Confusion Over Distributive Property in Tensor and External Tensor Products

Question

I've been delving into the properties of tensor ($\otimes$) and external tensor products ($\boxtimes$) within the context of coalgebra, particularly examining how the coproduct $\Delta$ applies to tensor products of elements from a vector space $V$ and its extension. The operation $\otimes$ is understood as the tensor product within the tensor algebra $T(V)$, whereas $\boxtimes$ represents an external tensor product, combining elements from two separate spaces or algebras, $T(V) \boxtimes T(W)$.

According to a Wikipedia article on tensor algebra, the coproduct $\Delta: V \to V \boxtimes V$ is defined for $v \in V$ as $\Delta(v) = v \boxtimes 1 + 1 \boxtimes v$, extending homomorphically over $T(V)$. My inquiry focuses on the expansion of $\Delta(v \otimes w)$ for $v, w \in V$, specifically through these steps:

1. Starting from $\Delta(v\otimes w)$, the definition of $\Delta$ is applied to both $v$ and $w$: $$\Delta(v\otimes w) = (v\boxtimes 1 + 1\boxtimes v) \otimes (w\boxtimes 1 + 1\boxtimes w)$$

2. Employing the distributive property of the tensor product over addition, we get: $$(v\boxtimes 1) \otimes (w\boxtimes 1) + (v\boxtimes 1) \otimes (1\boxtimes w) + (1\boxtimes v) \otimes (w\boxtimes 1) + (1\boxtimes v) \otimes (1\boxtimes w)$$

3. The next step of simplification leads to confusion: $$= (v\otimes w) \boxtimes 1 + v\boxtimes w + w\boxtimes v + 1 \boxtimes (v\otimes w)$$

My confusion lies in the last step: How does this property follow formally from the definitions of $\otimes$ and $\boxtimes$? It appears to make use of a property of $\otimes$ interacting with $\boxtimes$, which is not immediately evident from their standard definitions: it's unclear why $(v \boxtimes w) \otimes (v' \boxtimes w') = (v \otimes v') \boxtimes (w \otimes w')$, if this is what they are doing. The article mentions that $v \otimes 1$ is just $v$, as this represents scalar multiplication in the algebra, this is clear enough.

Could someone help clarify this step?

Answer 1

The confusion arises from misunderstanding how the tensor ($\otimes$) and external tensor products ($\boxtimes$) interact. The key is to define the product $\otimes_{T(V) \boxtimes T(V)}$ such that for pure tensors (elements that are not sums of other tensors), the operation behaves as follows:

$$ (a \boxtimes b) \otimes_{T(V) \boxtimes T(V)} (c \boxtimes d) := (a \otimes_{T(V)} c) \boxtimes (b \otimes_{T(V)} d) $$

This is then extended bilinearly to non-pure tensors. This definition allows for the simplification seen in the last step. Essentially, when you have products like $(v \boxtimes 1) \otimes (w \boxtimes 1)$, you're using the $\otimes_{T(V) \boxtimes T(V)}$ operation, which under our definition simplifies to $(v \otimes w) \boxtimes 1$. Similarly, the terms $v \boxtimes w$ and $1 \boxtimes (v \otimes w)$ follow from the bilinear extension of our definition to non-pure tensors. It is possible to verify that the above product satisfies the axiomsto make $T(V) \boxtimes T(V)$ into an algebra.

Answer 2

While that article is trying to make things easier to comprehend with its notational choices, I think it may actually make things more confusing.

In general, if $(A,\cdot_A)$ and $(B,\cdot_B)$ are two algebras (associative and unital in the present context), then we can define a multiplication $\cdot_{A\otimes B}$ on the tensor product vector space $A\otimes B$ which satisfies $$(a_1\otimes b_1)\cdot_{A\otimes B}(a_2\otimes b_2)=(a_1\cdot_A a_2)\otimes(b_1\cdot_B b_2)\tag{1}$$ and makes $(A\otimes B,\cdot_{A\otimes B})$ an algebra.

If as in the article we swap the regular tensor product symbol $\otimes$ here with $\boxtimes$, then (1) becomes $$(a_1\boxtimes b_1)\cdot_{A\boxtimes B}(a_2\boxtimes b_2)=(a_1\cdot_A a_2)\boxtimes(b_1\cdot_B b_2)\tag{1'}$$ which makes $(A\boxtimes B,\cdot_{A\boxtimes B})$ an algebra.

If we now take $A=T(V)$ with $\cdot_A=\otimes_{T(V)}$ and similarly $B=T(V)$ with $\cdot_B=\otimes_{T(V)}$, and also write $\cdot_{A\boxtimes B}$ as $\otimes_{T(V)\boxtimes T(V)}$ we get $$(a_1\boxtimes b_1)\otimes_{T(V)\boxtimes T(V)}(a_2\boxtimes b_2)=(a_1\otimes_{T(V)} a_2)\boxtimes(b_1\otimes_{T(V)} b_2)\tag{2}$$ from which the rest follows. Note (2) differs from what you wrote in your answer.

One thing that is confusing with this notation is that we're using the symbol $\otimes$ in $T(V)$ and also in $T(V)\boxtimes T(V)$ for the algebra products, and the article regularly refers to them as "tensor products", but if $V\ne 0$ they are not actually tensor products. (They are associative and unital, so not universal bilinear.)