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Adem double-power comparison

Statement

Assume AC. For the finite-cellular mod-two cyclic squares, let $x \in H_{c e l l}^{q} (K; F_{2})$ . With all operations and binomial coefficients outside their ordinary nonnegative ranges declared zero, the double-power coefficients satisfy

$D_{2 q - a, 2 q - ℓ} (x) = \sum_{i = 0}^{q} (\binom{q - i}{q + i - ℓ}) S q_{c y c}^{a + ℓ - q - i} S q_{c y c}^{i} (x),$

and row--column transposition gives the identity

$\sum_{i = 0}^{q} (\binom{q - i}{q + i - ℓ}) S q_{c y c}^{a + ℓ - q - i} S q_{c y c}^{i} (x) = \sum_{r = 0}^{q} (\binom{q - r}{q + r - a}) S q_{c y c}^{a + ℓ - q - r} S q_{c y c}^{r} (x) .$

Consequently, if $a, b$ are positive integers with $0 < a < 2 b$ , if $2^{s} > a$ , if $q = 2^{s} - 1 + b$ , and if $x$ has degree $q$ , then

$S q_{c y c}^{a} S q_{c y c}^{b} (x) = \sum_{r = 0}^{⌊ a / 2 ⌋} (\binom{b - r - 1}{a - 2 r}) S q_{c y c}^{a + b - r} S q_{c y c}^{r} (x) .$

This is the finite-cellular high-degree coefficient calculation. Descent to all degrees and identification with the singular cup- $i$ squares are not asserted here.

Facts & Assumptions

Given: AC, integers $a, ℓ$ , a nonnegative degree $q$ , a finite oriented regular cell complex $K$ , and a class $x \in H_{c e l l}^{q} (K; F_{2})$ .

[F1]

The finite-cellular cyclic squares obey the external Cartan formula (Finite-cellular cyclic squares, Cartan formula, and cyclic-basis action).

[F4]

They act on the cyclic basis by $S q_{c y c}^{j} (w_{r}) = (\binom{r}{j}) w_{r + j}$ (Finite-cellular cyclic squares, Cartan formula, and cyclic-basis action).

[F2]

At $p = 2$ , the iterated double-power coefficients are symmetric: $D_{j, k} (x) = D_{k, j} (x)$ (Wreath double-power comparison and coefficient transposition).

[F5]

The full cyclic-power expansion has additive coefficients $D_{r}$ , and the external class is natural; uniqueness of its cyclic-basis coordinates therefore makes every $D_{r}$ natural (Equivariant p-fold external power and diagonal decomposition).

[F3]

AC supplies a choice function for every set-indexed family of nonempty sets (The Axiom of Choice).

Proof

technique · expand one iterated total square in the two cyclic bases, use row--column symmetry, and perform the binary-digit calculation after a high-degree substitution

1.1givenF1F4F5

Expand the iterated class in both cyclic coordinates. First justify the truncation of the inner expansion. For $r > q$ , the class $D_{r} (x)$ has degree $2 q - r < q$ . Restriction to the cellular $q$ -skeleton is an isomorphism in that degree, and naturality in [F5] identifies the restriction with $D_{r}$ of the restricted class. Collapse the $(q - 1)$ -skeleton of the $q$ -skeleton. The restricted $x$ is the pullback of a class on the resulting wedge of $q$ -spheres, so additivity and naturality in [F5] reduce the calculation to one sphere. For $q < r < 2 q$ the target $H^{2 q - r} (S^{q}; F_{2})$ is zero. For $r = 2 q$ and $q > 0$ , restrict to one point in each path component of the finite complex. Restriction to these points is injective in degree zero, while the positive-degree input restricts to zero and additivity gives $D_{2 q} (0) = 0$ on every component. When $q = 0$ , every $r > q$ is already outside the range $0 \leq r \leq 2 q$ ; for $q > 0$ , every $r > 2 q$ is outside the same range. Hence $D_{r} (x) = 0$ for all $r > q$ .

The first cyclic diagonal of a degree- $q$ class is

$\sum_{r = 0}^{2 q} w_{r} \times D_{r} (x) = \sum_{i = 0}^{q} w_{q - i} \times S q_{c y c}^{i} (x) .$

For the finitely many nonnegative basis degrees used below, choose one of the finite regular projective models $Q_{N}$ supplied by [F4], with $N$ larger than their maximum. The $w_{q - i}$ factor means the corresponding restricted class on $Q_{N}$ . Thus the following external Cartan computation takes place on the finite regular complex $Q_{N} \times K$ ; no one-cell projective skeleton is used as an input. Naturality in [F4] makes the result independent of increasing $N$ .

Apply the outer cyclic square. Its coordinate $w_{2 q - a}$ is obtained by applying $S q_{c y c}^{a}$ to every displayed product. By Cartan from [F1] and the basis action from [F4],

$S q_{c y c}^{a} (w_{q - i} \times S q_{c y c}^{i} x) = \sum_{j} (\binom{q - i}{j}) w_{q - i + j} \times S q_{c y c}^{a - j} S q_{c y c}^{i} x .$

The second cyclic coordinate equals $w_{2 q - ℓ}$ exactly when $j = q + i - ℓ$ . Substitution gives the first displayed formula in the statement. The outside-range conventions make this a finite equality for arbitrary integer $a, ℓ$ .

