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One has that a normed vector space (of arbitrary dimension) is an inner product space if and only if it satisfies the parallelogram law. Likewise, a (finite-dimensional) normed vector space is an inner product space if and only if the isometry group acts transitively on the unit sphere.

A reference for the latter statement can be found, for example, in the section on Finsler geometry in Volume 2 of Spivak's Comprehensive Introduction to Differential Geometry.

It follows from the above two statements that, in finite dimensions, a normed vector space satisfies the parallelogram law if and only if it is isotropic/transitive/the isometry group acts transitively on the unit sphere/the unit sphere is homogeneous. Whether the two are equivalent in infinite dimensions appears to be an open problem, called the Banach-Mazur rotation problem.

Question: Is there a way to see intuitively that, in finite dimensions (e.g. in $\mathbb{R}^2$), these two conditions, the parallelogram law and isotropy, should be equivalent?

In particular, are there any examples which allow one to geometrically visualize how the two properties are related? To make their equivalence "geometrically intuitively plausible"?

For example, how could drawing parallelograms allow me to see that the isometry group of such a space acts transitively on the unit sphere?

A simple example in $\mathbb{R}^2$ would be greatly appreciated -- an example in any other space would probably strain my visualization capabilities, and as noted below, it seems that understanding the two-dimensional case for some reason is sufficient to understand any finite-dimensional case.

Note: Such an understanding seems implicit in this question:

Consider a normed vector space $V$. Suppose that for every pair of unit vectors $v,w$ there exists a linear isometry which sends $v$ to $w$ (and leaves the subspace spanned by $v$ and $w$ invariant)... Does it follow that $V$ is an inner product space? By the formulation and the parallelogram law the problem immediately reduces to the two-dimensional case.

My question above may then perhaps be answered indirectly by answering this question:

Why does the formulation of the problem and the parallelogram law immediately reduce the problem to the two-dimensional case?

Related questions (not a comprehensive list):
- A norm which is symmetric enough is induced by an inner product?
- Isometry group of a norm is always contained in some Isometry group of an inner product?
- Isometry group of a norm is always contained in some Isometry group of an inner product?
- A real vector space is an inner product space if every two dimensional subspace is an inner product space ?
- Is this geometric intuition for the relationship between curvatures correct?
- Does the notion of "rotation" depend on a choice of metric?
- Are there two different ways to generalize the orthogonal group?

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