Mixing
Steenrod squares on integer cohomology classes.
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I know that Steenrod squares are defined as maps $Sq^k colon H^n(X;mathbb Z_2) to H^n+k(X,mathbb Z_2)$. But I often read papers where the $Sq^2 colon H^n(X;mathbb Z) to H^n+2(X,mathbb Z_2)$. What does this mean? Does it mean that I have to precompose the mod $2$ homomorphism $H^n(X,mathbb Z) to H^n(X,mathbb Z_2)$?
algebraic-topology homotopy-theory
asked Jul 23 at 9:23
Wilhelm L.
30515
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up vote
4
down vote
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I know that Steenrod squares are defined as maps $Sq^k colon H^n(X;mathbb Z_2) to H^n+k(X,mathbb Z_2)$. But I often read papers where the $Sq^2 colon H^n(X;mathbb Z) to H^n+2(X,mathbb Z_2)$. What does this mean? Does it mean that I have to precompose the mod $2$ homomorphism $H^n(X,mathbb Z) to H^n(X,mathbb Z_2)$?
algebraic-topology homotopy-theory
asked Jul 23 at 9:23
Wilhelm L.
30515
4
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I know that Steenrod squares are defined as maps $Sq^k colon H^n(X;mathbb Z_2) to H^n+k(X,mathbb Z_2)$. But I often read papers where the $Sq^2 colon H^n(X;mathbb Z) to H^n+2(X,mathbb Z_2)$. What does this mean? Does it mean that I have to precompose the mod $2$ homomorphism $H^n(X,mathbb Z) to H^n(X,mathbb Z_2)$?
algebraic-topology homotopy-theory
asked Jul 23 at 9:23
Wilhelm L.
30515
I know that Steenrod squares are defined as maps $Sq^k colon H^n(X;mathbb Z_2) to H^n+k(X,mathbb Z_2)$. But I often read papers where the $Sq^2 colon H^n(X;mathbb Z) to H^n+2(X,mathbb Z_2)$. What does this mean? Does it mean that I have to precompose the mod $2$ homomorphism $H^n(X,mathbb Z) to H^n(X,mathbb Z_2)$?
algebraic-topology homotopy-theory
asked Jul 23 at 9:23
Wilhelm L.
30515
asked Jul 23 at 9:23
Wilhelm L.
30515
asked Jul 23 at 9:23
Wilhelm L.
30515
asked Jul 23 at 9:23
Wilhelm L.
30515
30515
4
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27
add a comment |Â
4
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27
4
4
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27
add a comment |Â
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4
It means reduce mod $2$ and then do the Steenrod square: $H^n(X;mathbbZ)xrightarrowred_2 H^n(X;mathbbZ_2)xrightarrowSq^2 H^n+2(X;mathbbZ_2)$.
– Tyrone
Jul 23 at 10:27