Projective and injective tensor product - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T17:47:36Z http://mathoverflow.net/feeds/question/74007 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74007/projective-and-injective-tensor-product Projective and injective tensor product Celeban 2011-08-29T22:17:06Z 2011-08-29T22:17:06Z <p>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 $p &lt; q^{\prime}&lt;\infty$ 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 $p&lt; q^{\prime}$</p> <p>$$(\ell^p \hat\otimes_\pi \ell^q)^{\star}= \ell^{p^\prime} \hat \otimes_\varepsilon \ell^{q^\prime}.$$ Moreover, for $p^{\prime}>q$ $$(\ell^p \hat\otimes_\varepsilon \ell^q)^{\star}=\mathcal{N}(\ell^p, \ell^{q^\prime})= \ell^{p^\prime} \hat \otimes_\pi \ell^{q^\prime}.$$</p> <p>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).</p> <p><strong>Corollary</strong></p> <p>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 $$(\ell^p \hat\otimes_\varepsilon \ell^q)^{\star}= \ell^{p^\prime} \hat \otimes_\pi \ell^{q^\prime}.$$</p> <p><strong>Fact 1</strong></p> <p>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 $\ell^p$.</p> <p><strong>Fact 2</strong> 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 $\ell^1$.</p> <p><strong>Fact 3</strong> We know only that </p> <p>$$\ell^2 \hat \otimes_\varepsilon \ell^2 \subset (\ell^2 \hat\otimes_\pi \ell^2)^{\star}= \mathcal{L}(\ell^2, \ell^2)$$</p> <p><strong>Question</strong></p> <p>What other interesting properties are different for $\ell^2 \hat \otimes_\varepsilon \ell^2$ and $\ell^2 \hat\otimes_\pi \ell^2$?</p> <p><strong>Notation</strong></p> <p>$\hat\otimes_\pi$ - projective tensor product of two Banach spaces.</p> <p>$\hat\otimes_\varepsilon$ - injective tensor product.</p> <p>$\mathcal{L}(X, Y)$ - the space of all linear and bounded operators from $X$ to $Y$.</p> <p>$\mathcal{K}(X, Y)$ - the space of all compact operators from $X$ to $Y$.</p> <p>$\mathcal{N}(X, Y)$ - the space of all nuclear operators from $X$ to $Y$.</p>