1 The main result and its proof

To state our result, first recall from [Reference Wang27] or even [Reference Birkhoff4, p. 470] that for an integer $n \geqslant 1$ , a metric space $(X,d)$ is n-point homogeneous if given two finite subsets $Y,Y' \subseteq X$ with $|Y| = |Y'| \leqslant n$ , any isometry $T : Y \to Y'$ extends to an isometry $: X \to X$ . We will require a somewhat more restrictive notion.

We now motivate our main result (and the above definition via onto isometries). It seems to be folklore that the Euclidean space $\mathbb {R}^k$ is n-point homogeneous for all n – and more strongly, satisfies that every isometry between finite subsets ${Y,Y' \subseteq \mathbb {R}^k}$ , upon pre- and post-composing with suitable translations in order to send $0$ to $0$ , extends to an orthogonal linear map $: \mathbb {R}^k \to \mathbb {R}^k$ .

Returning to our motivation: in fact all inner product spaces satisfy this “orthogonal extension property” for all pairs of isometric finite subsets $Y,Y'$ , as we explain below. Here, we are interested in the converse question, i.e.,

1. (a) if this “orthogonal extension property” (with $Y,Y'$ finite) characterizes inner product spaces (among normed linear spaces); and

2. (b) if yes, then how much can this property be weakened without disturbing the characterization – and if it can in fact be weakened to use metric geometry alone. (In an arbitrary normed linear space, we necessarily cannot use orthogonality or inner products; but we also want to not use the vector space operations either.)

This note shows that indeed (a) holds. Moreover, (b), we can indeed weaken the orthogonal extension property to (i) replacing the orthogonal linear map by merely an onto isometry – not necessarily linear a priori – and (ii) working with three-point subsets $Y,Y'$ . More precisely, we have the following theorem.

To the best of our ability – and that of a dozen experts – we were unable to find such a result proved in the literature. Before proceeding to its proof, we discuss the assertion for $n=1,2$ . If $n=1,$ then Theorem 4 fails to hold, since every normed linear space $\mathbb {B}$ is strongly one-point homogeneous: given $x, y \in \mathbb {B}$ , the translation $z \mapsto z + y-x$ is an onto isometry sending x to y.

If instead $n=2,$ then one is asking if there exists a (real) linear space $(\mathbb {B}, \| \cdot \|)$ with $\| \cdot \|^2$ not arising from an inner product, such that given any $f,f',g,g' \in \mathbb {B}$ with $\| f'-f \| = \| g'-g \|$ , every isometry sending $f,f'$ to $g,g'$ , respectively, extends to an onto isometry of $\mathbb {B}$ . By pre- and post-composing with translations, one can assume $f=g=0$ and $\| f' \| = \| g' \|$ ; now one is asking if every real normed linear space with transitive group of onto-isometries fixing $0$ (these are called rotations) is isometrically isomorphic to an inner product space. Thus, we come to the well-known Banach–Mazur problem [Reference Banach2] – which was affirmatively answered by Mazur for finite-dimensional $\mathbb {B}$ [Reference Mazur21], has counterexamples among non-separable $\mathbb {B}$ [Reference Pełczyński and Rolewicz24], and remains open for infinite-dimensional separable $\mathbb {B}$ . (This is also called the Mazur rotations problem; see the recent survey [Reference Cabello Sánchez, Ferenczi and Randrianantoanina6].) This is when $n=2$ ; and the $n \geqslant 3$ case is Theorem 4.

Proof of Theorem 4

We begin by proving the real case. We first explain the forward implication, starting with $\mathbb {B} \cong (\mathbb {R}^k, \| \cdot \|_2)$ for an integer $k \geqslant 1$ . As is asserted (without proof) on [Reference Birkhoff4, p. 470], $\mathbb {R}^k$ is n-point homogeneous for every $n \geqslant 1$ ; we now show that it is moreover strongly n-point homogeneous – in fact, that it satisfies the “orthogonal extension property” above.

Let $Y,Y' \subseteq \mathbb {R}^k$ , and let $T : Y \to Y'$ be an isometry. By translating Y and $Y'$ , assume $0 \in Y \cap Y'$ and $T(0) = 0$ . Then, T extends to an orthogonal self-isometry $\widetilde {T}$ of ${(\mathbb {R}^k, \| \cdot \|_2)}$ .

This claim can be proved from first principles and is likely folklore (e.g., see, a skeleton argument in [Reference Blumenthal5, Section 38]). However, for self-completeness, we present a detailed sketch via a “lurking isometry” argument, along with some supplementary remarks. For full details, see [Reference Khare19, Theorem 22.3] and its proof.

