A question about the tensor product

Question

I am reading the interesting paper written by M. Aguiar. On page 267, I don't know whether we can replace the tensor symbols in an identity by some elements such that the identity is still established.

Maybe we can ask this question by the following way. Let $\mathbf{k}$ be a commutative ring with unit and $A$ be a $\mathbf{k}$-algebra with unit. Given an identity $$u_1\otimes v_1=u_2\otimes v_2 \text { for some }u_1,u_2,v_1,v_2\in A.$$ If we replace the tensor symbols by $x\in A$, the identity $$u_1x v_1=u_2x v_2$$ is still established? My question is how to prove it?

Answer 1

I assume that $\otimes$ denotes the tensor product over $\mathbf{k}$, so that $\otimes = \otimes_\mathbf{k}$.

To answer the question as stated: The map $$ A \times A \to A, \quad (u,v) \mapsto uxv $$ is $\mathbf{k}$-bilinear, and thus induces a $\mathbf{k}$-linear map $$ h \colon A \otimes_\mathbf{k} A \to A, \quad u \otimes v \mapsto uxv. $$ It then follows for $\sum_i u_i \otimes v_i = \sum_j u'_j \otimes v'_j$ and $x \in A$ that $$ \sum_i u_i x v_i = h\left( \sum_i u_i \otimes v_i \right) = h\left( \sum_j u'_j \otimes v'_j \right) = \sum_j u'_j x v'_j. $$

But I think we actually want that for $\sum_i u_i \otimes v_i \otimes w_i = \sum_j u'_j \otimes v'_j \otimes w'_j$ and $x, y \in A$ we have that $$ \sum_i u_i x v_i y w_i = \sum_j u'_j x v'_j y w'_j. $$ This can be shown in the same way as above by considering the $\mathbf{k}$-trilinear map $$ A \times A \times A \to A, \quad (u,v,w) \mapsto uxvyw. $$ For $r = \sum_i u_i \otimes v_i$ and $\beta \colon A \to A$ given by $\beta(x) = \sum_i u_i x v_i$ we have that \begin{align*} \beta(x)\beta(y) &= \sum_{i,j} u_i x v_i u_j y v_j, \\ \beta(\beta(x) y) &= \sum_{i,j} u_i u_j x v_j y v_i, \\ \beta(x \beta(y)) &= \sum_{i,j} u_j x u_i y v_i v_j, \end{align*} so it then follows with $\mathbf{A}(r) = 0$ (by replacing the tensor symbols $\otimes$ in the given equation by $x$ and $y$) that $$ \beta(x)\beta(y) = \beta(\beta(x) y) + \beta(x \beta(y)). $$ From what I understand, this is what it means for $\beta$ to be a Baxter operator.

Answer 2

Sure because your binary operation on $A$ $\cdot: AxA\to A$ is a balanced product and so, by universal property of tensor product, there exists a unique map $\phi:A\otimes A\to A$ such that $\cdot=\phi\circ\otimes$ and so if you have that $u_1\otimes v_1=u_2\otimes v_2$ than $u_1v_1=\phi\circ\otimes (u_1,v_1)=\phi(u_1\otimes v_1)=$ $\phi(u_2\otimes v_2)= \phi\circ\otimes(u_2, v_2) =u_2v_2$