00footnotetext: ∗ Corresponding author.
E-mail addresses<EMAIL_ADDRESS>(Pengfei Guo<EMAIL_ADDRESS>(Junyao
Pan).

# Sign-Balanced Pattern-Avoiding Permutation Classes

Junyao Pan1, Pengfei Guo2,∗

$1.$ School of Sciences, Wuxi University, Wuxi, 214105, P. R. China
$2.$ School of Mathematics and Statistics, Hainan Normal University, Haikou
571158, P. R. China

Abstract: A set of permutations is called sign-balanced if the set contains
the same number of even permutations as odd permutations. Let
$S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ be the set of permutations in
the symmetric group $S_{n}$ which avoids patterns
$\sigma_{1},\sigma_{2},\ldots,\sigma_{r}$. The aim of this paper is to
investigate when, for certain patterns
$\sigma_{1},\sigma_{2},\ldots,\sigma_{r}$,
$S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ is sign-balanced for every
integer $n>1$. We prove that for any
$\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}\subseteq S_{3}$, if
$\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ is sign-balanced except
$\\{132,213,231,312\\}$, then $S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$
is sign-balanced for every integer $n>1$. In addition, we give some results in
the case of avoiding some patterns of length $4$.

Keywords: Permutation, Sign-balanced, Symmetric group, Avoid patterns.

Mathematics Subject Classification (2020): 05A05, 06A07.

##  1 Introduction

In this paper, $S_{n}$ is always the symmetric group of degree $n$, its
identity element is used by $id_{n}$, and the binomial coefficient is denoted
by $C^{k}_{n}$.

Fix a permutation $\sigma=\sigma_{1}\sigma_{2}\cdots\sigma_{k}\in S_{k}$ and
let $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in S_{n}$ with $k\leq n$. If there exists
a subset of indices $1\leq i_{1}<i_{2}<\cdots<i_{k}\leq n$ such that
$\pi_{i_{s}}>\pi_{i_{t}}$ if and only if $\sigma_{s}>\sigma_{t}$ for any
$1\leq s<t\leq k$, then we call that $\pi$ contains _the pattern_ $\sigma$,
and the subsequence $\pi_{i_{1}}\pi_{i_{2}}\cdots\pi_{i_{k}}$ is called an
_occurrence_ of $\sigma$ in $\pi$, and expressed by $\sigma\leq\pi$.
Otherwise, we call that $\pi$ _avoids_ $\sigma$. For instance, $132\leq 24153$
since $253$ is an occurrence of $132$ in $24153$, and $53412$ avoids $132$. In
fact, the investigation of pattern avoidance in permutations began in 1968,
when Knuth [9] introduced a stack-sorting machine and showed that a
permutation can be sorted to the identity permutation using this machine if
and only if it avoids the pattern $231$. Henceforth, pattern avoidance in
permutations has been studied by many scholars, for some details see [2, 3, 4,
5, 11, 14, 15].

Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}$ be a permutation in $S_{n}$. A pair of
indices $i<j$ forms an _inversion_ in the permutation $\pi$ if
$\pi_{i}>\pi_{j}$, otherwise $i<j$ forms a _noninversion_ in the permutation
$\pi$. Additionally, we denote by $\tau(\pi)$ ($\theta(\pi)$) the number of
inversions (noninversions) in $\pi$. If $\tau(\pi)$ is an even number, then we
say that $\pi$ is an _even permutation_ , and otherwise $\pi$ is an _odd
permutation_. In addition, we say that a set of permutations is _sign-
balanced_ if the number of even permutations equals the number of odd
permutations in this set. Let $S_{n}(\sigma)$ denote the set of permutations
in $S_{n}$ that avoids pattern $\sigma$. Simion and Schmidt [13] proved that
$S_{n}(321)$ is sign-balanced if $n$ is even, and the number of even
permutations in $S_{n}(321)$ exceeds the number of odd permutations by the
Catalan number $C_{\frac{1}{2}(n-1)}$ if $n$ is odd. Afterwards, the sign-
balance property of permutation class that avoids one pattern was studied
under various conditions in [7, 8, 12, 14], and further the sign-balance of
permutation class with respect to certain statistics was studied in [1, 10].

In this note, we consider the permutation class that avoids several patterns.
Let $S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ be the set of
permutations in $S_{n}$ which avoids all patterns
$\sigma_{1},\sigma_{2},\ldots,\sigma_{r}$. We focus on the following problem
and further obtain some results.

######  Question 1.1.

Are there some patterns $\sigma_{1},\sigma_{2},\ldots,\sigma_{r}$ such that
$S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ is sign-balanced for every
integer $n>1$?

######  Theorem 1.2.

Suppose $\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\in S_{3}$. Then,
$S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ is sign-balanced for every
integer $n>1$ if and only if
$\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}\neq\\{132$, $213$, $231$,
$312\\}$ and $\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ is sign-balanced.

######  Theorem 1.3.

All of $S_{n}(1234,3214)$, $S_{n}(4321,4123)$, $S_{n}(4321,2341)$,
$S_{n}(1234,1432)$, $S_{n}(1243,2143)$, $S_{n}(3421,3412)$,
$S_{n}(4312,3412)$, $S_{n}(2134,2143)$, $S_{n}(1423,1432)$,
$S_{n}(3241,2341)$, $S_{n}(4132,4123)$ and $S_{n}(2314,3214)$ are sign-
balanced for every integer $n>1$.

##  2 Preliminaries

Recall some notions and notations. Consider $\sigma\in S_{l}$ and $\pi\in
S_{m}$. The _direct sum_ of $\sigma$ and $\pi$ is denote by $\sigma\oplus\pi$,
that is,

$\sigma\oplus\pi(i)=\begin{cases}\sigma(i),&\text{if $1\leq i\leq l$;}\\\
\pi(i)+l,&\text{if $l+1\leq i\leq l+m$.}\end{cases}$

and the _skew sum_ of $\sigma$ and $\pi$ is denote by $\sigma\ominus\pi$, that
is,

$\sigma\ominus\pi(i)=\begin{cases}\sigma(i)+m,&\text{if $1\leq i\leq l$;}\\\
\pi(i-l),&\text{if $l+1\leq i\leq l+m$.}\end{cases}$

Moreover, for any $\sigma\in S_{n}$, its reversal $\overline{\sigma}$ is given
by $\overline{\sigma}(i)=\sigma(n+1-i)$; its complement $\sigma^{*}\in S_{n}$
is the permutation $\sigma^{*}(i)=n+1-\sigma(i)$; the inverse $\sigma^{-1}$ is
the usual group theoretic inverse permutation. Let
$R=\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ be a set of permutations in
$S_{n}$. We use $R^{-1}$ and $R^{*}$ and $\overline{R}$ to denote
$\\{\sigma^{-1}_{1},\sigma^{-1}_{2},\ldots,\sigma^{-1}_{r}\\}$ and
$\\{\sigma^{*}_{1},\sigma^{*}_{2},\ldots,\sigma^{*}_{r}\\}$ and
$\\{\overline{\sigma_{1}},\overline{\sigma_{2}},\ldots,\overline{\sigma_{r}}\\}$
respectively.

Next we give some results which are useful in proving Theorem 1.2 and Theorem
1.3 by inductive method.

######  Lemma 2.1.

Let $\pi\in S_{n}$ and $\mu$ be a permutation that is from exchanging the
numbers in two different positions of $\pi$. Then the parity of $\pi$ and
$\mu$ is opposite.

Proof It is well-known that the parity of $\tau(\pi)$ and $\tau(\mu)$ is
opposite. Thus, the parity of $\pi$ and $\mu$ is opposite. $\hfill{\Box}$

######  Lemma 2.2.

Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{n}\in S_{n}$ and
$\mu=\pi_{1}\pi_{2}\cdots\pi_{i}(n+1)\pi_{i+1}\cdots\pi_{n}\in S_{n+1}$ for
some $i=0,1,\ldots,n$. In particular, if $i=0$ then
$\mu=(n+1)\pi_{1}\pi_{2}\cdots\pi_{n}\in S_{n+1}$. If $n-i$ is an even number,
then $\pi$ and $\mu$ have the same parity; if $n-i$ is an odd number, then
$\pi$ and $\mu$ have the opposite parity.

Proof Clearly, $\tau(\mu)=\tau(\pi)+n-i$. Thus, if $n-i$ is an even number,
then $\pi$ and $\mu$ have the same parity; if $n-i$ is an odd number, then
$\pi$ and $\mu$ have the opposite parity. $\hfill{\Box}$

######  Lemma 2.3.

Let $R$ be a set of permutations. Then $S_{n}(R^{*})=S^{*}_{n}(R)$ and
$S_{n}(\overline{R})=\overline{S_{n}(R)}$.

Proof Let $R=\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ and $\pi\in
S^{*}_{n}(R)$. If $\pi\notin S_{n}(R^{*})$, then $\pi$ contains
$\sigma^{*}_{i}$ that is in $R^{*}$. Thus, $\pi^{*}$ contains $\sigma_{i}$, in
other words, $\pi^{*}\notin S_{n}(R)$. However, $\pi\in S^{*}_{n}(R)$ means
that $\pi^{*}\in S_{n}(R)$, a contradiction. Therefore, $S^{*}_{n}(R)\subseteq
S_{n}(R^{*})$. Since $R=(R^{*})^{*}$, it follows that
$S^{*}_{n}(R^{*})\subseteq S_{n}(R)$, and which implies that
$S_{n}(R^{*})\subseteq S^{*}_{n}(R)$. Hence, $S_{n}(R^{*})=S^{*}_{n}(R)$.
Similarly, we can infer that $S_{n}(\overline{R})=\overline{S_{n}(R)}$.
$\hfill{\Box}$

######  Lemma 2.4.

Let $\sigma\in S_{n}$. If $C^{2}_{n}$ is an even number, then $\sigma^{*}$,
$\overline{\sigma}$ and $\sigma$ have the same parity; otherwise, the parity
of $\sigma$ is opposite to both of $\sigma^{*}$ and $\overline{\sigma}$.

Proof Note that $\theta(\sigma)+\tau(\sigma)=C^{2}_{n}$ and
$\theta(\sigma)=\tau(\sigma^{*})=\tau(\overline{\sigma})$. Thus, if
$C^{2}_{n}$ is an even number, then $\tau(\sigma^{*})$,
$\tau(\overline{\sigma})$ and $\tau(\sigma)$ have the same parity, and
otherwise the parity of $\tau(\sigma)$ is opposite to both of
$\tau(\sigma^{*})$ and $\tau(\overline{\sigma})$, as desired. $\hfill{\Box}$

######  Corollary 2.5.

Let $R$ be a set of permutations such that $S_{n}(R)$ is sign-balanced. Then
$S_{n}(R^{*})$ and $S_{n}(\overline{R})$ are sign-balanced.