The first step is to note that vectors $y_0, \dots , y_n$ in an inner product space are linearly dependent if and only if their Gram matrix $G := ( \langle y_i, y_j \rangle )_{i,j=0}^n$ is singular, since

(1.1) $$ \begin{align} v := \sum_{j=0}^n c_j y_j = \mathbf{0} \quad \implies \quad G \mathbf{c} = \mathbf{0} \quad \implies \quad \mathbf{c}^* G \mathbf{c} = 0 \quad \implies \quad \| v \|^2 = 0. \end{align} $$

This simple fact yields several noteworthy consequences; we mention two here, without proof. The first is an 1841 result by Cayley [Reference Cayley7] (during his undergraduate days).

Lemma 7 Let $(X = \{ x_0, \dots , x_n \}, d)$ be a finite metric space, and $\Psi : X \to \ell ^2$ an isometry. Then, the vectors $\Psi (X)$ are affine linearly-dependent (i.e., lie in an $(n-1)$ -dimensional affine subspace) if and only if the Cayley–Menger matrix of X is singular:

(1.2) $$ \begin{align} CM(X)_{(n+2) \times (n+2)} := \begin{pmatrix} 0 & d_{01}^2 & d_{02}^2 & \cdots & d_{0n}^2 & 1\\ d_{10}^2 & 0 & d_{12}^2 & \cdots & d_{1n}^2 & 1\\ d_{20}^2 & d_{21}^2 & 0 & \cdots & d_{2n}^2 & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ d_{n0}^2 & d_{n1}^2 & d_{n2}^2 & \cdots & 0 & 1\\ 1 & 1 & 1 & \cdots & 1 & 0 \end{pmatrix}, \ \text{where} \ d_{ij} = d_X(x_i, x_j).\nonumber\\ \end{align} $$

The second consequence is used below, but also underlies the global positioning system (GPS) “trilateration” – i.e., that every point on say an Euclidean plane P is uniquely determined by its distances from three non-collinear points in P. More generally, we have the following.

(See, e.g., [Reference Khare19, Proposition 22.7].) Returning to the proof of the orthogonal extension property for $\mathbb {R}^k$ : fix a maximal linearly independent subset $\{ y_1, \dots , y_r \}$ in Y. Then, we claim, so is $\{ T(y_1), \dots , T(y_r) \}$ . Indeed, by polarization we have

(1.3) $$ \begin{align} 2 \langle y_i, y_j \rangle = &\ \| y_i - 0 \|_2^2 + \| y_j - 0 \|_2^2 - \| y_i - y_j \|_2^2 \nonumber\\= &\ \| T(y_i) - T(0) \|_2^2 + \| T(y_j) - T(0) \|_2^2 - \| T(y_i) - T(y_j) \|_2^2 = 2 \langle T(y_i), T(y_j) \rangle\nonumber\\ \end{align} $$

for $1 \leqslant i,j \leqslant r$ . Thus, the Gram matrix of the $T(y_i)$ equals that of the $y_i$ , and hence is invertible as well; so by (1.1), the $T(y_j)$ are linearly independent too. Moreover, for any other $y \in Y$ , the Gram matrix of $y, y_1, \dots , y_r$ is singular by (1.1), hence, so is the Gram matrix of $T(y), T(y_1), \dots , T(y_r)$ by (1.3).

Next, we claim that the linear extension $\widetilde {T}$ of T from $\{ y_1 \dots , y_r \}$ to $\mathrm {span}_{\mathbb {R}}(Y)$ (hence, mapping into $\mathrm {span}_{\mathbb {R}}(Y')$ ) agrees with T on Y. Indeed, by maximality one writes $y \in \mathrm {span}_{\mathbb {R}}(Y)$ as $\sum _{j=1}^r c_j(y) y_j$ and $T(y) \in \mathrm { span}_{\mathbb {R}}(Y')$ as $\sum _{j=1}^r c^{\prime }_j(y) T(y_j)$ ; then (1.3) and Proposition 8 show that $c_j \equiv c^{\prime }_j$ on Y (for all j). Hence, $\widetilde {T} \equiv T$ on Y.

It also follows from (1.3) that $\widetilde {T}$ preserves lengths, hence is injective. Finally, choose orthonormal bases of the orthocomplements in $\mathbb {R}^k$ of $\mathrm {span}_{\mathbb {R}}(Y)$ and of $\widetilde {T} \left ( \mathrm {span}_{\mathbb {R}}(Y) \right )$ , and map the first of these bases (within $\mathrm {span}_{\mathbb {R}}(Y)^\perp $ ) bijectively onto the second; then extend $\widetilde {T}$ to all of $\mathbb {R}^k$ by linearity. Direct-summing these two orthogonal linear maps yields the desired extension of T to a linear self-isometry $\widetilde {T}$ of $(\mathbb {R}^k, \| \cdot \|_2)$ .

