Tensor Coherence and Algebraic Descent — Examples

Since $u^{\mathrm{\prime }}=2u$ and $v^{\mathrm{\prime }}=\frac{1}{2}v$, bilinearity of the elementary tensor [F1] gives $u^{\mathrm{\prime }}\otimes v^{\mathrm{\prime }}=(2u)\otimes (\frac{1}{2}v)=2\cdot \frac{1}{2}(u\otimes v)=u\otimes v$ because $2\cdot \frac{1}{2}=1$ in $k$ by the characteristic hypothesis; and $(u,v)\mathrm{\ne }(u^{\mathrm{\prime }},v^{\mathrm{\prime }})$ because $u=(1,0)\mathrm{\ne }(2,0)=u^{\mathrm{\prime }}$; moreover $u\otimes v=e_1\otimes e_2\mathrm{\ne }0$ by [F5], so an empty presentation, whose sum is $0$, can never satisfy ${\sum }_l{x}_l\otimes y_l=u\otimes v$.

The form $B$ is $k$-bilinear [F2], hence balanced, so it descends to the unique group homomorphism $\bar{B}:k^2\otimes k^2\to k$ with $\bar{B}(x\otimes y)=B(x,y)$ by [F3] and [F4]. It is $k$-linear: for $c\in k$ and $t={\sum }_l{x}_l\otimes y_l$, [F1] and [F2] give $\bar{B}(ct)={\sum }_{l}B(c{x}_l,y_l)=c{\sum }_{l}B(x_l,y_l)=c\bar{B}(t)$. If ${\sum }_l{x}_l\otimes y_l=u\otimes v$ is any finite presentation of the tensor, linearity of $\bar{B}$ and step 1.1 give ${\sum }_{l}B(x_l,y_l)=\bar{B}({\sum }_l{x}_l\otimes y_l)=\bar{B}(u\otimes v)=B(u,v)=1$, so the contraction has the same value $1$ on every presentation and is independent of the presentation.

Expand the first factor by bilinearity [F1]: $(l\otimes m)\otimes n=(l\otimes e_1)\otimes n+(l\otimes e_2)\otimes n$, and $n=2e_2$ may be written $n=2\cdot e_2$; since $(\cdot )\otimes x$ and both composites are linear, it suffices by [F3] to evaluate the two pentagon composites on the elementary tensors $((l\otimes e_i)\otimes n)\otimes x$, $i=1,2$, where $l,e_i,n,x\in k^2$.

On such an elementary tensor the right composite $\alpha \circ \alpha ={\alpha }_{L,M,N\otimes X}\circ {\alpha }_{L\otimes M,N,X}$ gives first $(l\otimes e_i)\otimes (n\otimes x)$ and then $l\otimes (e_i\otimes (n\otimes x))$, and the left composite $(\mathrm{i}\mathrm{d}\otimes \alpha )\circ \alpha \circ (\alpha \otimes \mathrm{i}\mathrm{d})$ gives first $(l\otimes (e_i\otimes n))\otimes x$, then $l\otimes ((e_i\otimes n)\otimes x)$ and finally $l\otimes (e_i\otimes (n\otimes x))$, all by the elementary-tensor formula of [F2]; the two values agree.

By linearity of the two composites in the first factor (step 1.1) and their agreement on the two elementary summands (step 1.2), both composites send $((l\otimes m)\otimes n)\otimes x$ to $l\otimes (m\otimes (n\otimes x))$, which is $e_1\otimes ((e_1+e_2)\otimes (2e_2\otimes (e_1-e_2)))$ by the definitions of $l,m,n,x$; this is the pentagon identity of [F3] checked on the four named vectors.

By [F2] the four tensors $e_i\otimes f_j$ form a basis of $V\otimes W$, and the classes $[e_2]\in V\mathrm{/}U$ and $[f_1]\in W\mathrm{/}Z$ form bases of the one-dimensional quotients, so $[e_2]\otimes [f_1]$ is the product basis of $(V\mathrm{/}U)\otimes (W\mathrm{/}Z)\cong k\otimes k\cong k$ by [F2]. Since $p$ kills $e_1$ and fixes the class of $e_2$, while $q$ kills $f_2$ and fixes the class of $f_1$, the map $p\otimes q$ kills exactly those basis tensors with $e_1$ in the first factor or $f_2$ in the second, namely $e_1\otimes f_1$, $e_1\otimes f_2$ and $e_2\otimes f_2$, and sends $e_2\otimes f_1$ to the basis element $[e_2]\otimes [f_1]$; in particular $p\otimes q$ is surjective by [F1] and [F2].

Since $(e_i\otimes f_j)$ is a basis of $V\otimes W$ by [F2], a tensor ${\sum }_{i,j}{c}_{ij}{e}_i\otimes f_j$ lies in the kernel of the linear map $p\otimes q$ exactly when ${\sum }_{i,j}{c}_{ij}(p\otimes q)(e_i\otimes f_j)=0$; step 1.1 shows that $(p\otimes q)(e_i\otimes f_j)=0$ for the three basis tensors $e_1\otimes f_1$, $e_1\otimes f_2$, $e_2\otimes f_2$, while $(p\otimes q)(e_2\otimes f_1)=[e_2]\otimes [f_1]\mathrm{\ne }0$ is a nonzero product-basis element, so in the basis expansion the kernel condition is the vanishing of the single coefficient of $e_2\otimes f_1$ and the kernel is exactly $\mathrm{span}\{e_1\otimes f_1,e_1\otimes f_2,e_2\otimes f_2\}$. That span equals $U\otimes W+V\otimes Z$ inside $V\otimes W$ because $U\otimes W=\mathrm{span}\{e_1\otimes f_1,e_1\otimes f_2\}$ and $V\otimes Z=\mathrm{span}\{e_1\otimes f_2,e_2\otimes f_2\}$ as the product bases of the respective tensor products, and by the basis expansion of [F2] its dimension is $3$. The dimension formula reads $\dim U\dim W+\dim V\dim Z-\dim U\dim Z=1\cdot 2+2\cdot 1-1\cdot 1=3$, in agreement with the kernel just computed and with the prediction of [F3], whose complement argument is not needed for this explicit computation.

