If I have a topological space, I say that a homeomorphic map preserves the structure of this space. Thus, in order to preserve topological properties we want to have a continuous bijection with a continuous inverse. A inner product space has some additional structure due to the inner product. Thus, I guess it is meaningful to demand that the appropriate isomorphism preserves the inner-product and hence must be unitary and linear (due to the vector space structure).

A normed space has actually less structure than a inner-product space, but apparently the isomorphism are the same kind of things: We want to have linear isometric isomorphisms.

In metric spaces, we are no longer interested in linear isomorphisms, as we are not dealing with vector spaces anymore. Thus, we can no longer demand that our isometric isomorphisms shall be linear.

Thus, if we have let's say the real numbers $\mathbb{R}$, with the canonical euclidean metric, then a translation would be metric preserving, although not norm-preserving in general. this means that the isomorphisms for metric spaces are actually more general than the ones for normed spaces, as every norm-preserving iso is a metric preserving one, but not vice versa. What I am confused about is, that I would say that if we have $$||Tx||=||x||$$ in a normed space whose norm is induced by a inner product, the inner-product is automatically preserved, too. Thus, there is an equivalence between the two isomorphism types.

Could anybody explain to me where this conceptual difference comes from?

If anything is unclear, please let me know.


1 Answer 1


Perhaps an example will help?

Consider the "taxicab metric" on $\mathbb{R}^2$ which is defined by $$||x||_1 = |x_1|+|x_2| \,\, \text{where $x = (x_1,x_2)$} $$ The self-isomorphisms of $\mathbb{R}^2$ with respect to this metric are exactly the translations, and compositions of translations with one of the eight norm-preserving isomorphisms, each of which is either a rotation by a multiple of $\pi/2$ or a reflection across one of the axes or one of the $45^\circ$ degree lines. So clearly not all inner product preserving isomorphisms are isometries of this metric.

As this example shows, different norms can define different metrics, and different metrics can have different self-isomorphisms.

Sometimes, by coincidence, or perhaps for some deeper reason, you will get two different norms whose metrics just so happen to have the same self-isomorphisms.

For example, the self-isomorphisms of $\mathbb{R}^2$ with respect to the metric given by the $||x||_1$ just so happen to be identical with the self-isomorphisms with respect to the metric given by the norm $$||x||_\infty = \text{max}\{|x_1|,|x_2|\} $$


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