# Heisenberg group, convex analysis, alike or not?

I posted on mathoverflow a question, with the purpose of clarifying the feelings I have concerning the formal resemblance between Heisenberg group and cyclically monotone operators from convex analysis. The question got an answer which made me realize something trivial (but sometimes we need somebody else to point to the obvious). However, I still think there is something worthy of further consideration here, that’s why I post it.

Setting: Let $S$ be a semigroup (i.e. has an associative operation with neutral element $e$) and let $(A,+)$ be a commutative group (with neutral element $0$).

Let’s say that a semigroup extension of $S$ with $A$ is any operation on $S \times A$, of the form

$(s,a)(s',a') = (s s' , a+a'+ \lambda(s,s'))$

where $\lambda: S \times S \rightarrow A$ is a function such that $S \times A$ with the mentioned operation is a semigroup with neutral element $(e,0)$. Obviously, the operation is encoded by the function $\lambda$, which has to satisfy:

$\lambda(s s', s") + \lambda(s,s') = \lambda(s, s's") + \lambda(s',s")$  (from associativity)

$\lambda(e,s) = \lambda(s,e) = 0$  (from the neutral element condition).

Here are two examples. Both are written in the setting of bipotentials, namely   $X$ and $Y$ are topological, locally convex, real vector spaces of dual variables $x \in X$ and $y \in Y$, with the duality product $\langle \cdot , \cdot \rangle : X \times Y \rightarrow \mathbb{R}$.

The spaces $X, Y$ have topologies compatible with the duality product, in the sense that for any continuous linear functional on $X$ there is an $y \in Y$ which puts the functional into the form $x \mapsto \langle x,y\rangle$ (respectively any continuous linear functional on $Y$ has the form $y \mapsto \langle x,y\rangle$, for an $x \in X$).

Example 1: (Heisenberg group) Let $S = X \times Y$ with the operation of addition of pairs of vectors and let $A = \mathbb{R}$ with addition. We may define a Heisenberg group over the pair $(X,Y)$ as $H(X,Y) = S \times A$ with the operation

$(x,y,a)(x',y',b) = (x+x', y+y', a+a'+ \langle x, y'\rangle - \langle x', y \rangle)$
Fact: There is no injective morphism from $(X\times Y, +)$ to $H(X,Y)$.

Example 2: (convex analysis) Let this time $S = (X \times Y)^{*}$, the free semigroup generated by $X \times Y$, i.e. the collection of all finite words with letters from $X \times Y$, together with the empty word $e$, with the operation of concatenation of words.

Let $A$ be the set of bi-affine real functions on $X \times Y$, i.e. the collection of all functions $a: X \times Y \rightarrow \mathbb{R}$ which are affine and continuous in each argument. $A$ is a commutative group with the addition of real valued functions operation.

We define the function $\lambda: S \times S \rightarrow A$ by:

$\lambda(e, c)(x,y) = \lambda(c,e)(x,y)=0$ for any $c \in S$ and any $(x,y) \in X \times Y$.
– if $c, h \in S$ are words $c = (x_{1},y_{1})...(x_{n}, y_{n})$ and $h = (u_{1},v_{1})...(u_{m}, v_{m})$, with $m,n \geq 1$, then

$\lambda(c,h)(x,y) = \langle u_{1} - x , y_{n} - y \rangle$ for any $(x,y) \in X \times Y$.

This $\lambda$ induces a semigroup extension operation on $S \times A$.

Fact: there is an injective morphism $F: S \rightarrow S \times A$, with the expression $F(c) = (c, E(c))$.

Here, for any $c = (x_{1},y_{1})...(x_{n}, y_{n})$ the expression $E(c)(x,y)$ is the well known circular sum associated to the “dissipation during the discrete cycle” $(x_{1},y_{1})...(x_{n}, y_{n})(x,y)$, namely:

$E(c)(x,y) = \langle x_{1},y\rangle + \langle x,y_{n}\rangle - \langle x,y \rangle + \sum_{k=1}^{n-1}\langle x_{k+1}, y_{k}\rangle - \sum_{k=1}^{n} \langle x_{k},y_{k} \rangle$

which appears in convex analysis, related to cyclically monotone operators.

The main difference between those two examples is that the example 2. is a direct product of a semigroup with a group. There are many resemblances though.