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(Co)bordism invariant of Eilenberg–MacLane space becomes vanished
Clash Royale CLAN TAG#URR8PPP
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Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
Omega^5_O(K(mathbbZ/2,2)).
$$
Here $u in H^2(K(mathbbZ/2,2),mathbbZ_2)$. The $K(mathbbZ/2,2)$ is Eilenberg–MacLane space. The $mathbbZ/2$ is the finite group of order 2.
Question: Is it true that such a (co)bordism invariant $u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$ when being pulled back from the $Omega^5_O(K(mathbbZ/2,2))$ to a pulldback new (co)bordism group
$$
Omega^5_frac(Pin^pm times SU(2))mathbbZ/2
$$
(thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$Omega^5_O(K(mathbbZ/2,2))$ to $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$),
the (co)bordism invariant
$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
Namely the effective manifold generators (respect of to $
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $Omega^5_O(K(mathbbZ/2,2))$ becomes trivial or vanished in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
How to prove this?
- Here $frac(Pin^pm times SU(2))mathbbZ/2$ is similar to the generalization of Pin$^c$ structure where $U(1)$ is now replaced by $SU(2)$.
at.algebraic-topology gt.geometric-topology smooth-manifolds characteristic-classes cobordism
asked 2 hours ago
annie heart
8912
add a comment |Â
up vote
3
down vote
favorite
Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
Omega^5_O(K(mathbbZ/2,2)).
$$
Here $u in H^2(K(mathbbZ/2,2),mathbbZ_2)$. The $K(mathbbZ/2,2)$ is Eilenberg–MacLane space. The $mathbbZ/2$ is the finite group of order 2.
Question: Is it true that such a (co)bordism invariant $u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$ when being pulled back from the $Omega^5_O(K(mathbbZ/2,2))$ to a pulldback new (co)bordism group
$$
Omega^5_frac(Pin^pm times SU(2))mathbbZ/2
$$
(thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$Omega^5_O(K(mathbbZ/2,2))$ to $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$),
the (co)bordism invariant
$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
Namely the effective manifold generators (respect of to $
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $Omega^5_O(K(mathbbZ/2,2))$ becomes trivial or vanished in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
How to prove this?
- Here $frac(Pin^pm times SU(2))mathbbZ/2$ is similar to the generalization of Pin$^c$ structure where $U(1)$ is now replaced by $SU(2)$.
at.algebraic-topology gt.geometric-topology smooth-manifolds characteristic-classes cobordism
asked 2 hours ago
annie heart
8912
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
Omega^5_O(K(mathbbZ/2,2)).
$$
Here $u in H^2(K(mathbbZ/2,2),mathbbZ_2)$. The $K(mathbbZ/2,2)$ is Eilenberg–MacLane space. The $mathbbZ/2$ is the finite group of order 2.
Question: Is it true that such a (co)bordism invariant $u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$ when being pulled back from the $Omega^5_O(K(mathbbZ/2,2))$ to a pulldback new (co)bordism group
$$
Omega^5_frac(Pin^pm times SU(2))mathbbZ/2
$$
(thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$Omega^5_O(K(mathbbZ/2,2))$ to $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$),
the (co)bordism invariant
$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
Namely the effective manifold generators (respect of to $
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $Omega^5_O(K(mathbbZ/2,2))$ becomes trivial or vanished in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
How to prove this?
- Here $frac(Pin^pm times SU(2))mathbbZ/2$ is similar to the generalization of Pin$^c$ structure where $U(1)$ is now replaced by $SU(2)$.
at.algebraic-topology gt.geometric-topology smooth-manifolds characteristic-classes cobordism
asked 2 hours ago
annie heart
8912
Consider a (co)bordism invariant
$$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$$
obtained from
$$
Omega^5_O(K(mathbbZ/2,2)).
$$
Here $u in H^2(K(mathbbZ/2,2),mathbbZ_2)$. The $K(mathbbZ/2,2)$ is Eilenberg–MacLane space. The $mathbbZ/2$ is the finite group of order 2.
