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The conditions on the $p_j$ and $q_j$ that I believe hold do not in fact hold. See the edit for the linked question showing this.

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

Suppose that $p$ is a projection and $\nu\in \mathcal{S}(A)$ a state such that $\nu(p)=1$. Consider projections $p_j,\,q_j$ such that $p_j\leq p$ and $q_j\leq p$. Consider $$\sum_{j=1}^n q_j\otimes p_j,$$ so that the $p_j$ may be taken as linearly independent.

Suppose now that:

$$\sum_{j=1}^n\nu(q_j) p_j=p.$$

Does it follow that $$\sum_{j=1}^n q_j\otimes p_j=p\otimes p?$$

Edit: Perhaps there are more conditions on the $p_j$ and $q_j$ that come from the context.

The element $$T(q)=\sum_{j=1}^nq_j\otimes p_j$$ is the image of a projection $q$ under a *-homomorphism $T:A\rightarrow A\otimes A$. My understanding, therefore, is that it can be written as a sum of elementary tensors $q_j\otimes p_j$, with the $q_j$ and $p_j$ projections.

My understanding is that for such a general element of $A\otimes A$, we can choose the $p_j$ to be linearly independent projections. Is this incorrect? Furthermore, can we also take the $q_j$ to be linearly independent (or just the $p_j$)? Perhaps even then we can take the $q_j$ and $p_j$ to be orthogonal?

These questions are asked here.

A fuller context is here.

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If you put no restrictions on the $q_j$, in particular if you allow $\nu(q_j)=0$, then the answer is obviously no because you can choose $p_1,\ldots,p_n=p$, $q_1=p$ with $\nu(q_1)=1$ and $\nu(q_2)=0$ and $q_3=\cdots=q_n=q_2$. Then your condition is satisfied and $$ \sum_jq_j\otimes p_j=p\otimes p+\left(\sum_{j\geq2}q_j\right)\otimes p. $$

If you require $\nu(q_j)>0$ for all $j$ your condition can still fail. Suppose that $p=p_1+p_2$, with $\nu(p_1)=1$. Let $q_1=q_2=p_1$. Then $$ \sum_j\nu(q_j)p_j=p_1+p_2=p, $$ while $$ \sum_j q_j\otimes p_j=p_1\otimes p_1+p_1\otimes p_2=p_1\otimes p $$ is a proper subprojection of $p\otimes p$.

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  • $\begingroup$ Thank you for your answer. The first scenario doesn't apply because I want the $p_j$ to be linearly independent. I will in a moment slightly edit the question which may rule out the second scenario, I am not sure. $\endgroup$ Nov 25 '19 at 7:52

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