2.1F2step 1.1

Apply row--column transposition. At $p = 2$ , both signs in the transposition formula of [F2] equal one in $F_{2}$ . Thus

$D_{2 q - a, 2 q - ℓ} (x) = D_{2 q - ℓ, 2 q - a} (x) .$

Apply step 1.1 to the right side with $a$ and $ℓ$ exchanged, and rename its inner index $r$ . The outer exponent remains $a + ℓ - q - r$ , while the basis coefficient becomes $(\binom{q - r}{q + r - a})$ . This is precisely the asserted two-sum identity.

3.1step 2.1

Isolate the left-hand summand after the high-degree substitution. Assume now $0 < a < 2 b$ , choose $s$ with $2^{s} > a$ , put $q = 2^{s} - 1 + b$ , and put $ℓ = q + b$ . On the left of step 2.1 the binomial coefficient is

$(\binom{q - i}{q + i - ℓ}) = (\binom{2^{s} - 1 + b - i}{i - b}) .$

We first prove the binary coefficient criterion used twice below. If $n = \sum_{ν} n_{ν} 2^{ν}$ , then in $F_{2} [z]$ the Frobenius identity gives

$(1 + z)^{n} = \prod_{ν : n_{ν} = 1} (1 + z)^{2^{ν}} = \prod_{ν : n_{ν} = 1} (1 + z^{2^{ν}}) .$

Thus $(\binom{n}{d})$ is odd exactly when every nonzero binary digit of $d$ is also a nonzero digit of $n$ .

It is zero for $i < b$ by the negative-lower-index convention. If $i = b + h$ with $h > 0$ , then it is $(\binom{2^{s} - 1 - h}{h})$ . Here $0 < h < 2^{s}$ . Let $2^{v}$ be the lowest nonzero binary digit of $h$ . In the first $s$ binary digits, $2^{s} - 1 - h$ is the digitwise complement of $h$ , so its $v$ th digit is zero while the $v$ th digit of $h$ is one. The proved binary criterion therefore makes the coefficient zero.

For $i = b$ , the coefficient is $(\binom{2^{s} - 1}{0}) = 1$ , and the outer exponent is $a + ℓ - q - b = a$ . Hence the entire left side of step 2.1 is $S q_{c y c}^{a} S q_{c y c}^{b} (x)$ .

4.1step 2.1step 3.1

Reduce every right-hand coefficient. For the right side, complementing the lower index inside the upper one gives

$(\binom{q - r}{q + r - a}) = (\binom{q - r}{a - 2 r}) .$

A nonzero term must have $0 \leq a - 2 r$ , so $0 \leq r \leq ⌊ a / 2 ⌋$ . Since $a < 2 b$ , every such $r$ satisfies $r < b$ . Put $c = b - r - 1 \geq 0$ . Then $q - r = 2^{s} + c$ , while $0 \leq a - 2 r < 2^{s}$ . The binary criterion from step 3.1 sees only the lowest $s$ digits of the upper number, and adding $2^{s}$ does not change those digits. Therefore

$(\binom{q - r}{a - 2 r}) \equiv (\binom{b - r - 1}{a - 2 r}) (m o d 2) .$

The operation exponent on this summand is $a + ℓ - q - r = a + b - r$ . Substitution into the right side of step 2.1 yields exactly the finite sum in the statement.

5.1F3step 1.1step 2.1step 3.1step 4.1

Check ranges, models, and choice. If $K$ is empty, its cellular complex is zero, or $x = 0$ , both sides are zero. For a point, the required positive degree $q = 2^{s} - 1 + b$ has zero cohomology, so the identity is again zero. The strict hypotheses $0 < a < 2 b$ and $2^{s} > a$ are used respectively to obtain $r < b$ and to keep the lower binary index below the added $2^{s}$ digit. The endpoints $r = 0$ and $r = ⌊ a / 2 ⌋$ are retained, including the case of a zero lower binomial index. Every negative or oversized binomial and every outside-range square was declared zero before the calculation.

This is a finite regular cellular argument, so singular degeneracies are item-specifically inapplicable. Step 1.1 explicitly chooses a sufficiently large regular $Q_{N}$ for its finite set of basis degrees. AC from [F3] is propagated exactly through the cyclic-square and double-power suppliers used in steps 1.1 and 2.1, including the cyclic supplier's ordinary cup comparison and field duality. Choosing $s$ can be done by taking the least integer with $2^{s} > a$ , and every sum and binary-digit test is finite. No Adem theorem, degree-descent result, or singular cup- $i$ comparison is used. ∎

Depends on

Dependency tree · 2 levels

Sources

N. E. Steenrod and D. B. A. Epstein, Cohomology Operations