# Projective tensor product of injective operators

I've seen claims that it is known that for a pair of bounded injective linear operators $$T\colon X\to Y, S\colon W\to V$$, their tensor product $$T\otimes S\colon X \otimes_\pi W\to Y \otimes_\pi V$$ need not be injective. Here $$\otimes_\pi$$ stands for the projective tensor product of Banach spaces.

1. Can this happen when $$T = {\rm id}_X$$, the identity operator on some Banach space $$X$$?
2. If so, can it happen for $$T = {\rm id}_{L_1}$$, the identity operator on $$L_1$$?

Question 2 has negative answer when $$S$$ is an isomorphism onto its range. Not surprisingly, the answer would be always positive for the injective tensor product.

• For Q2, I guess we get $\operatorname{id}\otimes S : L_1 \otimes_\pi W \rightarrow L_1 \otimes_\pi V$ but then we know that $L_1 \otimes_\pi W \cong L_1(W)$ and for $V$ and so wouldn't "working with functions" show that Q2 has a negative answer (i.e. it is always injective)? May 5 at 10:16
• For Q1: Isn't $T\otimes S=(id_Y\otimes S)\circ (T\otimes id_W)$? If the composition isn't injective then so is one factor. May 5 at 10:29

$$\require{AMScd}\newcommand{\id}{\operatorname{id}}$$I use a common characterisation of the approximation property as found in e.g. Ryan's book Zbl 1090.46001.
A Banach space $$X$$ has the approximation property if and only if for each Banach space $$Y$$ (it is enough to take $$Y=X^*$$) the natural map $$X \widehat\otimes Y \rightarrow X \check\otimes Y$$ is injective.
Here I write $$\widehat\otimes$$ and $$\check\otimes$$ for the completed projective, respectively, injective tensor products.
We can now answer (2) in the negative. Let $$X$$ have the approximation property, and let $$S:W\rightarrow V$$ be injective. Consider the commutative diagram $$\begin{CD} X\widehat\otimes W @>>> X \check\otimes W \\ @V{\id\otimes S}VV @VV{\id\otimes S}V \\ X\widehat\otimes V @>>> X \check\otimes V \end{CD}$$ The map $$\id\otimes S: X \check\otimes W \rightarrow X \check\otimes V$$ is injective, and the horizontal arrows are injective as $$X$$ has AP, so $$\id\otimes S: X \widehat\otimes W \rightarrow X \widehat\otimes V$$ is injective. In particular $$X=L_1$$ has the AP, showing the negation of (2).
As Jochen Wengenroth noted, Q1 can be reduced to the $$T\otimes S$$ case which the OP stated has a positive answer. However, here is a concrete example, following Chapter 5, Corollary 4 of Defant and Floret Zbl 0774.46018. Let $$X$$ be any Banach space, and let $$B_{X^*}$$ be the unit ball of the dual space $$X^*$$, consider $$\ell_\infty(B_{X^*})$$ and define $$j:X\rightarrow \ell_\infty(B_{X^*})$$ by evaluation: $$j(x) = ( \phi(x) )_{\phi\in B_{X^*}}$$. Then $$j$$ is an isometry onto its range. We know that $$\ell_\infty(B_{X^*})$$ has AP so $$X^* \widehat\otimes \ell_\infty(B_{X^*}) \rightarrow X^* \check\otimes \ell_\infty(B_{X^*})$$ is injective. Consider now the commutative diagram $$\begin{CD} X^* \widehat\otimes X @>>> X^* \check\otimes X \\ @V{\id\otimes j}VV @VV{\id\otimes j}V \\ X^* \widehat\otimes \ell_\infty(B_{X^*}) @>>> X^* \check\otimes \ell_\infty(B_{X^*}) \\ \end{CD}$$ The bottom arrow is injective, and the right-hand down arrow is. If $$X$$ does not have AP then the top arrow is not injective, and so the left-hand down arrow must fail to be injective, which gives an example of (1). (There is nothing special about $$\ell_\infty$$ here: any Banach space $$F$$ with the AP and any injection $$j:X\rightarrow F$$ would work.)