Let $A$ be a finite dimensional $\mathrm{C}^*$-algebra.

Suppose that $q$ is a projection and $$T:A\rightarrow A\otimes A$$ is a *-homomorphism.

Suppose we write:

$$T(q)=\sum_{j=1}^n q_j\otimes p_j.$$

What properties do the $q_j$ and $p_j$ have? Is there some presentation of $T(q)$ such that

  1. The $p_j$ and $q_j$ are projections?
  2. The $(p_j)$ are linearly independent projections.
  3. The $(p_j)$ and $(q_j)$ are linearly independent projections.
  4. The $(p_j)$ are (mutually) orthogonal projections $p_ip_j=\delta_{i,j}p_j$.
  5. The $(p_j)$ and $(q_j)$ are (mutually) orthogonal projections.

This is a question additional to this question.


Already the answer to 1 is "no". Let $A=M_2(\mathbb C)$. You can see $A\otimes A$ canonically as $M_4(\mathbb C)$ where $a\otimes b=\begin{bmatrix} ab_{11}&ab_{12}\\ ab_{21}&ab_{22}\end{bmatrix}$. Let $T:A\to A\otimes A$ be given by $$ T\left(\begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{bmatrix} \right)=\begin{bmatrix} a_{11}&0&0&a_{12}\\ 0&0&0&0\\ 0&0&0&0\\ a_{21}&0&0&a_{22}\end{bmatrix}. $$ Now let $q=\tfrac12\,\begin{bmatrix} 1&1\\1&1\end{bmatrix}$. You can check in this answer that $T(q)$ cannot be written as a sum of tensors of positive elements, let alone projections.

  • $\begingroup$ Oh dear I am an idiot and have wasted your time on that other question. Thank you. $\endgroup$ Nov 25 '19 at 17:38
  • 2
    $\begingroup$ You are welcome, and you are not an idiot. There is little that is obvious about tensor products. $\endgroup$ Nov 25 '19 at 18:20

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