tensor product of *-algebra as a *-algebra: well-definedness of involution

Question

I have a question regarding the proof of the following proposition:

Proposition: Let $A$ and $B$ two $\mathbb{K}$-algebras with involution. Let $A\otimes B$ be the algebraic tensor product of $A$ and $B$. Then $A\otimes B$ is a involutive $\mathbb{K}$-algebra with $$(a\otimes b)(c\otimes d)=ac\otimes bd,$$ $$(a\otimes b)^*=a^*\otimes b^*.$$ Proof (only the passage where I'm stuck): I'm stuck with the well-definedness of $$\sum_i a_i\otimes b_i\mapsto \sum_i a_i^*\otimes b_i^*:$$ 1. Why one has to check for well-definedness the following: if $\sum_i a_i\otimes b_i=0 \Rightarrow \sum_i a_i^*\otimes b_i^*=0$ ?

2.how to prove that/ Is my solution for 2. (see below) correct?

My idea for 2. is to use the fact if $\{x_1,..,x_n\}\subseteq A$ is linearly independent and $\{y_1,...,y_n\}\subseteq B$ is arbitrary and such that $$\sum_{i=1}^nx_i\otimes y_i=0,$$ then $y_1=...y_n=0$. Now, let $\{x_k\}_{k\in I}$ be a basis for $A$ and $a_i=\sum_{k\in I}\lambda_{i,k}x_k$ (this sum is finite). Furthermore let $\sum_i a_i\otimes b_i=0$ Then we have $$0=\sum_i a_i\otimes b_i=\sum_i \sum_{k\in I}\lambda_{i,k}x_k \otimes b_i=\sum_{k\in I} x_k\otimes (\sum_i \lambda_{i,k}b_i)$$ i.e. (with the fact above) $\sum_i \lambda_{i,k}b_i=0$ for all $k\in I$. It follows $$\sum_i a_i^*\otimes b_i^*=\sum_i \sum_{k\in I}\overline{\lambda_{i,k}}x_k^* \otimes b_i^*=\sum_{k\in I} x_k^*\otimes (\sum_i \overline{\lambda_{i,k}}b_i^*)=\sum_{k\in I} x_k^*\otimes (\sum_i \lambda_{i,k}b_i)^*=0.$$

But I have no idea why 1. proves well-definedness of the involution.

Answer 1

If $$\sum_{i=1}^n a_i\otimes b_i=\sum_{j=1}^m c_j\otimes d_j,$$ then $$0=\sum a_i\otimes b_i-\sum c_j\otimes d_j=\sum_{k=1}^{m+n} r_k\otimes s_k,$$ where the $$ r_k=\begin{cases}a_k,&\ \text{ if }k=1,\ldots,n\\ -c_{k-m},&\ \text{ if } k=n+1,\ldots,n+m\end{cases} $$ and $$ s_k=\begin{cases}b_k,&\ \text{ if }k=1,\ldots,n\\ -d_{k-m},&\ \text{ if } k=n+1,\ldots,n+m\end{cases} $$ So, if $\sum a_i\otimes b_i=\sum_j c_j\otimes d_j$, then $\sum_kr_k\otimes s_k=0$, and by $1.$ you get that $\sum_kr_k^*\otimes s_k^*=0$, i.e. $\sum a_i^*\otimes b_i^*=\sum_j c_j^*\otimes d_j^*$.

Answer 2

Questions about well-definedness of alleged maps from (categorically genuine) tensor products are often quite daunting if/when one attempts to address them in a too elementary way.

That is, in reality, the universal mapping property characterization of tensor products, namely, that all bilinear maps $B:M\times N\to Z$ are (uniquely) converted to linear maps $\beta:M\otimes N\to Z$ such that $B(m,n)=\beta(m\otimes n)$ removes most of the apparent difficulty "by magic". That is, unambiguous specification of a bilinear map, that is, just on elements $m\times n$, gives a well-defined linear map on the tensor product... That is, in effect, if specification of the elements $\beta(m\otimes n)$ (where $m\otimes n$ is simply the image of $m\times n$ in the tensor product) induce a bilinear map on $M\times N$, then there's no ambiguity.

The example of an involution is a map of the tensor product to itself, so to be clear about the well-definedness we ask about the alleged corresponding map $M\times N\to M\otimes N$.

So, with an involution $\theta$, since $(m_1+m_2)^\theta\otimes n^\theta=m_1^\theta\otimes n^\theta + m_2^\theta\otimes n^\theta$, and symmetrically, and for scalars, it is well-defined.