Question: Is it true that such a (co)bordism invariant $u_2 Sq^1 u_2+Sq^2 Sq^1 u_2$ when being pulled back from the $Omega^5_O(K(mathbbZ/2,2))$ to a pulldback new (co)bordism group
$$
Omega^5_frac(Pin^pm times SU(2))mathbbZ/2
$$
(thus we are allowed to identify more wider classes of manifolds or more (co)bordism invariants by enlarging the cobordant structures of manifolds from
$Omega^5_O(K(mathbbZ/2,2))$ to $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$),
the (co)bordism invariant
$
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$ becomes 0 in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
Namely the effective manifold generators (respect of to $
u_2 Sq^1 u_2+Sq^2 Sq^1 u_2
$) in $Omega^5_O(K(mathbbZ/2,2))$ becomes trivial or vanished in $Omega^5_frac(Pin^pm times SU(2))mathbbZ/2$?
How to prove this?
- Here $frac(Pin^pm times SU(2))mathbbZ/2$ is similar to the generalization of Pin$^c$ structure where $U(1)$ is now replaced by $SU(2)$.
at.algebraic-topology gt.geometric-topology smooth-manifolds characteristic-classes cobordism
at.algebraic-topology gt.geometric-topology smooth-manifolds characteristic-classes cobordism
asked 2 hours ago
annie heart
8912
asked 2 hours ago
annie heart
8912
asked 2 hours ago
annie heart
8912
asked 2 hours ago
annie heart
8912
asked 2 hours ago
annie heart
8912
8912
add a comment |Â
add a comment |Â
1 Answer
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$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $Pin^h+$ and $Pin^h-$, and it's also necessary
to specify the maps $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;mathbb Z/2)$ determines a natural map to $K(mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $Omega_5^Pin^h+$. This group is isomorphic to $mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. It's known that $H^*(W;mathbb Z/2) =
mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $mathrmSO_3$-bundle is $Q$, and because $langle
w_2(Q)w_3(W), [W]rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$Omega_5^Pin^h+$.
Using the Wu formula, $Sq^1z_2 = z_3$ and $Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)Sq^1w_2(Q) + Sq^2Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)
An analogous argument should be possible in the pin$h-$ case, but since
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
answered 55 mins ago
Arun Debray
2,71911337
thanks +1, you are so professional
– annie heart
2 mins ago
add a comment |Â
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $Pin^h+$ and $Pin^h-$, and it's also necessary
to specify the maps $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;mathbb Z/2)$ determines a natural map to $K(mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $Omega_5^Pin^h+$. This group is isomorphic to $mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. It's known that $H^*(W;mathbb Z/2) =
mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $mathrmSO_3$-bundle is $Q$, and because $langle
w_2(Q)w_3(W), [W]rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$Omega_5^Pin^h+$.
Using the Wu formula, $Sq^1z_2 = z_3$ and $Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)Sq^1w_2(Q) + Sq^2Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)
An analogous argument should be possible in the pin$h-$ case, but since
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
answered 55 mins ago
Arun Debray
2,71911337
thanks +1, you are so professional
– annie heart
2 mins ago
add a comment |Â
up vote
4
down vote
$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $Pin^h+$ and $Pin^h-$, and it's also necessary
to specify the maps $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;mathbb Z/2)$ determines a natural map to $K(mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $Omega_5^Pin^h+$. This group is isomorphic to $mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. It's known that $H^*(W;mathbb Z/2) =
mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $mathrmSO_3$-bundle is $Q$, and because $langle
w_2(Q)w_3(W), [W]rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$Omega_5^Pin^h+$.
Using the Wu formula, $Sq^1z_2 = z_3$ and $Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)Sq^1w_2(Q) + Sq^2Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)
An analogous argument should be possible in the pin$h-$ case, but since
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
answered 55 mins ago
Arun Debray
2,71911337
thanks +1, you are so professional
– annie heart
2 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $Pin^h+$ and $Pin^h-$, and it's also necessary
to specify the maps $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;mathbb Z/2)$ determines a natural map to $K(mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $Omega_5^Pin^h+$. This group is isomorphic to $mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. It's known that $H^*(W;mathbb Z/2) =
mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $mathrmSO_3$-bundle is $Q$, and because $langle
w_2(Q)w_3(W), [W]rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$Omega_5^Pin^h+$.
