# Unoriented bordism

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## 1 Introduction

We denote the unoriented bordism groups by $\mathcal N_i$${{Authors|Matthias Kreck}}{{Stub}} == Introduction == ; We denote the unoriented bordism groups by \mathcal N_i. The sum of these groups \mathcal N_* := \sum _i\mathcal N_i forms a ring under cartesian products of manifolds. Thom {{cite|Thom1954}} has shown that this ring is a polynomial ring over \mathbb Z/2 in variables x_i for i \ne 2^k -1 and he has shown that for i even one can take \mathbb {RP}^i for x_i. Dold {{cite|Dold1956}} has constructed manifolds for x_i with i odd. == Construction and examples == ; Dold constructs certain bundles over \mathbb {RP}^m with fibre \mathbb {CP}^n denoted by P(m,n):= (S^m \times \mathbb {CP}^n)/\tau, where \tau is the involution mapping (x,[y]) to (-x, [\bar y]) and \bar y = (\bar y_0,...,\bar y_n) for y =(y_0,...y_n). These manifolds are now called Dold manifolds. Using the results by Thom {{cite|Thom1954}} Dold shows that these manifolds give ring generators of \mathcal N_*. {{beginthm|Theorem (Dold) {{cite|Dold1956}}}} For i even set x_i:= [P(i,0) ]= [\mathbb {RP}^i] and for i = 2^r(2s+1)-1 set x_i:=[ P(2^r-1,s2^r)]. Then for i \ne 2^k-1 x_2,x_4,x_5,x_6,x_8,... are polynomial generators of \mathcal N_* olver \mathbb Z/2: \mathcal N_* \cong \mathbb Z/2[x_2,x_4,x_5,x_6,x_8...]. {{endthm}} == Invariants == ; To prove the Theorem Dold has to compute the characteristic numbers which according to Thom's theorem determine the bordism class. As a first step Dold computes the cohomology ring with \mathbb Z/2-coeffcients. The fibre bundle p:P(m,n) \to \mathbb {RP}^m has a section s([x]) := [(x,[1,...,1])] and we consider the cohomology classes (always with \mathbb Z/2-coefficients) c:= p^*(x) \in H^1(P(m,n)), where x is a generator of H^1(\mathbb {RP}^m), and d \in H^2(P(m,n)), which is characterized by the property that the restriction to a fibre is non-trivial and s^*(d)=0. {{beginthm|Theorem {{cite|Dold1956}}}} The classes c \in H^1(P(m,n)) and d\in H^2(P(m,n)) generate H^*(P(m,n);\mathbb Z/2) with only the relations c^{m+1} =0 and d^{n+1} =0. The Steenrod squares act by Sq^0 =id, \,\, Sq^1(c) = c^2,\,\, Sq^1(d) = cd,\,\, Sq^2(d) =d^2, and all other Squares Sq^i act trivially on c and d. On the decomposable classes the action is given by the Cartan formula. The total Stiefel-Whitney class of the tangent bundle is w(P(m,n)) = (1+c)^{m+1}(1+d)^{n+1}. {{endthm}} == Classification == ; To give explicit polynomial generators is useful information, if one wants to prove a formula like for example that = e(M)\,\, mod\,\, 2, where w_n is the n-th Stiefel-Whitney class of an m-dimensional manifold and e(M) is the Euler characteristic, which one has to check on generators. But it does not help to classify manifolds up to bordism. There is an explicit answer to this question by Thom: {{beginthm|Theorem {{cite|Thom1954}}}} Two closed n-manifolds M and N are bordant if and only if all Stiefel-Whitney numbers agree: = for all partitions i_1+...+i_k =n. {{endthm}} == Further discussion == ; For i= 2^r(2s+1)-1 odd i \ne 2^k-1 the manifolds P_i:= P(2^r-1,s2^r) are orientable and thus after choosing an orientation give an element in the oriented bordism group \Omega_i. Since P_i admits an obvious orientation reversing diffeomorphism, these elements are -torsion. Thus we obtain a subring in \Omega_* isomorphic to \mathbb Z/2[x_5, x_9, x_{11},...]. For more information about \Omega _* see the page on oriented bordism. == References == {{#RefList:}} [[Category:Manifolds]] [[Category:Bordism]]\mathcal N_i$. The sum of these groups

$\displaystyle \mathcal N_* := \sum _i\mathcal N_i$

forms a ring under cartesian products of manifolds. Thom [Thom1954] has shown that this ring is a polynomial ring over $\mathbb Z/2$$\mathbb Z/2$ in variables $x_i$$x_i$ for $i \ne 2^k -1$$i \ne 2^k -1$ and he has shown that for $i$$i$ even one can take $\mathbb {RP}^i$$\mathbb {RP}^i$ for $x_i$$x_i$. Dold [Dold1956] has constructed manifolds for $x_i$$x_i$ with $i$$i$ odd.

