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The pentagon on four named vectors in $k^{2}$

Example

Let $e_{1}, e_{2}$ be the standard basis of $k^{2}$ and put $l = e_{1}$ , $m = e_{1} + e_{2}$ , $n = 2 e_{2}$ , $x = e_{1} - e_{2}$ . Then in $k^{2} \otimes k^{2} \otimes k^{2} \otimes k^{2}$ the pentagon of Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors holds on $((l \otimes m) \otimes n) \otimes x$ : the two composites $(i d \otimes α) \circ α \circ (α \otimes i d)$ and $α \circ α$ both send that element to $l \otimes (m \otimes (n \otimes x))$ , i.e. to $e_{1} \otimes ((e_{1} + e_{2}) \otimes (2 e_{2} \otimes (e_{1} - e_{2})))$ .

Facts & Assumptions

Given: A field $k$ , the standard basis $e_{1}, e_{2}$ of $k^{2}$ , and $l = e_{1}$ , $m = e_{1} + e_{2}$ , $n = 2 e_{2}$ , $x = e_{1} - e_{2}$ in $k^{2}$ .

[F1]

The tensor product conventions: iterated tensor powers are left-associated, every element is a finite sum of elementary tensors, and the elementary tensor is bilinear, so $l \otimes m = l \otimes e_{1} + l \otimes e_{2}$ and scalar factors may be moved across the tensor sign (Scalars, tensor powers, the empty tensor, opposite algebras and finite sums).

[F2]

The associator is the unique linear isomorphism with $α ((u \otimes v) \otimes w) = u \otimes (v \otimes w)$ on elementary tensors (Symmetry and associativity isomorphisms for tensor products over a commutative ring).

[F3]

The pentagon identity $(i d \otimes α) \circ α \circ (α \otimes i d) = α \circ α$ holds as an equality of linear maps on $k^{2} \otimes k^{2} \otimes k^{2} \otimes k^{2}$ (Associator naturality, pentagon, unit triangle and symmetry hexagons on elementary tensors).

Verification

technique · direct

1.1givenF1F3algebra

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 $n = 2 e_{2}$ 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 $l, e_{i}, n, x \in k^{2}$ .

1.2givenF1F2algebra

On such an elementary tensor the right composite $α \circ α = α_{L, M, N \otimes X} \circ α_{L \otimes M, N, X}$ gives first $(l \otimes e_{i}) \otimes (n \otimes x)$ and then $l \otimes (e_{i} \otimes (n \otimes x))$ , and the left composite $(i d \otimes α) \circ α \circ (α \otimes i d)$ 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.

2.1step 1.1step 1.2F1F3∎

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 $e_{1} \otimes ((e_{1} + e_{2}) \otimes (2 e_{2} \otimes (e_{1} - e_{2})))$ by the definitions of $l, m, n, x$ ; this is the pentagon identity of [F3] checked on the four named vectors.

Depends on

Used by

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Sources

Keith Conrad, Tensor products (University of Connecticut expository notes, 60 pp.) (standard reference, not scraped)

The pentagon on four named vectors in k2k2