First, $b_1^{\ast },b_2^{\ast }$ as displayed are the dual basis of $b_1,b_2$: using $e_i^{\ast }(e_j)={\delta }_{ij}$ one computes $b_1^{\ast }(b_1)=\frac{1}{2}(1+1)=1$, $b_1^{\ast }(b_2)=\frac{1}{2}(1-1)=0$, $b_2^{\ast }(b_1)=\frac{1}{2}(1-1)=0$ and $b_2^{\ast }(b_2)=\frac{1}{2}(1+1)=1$, so by the characterization of [F2] the displayed functionals are $b_1^{\ast },b_2^{\ast }$. Then [F1] gives ${\sum }_j{b}_j\otimes b_j^{\ast }=\frac{1}{2}((e_1+e_2)\otimes (e_1^{\ast }+e_2^{\ast })+(e_1-e_2)\otimes (e_1^{\ast }-e_2^{\ast }))=\frac{1}{2}(2e_1\otimes e_1^{\ast }+2e_2\otimes e_2^{\ast })=e_1\otimes e_1^{\ast }+e_2\otimes e_2^{\ast }$, the two cross terms cancelling and the factor $\frac{1}{2}\cdot 2=1$ being legitimate because $\mathrm{char}k\mathrm{\ne }2$.

By step 1.1 the same element of $V\otimes V^{\ast }$ is computed by the sum over the standard basis and by the sum over the second basis, which is exactly the basis-independence asserted in [F3]; since [F3] identifies this element with the image of $1$ under the coevaluation, the coevaluation is computed by the same element in both bases, and the zigzag identities of [F3] hold for both ordered bases.

For $f,g\in V^{\ast }$, the bilinear pairing $(v,w)\mapsto f(v)g(w)$ defines a functional $\theta (f,g)\in (V\otimes V{)}^{\ast }$ by [F6]; the assignment $(f,g)\mapsto \theta (f,g)$ is itself bilinear, so [F6] gives a linear map $\theta :V^{\ast }\otimes V^{\ast }\to (V\otimes V{)}^{\ast }$ with $\theta (f\otimes g)(v\otimes w)=f(v)g(w)$, without a finite-dimensional hypothesis. The product basis $(e_i\otimes e_j{)}_{(i,j)\in I\times I}$ of [F2] is a basis of $V\otimes V$, so prescribing the values $L(e_i\otimes e_j)={\delta }_{ij}$ on it defines a unique linear functional $L\in (V\otimes V{)}^{\ast }$ by [F1]; in particular $L$ is well defined and $I\times I$, hence $I$, is infinite. Likewise the coordinate functionals $e_j^{\ast }$ of [F3] are well defined, and the set $S:=\{e_j^{\ast }:j\in I\}\subseteq V^{\ast }$ is infinite because $j\mapsto e_j^{\ast }$ is injective (they take different values at the single vector $e_j$), and it is linearly independent: if ${\sum }_{l=1}^r{\lambda }_l{e}_{j_l}^{\ast }=0$ is a finite relation with distinct indices $j_1,\dots ,j_r$, evaluating at $e_{j_l}$ gives ${\lambda }_l=0$.

Suppose $L$ is the image of an element of $V^{\ast }\otimes V^{\ast }$, written as a finite sum ${\sum }_{s=1}^m{f}_s\otimes g_s$ of elementary tensors by [F6]. For fixed $j\in I$ the functional $v\mapsto L(v\otimes e_j)$ is the coordinate functional $e_j^{\ast }$, because $L(e_i\otimes e_j)={\delta }_{ij}$, and by the canonical-map formula in step 1.1 it is also ${\sum }_{s=1}^m{f}_s(\cdot )g_s(e_j)={\sum }_{s=1}^m{g}_s(e_j)f_s$; hence every $e_j^{\ast }$ lies in the finite-dimensional span of $f_1,\dots ,f_m$. Thus the infinite linearly independent set $S$ of step 1.1 lies in a space spanned by $m$ elements, contradicting [F5], which forbids an infinite linearly independent subset in such a space.

No finite sum ${\sum }_s{f}_s\otimes g_s$ can have image $L$, so $L$ lies outside the image of the canonical map and the map is not surjective for this infinite-dimensional $V$: surjectivity of $V^{\ast }\otimes V^{\ast }\to (V\otimes V{)}^{\ast }$ is a finite-dimensional phenomenon, as claimed. For $I=\mathbb{N}$ and $v={\sum }_n{a}_n{e}_n$, $w={\sum }_m{b}_m{e}_m$ (finite sums) one computes $L(v\otimes w)={\sum }_{n,m}{a}_n{b}_{m}L(e_n\otimes e_m)={\sum }_n{a}_n{b}_n$, so the outside functional is exactly the coefficient pairing, which is not a finite sum of products of functionals by step 2.1.