{\textcolor{RedOrange}{\Z/3}}.
\end{gather}

We will address (<ref>) in <ref>, (<ref>) in <ref>, and (<ref>) in
As input to our computations, we need $\Ext_{\cA^{\tmf}}(\Z/3)$; this acts on $\Ext_{\cA^{\tmf}}(V)$ for any
$\cA^{\tmf}$-module $V$ by the Yoneda product (see <cit.>). The boundary maps in the long exact
sequences of Ext groups induced by short exact sequences of $\cA^{\tmf}$-modules are linear for this
$\Ext_{\cA^{\tmf}}(\Z/3)$-action, which we will use in <ref>.

$\Ext_{\cA^{\tmf}}(\Z/3)$ is generated by the classes
$c_6\in\Ext^{3,15}$, and
$\Delta\in \Ext^{3,27}$, modulo the relations $\alpha^2 = 0$, $h_0\alpha = 0$, $h_0\beta = 0$, $\alpha c_4 =
0$, $\beta c_4 = 0$, $\alpha c_6 = 0$, $\beta c_6 = 0$, and $c_4^3 - c_6^2 = h_0^3\Delta$.

Our notation differs from that of some authors who study $\Ext_{\cA^{\tmf}}(\Z/3)$. Compared with Hill <cit.>, our names for generators agree except that what we call $h_0$ Hill calls $v_0$. Comparing with
Bruner-Rognes <cit.>: our $h_0$ is their $a_0$, our $\alpha$ is
their $h_0$, and our $\beta$ is their $b_0$, and other names of generators agree.

The action of $h_0$ on the $E_\infty$-page of this Adams spectral sequence lifts to multiplying by $3$ on the
twisted $\tmf$-homology groups that the spectral sequence converges to.

In the long exact sequence in Ext corresponding to (<ref>), let $x\in\Ext^{0,0}$ be either generator of
$\Ext_{\cA^{\tmf}}(\textcolor{Periwinkle}{\Z/3})$ and $y\in\Ext^{0,4}$ be either generator of
$\Ext_{\cA^{\tmf}}(\textcolor{RubineRed}{\Sigma^4\Z/3})$, both as modules over $\Ext_{\cA^{\tmf}}(\Z/3)$. In both
cases, there are exactly two generators and they differ by a sign.

In the long exact sequence in Ext associated to (<ref>), $\partial(y) = \pm \alpha x$, $\partial(\beta y) =
\pm \alpha\beta x$, and the boundary map vanishes on all other elements in degrees $14$ and below (except for
$-c$ where $c$ was a class already listed).

We draw this in <ref>, bottom left.

Apart from on $\pm y$ and $\pm \beta y$, the boundary map vanishes for degree reasons; since $\partial$ commutes
with the action of $\Ext_{\cA^{\tmf}}(\Z/3)$, once we show $\partial(y) = \pm \alpha x$, $\partial(\beta y) = \pm
\alpha\beta x$ follows. Since $\Ext^{1,4}(\textcolor{Periwinkle}{\Z/3})\cong\Z/3$, if we show $\partial(y)\ne 0$
the only options for $\partial y$ are $\pm \alpha x$.

Since $y$ and $-y$ are the only nonzero elements in $\Ext^{4,0}$ of both $\textcolor{Periwinkle}{\Z/3}$ and
$\textcolor{RubineRed}{\Sigma^4\Z/3}$, $\partial(y) = 0$ if and only if $\Ext_{\cA^{\tmf}}^{0,4}(C\nu) = 0$. And
this Ext group is $\Hom_{\cA^{\tmf}}(C\nu, \Sigma^4 \Z/3) = 0$.

In $\Ext_{\cA^{\tmf}}(C\nu)$, $\alpha(\alpha y) = \beta x$,[This does not contradict the relation
$\alpha^2 = 0$ from <ref>: since $y$ was killed in the long exact sequence computing $\Ext(C\nu)$,
the class $\alpha y\in \Ext(C\nu)$ is not $\alpha$ times anything, so $\alpha(\alpha y)$ need not vanish.] but this
is not detected by the long exact sequence in Ext. This action is denoted with a dashed gray line in
<ref>, bottom right. We do not need this hidden $\alpha$-action, so we will not prove it;[We do
use this $\alpha$-action in the proof of <ref>, but only to determine Ext groups that will be in too high
of a degree to matter in the final computation, so that part of the proof can be left out.] one way to check it is
to compute $\Ext_{\cA_3}(C\nu)$ using the software developed by Bruner [30] or by Chatham-Chua [34],
obtain the hidden $\alpha$-action in $\Ext_{\cA_3}(C\nu)$, and chase it across the map of Ext groups induced by

Thus we obtain $\Ext(C\nu)$ in <ref>, bottom right.