Proof Let $\Omega_{+}=\\{\pi\in S_{n}(R):$ $\pi$ is an even permutation$\\}$
and $\Omega_{-}=\\{\pi\in S_{n}(R):$ $\pi$ is an odd permutation$\\}$. Since
$S_{n}(R)$ is sign-balanced, we deduce that $|\Omega_{+}|=|\Omega_{-}|$.
Applying Lemma 2.3, it follows that
$S_{n}(R^{*})=\Omega^{*}_{+}\cup\Omega^{*}_{-}$. Set $\Delta_{+}=\\{\pi\in
S_{n}(R):$ $\pi$ is an even permutation$\\}$ and $\Delta_{-}=\\{\pi\in
S_{n}(R):$ $\pi$ is an odd permutation$\\}$. Then by Lemma 2.4, we see that
$\Delta_{+}=\Omega^{*}_{+}$ and $\Delta_{-}=\Omega^{*}_{-}$ if $C^{2}_{n}$ is
an even number, and $\Delta_{-}=\Omega^{*}_{+}$ and
$\Delta_{+}=\Omega^{*}_{-}$ if $C^{2}_{n}$ is an odd number. In either case,
$|\Delta_{+}|=|\Delta_{-}|$ holds. Similarly, we can deduce that
$S_{n}(\overline{R})$ is also sign-balanced. $\hfill{\Box}$

######  Lemma 2.6.

Let $\sigma\in S_{l}$ and $\pi\in S_{m}$. If $ml$ is an even number, then
$\tau(\sigma\ominus\pi)$ and $\tau(\sigma)+\tau(\pi)$ have the same parity; if
$ml$ is an odd number, then the parity of $\tau(\sigma\ominus\pi)$ is opposite
to that of $\tau(\sigma)+\tau(\pi)$.

Proof The lemma follows from the fact that
$\tau(\sigma\ominus\pi)=\tau(\sigma)+\tau(\pi)+ml$. $\hfill{\Box}$

######  Lemma 2.7.

[13, Proposition 17] If $R\subseteq S_{3}$ and $|R|=4$, then
(a) $|S_{n}(R)|=0$ if $R\supset\\{123,321\\}$ and $n\geq 5$;
(b) $|S_{n}(R)|=2$ if $R\not\supset\\{123,321\\}$, $n\geq 2$.
More precisely, the permutations counted in (b) are the appropriate two,
depending on $R$, from among the identity, $23\cdot\cdot\cdot n1$,
$n(n-1)\cdot\cdot\cdot 312$, their reversals and complements. For the values
of $n$ omitted in part (a), we have: $S_{1}(R)=1$, $S_{2}(R)=2$, $S_{4}(R)=0$
except for $S_{4}(123,321,132,213)=S_{4}(123,321,231,312)=1$.

######  Lemma 2.8.

[6, Erdős-Szekeres Theorem] Any sequence of $mk+1$ distinct real numbers
contains either an increasing subsequence of $m+1$ terms or a decreasing
subsequence of $k+1$ terms.

##  3 Main results

We first consider Question 1.1 under the assumption of
$\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\in S_{3}$. In this case, we see that
if $S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ is sign-balanced for every
integer $n>1$, then $\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ is sign-
balanced because
$S_{3}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})=S_{3}\setminus\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$.
Thus for Theorem 1.2, it suffices to check that
$S_{n}(\sigma_{1},\sigma_{2},\ldots,\sigma_{r})$ is sign-balanced for every
integer $n>1$ if
$\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}\neq\\{132,213,231,312\\}$ and
$\\{\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\\}$ is sign-balanced, and further
$S_{n}(132,213,231,312)$ is not sign-balanced for some integer $n>1$. Now we
start from the case that $r=2$.

######  Proposition 3.1.

All of $S_{n}(132,231)$, $S_{n}(312,213)$, $S_{n}(132,312)$ and
$S_{n}(231,213)$ are sign-balanced for every integer $n>1$.

Proof Since $132$ is an odd permutation while $231$ is an even permutation, we
infer that $S_{n}(132,231)$ is sign-balanced for $n=2,3$. Next we prove this
proposition by induction on $n$. Assume that $S_{n}(132,231)$ is sign-balanced
for $n=k>1$. Consider $\pi=\pi_{1}\pi_{2}\cdots\pi_{k}\pi_{k+1}\in S_{k+1}$.
Note that if $\pi\in S_{k+1}(132,231)$, then $\pi_{1}=k+1$ or $\pi_{k+1}=k+1$,
otherwise $\pi$ contains either $132$ or $231$. Suppose $\pi_{1}=k+1$. It is
clear that $\pi\in S_{k+1}(132,231)$ if and only if
$\pi^{\prime}=\pi_{2}\pi_{3}\cdots\pi_{k+1}\in S_{k}(132,231)$. Also, if
$\pi_{k+1}=k+1$, then $\pi\in S_{k+1}(132,231)$ if and only if
$\pi^{\prime\prime}=\pi_{1}\pi_{2}\cdots\pi_{k}\in S_{k}(132,231)$. Therefore,
we deduce that

$S_{k+1}(132,231)=\\{1\ominus\mu|\mu\in S_{k}(132,231)\\}\cup\\{\mu\oplus
1|\mu\in S_{k}(132,231)\\}.$

Let $\Omega_{+}=\\{1\ominus\mu|\mu\in S_{k}(132,231),\mu$ is an even
permutation$\\}$ and $\Omega_{-}=\\{1\ominus\mu|\mu\in S_{k}(132,231),\mu$ is
an odd permutation$\\}$. Then by Lemma 2.2 we infer that if $k$ is even,
$\Omega_{+}$ is an even permutation set and $\Omega_{-}$ is an odd permutation
set; and if $k$ is odd, $\Omega_{+}$ is an odd permutation set and
$\Omega_{-}$ is an even permutation set. Applying inductive hypothesis, we
deduce that $\\{1\ominus\mu|\mu\in S_{k}(132,231)\\}$ is sign-balanced no
matter what the parity of $k$ is. Similarly, we see that $\\{\mu\oplus
1|\mu\in S_{k}(132,231)\\}$ is sign-balanced. Therefore, $S_{k+1}(132,231)$ is
sign-balanced, and so $S_{n}(132,231)$ is sign-balanced for every integer
$n>1$.

Note $\\{312,213\\}=\\{132^{*},231^{*}\\}$. It follows from Corollary 2.5 that
$S_{n}(312,213)$ is sign-balanced for every integer $n>1$. Additionally, we
see that $\\{132,312\\}=\\{132^{-1},231^{-1}\\}$ and
$\\{231,213\\}=\\{312^{-1},213^{-1}\\}$. Using a result in [13, Lemma 1], we
deduce that $S_{n}(132,312)=S^{-1}_{n}(132,231)$ and
$S_{n}(231,213)=S^{-1}_{n}(312,213)$. Since the permutations $\sigma$ and
$\sigma^{-1}$ have the same parity, it follows that $S_{n}(132,312)$ and
$S_{n}(231,213)$ are sign-balanced for every integer $n>1$. $\hfill{\Box}$

######  Proposition 3.2.

All of $S_{n}(123,213)$, $S_{n}(321,231)$, $S_{n}(123,132)$ and
$S_{n}(321,312)$ are sign-balanced for every integer $n>1$.

Proof It is clear that $S_{n}(123,213)$ is sign-balanced for $n=2,3$ because
$123$ is an even permutation while $213$ is an odd permutation. By induction
on $n$, we assume that $S_{n}(123,213)$ is sign-balanced for $n=k>1$. Consider
$\pi=\pi_{1}\pi_{2}\cdots\pi_{k}\pi_{k+1}\in S_{k+1}$. We see that if $\pi\in
S_{k+1}(123,213)$ then $\pi_{1}=k+1$ or $\pi_{2}=k+1$, otherwise $\pi$
contains either $123$ or $213$. In addition, it is straightforward to show
that if $\pi_{1}=k+1$ then $\pi\in S_{k+1}(123,213)$ if and only if
$\pi^{\prime}=\pi_{2}\pi_{3}\cdots\pi_{k+1}\in S_{k}(123,213)$, and if
$\pi_{2}=k+1$ then $\pi\in S_{k+1}(132,231)$ if and only if
$\pi^{\prime\prime}=\pi_{1}\pi_{3}\cdots\pi_{k+1}\in S_{k}(123,213)$.
Proceeding as in the proof of Proposition 3.1, we deduce that
$S_{k+1}(123,213)$ is sign-balanced from Lemma 2.2 and the inductive
assumption. Therefore, $S_{n}(123,213)$ is sign-balanced for every integer
$n>1$.

In addition, we see that $\\{321,231\\}=\\{123^{*},213^{*}\\}$,
$\\{123,132\\}=\\{321^{-1},231^{-1}\\}$ and
$\\{321,312\\}=\\{123^{-1},213^{-1}\\}$. An argument similar to the one used
in the proof of Proposition 3.1 shows that all of $S_{n}(321,231)$,
$S_{n}(123,132)$ and $S_{n}(321,312)$ are sign-balanced for every integer
$n>1$. $\hfill{\Box}$

######  Proposition 3.3.

$S_{n}(123,321)$ is sign-balanced for every integer $n>1$.

Proof It is obvious that $321$ is an odd permutation while $123$ is an even
permutation, and thus $S_{n}(123,321)$ is sign-balanced for $n=2,3$.
Additionally, it follows from Lemma 2.8 that it suffices to consider
$S_{4}(123,321)$. One easily checks that $S_{4}(123,321)\cap
A_{4}=\\{2143,3412\\}$ and $S_{4}(123,321)\cap(S_{4}\setminus
A_{4})=\\{3142,2413\\}$, where $A_{4}$ is the alternating group of degree $4$.
The completes the proof. $\hfill{\Box}$

So far, we have verified all situations of $S_{n}(\sigma_{1},\sigma_{2})$
under the assumption of $\sigma_{1},\sigma_{2}\in S_{3}$. Secondly, we
consider $S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ in case when
$\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\in S_{3}$, and obtain the
following result.

######  Proposition 3.4.

Let $\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\subseteq S_{3}$. Then,
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is sign-balanced for
every integer $n>1$ if and only if
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\neq\\{132,213,231,312\\}$
and $\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$ is sign-balanced.

Proof If
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}=\\{132,213,231,312\\}$,
then
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\not\supset\\{123,321\\}$
and further
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\cap\\{123,321\\}=\emptyset$.
Applying Lemma 2.7, we deduce that $S_{n}(132,213,231,312)=\\{12\cdots
n,n\cdots 21\\}$. However, Lemma 2.4 shows that $12\cdots n$ and $n\cdots 21$
have the same parity if $C^{2}_{n}$ is an even number, and so
$S_{n}(132,213,231,312)$ is not sign-balanced when $C^{2}_{n}$ is an even
number.

Assume that
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\cap\\{123,321\\}\neq\emptyset$
and $\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$ is sign-balanced. It
suffices to prove that $S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is
sign-balanced for every integer $n>1$. It is obvious that
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is sign-balanced when
$n=2,3$. Next we consider $S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$
for $n>3$.

Consider
$|\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\cap\\{123,321\\}|=2$.
Namely
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\supset\\{123,321\\}$.
According to Lemma 2.7, we deduce that
$|S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})|=0$ for every integer
$n\geq 5$. In addition, Lemma 2.7 shows that
$|S_{4}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})|\neq 0$ if and only if
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}=\\{123,321,132,213\\}$ or
$\\{123,321,231,312\\}$. However, we see that $\\{123,321,132,213\\}$ and
$\\{123,321,231,312\\}$ are not sign-balanced. Thus, we deduce that
$|S_{4}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})|=0$. Therefore,
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is sign-balanced for
every integer $n>1$.

