External tensor product of Hilbert modules

Question

I am reading Lance's book "Hilbert $C^*$-modules". In particular, I want to understand how to construct the (external) tensor product of Hilbert $C^*$-modules. Consider the following fragment from Lance's book (on p62) in which Kasparov's absorption theorem is used:

I understand everything in here, except the boxed equation. When I do the calculation, I get a double summation $\sum_{i,j,m,n}$ instead of a single summation $\sum_{i,j,n}$. This implies that the argument after that also fails.

What am I missing here?

It is worth noting that in the book "Elements of KK-theory" by Jensen and Thomsen, in section 1.2.4 (in which the external tensor product is defined), it is said that the implication $$\langle z,z\rangle = 0\implies z= 0$$ may fail (so that we have a semi-inner product and we need to pass through a quotient first to get a Hilbert module). So, maybe the argument in Lance's book doesn't work?

Thanks in advance for your insights.

Answer 1

I agree that something looks off here. However, I think we can repair the argument like this (written out in some detail to be sure): \begin{align*} \Big\langle \sum_i x_i\otimes y_i, \sum_j x_j\otimes y_j \Big\rangle &= \sum_{i,j} \langle x_i, x_j \rangle \otimes \langle y_i, y_j \rangle \\ &= \sum_{i,j,n,m} a_{in}^*a_{jn} \otimes b_{im}^*b_{jm} \\ &= \sum_{i,j,n,m} (a_{in}^*\otimes b_{im}^*)(a_{jn} \otimes b_{jm}) \\ &= \sum_{n,m} \Big(\sum_i a_{in}\otimes b_{im}\Big)^*\Big(\sum_j a_{jn} \otimes b_{jm}\Big) \end{align*} If this is equal to $0$ then as the final sum is of positive elements, each element must be $0$, and so $$ \sum_i a_{in} \otimes b_{im} = 0 \qquad (n,m\in\mathbb N). $$ This still shows that $z=0$.