# Projective and injective tensor product

It is well known that for arbitrary Banach spaces $X$ and $Y$ we have that the dual space $(X \hat{\otimes}_{\pi} Y)^* = \mathcal{L}(X, Y^*)$. If we take $\ell^p$ and $\ell^q$ such that $p < q^{\prime}<\infty$ we have that $$(\ell^p \hat{\otimes}_{\pi} \ell^q)^*= \mathcal{L}(\ell^p, \ell^{q^{\prime}})= \mathcal{K}(\ell^p, \ell^{q^{\prime}})$$ by virtue of Pitt's theorem. Since all $\ell^p$ for $p \in [1, \infty)$ have approximation property then $$\mathcal{K}(\ell^p, \ell^{q^{\prime}})= \ell^{p^{\prime}} \hat {\otimes}_{\varepsilon} \ell^{q^{\prime}}$$ and thus for $p< q^{\prime}$

$$(\ell^p \hat\otimes_\pi \ell^q)^{\star}= \ell^{p^\prime} \hat \otimes_\varepsilon \ell^{q^\prime}.$$ Moreover, for $p^{\prime}>q$ $$(\ell^p \hat\otimes_\varepsilon \ell^q)^{\star}=\mathcal{N}(\ell^p, \ell^{q^\prime})= \ell^{p^\prime} \hat \otimes_\pi \ell^{q^\prime}.$$

and it implies that $\ell^p \hat{\otimes}_{\pi} \ell^q$ is reflexive (for $p^{\prime}>q$). Of course the dual of reflexive is reflexive as well so we conclude that for $p^{\prime} > q$ the injective tensor product of $\ell^p$ and $\ell^q$ is reflexive as well (Here I had some LaTeX issues and this tensor product wasn't displayed properly).

Corollary

Therefore we have the condition that $\ell^p \hat\otimes_\pi \ell^q$ is reflexive iff $p>q^{\prime}$ and it so then $$(\ell^p \hat\otimes_\pi \ell^q)^{\star}= \ell^{p^\prime} \hat \otimes_\varepsilon \ell^{q^\prime}$$ and $\ell^p \hat\otimes_\varepsilon \ell^q$ is reflexive iff $p^{\prime}>q$ and it so then $$(\ell^p \hat\otimes_\varepsilon \ell^q)^{\star}= \ell^{p^\prime} \hat \otimes_\pi \ell^{q^\prime}.$$

Fact 1

Note that for $p=q=2$ the statement $$(\ell^2 \hat\otimes_\varepsilon \ell^2)^{\star}=\mathcal{N}(\ell^2, \ell^2)= \ell^2 \hat \otimes_\pi \ell^2,$$ is still true, because we use only the approximation property of $\ell^p$.

Fact 2 The converse is false i.e. $$(\ell^2 \hat\otimes_\pi \ell^2)^{\star} \neq \ell^2 \hat \otimes_\varepsilon \ell^2,$$ the space $\ell^2 \hat\otimes_\pi \ell^2$ is not reflexive since it contains a complemented isomorphic copy of $\ell^1$.

Fact 3 We know only that

$$\ell^2 \hat \otimes_\varepsilon \ell^2 \subset (\ell^2 \hat\otimes_\pi \ell^2)^{\star}= \mathcal{L}(\ell^2, \ell^2)$$

Question

What other interesting properties are different for $\ell^2 \hat \otimes_\varepsilon \ell^2$ and $\ell^2 \hat\otimes_\pi \ell^2$?

Notation

$\hat\otimes_\pi$ - projective tensor product of two Banach spaces.

$\hat\otimes_\varepsilon$ - injective tensor product.

$\mathcal{L}(X, Y)$ - the space of all linear and bounded operators from $X$ to $Y$.

$\mathcal{K}(X, Y)$ - the space of all compact operators from $X$ to $Y$.

$\mathcal{N}(X, Y)$ - the space of all nuclear operators from $X$ to $Y$.

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Since your question is open-ended ("what other interesting properties") I suggest that this question be made "community wiki", as its current formulation does not admit a single "best" answer. – Yemon Choi Aug 29 '11 at 23:42
Moreover: could you try and make more precise what you mean by "interesting properties"? The Banach spaces you mention are not isomorphic, but presumably you are after a less facile answer. Are you asking for e.g. type and cotype properties which are different? – Yemon Choi Aug 29 '11 at 23:44
Thanks for your comments. I think that a question about a type and cotype is interesting. The 2nd question can be e.g. projective tensor product of $\ell^2$ spaces is not WCG, what about injective tensor product of these spaces? ($\ell^2$ is of course WCG). Your suggestion to make it a "community wiki" is also good, as it seems to be open-ended. – Celeban Aug 30 '11 at 0:25
In above comment I was wrong the projective tensor product of $\ell^2$ spaces is WCG since it is a separable space, but changing $\ell^2$ to $\ell^2(\Gamma),$ where $\Gamma$ is an uncountable set the above statement is true. – Celeban Sep 5 '11 at 18:16