Consider the case that
$|\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}\cap\\{123,321\\}|=1$. Note
$321=123^{*}$. By Corollary 2.5, we can assume that
$123\notin\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$ and
$321\in\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$. Since
$\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$ is sign-balanced, it
follows that either
$123,132\notin\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$ or
$123,213\notin\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$. If
$123,132\notin\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$, then
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})=\\{1234\cdots
n,1234\cdots(n-2)n(n-1)\\}$ by Lemma 2.7. Clearly,
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is sign-balanced in this
case. If $123,213\notin\\{\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4}\\}$,
then $S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})=\\{1234\cdots
n,2134\cdots n\\}$ by Lemma 2.7. In this case,
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4})$ is also sign-balanced.
The proof of this proposition is completed. $\hfill{\Box}$

Proof of Theorem 1.2 Consider
$S_{n}(\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4},\sigma_{5},\sigma_{6})=S_{n}(S_{3})$.
It is obvious that $S_{n}(S_{3})$ is sign-balanced for every integer $n>1$
because $S_{n}(S_{3})=\emptyset$ when $n\geq 3$. So we have solved Question
1.1 under the assumption of $\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\in
S_{3}$. At the same time, we derive Theorem 1.2 immediately by Propositions
3.1-3.4. $\hfill{\Box}$

Thirdly, we solve Question 1.1 in case when
$\sigma_{1},\sigma_{2},\ldots,\sigma_{r}\in S_{4}$, and obtain some results as
follows.

######  Proposition 3.5.

Let $R=\\{\\{1234,3214\\},\\{4321,4123\\},\\{4321,2341\\},\\{1234,1432\\}\\}$.
Then for any $\\{\sigma_{1},\sigma_{2}\\}\in R$,
$S_{n}(\sigma_{1},\sigma_{2})$ is sign-balanced for every integer $n>1$.

Proof Note that $\\{4321,4123\\}=\\{\overline{1234},\overline{3214}\\}$,
$\\{4321,2341\\}=$ $\\{1234^{*},3214^{*}\\}$ and $\\{1234$, $1432\\}=$
$\\{\overline{4321},\overline{2341}\\}$. Then by Corollary 2.5, it suffices to
prove that $S_{n}(1234,3214)$ is sign-balanced for every integer $n>1$. Since
$1234$ is an even permutation while $3214$ is an odd permutation, it follows
that $S_{n}(1234,3214)$ is sign-balanced for $n=2,3,4$. Assume that
$S_{n}(1234,3214)$ is sign-balanced for $1<n\leq k$ with $k>3$. Let
$X^{(i)}=\\{\pi\in{S_{k+1}(1234,3214)}:\pi(i)=k+1\\}$. It is clear that
$S_{k+1}(1234,3214)$ is the disjoint union $\bigcup^{k+1}_{i=1}X^{(i)}$. It
follows from Lemma 2.8 that $X^{(i)}=\emptyset$ if $i>5$. Therefore, we deduce
that

$S_{k+1}(1234,3214)=\bigcup^{5}_{i=1}X^{(i)}.$

Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in S_{k+1}$ with $\pi_{i}=k+1$ and
$1\leq i\leq 3$. Clearly, $\pi\in X^{(i)}$ if and only if
$\pi^{\prime}=\pi_{1}\pi_{2}\cdots\pi_{i-1}\pi_{i+1}\cdots\pi_{k+1}\in
S_{k}(1234,3214)$. In particular, if $i=1$ then
$\pi^{\prime}=\pi_{2}\pi_{3}\cdots\pi_{k+1}$. Proceeding as in the proof of
Proposition 3.1, we deduce that all of $X^{(1)},X^{(2)}$ and $X^{(3)}$ are
sign-balanced by Lemma 2.2 and the inductive assumption. Next we prove that
$X^{(4)}$ and $X^{(5)}$ are also sign-balanced.

Consider $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in S_{k+1}$ with $\pi_{4}=k+1$.
Let $\pi^{\prime}=\pi_{1}\pi_{2}\pi_{3}\pi_{5}\cdots\pi_{k+1}$. Clearly, if
$\pi\in X^{(4)}$ then $\pi^{\prime}\in S_{k}(1234,3214)$. For convenience, we
set

$A^{(4)}=\\{\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
S_{k+1}:\pi_{4}=k+1,\pi_{1}\pi_{2}\pi_{3}\pi_{5}\cdots\pi_{k+1}\in
S_{k}(1234,3214)\\}.$

Proceeding as in the proof of Proposition 3.1, we deduce that $A^{(4)}$ is
sign-balanced by Lemma 2.2 and the inductive assumption. Suppose that
$\pi\notin X^{(4)}$ and $\pi^{\prime}\in S_{k}(1234,3214)$. It is simple to
see that $\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ is an occurrence of $1234$ or $3214$.
Moreover, if $\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ is an occurrence of $1234$, then
$\pi_{1}<\pi_{2}<\pi_{3}=k$ otherwise $\pi^{\prime}$ contains $1234$; if
$\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ is an occurrence of $3214$, then
$\pi_{1}=k>\pi_{2}>\pi_{3}$ otherwise $\pi^{\prime}$ contains an occurrence of
$3214$. Indeed, if $\pi^{\prime}\in S_{k}(1234,3214)$, then $\pi\notin
X^{(4)}$ if and only if $\pi_{1}<\pi_{2}<\pi_{3}=k$ or
$\pi_{1}=k>\pi_{2}>\pi_{3}$. Let
$B^{(4)}=\\{\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
A^{(4)}:\pi_{1}=k>\pi_{2}>\pi_{3}\\}$ and
$C^{(4)}=\\{\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
A^{(4)}:\pi_{1}<\pi_{2}<\pi_{3}=k\\}$. Thus, we have

$X^{(4)}=A^{(4)}\setminus(B^{(4)}\cup C^{(4)}).$

Consider $B^{(4)}\cup C^{(4)}$. We observe that
$B^{(4)}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
S_{k+1}:\pi_{1}=k,\pi_{2}>\pi_{3},\pi_{4}=k+1,\pi_{2}\pi_{3}\pi_{5}\cdots\pi_{k+1}\in
S_{k-1}(1234,3214)\big{\\}}$ and
$C^{(4)}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}:\pi_{1}<\pi_{2},\pi_{3}=k,\pi_{4}=k+1,\pi_{1}\pi_{2}\pi_{5}\cdots\pi_{k+1}\in
S_{k-1}(1234,3214)\big{\\}}$. Since $k-1$ and $k-3$ have the same parity, it
follows that $B^{(4)}\cup C^{(4)}$ is sign-balanced from Lemma 2.2 and the
inductive assumption. Therefore, $X^{(4)}$ is sign-balanced.

Suppose that $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in X^{(5)}$ with
$\pi_{5}=k+1$. Clearly, $\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ avoids $123$ and $321$.
Therefore, we deduce that either
$\pi_{1}>\pi_{2},\pi_{3}>\pi_{4},\pi_{3}>\pi_{1},\pi_{4}>\pi_{2}$ or
$\pi_{1}<\pi_{2},\pi_{3}<\pi_{4},\pi_{3}<\pi_{1},\pi_{4}<\pi_{2}$, otherwise
$\pi_{1}\pi_{2}\pi_{3}\pi_{4}$ contains an occurrence of $123$ or $321$.
Moreover, for any $\pi=\pi_{1}\pi_{2}\pi_{3}\pi_{4}\pi_{5}\cdots\pi_{k+1}\in
X^{(5)}$, we observe that
$\pi_{1}\pi_{3}\pi_{2}\pi_{4}\pi_{5}\cdots\pi_{k+1}\in X^{(5)}$. In other
words, exchanging the entries $2$ and $3$ is a bijection from $X^{(5)}$ to
$X^{(5)}$ and further the parity of the image is opposite to that of the
original image under this bijection. Thus, $X^{(5)}$ is sign-balanced.
According to the above arguments, it follows that $S_{k+1}(1234,3214)$ is
sign-balanced. The proof of this proposition is completed. $\hfill{\Box}$

######  Proposition 3.6.

Let $R=\\{\\{1243,2143\\},\\{3421,3412\\},\\{4312,3412\\},\\{2134,2143\\}\\}$.
Then for any $\\{\sigma_{1},\sigma_{2}\\}\in R$,
$S_{n}(\sigma_{1},\sigma_{2})$ is sign-balanced for every integer $n>1$.

Proof Note that $\\{3421,3412\\}=\\{\overline{1243},\overline{2143}\\}$,
$\\{4312,3412\\}=\\{1243^{*},2143^{*}\\}$ and
$\\{2134,2143\\}=\\{3421^{*},3412^{*}\\}$. According to Corollary 2.5, it
suffices to confirm that $S_{n}(1243,2143)$ is sign-balanced for every integer
$n>1$. Since $2143$ is an even permutation while $1243$ is an odd permutation,
we have that $S_{n}(1234,3214)$ is sign-balanced for $n=2,3,4$. By induction
on $n$, we assume that $S_{n}(1243,2143)$ is sign-balanced for $1<n\leq k$
with $k>3$. Now we consider $S_{k+1}(1243,2143)$. Let
$X^{(i)}=\\{\pi\in{S_{k+1}(1243,2143)}:\pi(i)=k+1\\}$. It is easy to see that
$S_{k+1}(1243,2143)$ is the disjoint union $\bigcup^{k+1}_{i=1}X^{(i)}$. We
claim that $X^{(i)}$ is sign-balanced for $i=1,2,...,k+1$.

Consider $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in S_{k+1}$ with $\pi_{1}=k+1$.
Note that $\pi\in X^{(1)}$ if and only if $\pi_{2}\cdots\pi_{k}\in
S_{k}(1243,2143)$. By Lemma 2.2 and the inductive assumption, we obtain that
$X^{(1)}$ is sign-balanced. Similarly, we can infer that $X^{(2)}$ and
$X^{(k+1)}$ are sign-balanced.

Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in X^{(i)}$ with $2<i<k+1$. Then there
at most one number in $\\{\pi_{1},\pi_{2},\ldots,\pi_{i-1}\\}$ which is
smaller than the biggest number in $\\{\pi_{i+1},\ldots,\pi_{k+1}\\}$,
otherwise $\pi$ contains an occurrence of $1243$ or $2143$. Thus,
$\\{\pi_{1},\pi_{2},...,\pi_{i-1}\\}$ is either $\\{k-i+2,k-i+3,...,k\\}$ or
$\\{m,k-i+3,...,k\\}$ for some $m\in\\{1,2,...,k-i+1\\}$. In the case that
$\\{\pi_{1},\pi_{2},...,\pi_{i-1}\\}=\\{k-i+2,k-i+3,...,k\\}$, set

$\Omega_{1}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
X^{(i)}:\\{\pi_{1},\pi_{2},...,\pi_{i-1}\\}=\\{k-i+2,k-i+3,...,k\\}\big{\\}}.$

We see that $\Omega_{1}=\\{(\sigma\oplus 1)\ominus\rho:\sigma\in
S_{i-1}(1243,2143),\rho\in S_{k-i+1}(1243,2143)\\}$. Proceeding as in the
proof of Proposition 3.1, we deduce that $\Omega_{1}$ is sign-balanced by
Lemma 2.2, Lemma 2.6 and the inductive assumption. In the case that
$\\{\pi_{1},\pi_{2},...,\pi_{i-1}\\}=\\{m,k-i+3,...,k\\}$ for every
$m\in\\{1,2,...,k-i+1\\}$, we set

$\Omega^{m}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
X^{(i)}:\\{\pi_{1},\pi_{2},...,\pi_{i-1}\\}=\\{m,k-i+3,...,k\\}\big{\\}}.$

Notice that exchanging the entries that are $k-i+2$ and $m$ is a bijection
from $\Omega^{m}$ to $\Omega_{1}$, and further the parity of the image is
opposite to that of the original image under this bijection. Since
$\Omega_{1}$ is sign-balanced, it follows that $\Omega^{m}$ is sign-balanced.
Therefore, $X^{(i)}$ is sign-balanced by the fact that $X^{(i)}$ is the
disjoint union $\Omega_{1}\bigcup(\bigcup^{k-i+1}_{m=1}\Omega^{m})$. According
to the above arguments, it follows that $S_{k+1}(1243,2143)$ is sign-balanced.
The proof of this proposition is completed. $\hfill{\Box}$

######  Proposition 3.7.

Let $R=\\{\\{1423,1432\\},\\{3241,2341\\},\\{4132,4123\\},\\{2314,3214\\}\\}$.
Then for any $\\{\sigma_{1},\sigma_{2}\\}\in R$,
$S_{n}(\sigma_{1},\sigma_{2})$ is sign-balanced for every integer $n>1$.