This linear orthogonal map $\widetilde {T}$ is necessarily injective on $\mathbb {R}^k$ , hence surjective as well. This shows the orthogonal extension property for all (possibly infinite) isometric subsets $Y,Y' \subseteq \mathbb {R}^k$ , and hence the forward implication in the main result for $\mathbb {B} \cong \mathbb {R}^k$ .

If instead $\mathbb {B}$ is an infinite-dimensional inner product space, and T an isometry between $Y, Y' \subset \mathbb {B}$ of common size at most n, first pre- and post-compose by translations to assume $0 \in Y \cap Y'$ and $T(0) = 0$ . Now let $\mathbb {B}_0 \cong (\mathbb {R}^k, \| \cdot \|_2)$ be the span of $Y \cup Y'$ . By the above analysis, $T : Y \to Y'$ extends to an orthogonal operator on $\mathbb {B}_0$ , which we still denote by T; as $\mathbb {B}_0$ is a complete subspace of $\mathbb {B}$ , by the “projection theorem” (or from first principles) we get $\mathbb {B} = \mathbb {B}_0 \oplus \mathbb {B}_0^\perp $ . Now the bijective orthogonal map $T|_{\mathbb {B}_0} \oplus \mathrm {Id}|_{\mathbb {B}_0^\perp }$ completes the proof of the forward implication – for any $n \geqslant 1$ .

We next come to the reverse implication; now $n \geqslant 3$ . From the definitions, it suffices to work with $n=3$ . Moreover, the cases of $\dim \mathbb {B} = 0,1$ are trivial since $\mathbb {B}$ is then an inner product space, so the reader may also assume $\dim \mathbb {B} \in [2,\infty ]$ in the sequel, if required.

We work via contradiction: let $(\mathbb {B}, \| \cdot \|)$ be a nonzero three-point homogeneous normed linear space, with $\| \cdot \|^2$ not induced by an inner product. Then, Lorch’s condition (IP5) fails to hold for any $\gamma ' \in \mathbb {R} \setminus \{ 0, \pm 1 \}$ . Fix such a scalar $\gamma '$ ; then there exist vectors $f',g' \in \mathbb {B}$ with $\| f' \| = \| g' \|$ but $\| f' + \gamma ' g' \| \neq \| g' + \gamma ' f' \|$ . In particular, $0, f', g'$ are distinct. Now let $Y = Y' = \{ 0, f', g' \}$ , and consider the isometry ${T : Y \to Y'}$ which fixes $0$ and interchanges $f',g'$ . By hypothesis, T extends to an onto isometry $: \mathbb {B} \to \mathbb {B}$ , which we also denote by T. But then T is affine-linear by the Mazur–Ulam theorem [Reference Mazur and Ulam22], hence is a linear isometry as $T(0) = 0$ . This yields

$$\begin{align*}\| f' + \gamma' g' \| = \| f' - (- \gamma') g' \| = \| T(f') - T(- \gamma' g') \| = \| g' + \gamma' f' \|, \end{align*}$$

which provides the desired contradiction, and proves the reverse implication.

The above analysis proves the real case; now suppose $(\mathbb {B}, \| \cdot \|)$ is a nonzero complex normed linear space. The forward implication follows from the real case, since ${(\mathbb {B}, \| \cdot \|)}$ is also a real normed space – which we denote by $\mathbb {B}|_{\mathbb {R}}$ . For the same reason, in the reverse direction, we obtain that $\mathbb {B}|_{\mathbb {R}}$ is isometrically isomorphic to a real inner product space: $\| f \| = \sqrt { \langle f, f \rangle _{\mathbb {R}} }$ for a real inner product $\langle \cdot , \cdot \rangle _{\mathbb {R}}$ on $\mathbb {B}|_{\mathbb {R}}$ and all $f \in \mathbb {B}|_{\mathbb {R}}$ . Now the complex polarization trick of Jordan–von Neumann [Reference Jordan and von Neumann16] gives that

(1.4) $$ \begin{align} \langle f, g \rangle := \langle f, g \rangle_{\mathbb{R}} - i \langle if, g \rangle_{\mathbb{R}} \end{align} $$

is indeed a complex inner product on $\mathbb {B}$ satisfying: $\| f \| = \sqrt {\langle f, f \rangle }$ for all f.

