What is the result of $\ell^1 \otimes \ell^1$?

Question

I recently stumbled upon the concept of a tensor product while studying quantum computing, and felt that my understanding of it from preliminary readings was incomplete. I challenged myself to determine the answer to this question in order to show that I had a full understanding of tensor products. I believe the answer should be $\mathbb{R}^\mathbb{N}$ (excuse the abuse of notation, I'm unfamiliar with a better way of denoting a countably infinite real space) or in fact $\ell^1$ itself once again. Any hints, ideas or explanations are appreciated!

Answer 1

Upon further thought and studying, I believe the answer here to be $\mathbb{R}^\infty$. Recall the definition of tensor product between two vector spaces $V$ and $W$: given a basis $\mathcal{B}_V$ for $V$ and $\mathcal{B}_W$ for $W$, $V \otimes W$ is the span of $\mathcal{B}_V \otimes \mathcal{B}_W$, where this basis uses the tensor product only symbolicly; it contains dim$(V) \times$ dim$(W)$ basis vectors.

Since $\ell^1$ has an uncountable basis, we have that $\ell^1 \otimes \ell^1$ is the (real) span of uncountably many basis vectors, hence $\mathbb{R}^\infty$.

Answer 2

Tensor products are not straightforward.

You say you want to consider the algebraic tensor product (that is, the product as vector spaces). That's doable of course, and easiest. But it also gives you basically nothing. If you treat $\ell^1$ as an algebraic vector space, then you are already working on a vector space of uncountable dimension (since $\ell^1$ has uncountable dimension as a vector space, this is a consequence of Baire's Category Theorem). It is then a trivial cardinality argument to say that the tensor product of two vector spaces of uncountable dimension is a vector space of (the same) uncountable dimension.

But your question is tagged "Functional Analysis", and then the above is entirely irrelevant. Usually one considers $\ell^1$ not as a vector space but as a Banach space. This means that one considers its vector space structure, together with a normed space structure that makes it complete. If you now want to make $\ell^1\otimes\ell^1$ a Banach space, what needs to be done is not trivial. The algebraic product of $\ell^1$ with itself is not complete. So to make it a Banach space one needs to complete it. And it turns out that in general there is more than one way in which to complete the algebraic product $\ell^1\otimes\ell^1$ such that the norm, restricted to the elementary tensors, satisfies $\|x\otimes y\|=\|x\|\,\|y\|$ (that is, a cross norm).

The theory is very deep. Grothendieck's thesis was a milestone in the topic sixty-something years ago. It is possible to prove that there exist two norms, called the injective and projective cross norms, that are respectively minimal and maximal among cross norms. In general they are not equal, which means that there are usually many different ways to complete the algebraic tensor product.