## 2 Construction and examples

Dold constructs certain bundles over $\mathbb {RP}^m$$\mathbb {RP}^m$ with fibre $\mathbb {CP}^n$$\mathbb {CP}^n$ denoted by

$\displaystyle P(m,n):= (S^m \times \mathbb {CP}^n)/\tau,$
where $\tau$$\tau$ is the involution mapping $(x,[y])$$(x,[y])$ to $(-x, [\bar y])$$(-x, [\bar y])$ and $\bar y = (\bar y_0,...,\bar y_n)$$\bar y = (\bar y_0,...,\bar y_n)$ for $y =(y_0,...y_n)$$y =(y_0,...y_n)$. These manifolds are now called Dold manifolds.

Using the results by Thom [Thom1954] Dold shows that these manifolds give ring generators of $\mathcal N_*$$\mathcal N_*$.

Theorem (Dold) [Dold1956] 2.1. For $i$$i$ even set $x_i:= [P(i,0) ]= [\mathbb {RP}^i]$$x_i:= [P(i,0) ]= [\mathbb {RP}^i]$ and for $i = 2^r(2s+1)-1$$i = 2^r(2s+1)-1$ set $x_i:=[ P(2^r-1,s2^r)]$$x_i:=[ P(2^r-1,s2^r)]$. Then for $i \ne 2^k-1$$i \ne 2^k-1$

$\displaystyle x_2,x_4,x_5,x_6,x_8,...$

are polynomial generators of $\mathcal N_*$$\mathcal N_*$ olver $\mathbb Z/2$$\mathbb Z/2$:

$\displaystyle \mathcal N_* \cong \mathbb Z/2[x_2,x_4,x_5,x_6,x_8...].$

## 3 Invariants

To prove the Theorem Dold has to compute the characteristic numbers which according to Thom's theorem determine the bordism class. As a first step Dold computes the cohomology ring with $\mathbb Z/2$$\mathbb Z/2$-coeffcients. The fibre bundle $p:P(m,n) \to \mathbb {RP}^m$$p:P(m,n) \to \mathbb {RP}^m$ has a section $s([x]) := [(x,[1,...,1])]$$s([x]) := [(x,[1,...,1])]$ and we consider the cohomology classes (always with $\mathbb Z/2$$\mathbb Z/2$-coefficients)

$\displaystyle c:= p^*(x) \in H^1(P(m,n)),$

where $x$$x$ is a generator of $H^1(\mathbb {RP}^m)$$H^1(\mathbb {RP}^m)$, and

$\displaystyle d \in H^2(P(m,n)),$

which is characterized by the property that the restriction to a fibre is non-trivial and $s^*(d)=0$$s^*(d)=0$.

Theorem [Dold1956] 3.1. The classes $c \in H^1(P(m,n))$$c \in H^1(P(m,n))$ and $d\in H^2(P(m,n))$$d\in H^2(P(m,n))$ generate $H^*(P(m,n);\mathbb Z/2)$$H^*(P(m,n);\mathbb Z/2)$ with only the relations

$\displaystyle c^{m+1} =0$

and

$\displaystyle d^{n+1} =0.$

The Steenrod squares act by

$\displaystyle Sq^0 =id, \,\, Sq^1(c) = c^2,\,\, Sq^1(d) = cd,\,\, Sq^2(d) =d^2,$

and all other Squares $Sq^i$$Sq^i$ act trivially on $c$$c$ and $d$$d$. On the decomposable classes the action is given by the Cartan formula.

The total Stiefel-Whitney class of the tangent bundle is

$\displaystyle w(P(m,n)) = (1+c)^{m+1}(1+d)^{n+1}.$

## 4 Classification

To give explicit polynomial generators is useful information, if one wants to prove a formula like for example that $ = e(M)\,\, mod\,\, 2$$ = e(M)\,\, mod\,\, 2$, where $w_n$$w_n$ is the $n$$n$-th Stiefel-Whitney class of an $m$$m$-dimensional manifold and $e(M)$$e(M)$ is the Euler characteristic, which one has to check on generators. But it does not help to classify manifolds up to bordism. There is an explicit answer to this question by Thom:

Theorem [Thom1954] 4.1. Two closed $n$$n$-manifolds
Tex syntax error
$M$ and $N$$N$ are bordant if and only if all Stiefel-Whitney numbers agree:
$\displaystyle = $

for all partitions $i_1+...+i_k =n$$i_1+...+i_k =n$.

## 5 Further discussion

For $i= 2^r(2s+1)-1$$i= 2^r(2s+1)-1$ odd $i \ne 2^k-1$$i \ne 2^k-1$ the manifolds $P_i:= P(2^r-1,s2^r)$$P_i:= P(2^r-1,s2^r)$ are orientable and thus after choosing an orientation give an element in the oriented bordism group $\Omega_i$$\Omega_i$. Since $P_i$$P_i$ admits an obvious orientation reversing diffeomorphism, these elements are $2$$2$-torsion. Thus we obtain a subring in $\Omega_*$$\Omega_*$ isomorphic to $\mathbb Z/2[x_5, x_9, x_{11},...]$$\mathbb Z/2[x_5, x_9, x_{11},...]$. For more information about $\Omega _*$$\Omega _*$ see the page on oriented bordism.