2 A second proof: Lorch’s characterizations for complex linear spaces

We next provide a second proof of the “reverse implication” of Theorem 4 for complex linear spaces $\mathbb {B} = \mathbb {B}|_{\mathbb {C}}$ , which requires the complex version of Lorch’s condition (IP5) above. Note that the above argument over $\mathbb {R}$ cannot immediately proceed verbatim over $\mathbb {C}$ for two reasons:

1. (a) Lorch’s condition (IP5) needs to be verified as characterizing a complex inner product.

2. (b) The Mazur–Ulam theorem does not go through over $\mathbb {C}$ – e.g., the isometry $\eta : (z,w) \mapsto (\overline {z},w)$ of the Hilbert space $(\mathbb {C}^2, \| \cdot \|_2)$ sends $(0,0)$ to itself, but is neither $\mathbb {C}$ -linear nor $\mathbb {C}$ -antilinear.

However, since $\eta $ is $\mathbb {R}$ -linear on $\mathbb {C}^2$ , this reveals how to potentially fix (b) for a second proof of the reverse implication: it suffices to use Lorch’s condition (IP5) for $\gamma '$ still real – and avoiding $0, \pm 1$ as above – now for all vectors $f',g' \in \mathbb {B}|_{\mathbb {C}}$ . If this still characterizes a complex inner product, then one could continue the above proof verbatim, using the Mazur–Ulam theorem for real normed linear spaces and replacing the word “linear” twice by “ $\mathbb {R}$ -linear” in the proof.

Thus, we need to verify if Lorch’s condition (IP5) characterizes a complex inner product. More broadly, one can ask which of Lorch’s conditions $(I_1)$ – $(I_6)$ and $(I^{\prime }_1) =$ (IP5) in [Reference Lorch20] – which characterized an inner product in a real normed linear space – now do the same over $\mathbb {C}$ .

In fact, these characterizations have since been cited and applied in many papers that work over real linear spaces. The need to work over complex normed linear spaces – as well as the fact that several characterizations in [Reference Day9, Reference Ficken11, Reference Jordan and von Neumann16] before Lorch held uniformly over both $\mathbb {R}$ and $\mathbb {C}$ – provide natural additional reasons to ask if Lorch’s conditions also characterize complex inner products.

We quickly explain why this does hold. Indeed, Lorch’s characterizations themselves involve real scalars, so even if one starts with a complex normed linear space $\mathbb {B}$ , one still obtains a real inner product $\langle \cdot , \cdot \rangle _{\mathbb {R}}$ on $\mathbb {B} = \mathbb {B}|_{\mathbb {R}}$ . Now the discussion around (1.4) recovers the complex inner product from $\langle \cdot , \cdot \rangle _{\mathbb {R}}$ . This yields the following.

In particular, this provides a second proof of one implication in Theorem 4 over  $\mathbb {C}$ .

3 Isosceles triangle characterization:weaker notions of homogeneity

Note that the arguments in Section 1 in fact show a strengthening of Theorem 4. Namely, continuing the discussion preceding Theorem 4, the “orthogonal extension property” can be weakened to strong three-point homogeneity, and even weaker – wherein one only works with $Y = Y'$ the vertices of an isosceles triangle in $\mathbb {B}$ (with the two equal sides of specified length).

Indeed, we showed above that $(4) \implies (1) \implies (2)$ (as the $f',g'$ in Lorch’s condition (IP5) can be simultaneously rescaled), while $(2) \implies (3) \implies (4)$ is trivial. To see why one cannot assert $\mathbb {C}$ -linearity in the second statement, let

$$\begin{align*}\mathbb{B} = (\mathbb{C}^2, \| \cdot \|_2), \qquad Y = \{ (0,0), (1,0), (i,0) \}, \qquad Y' = \{ (0,0), (1,0), (0,1) \}, \end{align*}$$

and let $T : Y \to Y'$ fix $(0,0)$ and $(1,0)$ , and send $(i,0)$ to $(0,1)$ . Then, T necessarily cannot extend to a $\mathbb {C}$ -linear map. Also note that akin to Theorem 4, here too the final three assertions each characterize inner products using (isosceles) metric geometry alone, and without appealing to the vector space structure in $\mathbb {B}$ .

