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As implicit in Bill's answer and prooved in Konrad's the answer to http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces, any isometry group of a normed space is a subgroup $G$ of $O(n)$ that contains $-id$.

It may be a good starting point.

Does the positive dimension imply that the space contains an Euclidean plane?

Forget the original norm and consider Euclidean $n$-space with a closed subgroup of $O(n)$ acting on it.

Let $A$ be any non zero element of the Lie algebra tangent to $G$. Since $exp(\epsilon A)$ is a rotation, and every rotation is equivalent to a matrix that has $2\times 2$ rotation blocks in the diagonal and zeros elsewhere. Expressed in the same (or other?) coordinates, $A$ probably had $2\times 2$ antisymmetric blocks in the diagonal and zeros elsewhere (although I don't know right now how to prove this fact without abandoning the algebra). The point having first coordinate $1$ and then zeros moves in a circle around the origin. Hence the spaces contains an Euclidean plane.

Related stuff:

http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean http://mathoverflow.net/questions/12452/maximal-ellipsoid http://mathoverflow.net/questions/64269/towards-a-metric-characterization-of-euclidean-spaces (not too related but I'm trying to promote it).

2 added 265 characters in body

As implicit in Bill's answer and prooved in Konrad's the answer to http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces, any isometry group of a normed space is a subgroup $G$ of $O(n)$ that contains $-id$.

It may be a good starting point.

Does the positive dimension imply that the space contains an Euclidean plane?

Forget the original norm and consider Euclidean $n$-space with a closed subgroup of $O(n)$ acting on it.

Let $A$ be any non zero element of the Lie algebra tangent to $G$. Since $exp(\epsilon A)$ is a rotation, and every rotation is equivalent to a matrix that has $2x2$ 2\times 2$rotation blocks in the diagonal (each characterized by an angle$\phi_i$), and zeros elsewhere. Expressed in the same (or other?) coordinates,$A$probably had$2x2$2\times 2$ antisymmetric blocks with the same numbers $\phi_i$ over the diagonal (although I don't know right now how to prove this fact without abandoning the algebra). The point having first coordinate $1$ and then zeros moves in a circle around the origin. Hence the spaces contains an Euclidean plane.

Related stuff:

http://mathoverflow.net/questions/41211/easy-proof-of-the-fact-that-isotropic-spaces-are-euclidean http://mathoverflow.net/questions/12452/maximal-ellipsoid http://mathoverflow.net/questions/64269/towards-a-metric-characterization-of-euclidean-spaces (not too related but I'm trying to promote it).

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As implicit in Bill's answer and prooved in Konrad's the answer to http://mathoverflow.net/questions/13596/continuous-automorphism-groups-of-normed-vector-spaces, any isometry group of a normed space is a subgroup $G$ of $O(n)$ that contains $-id$.

It may be a good starting point.

Does the positive dimension imply that the space contains an Euclidean plane?

Forget the original norm and consider Euclidean $n$-space with a closed subgroup of $O(n)$ acting on it.

Let $A$ be any non zero element of the Lie algebra tangent to $G$. $exp(\epsilon A)$ is a rotation, and every rotation is equivalent to a matrix that has $2x2$ rotation blocks in the diagonal (each characterized by an angle $\phi_i$), and zeros elsewhere. Expressed in the same coordinates, $A$ probably had $2x2$ antisymmetric blocks with the same numbers $\phi_i$ over the diagonal (I don't know right now how to prove this fact without abandoning the algebra). The point having first coordinate $1$ and then zeros moves in a circle around the origin. Hence the spaces contains an Euclidean plane.