algebraic theories in functional analysis

* **Functional Analysis**
## Overview diagrams
* topological vector spaces
* locally convex topological vector spaces
## Basic concepts
* topological vector space
* locally convex topological vector space
* Banach Spaces
* reflexive
* Hilbert Spaces, Fréchet Spaces, Sobolev spaces, Lebesgue Spaces
* Bornological Vector Spaces
* Barrelled Vector Spaces
* linear operator
* bounded, unbounded, self-adjoint, compact, Fredholm
* spectrum of an operator
* operator algebras
* functional calculus
## Theorems
* Stone-Weierstrass theorem
* spectral theory
* spectral theorem
* Gelfand duality
* functional calculus
* Riesz representation theorem
* measure theory
## Topics in Functional Analysis
* Bases
* Algebraic Theories in Functional Analysis
* An Elementary Treatment of Hilbert Spaces
* When are two Banach spaces isomorphic?

At the moment, this is a “place holder” page. I want to learn about the appearance of algebraic theories in functional analysis and shall record what I learn here. A preliminary outline is to find out about the following statements:

- The category of Banach spaces with linear short maps is not monadic over Set. The “nearest” algebraic theory is that of totally convex spaces.
- The category of Banach algebras is also not algebraic.
- The category of $C^*$-algebras is algebraic.

We consider the category of Banach spaces with linear short maps. That is, this is the category $\operatorname{Ban}$ with:

- Objects
- Banach spaces over $\mathbb{R}$
- Morphisms $E \to F$
- Linear short maps. That is, bounded linear transformations $T \colon E \to F$ such that $\|T\| \le 1$

We define a functor $B \colon \operatorname{Ban} \to \operatorname{Set}$ sending a Banach space to its unit ball. Since linear short maps $E \to F$ take the unit ball of $E$ into the unit ball of $F$, this is well-defined.

There is a functor in the opposite direction which assigns to a set the “free” Banach space on that set. That is, it assigns to a set $X$ the Banach space $\ell^1(X)$ of all absolutely summable sequences indexed by elements of $X$. It is a standard result that such a sequence must have countable support, no matter how large $X$ is.

$\ell^1$ is left adjoint to $B$.

We need to define the adjunction natural transformations: $\eta_X \colon X \to B \ell^1(X)$ and $\epsilon_E \colon \ell^1(B E) \to E$. The first is the map which assigns to $x$ the sequence $(\delta_{x y})$ which is $1$ at $x$ and $0$ elsewhere. The second is the summation map which assigns to an absolutely summable sequence $(a_e)$ indexed by $e \in B E$ its sum, $\sum a_e e$.

This adjunction defines a monad over $\operatorname{Set}$. Let us spell out the details. The functor $T \colon \operatorname{Set} \to \operatorname{Set}$ sends a set $X$ to the unit ball of $\ell^1(X)$. That is, an element of $T(X)$ is a weighted (formal) sum of elements of $X$, $\sum a_x$, such that $\sum |a_x| \le 1$. The unit for the monad sends an element $x \in X$ to the delta sequences in $T(X)$. The product, $\mu$, takes a “sum of sums” and evaluates them. That is, given a formal sum $\sum a_s$ where each $s$ is of the form $\sum s_x$, $\mu(\sum a_s) = \sum b_x$ where $b_x = \sum_s s_x$.

AS: I think! I need to check exactly how the product works in this example but I’m just getting the basic sketch down first.

The key question is whether or not $\operatorname{Ban}$ is (equivalent to) the category of algebras for this monad. That is, is $B \colon \operatorname{Ban} \to \operatorname{Set}$ *tripleable*? If not (as it will turn out), how close is it?

Beck's tripleability theorem gives three conditions for a functor to be tripleable. We already have one (the adjunction), let us show that the second also holds.

$B \colon \operatorname{Ban} \to \operatorname{Set}$ reflects isomorphisms.

Let $T \colon E \to F$ be a linear short map which induces an isomorphism on the unit balls of $E$ and $F$. It is evident that it is therefore a bijection from the underlying set of $E$ to that of $F$. Hence, by the open mapping theorem, it is a linear homeomorphism. It remains to show that $\|T(x)\| = \|x\|$ (so that its inverse is a short map as well). This is simple to show: if we had some $x \in E$ with $\|x\| = 1$ but $\|T(x)\| \lt 1$ (if it fails, it must fail that way as $T$ is short) then there would be some $\lambda \gt 1$ such that $\|T(\lambda x)\| \le 1$. As $B(T) \colon B E \to B F$ is surjective, there is some $y \in B E$ such that $T(y) = T(\lambda x)$. But as $\lambda \gt 1$, $\lambda x \notin B E$ so $\lambda x \ne y$, contradicting the injectivity of $T$. (Incidentally, this argument is valid constructively; it is a property of located real numbers that any number that is neither greater nor smaller than $1$ must equal $1$.)

AS: To be continued …

The above is essentially my “notes” on reading the following (and whatever necessary to understand the following):

Section 4.4 of *Toposes, Triples, and Theories* by Barr and Wells

*On the equational theory of $C^*$-algebras*, Pelletier, J. Wick and Rosický, J., MR1223636- Any more suggested by this question on mathoverflow

Revised on May 14, 2010 15:35:53
by Urs Schreiber
(131.211.233.156)