(0, 0);
in -5, 0

[->, thick] (-4.5, 2) – (-0.5, 2);

in 0, 5

[->, thick] (0.5, 0) – (4.5, 0);

[below=2pt] at (-5, 0) $\Sigma^4\Z/3$;
[below=2pt] at (0, 0) $C\nu$;
[below=2pt] at (5, 0) $\Z/3$;

Top: the short exact sequence (<ref>) of $\cA^{\tmf}$-modules. Lower left: the induced long exact
sequence in Ext; we compute the pictured boundary maps in <ref>. Lower right: $\Ext_{\cA^{\tmf}}(C\nu)$ as
computed by the long exact sequence. The dashed line is a nonzero $\alpha$-action not visible to this computation;
see <ref>.

Now we turn to (<ref>) and its long exact sequence in Ext, depicted in <ref>. We keep the notation
for elements of $\Ext(\textcolor{PineGreen}{C\nu})$ from above, so elements are specified by products of classes in
$\Ext(\Z/3)$ with $x$ or $y$. In the long exact sequence induced by (<ref>), let $z\in\Ext^{0,5}$ be a
generator of $\Ext(\textcolor{Cerulean}{\Sigma^5\Z/3})$ as a module over $\Ext(\Z/3)$ (again, there is exactly one
other generator, which is $-z$).

In the long exact sequence in Ext associated to (<ref>), $\partial(h_0^iz) = \pm h_0^iy$, $\partial(h_0c_4z) =
\pm h_0^i c_4y$, and the boundary map vanishes on all other elements in degrees $14$ and below (except for $-c$
where $c$ was a class already listed).

We draw this in <ref>, bottom left.

The proof is essentially the same as for <ref>: all boundary maps other than the ones in the theorem
statement vanish for degree reasons; then, $\Ext(\Z/3)$-linearity of boundary maps reduces the theorem statement to
the computation of $\partial(z)$, which must be $\pm h_0y$ because $\Ext_{\cA^{\tmf}}^{0,5}(N_1) =
\Hom_{\cA^{\tmf}}(N_1, \Sigma^5\Z/3) = 0$.

Like in <ref>, the long exact sequence does not fully specify the $\Ext(\Z/3)$-action on $\Ext(N_1)$.
One can show that $h_0\cdot\alpha z = \pm c_4x$, but this is missed by our long exact sequence calculation. We do not need this relation in our proof of <ref>, so we do not
prove it; one way to see $h_0\cdot\alpha z = \pm c_4x$ would be to deduce it from the analogous $h_0$-action in
$\Ext(N_2)$ via the long exact sequence in Ext induced from (<ref>). To see the corresponding $h_0$-action
in $\Ext(N_2)$, let $N_3$ be a nonsplit $\cA^{\tmf}$-module extension of $C\nu$ by $\Sigma^8\Z/3$; this
characterizes $N_3$ up to isomorphism. Then there is a short exact sequence $\Sigma^9\Z/3\to N_2\to N_3$, and the $h_0$-action we want to detect is visible to the corresponding long exact sequence in Ext.

Thus we have $\Ext(N_1)$ in <ref>, bottom right.

(0, 2);
in -5, 0

[->, thick] (-4.5, 2.5) – (-0.5, 2.5);

in 0, 5
(, 0);

[->, thick] (0.5, 0) – (4.5, 0);

[below=2pt] at (-5, 0) $\Sigma^5\Z/3$;
[below=2pt] at (0, 0) $N_1$;
[below=2pt] at (5, 0) $C\nu$;

Top: the short exact sequence (<ref>) of $\cA^{\tmf}$-modules. Lower left: the induced long exact
sequence in Ext. We compute the pictured boundary maps in <ref>. Lower right: $\Ext_{\cA^{\tmf}}(N_1)$ as
computed by the long exact sequence. The gray line joining $\alpha z$ and $c_4x$ indicates a
nonzero $h_0$-action not visible to this computation; see <ref>.

