Math 2.0

Algebraic topology Definition

Steenrod square

Builds on Cohomology operationSingular cohomologyCup product

Definition

Write H∗(X,A)H^*(X, A) for singular cohomology of a pair of spaces with coefficients in ℤ/2ℤ\Z/2\Z. The Steenrod squares are the cohomology operations Sq⁡i ⁣:Hn(X,A)→Hn+i(X,A),i≥0, n≥0,\operatorname{Sq}^i \colon H^n(X, A) \to H^{n+i}(X, A), \quad i \ge 0, \ n \ge 0, defined for all pairs (X,A)(X, A), that satisfy:

  1. Naturality: f∗Sq⁡i=Sq⁡if∗f^* \operatorname{Sq}^i = \operatorname{Sq}^i f^* for every map of pairs ff.
  2. Additivity: Sq⁡i(x+y)=Sq⁡ix+Sq⁡iy\operatorname{Sq}^i(x + y) = \operatorname{Sq}^i x + \operatorname{Sq}^i y.
  3. Cartan formula: Sq⁡k(x⌣y)=∑i+j=kSq⁡ix⌣Sq⁡jy\operatorname{Sq}^k(x \smile y) = \sum_{i + j = k} \operatorname{Sq}^i x \smile \operatorname{Sq}^j y, with ⌣\smile the cup product.
  4. Stability: Sq⁡i\operatorname{Sq}^i commutes with the connecting homomorphism δ ⁣:Hn(A)→Hn+1(X,A)\delta \colon H^n(A) \to H^{n+1}(X, A).
  5. Sq⁡0\operatorname{Sq}^0 is the identity.
  6. If xx has degree nn, then Sq⁡nx=x⌣x\operatorname{Sq}^n x = x \smile x, and Sq⁡ix=0\operatorname{Sq}^i x = 0 for i>ni > n.

There is exactly one family of operations with these properties.

Remarks

The total square is Sq⁡=Sq⁡0+Sq⁡1+Sq⁡2+⋯\operatorname{Sq} = \operatorname{Sq}^0 + \operatorname{Sq}^1 + \operatorname{Sq}^2 + \cdots; on each class only finitely many terms are nonzero. The square Sq⁡1\operatorname{Sq}^1 is the Bockstein homomorphism of the sequence 0→ℤ/2ℤ→ℤ/4ℤ→ℤ/2ℤ→00 \to \Z/2\Z \to \Z/4\Z \to \Z/2\Z \to 0.

Example: in H∗(ℝP∞)=(ℤ/2ℤ)[a]H^*(\R P^\infty) = (\Z/2\Z)[a] with deg⁡a=1\deg a = 1, one has Sq⁡a=a+a2\operatorname{Sq} a = a + a^2 by clauses 5 and 6. By the Cartan formula, Sq⁡(an)=(a+a2)n\operatorname{Sq}(a^n) = (a + a^2)^n, so Sq⁡i(an)=(ni)an+i\operatorname{Sq}^i(a^n) = \binom{n}{i} a^{n+i}. For instance Sq⁡1(a2)=2a3=0\operatorname{Sq}^1(a^2) = 2a^3 = 0 and Sq⁡1(a3)=a4\operatorname{Sq}^1(a^3) = a^4.

Non-example: the family with Sq⁡0=id⁡\operatorname{Sq}^0 = \id and Sq⁡i=0\operatorname{Sq}^i = 0 for all i≥1i \ge 1 satisfies clauses 1 to 5 but fails clause 6, since Sq⁡1a\operatorname{Sq}^1 a would be 00 instead of a2a^2 in H2(ℝP∞)H^2(\R P^\infty).

Reading path

To understand this page from the ground up, read these 4 pages in order. The foundations come first.

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Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
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  1. Cohomology operation concept
  2. Cup product concept
  3. Singular cohomology concept
  4. Steenrod square definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Steenrod_algebra
About this page
TagH0T4RFG2AD
URIm2://H0T4RFG2AD
Kinddefinition
WritingW0 stub
MSC55S10, 55S05
DefinesSteenrod square
UsesCohomology operation, Singular cohomology, Cup product
Changed2 November 2026

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