In Ryan Introduction to Tensor Products of Banach spaces, Example 2.10, it is shown that the space $\ell^2 \hat{\otimes}_\pi \ell^2$, where $\hat{\otimes}_\pi$ denotes the projective tensor product and $\ell^2$ the usual space of square-summable sequences, contains $\ell^1$ as a subspace, and that $\ell^1$ is complemented in $\ell^2 \hat{\otimes}_\pi \ell^2$, that is, a subspace $M$ exists such that $\ell^2 \hat{\otimes}_\pi \ell^2 = \ell^1 \oplus M$. Is anything known about this space $M$? I'd be also intersted if there are other references that discuss the space $\ell^2 \hat{\otimes}_\pi \ell^2$ in more detail.
Projective tensor product $\ell^2 \hat{\otimes}_\pi \ell^2$
banach-spacesfunctional-analysistensor-products
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Your question is a very natural one (if I understand it correctly) and at the same time it raises a rather difficult question.
As you say, it makes sense to identify the operators $A : X \to Y$ of finite rank with elements of $X^{\ast} \otimes Y$. Now you want the operator norm of $A$ to coincide with its norm in $X^{\ast} \otimes Y$ with respect to some tensor norm. I leave it to you to check that the injective tensor norm defined for $\omega = \sum x_{i}^{\ast} \otimes y_{i}$ by $$\Vert \omega \Vert_{\varepsilon} = \sup_{\substack{\Vert \phi \Vert_{X^{\ast\ast}} \leq 1 \\\ \Vert \psi \Vert_{Y^{\ast}} \leq 1}}{\left\vert \sum \phi(x_{i}^{\ast}) \, \psi(y_{i})\right\vert}$$ does what you want (it is independent of the representation of $\omega$ as a finite sum of elementary tensors and $\|A\| = \|A\|_{\varepsilon}$ for operators of finite rank). Edit: To see the second claim, use Goldstine's theorem that allows you to replace the supremum over $\phi \in X^{\ast\ast}$ with $\Vert\phi\Vert_{X^{\ast\ast}} \leq 1$ by the supremum over $\operatorname{ev}_{x}$ with $x \in X$ and $\Vert x \Vert_{X} \leq 1$.
The projective tensor norm of a finite rank operator is usually much larger than its operator norm (see the discussion on pp.41ff in Ryan, for example).
Given this, we can identify the completion $X^{\ast} \otimes_{\varepsilon} Y$ of $X^{\ast} \otimes Y$ with respect to the injective tensor norm with a space $K_{0}(X,Y) \subset L(X,Y)$ of operators $X \to Y$ and we will freely do so from now on. Note that $K_{0}(X,Y)$ is nothing but the closure of the operators of finite rank in $L(X,Y)$.
Now the question is: What are the operators lying in $K_{0}(X,Y)$ ?
This is really difficult and I'll outline the closest I know to an answer to that.
As a first observation note that the compact operators $K(X,Y) \subset L(X,Y)$ are a closed subspace of $L(X,Y)$ containing $X^{\ast} \otimes_{\varepsilon} Y = K_{0}(X,Y)$. Looking at the examples of Hilbert spaces or the classical Banach spaces one finds out that quite often $K(X,Y) = K_{0}(X,Y)$ holds. However, it may fail in general, and that's where the famous Approximation Property comes in. I'll refrain from delving into the numerous equivalent formulations and use it as a black box. We have the following theorem due to Grothendieck:
Theorem. The Banach space $X^{\ast}$ has the approximation property if and only if $K_{0}(X,Y) = K(X,Y)$ holds for all Banach spaces $Y$.
Edit 2: (in response to a comment of the OP) It follows that for a reflexive Banach space $X$ with the approximation property we have $K(X,Y) = X^{\ast} \otimes_{\varepsilon} Y$ for all Banach spaces $Y$.
Now most of the Banach spaces you'll run into have the approximation property, e.g. $L^{p}$, $C(X)$ and so on. However, P. Enflo (in a veritable tour de force) has shown that there exist Banach spaces failing the approximation property. An explicit example (identified by Szankowski) is the space $L(H,H)$ of a separable Hilbert space. Note that this space is the dual space of the trace class operators. A famously open question is whether the space $H^{\infty}(D)$ of bounded holomorphic functions on the open unit disk has the approximation property.
I hope this answers your question. The approximation property is discussed in detail in any book that treats the tensor products of Banach spaces. In particular, this is well treated in Ryan's book.
Here's a sketch of the argument based on the hint from Stefan Walter.
Start by fixing $x\in X$, and define the bilinear map $Y\times Z\to (X\otimes Y)\otimes Z$ by $$(y,z)\mapsto (x\otimes y)\otimes z$$
By the universal property stated in my question, this induces a linear map $$A_{x}:Y\otimes Z\to (X\otimes Y)\otimes Z\text{ such that }A_{x}(y\otimes z) = (x\otimes y)\otimes z\text{ for every }y\in Y,z\in Z$$
Next define the bilinear map $X\times (Y\otimes Z)\to (X\otimes Y)\otimes Z$ by $$(x,\sum_{i=1}^{n}y_{i}\otimes z_{i})\mapsto A_{x}(\sum_{i=1}^{n}y_{i}\otimes z_{i})$$
Passing this map through the universal property yields the isomorphism.
Best Answer
$\ell^2$ has approximation property (AP) since it possess a Schauder basis. $\ell^2$ also has the Radon-Nikodym property (RNP) since every reflexive Banach space has RNP. You may deduce the following from the more general theorems in Ryan's book: If a Banach space $X$ is reflexive and has AP, then
Thus, $\ell^2\hat{\otimes}_{\epsilon}\ell^2 = K(\ell^2)$, and $\ell^2\hat{\otimes}_{\pi}\ell^2 = N(\ell^2)$, which is the space of the trace class operators on $\ell^2$.
For the other question, given a complemented subspace $V$ of a Banach space $X$, $V$ may not have a unique complement. However, the complemented $\ell^1$ subspace given in Ryan's book is the closed linear span of the tensor diagonal $(e_k\otimes e_k)_{k\in\mathbb{N}}$, where $(e_k)_{k\in\mathbb{N}}$ can be chosen any orthonormal basis for $\ell^2$. Also, a particular projection (call it $P$) is also given in the book. One may simply define $M$ to be the range of the projection $I-P$, where $I$ is the identity map.