Admissible square actions have distinct leading monomials

The published projective-space lemma gives the polynomial ring of ${\mathbb{R}\mathbb{P}}^{\mathrm{\infty }}$ and skeletal restriction isomorphisms through degree $L$. Finite projective space has one mod-two cellular generator in every degree $0,\dots ,L$, zero cellular differential, and no cells above $L$; the cellular comparison gives degreewise finite-free homology, and field evaluation duality makes cohomology vanish above $L$. Thus restriction computes its ring as ${\mathbb{F}}_2[x]\mathrm{/}(x^{L+1})$, and iterative application of the published finite-free Künneth theorem gives the displayed product ring. Its distinct surviving monomials are linearly independent.

For a degree-one generator, normalization gives $Sq(x)=x+x^2$. Cartan on a product of $m$ copies gives, by the finite binomial expansion, $S{q}^a(x^m)=\left(\genfrac{}{}{0px}{}{m}{a}\right)x^{m+a}\mathrm{.}$ In particular, for $m=2^h$ the polynomial identity $(1+t{)}^{2^h}=1+t^{2^h}$ over ${\mathbb{F}}_2$ gives $S{q}^0(x^{2^h})=x^{2^h}$, $S{q}^{2^h}(x^{2^h})=x^{2^{h+1}}$, and $S{q}^a(x^{2^h})=0$ for all other nonnegative $a$.

Consequently every term in an iterated action on $P$ arises by a finite schedule of doublings. At the step labelled $i_j$, a set of variables is doubled whose current exponents sum to $i_j$. Different schedules may give the same final monomial, so their parity must be considered; their supports are not disjoint.

We prove the largest-monomial assertion by induction on $k$. For $k=1$, exactly $i_1$ of the variables are doubled. Since $r\ge e(I)=i_1$, the largest term doubles the first $i_1$ variables, uniquely. It has coefficient one.

For $k>1$, a variable reaches exponent $2^k$ only if it is doubled at every one of the $k$ steps. The first step to act is $i_k$, when all exponents equal one; hence at most $i_k=d_k$ variables can reach $2^k$. Achieving $i_k$ such variables exhausts that first-step budget, and their later costs are forced to be $2^{k-j}{i}_k$ at step $j$. Subtract those costs from the earlier budgets and discard the now exhausted final step. The remaining sequence is $I^{\mathrm{\prime }}=(i_1-2^{k-1}{i}_k,i_2-2^{k-2}{i}_k,\dots ,i_{k-1}-2i_k)\mathrm{.}$ All its entries are nonnegative by admissibility, and its successive admissibility differences are $d_1,\dots ,d_{k-1}$. Remove any terminal zeros. Its excess is $e(I)-i_k$, and $r-i_k\ge e(I^{\mathrm{\prime }})$. Thus the induction hypothesis constructs the largest remaining term on the remaining variables. Taking the first $i_k$ variables for the full doubling chains constructs a nonzero term with the stated exponent vector.

A monomial with fewer than $i_k$ variables of exponent $2^k$ is lexicographically smaller than that term after placing its exponents in decreasing order. Symmetry of $P$ and of its square action ensures that arranging exponents in decreasing order maximizes lexicographic order. If precisely $i_k$ variables attain exponent $2^k$, their forced all-step chains consume exactly the costs above, so comparison of the remaining exponents is the induction problem for $I^{\mathrm{\prime }}$. Therefore none is larger. For the specified leading monomial, its first $i_k$ variables have uniquely forced chains and the residual leading term has coefficient one by induction; hence the full coefficient is one, not an unproved parity assertion. No displayed term is truncated: a variable's exponent cannot exceed $1+\mathrm{\mid }I\mathrm{\mid }$, because every doubling cost contributes to the total increase $\mathrm{\mid }I\mathrm{\mid }$. Thus $L\ge \mathrm{\mid }I\mathrm{\mid }+1$ suffices. Finally the multiplicities $d_j$ recover the sequence by the backward recursion $i_j=d_j+2i_{j+1}$. Since a normalized sequence has $d_k=i_k>0$, the largest exponent also recovers its length. Distinct admissible sequences therefore have distinct leading monomials. Empty sequences and $r=0$ give the identity action on the unit.

On $P=xyz$, $S{q}^3(P)=x^2{y}^2{z}^2,S{q}^{2}S{q}^1(P)=x^{4}yz+x{y}^{4}z+xy{z}^4+x^2{y}^2{z}^2\mathrm{.}$ Both composites are admissible and their supports overlap. Their leading monomials differ, which is exactly the property needed for independence.