# How general is the convergence of the exponential function's power series?

I asked essentially this over two weeks ago on MSE, and nothing was else was posted to that question.

Let $\mathbf{V}$ be a Fréchet space whose underlying set is $V$.
Let $\;\; \beta \: : \: V\times V \: \to \: V \;\;$ be a continuous bilinear map
that has an identity element and is power-associative.
For vectors $v$ and non-negative integers $n$, define $\hspace{.02 in}v^{\hspace{.02 in}n}\hspace{.02 in}$ in the obvious way.

Does it follow that for all vectors $\hspace{.02 in}v$, $\;\;\; \displaystyle\sum_{n=0}^{\infty} \; \left(\hspace{-0.03 in}\frac1{n!}\hspace{-0.05 in}\cdot \hspace{-0.02 in}v^{\hspace{.02 in}n}\hspace{-0.05 in}\right) \;\;\;$ exists?

If no, what if we additionally assume that $\hspace{.02 in}\beta\hspace{.02 in}$ is associative
and/or commutative and/or every vector has an inverse?

• Have you tried looking into the theory of analyticity in Fréchet spaces? There are a list of criterion for convergence of such "power series". Only a passing thought, though. Oct 18, 2013 at 6:24

Let me give you a counter-example with an associative $\beta$, i.e. a Frechet algebra for which the exponentials do not exist in general: The main reason is that a Frechet algebra needs not to be locally multiplicatively convex. On the Weyl algebra with two generators $Q$ and $P$ subject to the commutation relations $[Q, P] = 1$ there are several locally convex topologies possible such that the completion yields a Frechet algebra. Now it is well-known that none of them can be locally multiplicatively convex. In fact, one can show that e.g. the exponentials of quadratic expressions in the generators like $Q^2$ do not converge.

If of course you have a Frechet algebra which is locally multiplicatively convex then you have an entire calculus and hence in particular exponentials of all elements.

Edit: here is just simple argument that in an algebra with elements satisfying ccr one can not have submultiplicative seminorms: you compute $[Q, \cdots [Q, P^n] \cdot]$ directly (with $n$ commutators) and get essentially $n!$ times the identity of the algebra. On the other hand, if there would be a submultiplicative seminorm then using the submultiplicativity and a simple counting of all terms gives that the seminorm on this expression grows like $2^n$ times the seminorm of the identity. Thus the seminorm has to vanish on the identity and hence it is identically zero. It is essentially the same proof as the one to show that there is no normed algebra with canonical commutation relations (ccr).

In case you are interested in, I have a recent preprint on the arXiv where several locally convex topologies for the Weyl algebra are discussed, none of course mutliplicatively convex ;)

• Stefan, what is a good reference for this failure of locally multiplicative convexity? Thanks! Oct 18, 2013 at 18:04
• What is "ccr"? $\:$ (Also, I find "mutliplocatively" funny.) $\;\;\;$
– user5810
Oct 19, 2013 at 10:00
• @Ricky Demer see edit ;) Oct 19, 2013 at 10:51
• @Stefan : $\:$ You haven't quite corrected that word. $\;\;\;$
– user5810
Oct 21, 2013 at 1:21

Suppose that $N$ is the set of all continuous semi-norms in $V$. For every semi-norm $p\in N$ there exists semi-norm $q \in N$ such that $p(\beta(a,b)) \leq q(a)q(b)$ --- this is a consequence of the continuity of $\beta$. Therefore $p(v^n)\leq q(v)^n$ and hence the series converges in any semi-norm which is enough.

• This is not possible: for $v^3$ you have two $\beta$'s you have to estimate. Thus estimating the first one with the seminorm $q$ requires estimating the second one by yet another seminorm. This way, we need more and more seminorms and you can not control the convergence. Oct 18, 2013 at 11:11