Proof Note that $\\{3241,2341\\}=\\{\overline{1423},\overline{1432}\\}$,
$\\{4132,4123\\}=\\{1423^{*},1432^{*}\\}$ and
$\\{2314,3214\\}=\\{3241^{*},2341^{*}\\}$. So it suffices to prove that
$S_{n}(1423,1432)$ is sign-balanced for every integer $1<n\leq 4$. Since
$1423$ is an even permutation while $1432$ is an odd permutation, we have that
$S_{n}(1423,1432)$ is sign-balanced for $n=2,3,4$. By induction on $n$, we
assume that $S_{n}(1423,1432)$ is sign-balanced for $1<n\leq k$ with $k>3$.
Now we consider $S_{k+1}(1423,1432)$. Let
$X^{(i)}=\\{\pi\in{S_{k+1}(1423,1432)}:\pi(i)=k+1\\}$. It is easy to see that
$S_{k+1}(1423,1432)$ is the disjoint union $\bigcup^{k+1}_{i=1}X^{(i)}$. We
claim that $X^{(i)}$ is sign-balanced for $i=1,2,...,k+1$.

Consider $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in S_{k+1}$ with $\pi_{1}=k+1$.
Note that $\pi\in X^{(1)}$ if and only if $\pi_{2}\cdots\pi_{k}\in
S_{k}(1423,1432)$. Applying Corollary 2.2 and the inductive assumption, we
obtain that $X^{(1)}$ is sign-balanced. Similarly, we can infer that $X^{(k)}$
and $X^{(k+1)}$ are sign-balanced.

Let $\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in X^{(i)}$ with $1<i<k$. Then there at
most one number in $\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}$ which is
bigger than the smallest number in $\\{\pi_{1},\ldots,\pi_{i-1}\\}$, otherwise
$\pi$ contains an occurrence of $1423$ or $1432$. Therefore,
$\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}$ is either $\\{1,2....,k-i+1\\}$
or $\\{1,2....,k-i,m\\}$ for some $m\in\\{k-i+2,k-i+3,...,k\\}$. In the case
that $\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}=\\{1,2....,k-i+1\\}$, set

$\Omega_{1}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
X^{(i)}:\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}=\\{1,2....,k-i+1\\}\big{\\}}.$

We see that $\Omega_{1}=\\{(\sigma\oplus 1)\ominus\rho:\sigma\in
S_{i-1}(1243,2143),\rho\in S_{k-i+1}(1243,2143)\\}$. Proceeding as in the
proof of Proposition 3.1, we deduce that $\Omega_{1}$ is sign-balanced by
Lemma 2.2 and Lemma 2.6 and the inductive assumption. In the case that
$\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}=\\{1,2....,k-i,m\\}$ for every
$m\in\\{k-i+2,k-i+3,...,k\\}$, we set

$\Omega^{m}=\big{\\{}\pi=\pi_{1}\pi_{2}\cdots\pi_{k+1}\in
X^{(i)}:\\{\pi_{i+1},\pi_{i+2},\ldots,\pi_{k+1}\\}=\\{1,2....,k-i,m\\}\big{\\}}.$

Notice that exchanging the entries that are $k-i+1$ and $m$ is a bijection
from $\Omega^{m}$ to $\Omega_{1}$, and further the parity of the image is
opposite to that of the original image under this bijection. Since
$\Omega_{1}$ is sign-balanced, it follows that $\Omega^{m}$ is sign-balanced.
Therefore, $X^{(i)}$ is sign-balanced by the fact that $X^{(i)}$ is the
disjoint union $\Omega_{1}\bigcup(\bigcup^{k-i+1}_{m=1}\Omega^{m})$. According
to the above arguments, it follows that $S_{k+1}(1243,2143)$ is sign-balanced.
The proof of this proposition is completed. $\hfill{\Box}$

Proof of Theorem 1.3 Applying Propositions 3.5-3.7, we have that Theorem 1.3
holds immediately. $\hfill{\Box}$

However, we observe that $S_{n}(\sigma_{1},\sigma_{2})$ is not always sign-
balanced for $\sigma_{1},\sigma_{2}\in S_{4}$. Finally, we give an example
which seems to mean that Question 1.1 is more and more complicated as the
length of avoiding pattern increases.

######  Example 3.8.

Both $S_{n}(1324,2143)$ and $S_{n}(4231,3412)$ are not sign-balanced for some
integer $n>1$.

Proof Note that $\\{4231,3412\\}=\\{\overline{1324},\overline{2143}\\}$. So it
suffices to verify $S_{n}(1324,2143)$. In addition, it is clear that
$S_{n}(1324,2143)$ is sign-balanced for all integer $1<n\leq 4$. Next we
observe that $S_{5}(2341,2143)$ is not sign-balanced.

Let $X^{(i)}=\\{\pi\in{S_{5}(1324,2143)}:\pi(i)=5\\}$. It is clear that
$5\pi_{1}\pi_{2}\pi_{3}\pi_{4}\in X^{(5)}$ if and only if
$\pi_{1}\pi_{2}\pi_{3}\pi_{4}\in{S_{4}(1324,2143)}$. Thus $X^{(1)}$ is sign-
balanced. Similarly, we deduce that $X^{(2)}$ is sign-balanced. By the
computation, it follows that $X^{(3)}=\\{12534$, $13542$, $14523$, $23514$,
$24531$, $34512$, $41532$, $42513$, $43521\\}\cup\\{12543$, $14532$, $23541$,
$24513$, $34521$, $41523$, $42531$, $43512\\}$. Note that the number of even
permutations is one more than that of odd permutations in $X^{(3)}$.
Similarly, $X^{(4)}=\\{12453$, $23451,31452$, $34251,41253$,
$42351,43152\\}\cup\\{12354$, $13452,32451$, $34152,41352,43251\\}$ and the
number of even permutations is also one more than that of odd permutations in
$X^{(4)}$; $X^{(5)}=\\{12345$, $23145,31245$, $32415,34125$,
$42135,43215\\}\cup\\{21345$, $23415,32145$, $34215,41235$, $42315,43125\\}$
and the number of even permutations is equal to that of odd permutations.
Hence, $S_{5}(1324,2143)$ is not sign-balanced. $\hfill{\Box}$

Acknowledgments

The work was supported by the National Natural Science Foundation of China
(No. 12061030) and Hainan Provincial Natural Science Foundation of China (No.
122RC652).

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# Cobordism Obstructions to Complex Sections II: Torsion Obstructions

Dennis Nguyen  Department of Mathematics, University of Oregon
<EMAIL_ADDRESS>
###### Abstract.

In the previous paper, we studied the existence of complex sections on almost
complex manifolds up to cobordism. We proved a necessary and sufficient
condition for their existence in $\Omega_{*}^{U}\otimes\mathbb{Q}$ in terms of
the Chern classes. In this paper we study the existence question in
$\Omega^{U}_{*}$. We call corresponding obstructions torsion and we determine
them for small values of $r$. Calculations with the Adams-Novikov spectral
sequence for the Thom spectra defined in the previous paper allow us to show
the vanishing of the torsion obstruction for low $r$. For prime $p$, we show
the vanishing of $p$ torsion obstructions for finding $r$ complex sections for
$r<p^{2}-p$.

##  1\. Introduction

There is a classical problem to determine whether a manifold admits $r$
linearly independent tangent vector fields. In the work of Bokstedt, Dupont
and Svane [4] and in the previous paper [9], we approached this problem by
instead determining the obstruction to finding a cobordant manifold with $r$
vector fields. We were able to describe the rational obstruction to $r$
linearly independent complex sections of the tangent bundle of an almost
complex manifold up to cobordism. This obstruction is given in terms of Chern
characteristic numbers as follows:

######  Theorem 1.1.

Let $M^{2d}$ be a $d$-dimensional almost-complex manifold. The Chern
characteristic classes $s_{\omega}(M^{2d})=0$ for all $\omega$ of length
greater than or equal to $d-r+1$ if and only if there exists integer $c$ such
that the cobordism class $c[M^{2d}]$ contains a manifold $N^{2d}$ with $r$
complex sections on $TN$.

For any cobordism class $[M^{2d}]$ with vanishing rational obstruction, we can
ask whether there are any divisibility conditions for the constant $c$ in the
above theorem. If, for all $[M^{2d}]\in\Omega_{2d}^{U}$ with vanishing
rational obstruction, the constant $c$ may be chosen to be $1$ mod $p$ for
prime $p$, we say that the $p$ torsion obstruction vanishes. This is
equivalent to a particular map having no $p$-torsion in its image, see Theorem
4.6 in [9]. We will show in this paper, Proposition 3.6, that any $p$-torsion
in the image of this map is equivalent to a non-vanishing differential in the
$p$-primary Adams-Novikov spectral sequence for $\mathbf{MTU}(d-r)$. The prime
$p>2$ case is tackled first:

######  Theorem 1.2.

For $p>2$, the $p$-torsion obstruction to the existence of $r$ complex
sections vanishes when $r<p^{2}-p$.

We can further show that the $2$-torsion obstruction vanishes for $r=2$ or
$r=3$ complex sections. Thus we can prove the main result of the paper:

######  Theorem 1.3.

Let $M^{2d}$ be a $d$-dimensional almost-complex manifold with $d\geq 6$.
There is a manifold $N^{2d}\in[M^{2d}]$ with $2$ complex sections of $TN$ if
and only if $\chi([M^{2d}])=0$ and $s_{2,1,...,1}([M^{2d}])=0$.

There is a manifold $N^{2d}\in[M^{2d}]$ with $3$ complex sections of $TN$ if
and only if $s_{2,2,1...,1}([M^{2d}])=0$, $s_{3,1,...,1}([M^{2d}])=0$,
$s_{2,1,...,1}([M^{2d}])=0$, and $\chi([M^{2d}])=0$.

We can make a further improvement to the above result when $d$ is odd.

######  Theorem 1.4.

If $d>6$ is odd, there is a manifold $N^{2d}\in[M^{2d}]$ with $4$ complex
sections of $TN$ if and only if $s_{4,1...,1}([M^{2d}])=0$,
$s_{3,2,1...,1}([M^{2d}])=0$, $s_{2,2,2,...,1}([M^{2d}])=0$,
$s_{2,2,1...,1}([M^{2d}])=0$, $s_{3,1,...,1}([M^{2d}])=0$,
$s_{2,1,...,1}([M^{2d}])=0$, and $\chi([M^{2d}])=0$.

##  2\. Construction of Spectra

###  2.1. The spectra MTU(d) and MTU(d,r)

Here we introduce the spectra $\mathbf{MTU}(d)$ and $\mathbf{MTU}(d,r)$. The
homotopy groups of these spectra will be the natural objects of study. The
real case is described in [4] and [7]. Most of this work is repeated from our
previous paper [9]. It should be noted that there is some disagreement about
the proper indexing of these spectra in [4] and [7]. We follow the convention
given in Bokstedt, Dupont and Svane. [4] Write $G_{\mathbb{C}}(d,n)$ for the
complex Grassmannian of $d$ dimensional complex subspaces of
$\mathbb{C}^{d+n}$. Let $U_{\mathbb{C},d,n}\rightarrow G_{\mathbb{C}}(d,n)$ be
the tautological $d$ dimensional complex vector bundle and let
$U_{\mathbb{C},d,n}^{\perp}\rightarrow G_{\mathbb{C}}(d,n)$ be the $n$
dimensional orthogonal complement of $U_{\mathbb{C},d,n}$.

