Eq. (11) — Tensor-product distance upper bound

For every pair of finite-length based binary chain complexes $A$ and $B$ over $\mathbb F_2=\mathbb Z/2\mathbb Z$, and every natural number $j$, the declaration asserts a single non-strict inequality in $\operatorname{WithTop}\mathbb N=\mathbb N\cup\{\infty\}$:

These are the theorem's only quantified arguments and it has no additional hypotheses. Concretely, a complex $X\in\{A,B\}$ consists of a function $n\mapsto\dim_X(n)\in\mathbb N$, binary word spaces $X_n=\bigl(\operatorname{Fin}(\dim_X(n))\to\mathbb F_2\bigr)$, and boundary maps indexed by their domains: $\partial^X_0:X_0\to(\mathrm{Empty}\to\mathbb F_2)$, necessarily the zero map into the one-element zero vector space, and $\partial^X_{n+1}:X_{n+1}\to X_n$ for $n\ge0$. They satisfy $\partial^X_n\circ\partial^X_{n+1}=0$ for every $n$. The structure also supplies a natural number $L_X$ such that $\dim_X(n)=0$ whenever $L_X<n$; it does not require $L_X$ to be minimal or require any dimension at or below $L_X$ to be nonzero. The component distance $d_n(X)$ is the infimum, in $\mathbb N\cup\{\infty\}$, of the exact set of Hamming weights

Thus at $n=0$, because $\partial^X_0$ maps to the zero space, every element of $X_0$ satisfies the kernel condition and the candidates are precisely the words outside $\operatorname{range}(\partial^X_1)$. If the displayed candidate set is empty, its infimum is $\infty$; otherwise, since $X_n$ is finite, the value is the least occurring finite Hamming weight. It cannot be $0$, because weight zero forces $x=0$, while zero belongs to the range of every linear map. The degree-$n$ tensor word space is indexed by the dependent disjoint union

where $(p,q)$ ranges over pairs whose two entries lie in $\operatorname{Fin}(n+1)$ and whose natural-number values satisfy exactly $p+q=n$. Thus only nonnegative complementary degrees occur, each of the $n+1$ pairs $(0,n),(1,n-1),\ldots,(n,0)$ is represented, and different degree-pair tags give distinct tensor coordinates; a fiber is empty when either corresponding dimension is zero, and the whole coordinate type can be empty. Write $T_n=I_n(A,B)\to\mathbb F_2$. The tensor boundary indexed by its domain is $\Delta_0:T_0\to(\mathrm{Empty}\to\mathbb F_2)$, defined to be zero, and, for $n\ge0$, $\Delta_{n+1}:T_{n+1}\to T_n$. At a target coordinate tagged by $(p,q)$ with $p+q=n$, with basis coordinates $a\in\operatorname{Fin}(\dim_A(p))$ and $b\in\operatorname{Fin}(\dim_B(q))$, its value on $e\in T_{n+1}$ is exactly

The first source coordinate is tagged by the degree pair $(p+1,q)$, the second by $(p,q+1)$, and the two contributions are added in $\mathbb F_2$, with no sign factor. The imported declarations prove $\Delta_n\circ\Delta_{n+1}=0$ for every $n$, including the $n=0$ endpoint through the zero map $\Delta_0$. The tensor-product distance on the left is therefore the infimum

At $j=0$, $I_0(A,B)$ has only the degree pair $(0,0)$, $\Delta_0=0$, every tensor word is a cycle, and the elements considered are exactly those outside $\operatorname{range}(\Delta_1)$. If this candidate set is empty, or if $T_j$ is the zero vector space and hence contains only the zero word, then $D_j(A\otimes B)=\infty$; otherwise it is a positive finite minimum. The right-hand side is defined as the infimum of

Its index is exactly $i=0,1,\ldots,j$. Because $i<j+1$ implies $i\le j$, the natural-number subtraction $j-i$ never underflows, and because $\operatorname{Fin}(j+1)$ is finite and nonempty, this infimum is an actual minimum of $j+1$ terms and is never $\infty$ merely because its indexing set is empty. At $j=0$, it is the single product $d_0(A)d_0(B)$. Since component distances are never zero, such a product is finite exactly when both factors are finite and is $\infty$ when at least one factor is $\infty$; consequently the right-hand side is $\infty$ exactly when, for every $i=0,\ldots,j$, at least one of $d_i(A)$ and $d_{j-i}(B)$ is $\infty$. In that case the asserted upper bound holds automatically because every value is at most $\infty$. If the right-hand side is finite, the assertion forces the tensor-product distance to be finite and no larger than that finite minimum; if the tensor-product distance is $\infty$, the inequality can hold only when the right-hand side is also $\infty$. The quantification includes complexes with zero-dimensional groups, length zero, overestimated lengths, trivial homology in any or all degrees, and values of $j$ larger than either length or their sum. In particular, if $j>L_A+L_B$, every complementary pair has at least one degree above its complex's length, all corresponding products are $\infty$, the tensor coordinate type is empty, and the displayed inequality specializes to $\infty\le\infty$. No negative degrees occur, no nontrivial-homology or finiteness-of-distance premise is imposed, and the declaration states only this upper inequality, not equality, a converse, or a lower bound.