The last long exact sequence we have to run is the one induced by (<ref>). We keep the notation for
elements of $\Ext(\textcolor{RawSienna}{N_1})$ from above — classes in $\Ext(\Z/3)$ times $x$, $y$, or $z$. We let
$w$ denote a generator of $\Ext(\textcolor{RedOrange}{\Z/3})$ as an $\Ext(\Z/3)$-module; like before, the two
generators are $w$ and $-w$.

In the long exact sequence in Ext associated to (<ref>), the boundary map takes the values $\partial(x) =
\pm \alpha w$, $\partial(\alpha y) = \pm \beta w$, and $\partial(\beta x) = \pm \alpha\beta w$, and vanishes on all
other classes in degrees $14$ and below (except for $-c$ where $c$ was a class already listed).

We draw this in <ref>, bottom left.

As in <ref>, apart from $\partial(\pm x)$, $\partial(\pm \alpha y)$, and $\partial(\pm \beta x)$,
the boundary map vanishes for degree reasons, and we infer $\partial(x) = \pm \alpha w$ because this is the only
way for $\Ext^{0,4}(N_2) = \Hom(N_2, \Sigma^4 \Z/3)$ to vanish. And since $\alpha(\alpha y) = \beta x$, as we
discussed in <ref>, it remains only to prove $\partial(\alpha y) = \pm\beta w$; then $\partial(\beta
x) = \alpha\beta w$ follows from $\Ext(\Z/3)$-linearity; and since
$\Ext_{\cA^{\tmf}}^{2,12}(\textcolor{RedOrange}{\Z/3})$ is one-dimensional, to show $\partial(\alpha y) = \pm \beta
w$ it suffices to show $\partial(\alpha y)$ is nonzero.

To compute $\partial(\alpha y)$, we use the characterization of
$\Ext_{\cA^\tmf}^{1, t}(M, N)$ as a set of equivalence classes of $\cA^{\tmf}$-module extensions $0\to \Sigma^t
N\to L\to M\to 0$. We will represent $\alpha y$ as an explicit extension of $\Sigma^4 N_1$ by $\Sigma^{12}\Z/3$ and
then show this extension cannot be the pullback of an extension of $N_2$ by $\Sigma^{12}\Z/3$, which implies
$\partial(\alpha y)\ne 0$ by exactness. Up to isomorphism, there is only one non-split extension of $\Sigma^4N_1$
by $\Sigma^{12}\Z/3$, with $\alpha y$ and $-\alpha y$ distinguished by a sign in the extension maps; we draw this
extension in <ref>, left. In <ref>, right, we illustrate what goes wrong if we
try to obtain this extension as the pullback of an extension of $N_2$: the relation $(\cP^1)^3 = 0$ in $\cA^{\tmf}$
is violated. Thus $\partial(\alpha y)\ne 0$.

(0, 4);
in -5, 0

[->, thick] (-4.5, 6) – (-0.5, 6);

in 0, 5
(, 2);
(, 4);

[->, thick] (0.5, 2) – (4.5, 2);

[below=2pt] at (-5, 0) $\Sigma^{12}\Z/3$;
[below=2pt] at (5, 0) $\Sigma^4 N_1$;

(0, 4);
in -5, 0

[->, thick] (-4.5, 6) – (-0.5, 6);

in 0, 5
(, 2);
(, 4);

[->, thick] (0.5, 2) – (4.5, 2);
(0, 0);

(5, 0);

[below=2pt] at (-5, 0) $\Sigma^{12}\Z/3$;
[below=2pt] at (5, 0) $N_2$;

Left: an extension of $\cA(1)$-modules representing the class $\alpha
y\in\Ext_{\cA^{\tmf}}^{1,12}(\Sigma^4 N_1)$. Right: if we try to form an analogous extension of $N_2$, we are
obstructed by the fact that $(\cP^1)^3 = 0$ in $\cA^{\tmf}$. This is part of the proof of <ref>.

(0, 0);
in -5, 0
(, 2);
(, 4);

[->, thick] (-4.5, 2) – (-0.5, 2);

in 0, 5

[->, thick] (0.5, 0) – (4.5, 0);

[below=2pt] at (-5, 0) $\Sigma^4 N_1$;
[below=2pt] at (0, 0) $N_2$;
[below=2pt] at (5, 0) $\Z/3$;

Top: the short exact sequence (<ref>) of $\cA^{\tmf}$-modules. Lower left: the induced long exact
sequence in Ext. We compute the boundary maps in <ref>. Lower right: $\Ext_{\cA^{\tmf}}(N_2)$ as computed by the long exact sequence.