Using the Wu formula, $Sq^1z_2 = z_3$ and $Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)Sq^1w_2(Q) + Sq^2Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)
An analogous argument should be possible in the pin$h-$ case, but since
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
answered 55 mins ago
Arun Debray
2,71911337
$newcommandPinmathrmPinnewcommandSqmathrmSq$
To simplify notation, I'll denote $(Pin_n^pmtimesmathrmSU_2)/(mathbb Z/2)$ by $Pin_n^hpm$, since these are
analogues of the groups $mathrmSpin_n^h = (mathrmSpin_ntimesmathrmSU_2)/(mathbb Z/2)$. Analogously to how a
spin$c$-structure on a manifold $M$ determines a line bundle $L$ with $w_2(L) = w_2(M)$, a
pin$hpm$-structure on $M$ determines a principal $mathrmSO_3$-bundle $Q$ with $w_2(Q) = w_2(M)$ (for
pin$h+$) or $w_2(Q) = w_2(M) + w_1^2(M)$ (for pin$h-$).
This question is actually several questions in one: the argument differs for $Pin^h+$ and $Pin^h-$, and it's also necessary
to specify the maps $Omega_5^Pin^hpmtoOmega_5^O(K(mathbb Z/2, 2))$. Any characteristic class of $M$ or
$Q$ in $H^2(M;mathbb Z/2)$ determines a natural map to $K(mathbb Z/2,2)$, and hence a map between these bordism
groups. There are several examples, such as $w_2(Q)$ and $w_2(Q) + w_1^2(M)$, and the answer could in principle differ depending
on which class one chooses.
As such, I have a partial answer: in the case of pin$h+$ 5-manifolds where the map to $K(mathbb Z/2,2)$
is $w_2(Q)$ or $w_2(Q) + w_1^2(M)$, the answer is yes, the invariant vanishes.
Since $u_2mathrmSq^1 u_2 + mathrmSq^2mathrmSq^1u_2$ is a cobordism invariant, it suffices to check this on
a generator of $Omega_5^Pin^h+$. This group is isomorphic to $mathbb Z/2$ (this is due to
Freed-Hopkins, Theorem 9.89), so we need only to find a single nonbounding
pin$h+$ 5-manifold and check there.
This generator is the Wu manifold $W := mathrmSU_3/mathrmSO_3$, with the principal $mathrmSO_3$-bundle $Q$
given by the quotient map $mathrmSU_3twoheadrightarrow W$. It's known that $H^*(W;mathbb Z/2) =
mathbb Z/2[z_2,z_3]/(z_2^2, z_3^2)$ with $|z_i| = i$; its nonzero Stiefel-Whitney classes are $w_2(W) = z_2$ and $w_3(W)
= z_3$, and that $w_2(Q) = z_2$ (this is discussed, for example, at the end of section 6 of Xuan Chen's
thesis). In particular, $w_2(W) = w_2(Q)$, so $W$
admits a pin$h+$-structure whose corresponding $mathrmSO_3$-bundle is $Q$, and because $langle
w_2(Q)w_3(W), [W]rangle = 1$, it doesn't bound as a pin$h+$ manifold. Therefore it generates
$Omega_5^Pin^h+$.
Using the Wu formula, $Sq^1z_2 = z_3$ and $Sq^2z_3 = z_2z_3$, so
$$ w_2(Q)Sq^1w_2(Q) + Sq^2Sq^1w_2(Q) = 2z_2z_3 = 0,$$
so we conclude that this invariant vanishes for all pin$h+$ 5-manifolds. (Since $w_1(W)^2 = 0$, we can
also conclude this for the map to $K(mathbb Z/2,2)$ given by $w_2(Q) + w_1(M)^2$.)
An analogous argument should be possible in the pin$h-$ case, but since
$Omega_5^Pin^h-congmathbb Z/2oplusmathbb Z/2$ (this is again Freed-Hopkins, Theorem 9.89), one would have
to find two linearly independent generators and check on them. $W$ also admits a pin$h-$-structure with
corresponding bundle $Q$, but I wasn't able to figure out what the other generator is.
answered 55 mins ago
Arun Debray
2,71911337
edited 41 mins ago
answered 55 mins ago
Arun Debray
2,71911337
answered 55 mins ago
Arun Debray
2,71911337
answered 55 mins ago
Arun Debray
2,71911337
2,71911337
thanks +1, you are so professional
– annie heart
2 mins ago
add a comment |Â
thanks +1, you are so professional
– annie heart
2 mins ago
thanks +1, you are so professional
– annie heart
2 mins ago
thanks +1, you are so professional
– annie heart
2 mins ago
add a comment |Â
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