######  Definition 2.1.

Define $\mathbf{MTU}(d)$ to be the spectrum whose $2n$-th space is
$\mathbf{MTU}(d)_{2n}=Th(U_{\mathbb{C},d,n}^{\perp})$.

There exists a canonical map $G_{\mathbb{C}}(d,n)\rightarrow
G_{\mathbb{C}}(d,n+1)$ defined using the composition:

$\mathbb{C}^{d}\rightarrow\mathbb{C}^{d+n}\hookrightarrow\mathbb{C}\oplus\mathbb{C}^{d+n}\cong\mathbb{C}^{d+n+1}$

The restriction of the bundle $U_{\mathbb{C},d,n+1}^{\perp}$ to
$G_{\mathbb{C}}(d,n)$ under this map is
$U_{\mathbb{C},d,n}^{\perp}\oplus\mathbb{C}$. (Here $\mathbb{C}$ is a one
dimensional complex trivial bundle.) In other words, there is a bundle map
$U_{\mathbb{C},d,n}^{\perp}\oplus\mathbb{C}\rightarrow
U_{\mathbb{C},d,n+1}^{\perp}$ covering the map $G(d,n)\rightarrow G(d,n+1)$.
This map induces a map of Thom spaces and the following composition gives the
spectrum map
$\Sigma^{2}\mathbf{MTU}(d)_{2n}\rightarrow\mathbf{MTU}(d)_{2n+2}$:

$\Sigma^{2}(Th(U_{\mathbb{C},d,n}^{\perp}))\cong S^{2}\wedge
Th(U_{\mathbb{C},d,n}^{\perp})\cong Th(\mathbb{C}\oplus
U_{\mathbb{C},d,n}^{\perp})\rightarrow Th(U_{\mathbb{C},d,n+1}^{\perp})$

There is also a map $G(d-r,n)\rightarrow G(d,n)$ which takes a $(d-r)$-complex
plane $P\subset\mathbb{C}^{d-r+n}$ to the $d$-plane
$P\oplus\mathbb{C}^{r}\subset\mathbb{C}^{d-r+n}\oplus\mathbb{C}^{r}$. Under
this map, the pullback of $U_{\mathbb{C},d,n}^{\perp}$ is
$U_{\mathbb{C},d-r,n}^{\perp}$. Thus, there are maps of Thom spaces
$\mathbf{MTU}(d-r)_{2n}\rightarrow\mathbf{MTU}(d)_{2n}$. Since this map
commutes with the spectrum map, it defines a map of spectra
$\mathbf{MTU}(d-r)\rightarrow\mathbf{MTU}(d)$.

######  Definition 2.2.

Let $\mathbf{MTU}(d,r)$ be the cofiber of the map
$\mathbf{MTU}(d-r)\rightarrow\mathbf{MTU}(d)$.

We note that, having defined $\mathbf{MTU}(d,r)$, we immediately get a
cofibration for $k\leq d-r$.

$\mathbf{MTU}(d-r,k)\rightarrow\mathbf{MTU}(d,r+k)\rightarrow\mathbf{MTU}(d,r)$

This cofibration reduces to the definition when $k=d-r$. There is a sequence
of spectra:

$\mathbf{MTU}(d)\rightarrow\mathbf{MTU}(d+1)\rightarrow\mathbf{MTU}(d+2)\rightarrow...$

Let us call the colimit of this spectrum $\mathbf{MTU}$; this spectrum is the
Thom spectrum of the bundle $U^{\perp}_{\mathbb{C}}$ over $BU$. There is an
inversion map $BU\rightarrow BU$ which takes a vector space to its complement.
This map is covered by a bundle map $U^{\perp}_{\mathbb{C}}\rightarrow
U_{\mathbb{C}}$ which defines a homotopy equivalence
$\mathbf{MTU}\rightarrow\mathbf{MU}$. See [6] for the real case of this
equivalence. In particular, there is a well defined map
$\mathbf{MTU}(d)\rightarrow\mathbf{MU}$.

######  Definition 2.3.

Let $\overline{\mathbf{MTU}}(d)$ be the cofiber of the map
$\mathbf{MTU}(d)\rightarrow\mathbf{MU}$.

There is an alternative interpretation of $\overline{\mathbf{MTU}}(d)$ as the
colimit of the sequence of spectra:

$\mathbf{MTU}(d+1,1)\rightarrow\mathbf{MTU}(d+2,2)\rightarrow...$

The following commutative diagram relates these two constructions and shows
that they are the same.

$\mathbf{MTU}(d)$$\mathbf{MTU}(d+1)$$\mathbf{MTU}(d+1,1)$$\mathbf{MTU}(d+2)$$\mathbf{MTU}(d+2,2)$⋮⋮$\mathbf{MU}$$\overline{\mathbf{MTU}}(d)$

######  Proposition 2.1.

There is a cofibration
$\mathbf{MTU}(d,r)\rightarrow\overline{\mathbf{MTU}}(d-r)\rightarrow\overline{\mathbf{MTU}}(d)$

###### Proof.

There is a commutative diagram as below. The columns are the sequences of
spectra used to define $\overline{\mathbf{MTU}}(d-r)$ and
$\overline{\mathbf{MTU}}(d)$. The rows are each cofibrations.

$\mathbf{MTU}(d,r)$$\mathbf{MTU}(d+1,r+1)$$\mathbf{MTU}(d+1,1)$$\mathbf{MTU}(d+2,r+2)$$\mathbf{MTU}(d+2,2)$⋮⋮$\overline{\mathbf{MTU}}(d-r)$$\overline{\mathbf{MTU}}(d)$

The colimit of the cofibrations defined by each row gives the desired
cofibration. ∎

There is a second construction of spectrum $\mathbf{MTU}(d)$ due to [4] which
connects these spectra to the structure of complex sections. For any
$d$-dimensional complex fiber bundle $E\rightarrow X$ equipped with a
Hermitian inner product there is a complex frame bundle
$V_{\mathbb{C},r}(E)\rightarrow X$ with fiber $V_{\mathbb{C},d,r}$. The space
$V_{\mathbb{C},d,r}$ is the frame manifold of $r$ ordered complex frames in
$\mathbb{C}^{d}$. Now consider the specific case where the bundle is
$U_{\mathbb{C},d,n}\rightarrow G_{\mathbb{C}}(d,n)$. We can pullback the
original construction:

${p_{V_{\mathbb{C},r}}^{*}U_{\mathbb{C},d,n}^{\perp}}$${U_{\mathbb{C},d,n}^{\perp}}$${V_{\mathbb{C},r}(U_{\mathbb{C},d,n})}$${G_{\mathbb{C}}(d,n)}$$\scriptstyle{p_{V_{\mathbb{C},r}}}$

We form the Thom spaces
$Th(p_{V_{\mathbb{C},r}}^{*}U_{\mathbb{C},d,n}^{\perp})$ into a Thom spectra
which we call $\mathbf{MTU}(d)_{V_{r}}$. Similarly let
$W_{\mathbb{C},r}(U_{\mathbb{C},d,n})$ be the bundle whose fiber is the space
of ordered $r$-tuples in $U_{\mathbb{C},d,n}$ which are (hermitian) orthogonal
and of the same length which may be between $0$ and $1$. The fiber
$W_{\mathbb{C},d,r}$ is the cone over $V_{\mathbb{C},d,r}$ by construction. We
can construct a spectrum $\mathbf{MTU}(d)_{W_{r}}$ in an analogous way. In
[9], we prove the following:

######  Proposition 2.2.

There is a commutative diagram where all vertical maps are homotopy
equivalences.

${\mathbf{MTU}(d-r)}$${\mathbf{MTU}(d)}$${\mathbf{MTU}(d)_{V_{r}}}$${\mathbf{MTU}(d)_{W_{r}}}$

By using these homotopy equivalences in [9], we show that:

######  Theorem 2.3.

$\pi_{2d}(\mathbf{MTU}(d-r))$ is the even complex section cobordism group of
$2d$ dimensional manifolds with $r$ complex sections. That is, elements of
$\pi_{2d}(\mathbf{MTU}(d-r))$ are classes of almost complex manifolds with $r$
complex sections. Two classes are the same if there is a complex cobordism
equipped with $r+1$ sections between them. The map
$\pi_{2d}(\mathbf{MTU}(d-r))\rightarrow\pi_{2d}(\mathbf{MU})$ is the forgetful
map. The image of the forgetful map is all cobordism classes which contain a
manifold which can be equipped with $r$ complex sections.

The space $\mathbf{MTU}(d,1)$ can be given a simpler description which will be
useful for computing the homotopy groups:

######  Theorem 2.4.

There is a homotopy equivalence:

$\mathbf{MTU}(d,1)\cong S^{\infty+2d}\vee\Sigma^{\infty+2d}BU(d)$

In particular,

$\pi_{q+2d}(\mathbf{MTU}(d,1))\cong\pi_{q}^{s}(S^{0})\oplus\pi_{q}^{s}(BU(d))$

###### Proof.

By the second construction, we observe that $\mathbf{MTU}(d,1)_{2n}\cong
Th(p^{*}_{W_{\mathbb{C},1}}U^{\perp}_{\mathbb{C},d,n})/Th(p^{*}_{V_{\mathbb{C},1}}U^{\perp}_{\mathbb{C},d,n})$
where $p^{*}_{V_{\mathbb{C},1}}:V_{1}(U_{\mathbb{C},d,n})\cong
S(U_{\mathbb{C},d,n})\rightarrow G_{\mathbb{C}}(d,n)$ and
$p^{*}_{V_{\mathbb{C},1}}:W_{1}(U_{\mathbb{C},d,n})\cong
D(U_{\mathbb{C},d,n})\rightarrow G_{\mathbb{C}}(d,n)$.
There is a property that for two vector bundles $V_{1}$ and $V_{2}$ over the
same base,

$Th(V_{1}\oplus V_{2})\cong Th(V_{2}|_{D(V_{1})})/Th(V_{2}|_{S(V_{1})})$

Thus,

$\mathbf{MTU}(d,1)_{2n}\cong Th(U_{\mathbb{C},d,n}\oplus
U^{\perp}_{\mathbb{C},d,n})\cong Th(\mathbb{R}^{2d+2n}\times
G_{\mathbb{C}}(d,n))\cong S^{2d+2n}\vee\Sigma^{2d+2n}G_{\mathbb{C}}(d,n)$

Letting $n\rightarrow\infty$ proves the result. ∎

We briefly describe the homology of the spectra $\mathbf{MTU}(d)$.

######  Proposition 2.5.

There is a Thom isomorphism:

$H_{*}(BU(d);\mathbb{Z})\cong H_{*}(\mathbf{MTU}(d);\mathbb{Z})$

and the map $H_{*}(\mathbf{MTU}(d);\mathbb{Z})\rightarrow
H_{*}(\mathbf{MU};\mathbb{Z})$ is an injection. Moreover, the image of
$H_{*}(\mathbf{MTU}(d))$ in
$H_{*}(\mathbf{MU};\mathbb{Z})\cong\mathbb{Z}[B_{1},B_{2},...]$ is generated
as a group by monomials of degree less than $d$.