[scale=0.6, every node/.style = font=]
in 0, 2, ..., 12
at (-2, /2) $\y$;

(0, 0);
(0, 2);
(0, 4);

(2, 4);
(2, 6);

(4, 4);
(4, 6);

Left: the $\cA^{\tmf}$-module structure on $H_{\tmf}^*(M)$ in low degrees; the pictured submodule contains
all elements in degrees $11$ and below. Right: the $E_2$-page of the Adams spectral sequence computing
$\pi_*(M)_3^\wedge$, which as we discuss in the proof of <ref> is isomorphic to the $3$-completion
of the twisted string bordism groups relevant for $E_8\times E_8$ heterotic string theory.

Now that we know the Ext groups of all $\cA^{\tmf}$-modules appearing in (<ref>), we can draw the
$E_2$-page of the Adams spectral sequence computing $\pi_*(M^\tmf f_{0,c_1+c_2})_3^\wedge$ in
<ref>, right. For degree reasons, this spectral sequence collapses at $E_2$ in degrees $t-s\le
11$; since $h_0$-actions lift to multiplication by $3$, there is no $3$-torsion in this range, and we conclude.

Other examples of twisted string structures appear in the math and physics literature; see
Dierigl-Oehlmann-Schimmanek <cit.> for another $3$-primary example.

Just as in <ref>, Kreck's modified surgery gives a classification of some closed, smooth $8$-manifolds up
to stable diffeomorphism in terms of twisted string bordism. There is work applying this in examples corresponding
to vector bundle twists [56, 53, 59, 118, 35]; it would be interesting to apply the $\tmf$-module Adams
spectral sequence to classes of manifolds where the twist is not given by a vector bundle.

§.§ $H\Z/2$ as a $\ku$-module Thom spectrum

Devalapurkar uses methods from chromatic homotopy theory to prove the following result. We will reprove it using
the tools in this paper.

There is a map $f\colon \U_2\to B\GL_1(\ku)$ and a $2$-local equivalence $Mf\simeq H\Z/2$.

Borel <cit.> proved that $H^*(\U_2;\Z)\cong \Z[b_1, b_3]/(b_1^2, b_3^2)$ and
$H^*(\U_2;\Z/2)\cong\Z/2[\overline b_1, \overline b_3]/(\overline b_1{}^2, \overline b_3{}^2)$, where $\overline
b_i = b_i\bmod 2$, $\abs{b_1} = \abs{\overline b_1} = 1$ and $\abs{b_3} = \abs{\overline b_3} = 3$; for degree
reasons, $\Sq^1$ and $\Sq^2$ act trivially on the mod $2$ cohomology ring.

Let $f\colon \U_2\to B\GL_1(\ku)$ be the fake vector bundle twist given by $(\overline b_1, b_3)$ (see
<ref> for the definition of this class of twists).

<Ref> shows that $H_\ku^*(Mf)$ is isomorphic to $H^*(\U_2;\Z/2)$ as $\Z/2$-vector spaces, and
that the $\cE(1)$-action is twisted by $Q_0(U) = U\overline b_1$ and $Q_1(U) = U\overline b_3$. This and the Cartan
rule imply $H_\ku^*(Mf)\cong\cE(1)$ as $\cE(1)$-modules, so $\Ext_{\cE(1)}(H_\ku^*(Mf), \Z/2)$ consists of a single
$\Z/2$ in bidegree $(0, 0)$ and vanishes elsewhere. Thus the $\ku$-module Adams spectral sequence immediately
collapses, and we learn $\pi_0(Mf)_2^\wedge\cong\Z/2$ and all other homotopy groups vanish. This property
characterizes $H\Z/2$ up to $2$-local equivalence (e.g. it implies $H^0(Mf;\Z/2)\cong\Z/2$, giving a map $Mf\to
H\Z/2$ which is an isomorphism on $2$-completed homotopy groups, allowing us to conclude by Whitehead).

Devalapurkar also shows that the equivalence in <ref> can be upgraded to an equivalence of $E_1$-ring
spectra; this is not accessible to our methods.

The statement and proof of <ref> can be upgraded to show that the Thom spectrum of the analogous map
$f_n\colon\U_n\to B\GL_1(\ku)$ is $2$-locally equivalent to a wedge sum of shifts of copies of $H\Z/2$ for all $n >

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