Skip to main content
MathOverflow

Return to Question

typo x instead of A; edited body
Source Link
Andi Bauer
Andi Bauer
  • 3k
  • 8
  • 16

There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula commutes with the derivative in the sense that $$d(A\cup B)=dA\cup B+(-1)^j A\cup dB$$ holds exactly on the level of cochains. This property ensures that if we apply the cup product to two cocycles, the result will again be a cocycle.

Another operation which can be defined on the cochain level is the $x$th Steenrod square of a $i$-cocycle $A$, $$Sq^x(A)=A\cup_{i-x} A$$ via higher order cup products, see, e.g., appendix B in this paper. Again, the formula applied to cocycles yields cocycles. However, the map does not commute with the derivative on the level of cochains. Specifically, using the first formula in the mentioned appendix, we get $$d Sq^x(A) = Sq^x(dA)+d(x\cup_{i-x+1} dx)$$$$d Sq^x(A) = Sq^x(dA)+d(A\cup_{i-x+1} dA)$$ up to some minus signs I'm potentially missing (I'm mostly thinking about $\mathbb{Z}_2$-valued cohomology anyways).

Question: Is there a local combinatorial formula for the Steenrod squares for which commutes with the derivative on the level of cochains such, such as the one for the cup product? Or is there some obstruction to this?

There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula commutes with the derivative in the sense that $$d(A\cup B)=dA\cup B+(-1)^j A\cup dB$$ holds exactly on the level of cochains. This property ensures that if we apply the cup product to two cocycles, the result will again be a cocycle.

Another operation which can be defined on the cochain level is the $x$th Steenrod square of a $i$-cocycle $A$, $$Sq^x(A)=A\cup_{i-x} A$$ via higher order cup products, see, e.g., appendix B in this paper. Again, the formula applied to cocycles yields cocycles. However, the map does not commute with the derivative on the level of cochains. Specifically, using the first formula in the mentioned appendix, we get $$d Sq^x(A) = Sq^x(dA)+d(x\cup_{i-x+1} dx)$$ up to some minus signs I'm potentially missing (I'm mostly thinking about $\mathbb{Z}_2$-valued cohomology anyways).

Question: Is there a local combinatorial formula for the Steenrod squares for which commutes with the derivative on the level of cochains such, as the one for the cup product? Or is there some obstruction to this?

There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula commutes with the derivative in the sense that $$d(A\cup B)=dA\cup B+(-1)^j A\cup dB$$ holds exactly on the level of cochains. This property ensures that if we apply the cup product to two cocycles, the result will again be a cocycle.

Another operation which can be defined on the cochain level is the $x$th Steenrod square of a $i$-cocycle $A$, $$Sq^x(A)=A\cup_{i-x} A$$ via higher order cup products, see, e.g., appendix B in this paper. Again, the formula applied to cocycles yields cocycles. However, the map does not commute with the derivative on the level of cochains. Specifically, using the first formula in the mentioned appendix, we get $$d Sq^x(A) = Sq^x(dA)+d(A\cup_{i-x+1} dA)$$ up to some minus signs I'm potentially missing (I'm mostly thinking about $\mathbb{Z}_2$-valued cohomology anyways).

Question: Is there a local combinatorial formula for the Steenrod squares which commutes with the derivative on the level of cochains, such as the one for the cup product? Or is there some obstruction to this?

Source Link
Andi Bauer
Andi Bauer
  • 3k
  • 8
  • 16

Is there a local simplicial formula for the Steenrod squares which commutes with the derivative on cochain level?

There is a well-known formula for the cup product of an $i$-cochain $A$ and $j$-cochain $B$ in simplicial homology given by $$(A\cup B)(0\ldots i+j) = A(0\ldots i) B(0\ldots j)\;.$$ This formula commutes with the derivative in the sense that $$d(A\cup B)=dA\cup B+(-1)^j A\cup dB$$ holds exactly on the level of cochains. This property ensures that if we apply the cup product to two cocycles, the result will again be a cocycle.

Another operation which can be defined on the cochain level is the $x$th Steenrod square of a $i$-cocycle $A$, $$Sq^x(A)=A\cup_{i-x} A$$ via higher order cup products, see, e.g., appendix B in this paper. Again, the formula applied to cocycles yields cocycles. However, the map does not commute with the derivative on the level of cochains. Specifically, using the first formula in the mentioned appendix, we get $$d Sq^x(A) = Sq^x(dA)+d(x\cup_{i-x+1} dx)$$ up to some minus signs I'm potentially missing (I'm mostly thinking about $\mathbb{Z}_2$-valued cohomology anyways).

Question: Is there a local combinatorial formula for the Steenrod squares for which commutes with the derivative on the level of cochains such, as the one for the cup product? Or is there some obstruction to this?