###### Proof.

This result follows by dualizing the corresponding result Proposition 2.2 in
cohomology from [9]. ∎

The lower homotopy groups of $\mathbf{MTU}(d)$ are trivial to compute [5],
[7].

######  Lemma 2.6.

The group $\pi_{q}(\mathbf{MTU}(d))\cong\pi_{q}(\mathbf{MU})$ for $q\leq 2d$.

There is an obstruction to lifting an element of $\pi_{2d}(\mathbf{MU})$ to
$\pi_{2d}(\mathbf{MTU}(d-r))$ which lands in the homotopy group
$\pi_{2d}(\mathbf{MTU}(d,r))$. In [9] we described the image of
$\pi_{*}(\mathbf{MTU}(d))\otimes\mathbb{Q}\rightarrow\pi_{*}(MU)\otimes\mathbb{Q}$
as follows:

######  Theorem 2.7.

Let $M^{2d}$ be a $d$-dimensional almost-complex manifold. The Chern
characteristic classes $s_{\omega}(M^{2d})=0$ for all $\omega$ of length
greater than or equal to $d-r+1$ if and only if the cobordism class $[M^{2d}]$
lifts to $\pi_{2d}(\mathbf{MTU}(d-r))\otimes\mathbb{Q}$. Geometrically, this
implies there exists integer $c$ such that the cobordism class $c[M^{2d}]$
contains a manifold $N^{2d}$ with $r$ complex sections on $TN$.

In this document, we begin to describe the image of the forgetful map
$\pi_{*}(\mathbf{MTU}(d))\rightarrow\pi_{*}(\mathbf{MU})$. As stated in the
introduction, we will call the torsion in the cokernel, the torsion
obstructions. Thus we determine which cobordism classes of manifolds can be
equipped with complex sections. To do so, we will study the $p$-local Adams-
Novikov spectral sequence for $\mathbf{MTU}(d)$.

##  3\. Computing the torsion obstruction

This section will be devoted to identifying which elements of
$\pi_{*}(\mathbf{MU})$ can be lifted to $\pi_{*}(\mathbf{MTU}(d))$. We will
use the following notational conventions. Then let $B_{i}\in
MU_{*}(\mathbf{MU})=\pi_{*}(\mathbf{MU})\otimes H_{*}(\mathbf{MU})$, be the
generators (as a $\pi_{*}(MU)$ algebra). These $B_{i}$ correspond to the
elements $b_{i}\in H_{*}(\mathbf{MU})\cong\mathbb{Z}[b_{1},b_{2},...]$. There
is an injective Hurewicz homomorphism $\pi_{*}(\mathbf{MU})\rightarrow
H_{*}(\mathbf{MU})$. By slight abuse of notation, we call the elements of
$\pi_{*}(\mathbf{MU})$ by their image in $H_{*}(\mathbf{MU})$. We begin by
stating the following results about $MU$ theory originally developed by [10]
and described in detail in [11] Appendix 2.1.

######  Theorem 3.1.

The pair $(\pi_{*}(\mathbf{MU}),MU_{*}(\mathbf{MU}))$ is a Hopf algebroid with
the following structures:

  * •

An augmentation map
$\varepsilon:MU_{*}(\mathbf{MU})\rightarrow\pi_{*}(\mathbf{MU})$, defined by
$\varepsilon(B_{i})=0$.

  * •

A left unit $\eta_{L}:\pi_{*}(\mathbf{MU})\rightarrow MU_{*}(\mathbf{MU})$,
defined as the standard inclusion.

  * •

A right unit corresponding to the Hurewicz homomorphism
$\eta_{R}:\pi_{*}(\mathbf{MU})\rightarrow MU_{*}(\mathbf{MU})$, defined by the
formula:

$\sum_{i}\eta_{R}(b_{i})x^{i+1}=\sum_{j}B_{j}\left(\sum_{k}b_{k}^{i+1}\right)^{j+1}$

  * •

A product inherited from $H_{*}(\mathbf{MU})$.

  * •

A coproduct defined by:

$\sum_{i}\Delta(B_{i})=\sum_{j}\left(\sum_{i}B_{i}\right)^{j+1}\otimes B_{j}$

It is typically convenient to localize at a prime $p$. The structure of
$\mathbf{MU}$ localized at $p$ is described as follows. See [11] Appendix 2.1
for more details.

######  Theorem 3.2.

For each prime $p$, there is an associative ring spectrum $\mathbf{BP}$ which
is a retract of $\mathbf{MU}_{(p)}$, moreover, $\mathbf{MU}_{(p)}$ splits as
copies of $\mathbf{BP}$. The homotopy group $\pi_{*}(\mathbf{BP})$ is
$\mathbb{Z}_{(p)}[v_{1},v_{2},...]$ where
$v_{i}\in\pi_{2(p^{i}-1)}(\mathbf{BP})$. The pair
$(\pi_{*}(\mathbf{BP}),BP_{*}(\mathbf{BP}))$ is a Hopf algebroid. We will
write $BP_{*}(\mathbf{BP})=\pi_{*}(\mathbf{BP})[t_{1},t_{2},...]$ with $t_{i}$
in degree $2(p^{i}-1)$. The algebroid has the structure:

  * •

An augmentation map
$\varepsilon:BP_{*}(\mathbf{BP})\rightarrow\pi_{*}(\mathbf{BP})$, defined by
$\varepsilon(t_{i})=0$ and $\varepsilon(v_{i})=v_{i}$.

  * •

A left unit $\eta_{L}:\pi_{*}(\mathbf{BP})\rightarrow BP_{*}(\mathbf{BP})$,
defined as the standard inclusion.

  * •

A right unit corresponding to the Hurewicz homomorphism
$\eta_{R}:\pi_{*}(\mathbf{BP})\rightarrow BP_{*}(\mathbf{BP})$, defined by the
formula:

$\eta_{R}(m_{i})=\sum_{j=1}^{n}m_{j}t_{i-j}^{p^{i}}$

  * •

A product inherited from $H_{*}(\mathbf{BP})$.

  * •

A coproduct determined by:

$\sum_{i,j}m_{i}\Delta(t_{j})^{p^{i}}=\sum_{i,j,k}\lambda_{i}t_{j}^{p^{i}}\otimes
t_{k}^{p^{i+j}}$

The basis $m_{i}$ comes from the elements $[\mathbb{C}P^{p^{i}-1}]/p$. The
generators $v_{i}$ are defined by

$pm_{n}=\sum_{i=0}^{n-1}m_{i}v_{n-i}^{p^{i}}$

We specifically note that $\eta_{R}(v_{1})=v_{1}+2t_{1}$, and
$\Delta(t_{1})=t_{1}\otimes 1+1\otimes t_{1}$.

######  Proposition 3.3.

Let $X$ be a spectrum with torsion-free homology concentrated in even degrees.
Let $E$ be either $\mathbf{MU}$ or $\mathbf{BP}$ for some prime $p$. Then,
$E_{*}(X)\cong\pi_{*}(E)\otimes_{\mathbb{Z}}E_{*}(X;\mathbb{Z})$ and is
torsion-free concentrated in even degrees.

###### Proof.

The Atiyah-Hirzebruch spectral sequence collapses, giving the stated result.
[3] ∎

Using Proposition 2.5, we immediately conclude:

######  Corollary 3.4.

Let $E$ be either $\mathbf{MU}$ or $\mathbf{BP}$ for some prime $p$. The map

$E_{*}(\mathbf{MTU}(d))\rightarrow
E_{*}(\mathbf{MU})=\pi_{*}(E)[B_{1},B_{2},...]$

is an injection, which is an isomorphism in degrees less than $2d+2$.
Moreover, both modules are concentrated in even degrees. The image of
$E_{*}(\mathbf{MTU}(d))$ maps onto polynomials of degree $d$.

$E_{*}(\mathbf{MU})\rightarrow E_{*}(\overline{\mathbf{MTU}}(d))$

is the quotient map killing all polynomials of degree $d$ or less. These maps
induce natural structures as $E_{*}(E)$ comodules.

We can now describe the $p$-local Adams-Novikov spectral sequence, see [10] or
[11].

######  Theorem 3.5.

For a spectra of finite type $X$, there is a spectral sequence
$\\{E^{s,t}_{k}(X),d_{k}\\}$ with differentials
$d_{k}:E^{s,t}_{k}(X)\rightarrow E^{s+r,t+r-1}_{k}(X)$: such that:

  1. (1)

$E^{s,t}_{2}(X)\cong\mathbf{Ext}^{s,t}_{BP_{*}(\mathbf{BP})}(\pi_{*}(\mathbf{BP}),BP_{*}(X))$

  2. (2)

If $X$ is connective and $p$ local, there is a filtration

$\pi_{t-s}(X)=F^{0,t-s}\supseteq...\supseteq F^{s,t}\supseteq
F^{s+1,t+1}\supseteq...$

such that $E^{s,t}_{\infty}(X)=F^{s,t}/F^{s+1,t+1}$.

For an element in $E_{q}^{s,t}$, we will refer to $s$ as the resolution of the
element and $t$ as the degree. The element, if it survives, corresponds to an
element of $\pi_{t-s}$. Consider the Adams-Novikov spectral sequence for
$\mathbf{MU}_{(p)}$. Since $\mathbf{MU}_{(p)}$ splits as a copy of
$\mathbf{BP}$’s, $E_{2}^{s,*}(\mathbf{MU}_{(p)})=0$ for $s>0$. So the spectral
sequence collapses, and
$E_{2}^{0,*}(\mathbf{MU}_{(p)})=\pi_{*}(\mathbf{MU}_{(p)})$. In particular, it
is a free $\mathbb{Z}_{(p)}$ module. Moreover,

$E_{2}^{0,*}(\mathbf{MTU}(d)_{(p)})=\mathbf{Hom}(BP_{*}(\mathbf{MTU}(d)_{(p)}),\pi_{*}(\mathbf{BP}))$

We are primarily concerned with the image of $BP_{*}(\mathbf{MTU}(d)_{(p)})$,
so we only need to know the image of the map
$E_{2}^{0,*}(\mathbf{MTU}(d))\rightarrow E_{2}^{0,*}(\mathbf{MU})$. This is
entirely described by the map

$\mathbf{Hom}_{BP_{*}(\mathbf{BP})}(\pi_{*}(\mathbf{BP}),BP_{*}(\mathbf{MTU}(d)_{(p)}))\rightarrow\mathbf{Hom}_{BP_{*}(\mathbf{BP})}(\pi_{*}(\mathbf{BP}),BP_{*}(\mathbf{MU}_{(p)}))=\pi_{*}(\mathbf{MU}_{(p)})$

This map has cokernel which is a free $\mathbb{Z}_{(p)}$ module. The following
proposition follows:

######  Proposition 3.6.

Fix prime $p$. Suppose $p[M]\in\pi_{*}(\mathbf{MU})$ lifts to
$\pi_{*}(\mathbf{MTU}(d))$. Then $[M]$ lifts to an element $x\in
E_{2}^{0,*}(\mathbf{MTU}(d))$ via the map:

$\mathbf{Hom}_{BP_{*}(\mathbf{BP})}(\pi_{*}(\mathbf{BP}),BP_{*}(\mathbf{MTU}(d)_{(p)}))\rightarrow\pi_{*}(\mathbf{MU}_{(p)})$

Moreover, $[M]$ lifts to $\pi_{*}(\mathbf{MTU}(d))$ if and only if there is no
Adams differential originating from $x$. Moreover, any free part of
$\pi_{*}(\mathbf{MTU}(d))$ comes from the zero line.

