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Stiefel-Whitney and Wu classes of $d$-manifold
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Is it true that Stiefel-Whitney $w_i$ and Wu classes $u_i$ of $d$-manifold, we always have the following:
$$
u_d-1=0, tag1
$$
$$
u_d=0, tag2
$$
$$
Sq^1(u_d-1)=0. tag3
$$
in any dimensions $d$?
Why are the above conditions true?
Do we have more conditions than the above in some dimensions $d$?
algebraic-topology differential-topology characteristic-classes
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
asked Jul 19 at 22:16
wonderich
1,65821226
add a comment |Â
up vote
0
down vote
favorite
Is it true that Stiefel-Whitney $w_i$ and Wu classes $u_i$ of $d$-manifold, we always have the following:
$$
u_d-1=0, tag1
$$
$$
u_d=0, tag2
$$
$$
Sq^1(u_d-1)=0. tag3
$$
in any dimensions $d$?
Why are the above conditions true?
Do we have more conditions than the above in some dimensions $d$?
algebraic-topology differential-topology characteristic-classes
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
asked Jul 19 at 22:16
wonderich
1,65821226
1
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is it true that Stiefel-Whitney $w_i$ and Wu classes $u_i$ of $d$-manifold, we always have the following:
$$
u_d-1=0, tag1
$$
$$
u_d=0, tag2
$$
$$
Sq^1(u_d-1)=0. tag3
$$
in any dimensions $d$?
Why are the above conditions true?
Do we have more conditions than the above in some dimensions $d$?
algebraic-topology differential-topology characteristic-classes
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
asked Jul 19 at 22:16
wonderich
1,65821226
Is it true that Stiefel-Whitney $w_i$ and Wu classes $u_i$ of $d$-manifold, we always have the following:
$$
u_d-1=0, tag1
$$
$$
u_d=0, tag2
$$
$$
Sq^1(u_d-1)=0. tag3
$$
in any dimensions $d$?
Why are the above conditions true?
Do we have more conditions than the above in some dimensions $d$?
algebraic-topology differential-topology characteristic-classes
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
asked Jul 19 at 22:16
wonderich
1,65821226
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
edited Jul 19 at 22:20
Ted Shifrin
59.5k44387
59.5k44387
asked Jul 19 at 22:16
wonderich
1,65821226
asked Jul 19 at 22:16
wonderich
1,65821226
asked Jul 19 at 22:16
wonderich
1,65821226
1,65821226
1
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50
add a comment |Â
1
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50
1
1
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $M$ be a closed $d$-dimensional manifold. Recall that the Wu classes $u_i$ satisfy $operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. One of the properties of Steenrod squares is that $operatornameSq^i(x) = 0$ for $i > deg x$. In particular, for $i > d - i$ (i.e. $i > fracd2$), we have $0 = operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. By Poincaré duality, we see that $u_i = 0$.
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $M$ be a closed $d$-dimensional manifold. Recall that the Wu classes $u_i$ satisfy $operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. One of the properties of Steenrod squares is that $operatornameSq^i(x) = 0$ for $i > deg x$. In particular, for $i > d - i$ (i.e. $i > fracd2$), we have $0 = operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. By Poincaré duality, we see that $u_i = 0$.
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
 |Â
show 1 more comment
up vote
2
down vote
accepted
Let $M$ be a closed $d$-dimensional manifold. Recall that the Wu classes $u_i$ satisfy $operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. One of the properties of Steenrod squares is that $operatornameSq^i(x) = 0$ for $i > deg x$. In particular, for $i > d - i$ (i.e. $i > fracd2$), we have $0 = operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. By Poincaré duality, we see that $u_i = 0$.
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $M$ be a closed $d$-dimensional manifold. Recall that the Wu classes $u_i$ satisfy $operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. One of the properties of Steenrod squares is that $operatornameSq^i(x) = 0$ for $i > deg x$. In particular, for $i > d - i$ (i.e. $i > fracd2$), we have $0 = operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. By Poincaré duality, we see that $u_i = 0$.
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
Let $M$ be a closed $d$-dimensional manifold. Recall that the Wu classes $u_i$ satisfy $operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. One of the properties of Steenrod squares is that $operatornameSq^i(x) = 0$ for $i > deg x$. In particular, for $i > d - i$ (i.e. $i > fracd2$), we have $0 = operatornameSq^i(x) = u_icup x$ for all $x in H^d-i(M; mathbbZ_2)$. By Poincaré duality, we see that $u_i = 0$.
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
answered Jul 20 at 13:16
Michael Albanese
61.2k1591289
61.2k1591289
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
 |Â
show 1 more comment
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
thanks +1, So the above criteria I listed are the all conditions for generic $d$-manifold --- are there more? (i.e. Without imposing addition structures on manifolds, such as spin, pin$^pm$, etc.)
– wonderich
Jul 20 at 14:27
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
First, the conditions you wrote do not hold for all $d$ (see Aleksandar Milivojevic's comment). Second, you only wrote that the last two Wu classes vanish, my answer shows that in general there are many more which vanish.
– Michael Albanese
Jul 20 at 18:31
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
Just to make sure, in 3d, as far as I understood before, I have $u_2=u_3=Sq^1(u_2)=0$, and in 4d, I have $u_3=u_4=Sq^1(u_3)=0$. But it looks that you wrote something different than what I thought?
– wonderich
Jul 21 at 0:10
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
For your notation, $Sq^i(x)$, I have $i=1$, but you wrote $i>d/2$. Thanks.
– wonderich
Jul 21 at 0:12
1
1
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
What you wrote for $d = 3$ and $4$ is correct. But for $d > 4$, what I wrote is stronger than what you wrote in your question. For example, when $d = 5$, not only do we have $u_4 = u_5 = 0$, and hence $operatornameSq^1(u_4) = operatornameSq^1(0) = 0$, we also have $u_3 = 0$.
– Michael Albanese
Jul 21 at 14:19
 |Â
show 1 more comment
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1
Note that in dimension 2 it is not necessarily true that $Sq^1(u_1) = 0$. For example, consider $mathbbRmathbbP^2$. Then $u_1 = w_1$ and $w_1^2 neq 0$.
– Aleksandar Milivojevic
Jul 20 at 14:39
that is a good point - how about general $d$? Do we have more conditions for generic manifolds?
– wonderich
Jul 20 at 15:26
For $d>2$ see Michael’s answer below.
– Aleksandar Milivojevic
Jul 20 at 15:50