Lee: If a space is an absolute -Lipschitz retract, with , can you say anything of its injective hull ? Answer: probably no. For instance, the injective hull of Euclidean is which seems very far from although is an absolute -Lipschitz retract.

**1. Injective hulls: basics **

** 1.1. Extremal functions **

Let be a metric space. Leopoldo Nachbin, in the 1950’s, already considered the set

Definition 1The minimal elements of are calledextremal functions. The set of extremal functions is denoted by .

Note that if is compact, is extremal iff for all , there exists such that . In general,

is extremal iff for all , .

Note that all belong to . This yields a map .

Lemma 2Extremal functions are -Lipschitz.

*Proof:* Given , define by and for . Then , so

Let

Then the -norm defines a metric on .

Proposition 3 (Dress)There is a map such that

- .
- .

*Proof:* For , set , so is extremal iff . Note that . By definition, for all , ,

so , and . For every ,

Thus

Define (pointwise limit). Then . For all , , so

which tends to , so .

Remark 1If , an infinite iteration is indeed needed to obtain .

** 1.2. Injectivity **

Proposition 4is injective.

*Proof:* Let be metric spaces, let be -Lipschitz. For , , put

Then . and is -Lipschitz. For , . We have

thus . Extend to by setting . Then the extended is -Lipschitz.

Theorem 5 (Isbell)

- is injective.
- If is -Lipschitz, and fixes pointwise, then is the identity.
- If is -Lipschitz and is an isometric embedding, then is an isometric embedding.
- If is an isometric embedding and is injective, then there is an isometric embedding such that .

*Proof:* 2. , so by minimality. 3. follows from 2. and 4. follows from 3.

**2. First examples **

** 2.1. Polyhedral structure on , finite **

If is bounded, then . If is compact, so is (Arzela-Ascoli).

If is finite, is a polyhedron in . Note that and are convex, but is not in general. The polyhedral structure can be detected by looking at “equality graphs”. For , let be the set of pairs such that . Then is a graph with loops: iff . Also

iff has no isolated vertices.

Say a graph is *admissible* if there exists such that . Every admissible graph corresponds to a polyhedral cell in , whose interior points are precisely the functions such that . is a face of iff contains .

** 2.2. Dimension of **

If two functions , have the same graph, then for all , , thus

Hence, if there is a path from to in of length , then

On connected components of containing an odd cycle, and coincide. On other (even) components, there is exactly one degree of freedom to vary without changing the graph.

Proposition 6Define therankas the number of even components. Then .

Example 1.

Then the possible graphs are an edge (rank one) and one edge with one loop (rank zero). This produces an interval.

Example 2.

Then is a tripod. Indeed, admissible graphs of extremal functions can have at most one even component. The complete graph is admissible but odd, producing a vertex. Two-edge admissible graphs contribute three segments. Hinges with a loop contribute three vertices.

The tripod lies in on a triangular face (standard simplex) at the bottom of polyhedron . It takes infinitely many steps (iterates of ) to map a constant function to . It is a bit stupid.

Lee: why don’t you consider the optimal such that belongs to ?

Example 3has 4 points with one distance equal to and all other equal to or .

The graph with two disjoint edges (one of length ) is admissible and has rank , it contributes a -cell, a square sitting diagonally in . The resulting polyhedron is the union of the -cell with two opposite protruding edges, each of length .