Thus, the differentials in the Adams-Novikov spectral sequence for
$\mathbf{MTU}(d)$ correspond with torsion obstructions. We will use the cobar
complex in order to compute the $E_{2}$ page. See Appendix 1 in [11] for more
details. Recall that $BP_{*}(\mathbf{BP})$ has an action of
$\pi_{*}(\mathbf{BP})$ given by multiplication by the image of $\eta_{R}$. Let
$\overline{\Gamma}$ be the quotient of $BP_{*}(\mathbf{BP})$ the image of
$\eta_{L}$.

######  Definition 3.1.

Let $(M,\psi)$ be a left $BP_{*}(\mathbf{BP})$ comodule. Define the cobar
complex as the tensor product $C_{BP}^{s}(M)=\overline{\Gamma}^{\otimes
s}\otimes_{\pi_{*}(\mathbf{BP})}M$ The differential
$d_{1}:C_{BP}^{s}(M)\rightarrow C_{BP}^{s+1}(M)$ is given by:

$d_{1}(\gamma_{1}\otimes...\otimes\gamma_{s}\otimes
m)=\sum_{i=1}^{s}(-1)^{i}\gamma_{1}\otimes...\otimes\Delta(\gamma_{i})\otimes...\otimes
m+(-1)^{s+1}\gamma_{1}\otimes...\otimes\gamma_{s}\otimes\psi(m)$

######  Proposition 3.7.

The homology of the cobar complex is the Ext groups of $M$, specifically:

$H(C_{BP}^{*}(M))=\mathbf{Ext}_{BP_{*}(\mathbf{BP})}(\pi_{*}(\mathbf{BP}),M)$

It will typically be convenient to view the cobar complex as the $E_{1}$ page
of the Adams-Novikov spectral sequence. We observe that the $0$ line of the
$E_{2}$ page of the Adams-Novikov spectral sequence is generated by
primitives, i.e. elements of the form $\psi(x)=1\otimes x$ where $\psi$ is the
coaction. The following proposition will be used to compute
$E_{2}(\mathbf{MTU}(d))$

######  Proposition 3.8.

There is an isomorphism for $s\geq 1$:

$E_{*}^{s,t}(\overline{\mathbf{MTU}}(d))\cong E_{*}^{s+1,t}(\mathbf{MTU}(d))$

###### Proof.

By Theorem 2.3.4 [11], there is an exact sequence of spectral sequences

$...\rightarrow E_{*}^{s,t}(\mathbf{MTU}(d))\rightarrow
E_{*}^{s,t}(\mathbf{MU})\rightarrow
E_{*}^{s,t}(\overline{\mathbf{MTU}}(d))\rightarrow
E_{*}^{s+1,t}(\mathbf{MTU}(d))\rightarrow...$

Since $E_{*}^{s,t}(\mathbf{MU})$ for $s>0$, the result follows. ∎

We can now prove Theorem 1.2.

######  Theorem 3.9.

The $p$-torsion obstruction to the existence of $r$ complex sections vanishes
when $r<p^{2}-p$.

###### Proof.

Fix prime $p>2$. Consider the cobar complexes for
$\overline{\mathbf{MTU}}(d)$. We will consider the cobar complex as the
$E_{1}$ page of the Adams-Novikov spectral sequence. Suppose we have an
element $\beta\in E_{2}^{s,t}(\overline{\mathbf{MTU}}(d))$. This element is
represented by $\sum_{k}\mu_{k}\otimes x_{k}\in\Gamma^{\otimes s}\otimes
BP_{*}(\mathbf{MTU}(d))$ such that $\sum_{k}\mu_{k}\otimes
x_{k}\in\textnormal{Ker}(d_{1})$. Suppose $t$ is such that
$E^{s,r}_{2}(\Sigma^{\infty}S^{0})=0$ for all $r\leq t-(2d+2)$. In particular
for $s=2$, this is equivalent to supposing $t<2(p^{2}-p+d+1)$. The element
$\sum_{k}\mu_{k}\otimes x_{k}$ has a term with maximal degree $x_{k}$. We will
call this term $\mu_{i}\otimes x_{i}$. Since $\sum_{k}\mu_{k}\otimes
x_{k}\in\textnormal{Ker}(d_{1})$, we see that $d_{1}(\mu_{i})=0$. We know
$\mu_{i}$ has degree less than or equal to $t-(2d+2)$ because $x_{i}$ has
degree greater than or equal to $(2d+2)$. Thus $\mu_{i}\otimes 1$ is in the
image of $\sum_{j}\tilde{\mu}_{j}\otimes\nu_{j}$ where
$\tilde{\mu}_{j}\in\overline{\Gamma}^{\otimes s-1}$ and
$\nu_{j}\in\pi_{*}(BP)$ because the corresponding element must vanish in
$E_{2}(S^{0})$. Then, we observe that
$d_{1}(\sum_{j}\tilde{\mu}_{j}\otimes\nu_{j}x_{i})=\mu_{i}\otimes x_{i}$.
Thus, $\sum_{k\neq i}\mu_{k}\otimes x_{k}$ also represents $\beta$. Repeating
this process, $\beta$ can be represented by $0$, and thus $\beta=0$. (We
choose the letter $\beta$ suggestively, as the first potentially non zero
$\beta$ corresponds to $\beta_{1}$ constructed in Definition 1.3.14 [11]). By
Proposition 3.8, $E_{2}^{3,t}(\mathbf{MTU}(d))=0$ and $t<2(p^{2}-p+d+1)$.
Thus, $d_{3}:E_{3}^{0,t}\rightarrow E_{3}^{3,t+2}$ is trivial for
$t<2(p^{2}-p+d)$. Similarly, the higher differentials must also vanish. By
Proposition 3.6, the torsion obstruction to $r<p^{2}-p$ complex sections
vanishes. ∎

######  Remark 3.1.

This result is not sharp, and we can improve it significantly by closer
analysis. However, this bound is sufficient for this paper, since it will show
the vanishing of all $p>2$ torsion obstructions for $r<6$.

We now focus on the case $p=2$. Thus for the rest of the paper all homotopy
groups are assumed to be 2-primary. We will need to compute the homotopy
groups of $\overline{\mathbf{MTU}}(d)$. To do so we remind ourselves of the
Adams-Novikov spectral sequence for the sphere spectrum: [14]

012$0$$2$$4$$\mathbb{Z}/2$$0$$\mathbb{Z}/2$3$\mathbb{Z}/2$$\mathbb{Z}/4$$\mathbb{Z}$$4$$\mathbb{Z}/2$$\mathbb{Z}/2$$s$$t-s$$\mathbb{Z}/2$05613

In particular, we will call the generators of $\pi_{1}^{s}(S^{0})$ and
$\pi_{3}^{s}(S^{0})$, $h_{1}$ and $h_{2}$ respectively. These can be
represented by the Hopf maps.

######  Theorem 3.10.

The homotopy groups of $\overline{\mathbf{MTU}}(d)$ for $d\geq 3$ are given
by:

$\begin{array}[]{cccccc}q&2d&2(d+1)&2(d+2)&2(d+3)&2(d+4)\\\ d\equiv 0\
(\textnormal{mod}\
2)&\mathbb{Z}&\mathbb{Z}/2&\mathbb{Z}/2\oplus\mathbb{Z}^{2}&?&\mathbb{Z}^{4}\\\
d\equiv 1\ (\textnormal{mod}\
2)&\mathbb{Z}&0&\mathbb{Z}^{2}&0&\mathbb{Z}^{4}\\\ \end{array}$

###### Proof.

The coaction $\psi:BP_{*}(\mathbf{MTU}(d))\rightarrow
BP_{*}(\mathbf{MU})\otimes_{\pi_{*}(\mathbf{BP})}BP_{*}(\mathbf{MTU}(d))$ is
the natural quotient of the coaction in $BP_{*}(\mathbf{MU})$ induced by the
map $\mathbf{MU}\rightarrow\mathbf{MTU}(d)$. Note that the right action of
$\pi_{*}(\mathbf{BP})$ on $BP_{*}(\mathbf{MU})$ is given by multiplication by
the image of the Hurewicz homomorphism $\eta_{R}$. We compute that
$\eta_{R}(v_{1})=v_{1}+2t_{1}$. (See [11], appendix 2.2.)

We start by computing the following coactions in $BP_{*}(\mathbf{MU})$.

$\displaystyle\psi(B_{1})$ $\displaystyle=1\otimes B_{1}+t_{1}\otimes 1$
$\displaystyle\psi(B_{2})$ $\displaystyle=1\otimes B_{2}+(2t_{1})\otimes
B_{1}+t_{1}^{2}\otimes 1$ $\displaystyle\psi(B_{3})$ $\displaystyle=1\otimes
B_{3}+3t_{1}\otimes B_{2}+(t_{1}^{2}-2v_{1}t_{1})\otimes B_{1}+t_{2}\otimes 1$

We can then compute:.

$\displaystyle\psi(B_{1}^{d+1})$ $\displaystyle=1\otimes B_{1}^{d+1}$
$\displaystyle\psi(B_{2}B_{1}^{d})$ $\displaystyle=2t_{1}\otimes
B_{1}^{d+1}+1\otimes B_{2}B_{1}^{d}$ $\displaystyle\psi(B_{1}^{d+2})$
$\displaystyle=1\otimes B_{1}^{d+2}+(d+2)t_{1}\otimes B_{1}^{d+1}$
$\displaystyle\psi(v_{1}B_{1}^{d+1})$ $\displaystyle=v_{1}\otimes
B_{1}^{d+1}=1\otimes v_{1}B_{1}^{d+1}-2t_{1}\otimes B_{1}^{d+1}$

We have two cases, if $d$ is even, then the cokernel of
$d_{1}:E_{1}^{0,2d+4}\rightarrow E_{1}^{1,2d+4}$ is $\mathbb{Z}/2$ generated
by $t_{1}\otimes B_{1}^{d+1}$ and if $d$ is odd, the cokernel is zero. Next we
compute the cokernel of $d_{1}:E_{1}^{1,2d+6}\rightarrow E_{1}^{2,2d+6}$.

$\displaystyle d_{1}(t_{1}\otimes B_{2}B_{1}^{d})=2t_{1}\otimes
t_{1}\otimes\beta_{1}^{d+1}$ $\displaystyle d_{1}$ $\displaystyle(t_{1}\otimes
B_{1}^{d+2})=(d+2)t_{1}\otimes t_{1}\otimes\beta_{1}^{d+1}$ $\displaystyle
d_{1}(t_{1}\otimes v_{1}B_{1}^{d+1})=-2t_{1}\otimes
t_{1}\otimes\beta_{1}^{d+1}$ $\displaystyle d_{1}$
$\displaystyle(t_{1}^{2}\otimes B_{1}^{d+1})=-2t_{1}\otimes
t_{1}\otimes\beta_{1}^{d+1}$

Once again, if $d$ is even, then the cokernel of
$d_{1}:E_{1}^{1,2d+6}\rightarrow E_{1}^{2,2d+6}$ is $\mathbb{Z}/2$, and if $d$
is odd, the cokernel is zero. We omit the details, but for similar reasons,
$d_{1}:E_{1}^{2,2d+8}\rightarrow E_{1}^{3,2d+8}$ has cokernels $\mathbb{Z}/2$
or $0$ when $d$ is even or odd respectively.

We now compute the image of $d_{1}:E_{1}^{0,2d+6}\rightarrow E_{1}^{1,2d+6}$.
We will write this as a table for brevity.