We conclude by exploring two weakenings of the notion of “strong” n-point homogeneity that was used to characterize when $\mathbb {B}$ is an inner product space. The first question is to ask if one can work with “usual” n-point homogeneity, i.e., if one can remove the “onto” part of that definition. Note that in the literature, the requirement of extending subset-isometries to onto-isometries of the whole (metric) space is perhaps more natural than merely into-isometries, given their appearance in well-known results and open problems in the Banach space literature – the Mazur–Ulam theorem [Reference Mazur and Ulam22], the Banach–Mazur problem [Reference Banach2, Reference Mazur21], and Tingley’s problem [Reference Tingley26] among others – and also given the folklore fact that this holds for all inner product spaces and in our results above. Now having studied the “onto” picture in detail, the first question (above) is natural.

To answer it, note that the proof of Theorem 4 required the “onto” hypothesis solely to invoke the Mazur–Ulam theorem. This hypothesis can thus be bypassed, at a cost.

Theorem 13 Suppose $(\mathbb {B}, \| \cdot \|)$ is a nonzero real or complex normed linear space that is moreover strictly convex:

$$\begin{align*}\| a+b \| = \|a\| + \|b\| \quad \implies \quad a,b \in \mathbb{B} \text{ are linearly dependent}. \end{align*}$$

Then, $\| \cdot \|^2$ arises from a (real or complex, respectively) inner product if and only if $\mathbb {B}$ is n-point homogeneous for any (every) $n \geqslant 3$ – equivalently, if the flip map on the vertices of each isosceles triangle extends to a self-isometry of $\mathbb {B}$ .

In particular, this reveals yet another equivalent condition to the inner product property: $\mathbb {B}$ is strictly convex and (“usual”) three-point homogeneous.

We formulate the natural question corresponding to the remaining case.

We end by answering a natural question that arises from Theorem 12(4). Note that (strong) n-point homogeneity concerns subsets $Y,Y'$ of size at most n. However, Theorem 12(4) shows that one can work with subsets of size exactly $n=3$ . Given Theorem 4, one can ask if subsets of size exactly $4$ , or any fixed integer $>4$ , will suffice to isolate an inner product. Our final result provides an affirmative answer, thereby adding to Theorem 12.

Proof Clearly $(2) \implies (5)$ . Conversely, it suffices to assume $(5)$ and show Theorem 12(1). We follow the proof of Theorem 4: if $\| \cdot \|$ is not induced by an inner product, there exist $\gamma ' \in \mathbb {R} \setminus \{ 0, \pm 1 \}$ and $f',g' \in \mathbb {B}$ with $\| f' \| = \| g' \|$ , but $\| f' + \gamma ' g' \| \neq \| g' + \gamma ' f' \|$ . Thus, $0, f', g'$ are distinct.

There are now two cases. First, if $g' = -f',$ then the flip map extends to the onto isometry $T \equiv - \mathrm {id}_{\mathbb {B}}$ . Else $f', g'$ are $\mathbb {R}$ -linearly independent, so their $\mathbb {R}$ -span is a real normed plane $P \subseteq \mathbb {B}$ , say. As all norms on $\mathbb {R}^2$ are equivalent, $(P, \| \cdot \|)$ is linearly bi-Lipschitz – hence homeomorphic – to $(\mathbb {R}^2, \| \cdot \|_2)$ .

We now claim that the “equidistant locus in P” – i.e., the locus $Z_{f',g'} := \{ h \in P : \| h - f' \| = \| h - g' \| \}$ is uncountable. Indeed, if this is not true then $Z_{f',g'}$ is at most countable, so $P \setminus Z_{f',g'}$ is homeomorphic to a countably-punctured Euclidean plane, hence is path-connected. As the function

$$\begin{align*}\varphi : P \to \mathbb{R}; \quad h \mapsto \| h - f' \| - \| h - g' \| \end{align*}$$

is continuous on $P \setminus Z_{f',g'}$ , and switches signs from $f'$ to $g'$ , it must vanish at an “intermediate” point in $P \setminus Z_{f',g'}$ . This contradiction shows the claim.

Finally, as $n \geqslant 3$ , choose distinct points $h_1, \dots , h_{n-3} \in Z_{f',g'} \setminus \{ 0 \}$ , and define $Y, Y'$ via:

$$\begin{align*}Y = Y' := \{ 0, f', g', \ h_1, \dots, h_{n-3} \}. \end{align*}$$

Let $T : Y \to Y'$ interchange $f',g'$ and fix all other points in Y. By hypothesis, T extends to an onto isometry of $\mathbb {B}$ . Now repeating the proof of Theorem 4, it follows that $\mathbb {B}$ is a real inner product space. Finally, if $\mathbb {B}$ was a complex linear space then the discussion around (1.4) reveals the complex inner product structure.