$\begin{array}[]{c|c|c|c|c}&t_{1}\otimes B_{2}B_{1}^{d},&t_{1}\otimes
B_{1}^{d+2}&t_{1}\otimes v_{1}B_{1}^{d+1}&t_{1}^{2}\otimes B_{1}^{d+1}\\\
B_{3}B_{1}^{d}&3&0&-2&5\\\ B_{2}^{2}B_{1}^{d-1}&4&0&0&4\\\
B_{2}B_{1}^{d+1}&(d+1)&2&0&2d+3\\\
B_{1}^{d+3}&0&(d+3)&0&\frac{1}{2}(d+3)(d+2)\\\
v_{1}B_{2}B_{1}^{d}&-2&0&2&-4\\\ v_{1}B_{1}^{d+2}&0&-2&(d+2)&2(d+2)\\\
v_{1}^{2}B_{1}^{d+1}&0&0&-4&4\\\ \end{array}$

The kernel of $d_{1}:E_{1}^{0,2d+6}\rightarrow E_{1}^{1,2d+6}$ is generated by
elements where $2a_{1}+(d+2)a_{2}-2(a_{3}+a_{4})=0$:

$a_{1}t_{1}\otimes B_{2}B_{1}^{d}+a_{2}t_{1}\otimes
B_{1}^{d+2}+a_{3}t_{1}\otimes v_{1}B_{1}^{d+1}+a_{4}t_{1}^{2}\otimes
B_{1}^{d+1}$

Thus, we have two cases for $E_{2}^{1,2d+4}$: if $d$ is even,
$E_{2}^{1,2d+4}=\mathbb{Z}/2$ and if $d$ is odd, $E_{2}^{1,2d+4}=0$. Similarly
we can compute $d_{1}:E_{1}^{1,2d+8}\rightarrow E_{1}^{2,2d+8}$. We find that
$E_{2}^{2,2d+6}$ is always zero.

$\begin{array}[]{c|c|c|c|c|c}&t_{1}\otimes t_{1}\otimes
B_{2}B_{1}^{d},&t_{1}\otimes t_{1}\otimes B_{1}^{d+2}&t_{1}\otimes
t_{1}\otimes v_{1}B_{1}^{d+1}&t_{1}\otimes t_{1}^{2}\otimes
B_{1}^{d+1}&t_{1}^{2}\otimes t_{1}\otimes B_{1}^{d+1}\\\ t_{1}\otimes
B_{3}B_{1}^{d}&3&0&-2&5&0\\\ t_{1}\otimes B_{2}^{2}B_{1}^{d-1}&4&0&0&4&0\\\
t_{1}\otimes B_{2}B_{1}^{d+1}&(d+1)&2&0&2d+3&0\\\ t_{1}\otimes
B_{1}^{d+3}&0&(d+3)&0&\frac{1}{2}(d+3)(d+2)&0\\\ t_{1}\otimes
v_{1}B_{2}B_{1}^{d}&-2&0&2&-4&0\\\ t_{1}\otimes
v_{1}B_{1}^{d+2}&0&-2&(d+2)&2(d+2)&0\\\ t_{1}\otimes
v_{1}^{2}B_{1}^{d+1}&0&0&-4&4&0\\\ t_{1}^{2}\otimes
B_{2}B_{1}^{d}&-2&0&0&0&2\\\ t_{1}^{2}\otimes B_{1}^{d+2}&0&-2&0&0&(d+2)\\\
t_{1}^{2}\otimes v_{1}B_{1}^{d+1}&0&0&-2&0&-2\\\ t_{2}\otimes
B_{1}^{d+1}&0&0&-1&3&2\\\ t_{1}^{3}\otimes B_{1}^{d+1}&0&0&0&3&3\\\
\end{array}$

Then, in the relevant range the Adams chart for $d$ even is:
012$2d+2$$2d+4$$2d+6$$\mathbb{Z}/2$$0$$\mathbb{Z}/2$3$\mathbb{Z}/2$$\mathbb{Z}/2$$\mathbb{Z}$$\mathbb{Z}^{\oplus
2}$$\mathbb{Z}^{\oplus 4}$$4$$\mathbb{Z}/2$$s$$t-s$

The non-trivial $d_{3}$ differential can be observed by looking at the map
$S^{\infty+2d}\rightarrow\overline{\mathbf{MTU}}(d)$ given by the generator of
$\pi_{2d}(\overline{\mathbf{MTU}}(d))$. There may be an additional
differential killing $E_{2}^{2,2d+8}$ but since this element is in odd degree,
we will not need to know it. The Adams-Novikov spectral sequence for $d$ odd
is trivial in this range. ∎

######  Remark 3.2.

The vanishing of the torsion obstruction to $r$ complex sections does not
imply the vanishing of the torsion obstruction to $r-1$ complex sections, thus
each result must be proved separately.

We use these Adams charts and homotopy groups to prove:

######  Theorem 3.11.

There is no torsion obstruction to the existence of 2 or 3 linearly
independent complex sections on $M^{2d}$ for $d\geq 6$.

###### Proof.

Since $\pi_{2d+6}(\overline{\mathbf{MTU}}(d))$ is free abelian, there is no
possible target for a torsion obstruction to $3$ complex sections. Similarly,
there is no possible target for a torsion obstruction to $2$ complex sections
when $d$ is odd. In the case $d$ is even, there could be an obstruction to $2$
complex sections. However, we recall from [8] that $S^{\infty+2d}$ splits off
$MTU(d)$ for $d$ even. The element in $E_{2}^{3,2d+3}$ comes from this split
$S^{\infty+2d}$ (specifically, it can be identified with $h_{1}^{3}$), and so
cannot be the target of a differential. ∎

######  Theorem 3.12.

There is no torsion obstruction to $4$ complex sections when $d>6$ is odd.

###### Proof.

Let $M\in\pi_{2d}(\mathbf{MU})$ with vanishing rational obstruction. Recall
that the image of $M$ in the group $\pi_{2d}(\overline{\mathbf{MTU}}(d-4))$ is
the obstruction to $4$ complex sections. Thus the image is torsion. By Theorem
3.11, there is no torsion obstruction to $3$ complex sections so the image of
$M$ in $\pi_{2d}(\overline{\mathbf{MTU}}(d-3))$ is zero. Thus the obstruction
to $4$ complex sections lifts to
$\pi_{2d}(\mathbf{MTU}(d-3,1))\cong\pi_{2d}^{s}(S^{2d-6})\oplus\pi_{6}^{s}(BU(d-3))$.

$\pi_{2d}(\mathbf{MTU}(d-4))$$\pi_{2d}(\mathbf{MU})$$\pi_{2d}(\overline{\mathbf{MTU}}(d-4))$$\pi_{2d}(\mathbf{MTU}(d-3))$$\pi_{2d}(\mathbf{MU})$$\pi_{2d}(\overline{\mathbf{MTU}}(d-3))$$\pi_{2d}(\mathbf{MTU}(d-3,1))$

We know that $\pi_{6}^{s}(BU(d-3))$ is torsion free. The torsion element in
$\pi_{6}^{s}(S^{0})\cong h_{2}\cdot h_{2}$. The map
$\pi_{2d}^{s}(S^{2d-6})\rightarrow\pi_{2d}(MTU(d-3))$ is a
$\pi_{*}^{s}(S^{0})$ module map. In particular, it must take $h_{2}$ to $0$ if
$d$ is odd and we conclude that $h_{2}\cdot h_{2}\mapsto h_{2}\cdot 0=0$. Thus
the torsion obstruction to $M$ having $4$ complex sections vanishes. ∎

######  Remark 3.3.

For $d$ even, there is a potential obstruction of order $2$. We expect that
this obstruction is non vanishing and, based on similar results in [2] and
[4], we expect that the obstruction is a divisibility condition on the
signature of the manifold.

We can give a bound on the possible torsion obstruction to the existence of
$r$ complex sections. Using the following theorem of Segal [12].

######  Theorem 3.13.

There are no non-zero differentials originating in the bottom row of the
Adams-Novikov spectral sequence for
$MTU(1)\cong\Sigma^{\infty-2}\mathbb{C}P^{\infty}$.

######  Corollary 3.14.

There is no torsion obstruction to the existence of $d-1$ sections on
$[M^{2d}]$.

In particular, we can find an element of $N^{2d}\in\pi_{2d}(\mathbf{MTU}(1))$
with all characteristic classes zero except $s_{d}(N^{2d})=(d+1)!/B_{2d}$
where $B_{2d}$ is the Bernoulli number [13]. From Milnor’s characterization of
the generators of the complex cobordism ring, the following proposition
immediately follows. These numbers are known to play a role in the theory of
characteristic classes cf. [1].

######  Proposition 3.15.

For every $d$, define

$a_{d}=\left\\{\begin{array}[]{cc}(d+1)!/B_{2d}&d\neq p^{i}-1\\\
(d+1)!/(pB_{2d})&d=p^{i}-1\\\ \end{array}\right.$

Then for any set of multiplicative generators $M^{2d}\in\Omega_{2d}^{U}$, we
can write $N^{2d}=\tilde{a}_{d}M^{2d}+\textnormal{decomposables}$

For any partition $I=\\{i_{1},...,i_{n}\\}$, define
$a_{I}=\tilde{a}_{i_{1}}...\tilde{a}_{i_{k}}$. Let $M^{2i}\in\Omega_{2i}^{U}$
be any collection of multiplicative generators and write
$M^{I}=M^{2i_{1}}\times...\times M^{2i_{k}}$. Define a partial order on
partitions of $d$ by

$\\{i_{1},...,i_{k}\\}>\\{j_{1,1},...,j_{1,l_{1}},j_{2,1},...,j_{2,l_{1}},....,j_{k,1},...,j_{k,l_{k}}\\}$

if $i_{n}=\sum_{m=1}^{l_{m}}j_{n,m}$.

######  Theorem 3.16.

Let $K^{2d}=M^{I}+\sum_{J<I}b_{J}M^{J}$ be a manifold with vanishing rational
obstruction to $r$ complex sections. Then $\textnormal{lcm}(a_{J}\mid
J<I)[K^{2d}]$ contains a manifold which can be equipped with $r$ complex
sections.

###### Proof.

Call $C=\textnormal{lcm}(a_{J}\mid J<I)$. By Proposition 3.15, we can write

$CK^{2d}=\frac{C}{a_{J}}N^{I}+C\sum_{J<I}b_{J}^{\prime}M^{J}$

for new constants $b_{J}^{\prime}$. Choose maximal $J$ with
$b_{J}^{\prime}\neq 0$. Then we can rewrite $CM^{J}$ as
$\frac{C}{a_{J}}\sum_{L<J}N^{J}$. By inducting down on the partial order, we
can write

$CK^{2d}=c_{I}N^{I}+\sum_{J<I}c_{J}N^{J}$

for some integers $c_{J}$. Note that $N^{J}$ has $s_{J}(N^{J})\neq 0$ and all
other characteristic classes equal to 0. So, in this sum, $c_{J}$ must be zero
for all $J$ of length greater than or equal to $d-r+1$ because we are assuming
the rational obstruction vanishes. Note that by construction $N^{2i}$ has
$i-1$ complex sections, so $N^{J}$ has $d-l(J)\geq r$ complex sections. Thus
$a_{I}K^{2d}$ also can be equipped with $r$ complex sections. ∎

######  Remark 3.4.

We expect that the Bernoulli numbers are measuring unexpected obstructions to
complex sections, by comparing the formula in Proposition 3.15 with the
formula for the primitives in [12]. Thus, we conjecture that they relate to
the presence of non trivial differentials.

This result is not sharp, but it allows us to construct cobordism classes
containing a manifold with $r$ complex sections for any $r$.

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