If 2 vector space are isometric then they are isomorphic

Question

We call a map between 2 normed vector spaces is isometric isomorphism if it is an isometry and an linear isomorphism. Then, these 2 vector spaces are called isometrically isomorphic.

Suppose $X,\,Y$ be 2 metric vector space of dimension $n$. Suppose $f:\ X\longrightarrow Y$ is a isometric, that is $$d_X(x,y)=d_Y\big(f(x),f(y)\big),\ \forall x,y\in X.$$ I want to know if $X,Y$ are linear isomorphic ?

What I know is if $X,Y$ are normed $n-$vector space, then every isometry $f:\ X\longrightarrow Y$ must be affine, which means \begin{align} f(x)=L(x)+u, \end{align} for some bijective linear $L:\ X\longrightarrow Y$ and $u\in Y$. Then $L$ will be a isometric isomorphism.

But is it true for the preceding case ? Moreover, if 2 metric vector spaces $n$ dimensional are isometric, then would they be isomorphic ?

I hope someone will help me to clarify this. Thanks

Assuming you meant $X$ and $Y$ over the same base field, then as long as they are both of dimension $n$, they are automatically linearly isomorphic, so in that sense this is trivially true. — David Gao, Dec 31 '23 at 19:55
@user114263 no, my definition of metric vector space is just simply that it is a vector space with a metric — PermQi, Dec 31 '23 at 20:11
Are you asking about the existence of a linear isometric isomoprhism or just a linear isomorphism? — user114263, Dec 31 '23 at 20:12
Is this the same as https://math.stackexchange.com/questions/3183539/are-any-two-isomorphic-normed-linear-spaces-homeomorphic ? — Gerry Myerson, Dec 31 '23 at 20:20
@user114263 my question is if 2 metric vector spaces of same dimension are isometry, then should it be linear isomorphic ? The same question is for the normed vector spaces, my guess is in the normed case, it is true as I stated above. But when I see the link that Gerry Myerson gives, I see that this is not true. So maybe, the true statement for the normed case should be "If $X$ is normed space and $f:\ X\to X$ preserves the norm (the same norm, not different norm) then $f$ is affine". — PermQi, Jan 01 '24 at 04:06
Any two vector spaces (over the same field) of the same dimension are linearly isomorphic. Norms and metrics aren't needed. — user114263, Jan 01 '24 at 13:47
@user114263 you're right, I almost forgot that. So for what reason they define the concept "isometrically isomorphic" ? Any 2 $K-$vector spaces of the same dimension are always linearly isomorphic regardless they are isometric or not, so didn't that mean isometric vector spaces implies isometrically isomorphic vector spaces ? — PermQi, Jan 01 '24 at 13:57
Linear maps preserve algebraic structure, isometries preserve distances. Linear isometries preserve both. The importance of the result you cited on normed vector spaces is that if there is an isometry between them, then there is actually a linear isometry between them, and in fact every isometry is almost already linear (it's just affine, which is linear $+$ a constant). — user114263, Jan 01 '24 at 14:01
@user114263 I have another question : if a metric space $E$ of $n$-dimension (as we can consider $E$ as a topological manifold) is isometric to Euclidean space $\mathbb R^n$ with Euclidean distance, then is $E$ a vector space ? — PermQi, Jan 01 '24 at 16:09
First, I want to emphasize that an isometry$F$ of a normed vector space is not necessarily linear because it's possible that $F(0) \ne 0$. By the Mazur-Ulam theorem, it is ncessarily affine, which implies means the map $G(x) = F(x) - F(0)$ is a linear isometry. Second, although the proof of Mazur-Ulam requires at least some cleverness, the proof in the case for a Euclidean vector space is a nice exercise using the dot product. — Deane, Jan 01 '24 at 18:51

score 0 · Answer 1 · answered Jan 01 '24 at 14:10

The result that you have cited about isometries between normed spaces being affine (which is linear plus a constant and subtracting the constant gives a linear isometry) is that the norm, by its definition, has some interplay with the linear structure. For example, homogeneity: $\|tx\|=|t|\|x\|$ connects the norm to scalar multiples (algebraic operation). The triangle inequality connects the norm to sums (algebraic operation).

If we drop this and ask about general metrics which don't have any particular connection to the linear structure of the space, we don't get any nice result. For example, take $X=Y=\mathbb{R}$. Define $$d_X(x,y)=\left\{\begin{array}{ll} 0 & : x=y \\ 1 & : x\neq y, \{x,y\}\cap \{0\}=\varnothing,\\ 2 & : x\neq y, \{0\}\subsetneq \{x,y\}.\end{array}\right.$$

So any two non-zero points are distance $1$ from each other, and $0$ is distance $2$ from everything else.

$$d_Y(x,y)=\left\{\begin{array}{ll} 0 & : x=y \\ 1 & : x\neq y, \{x,y\}\cap \{1\}=\varnothing,\\ 2 & : x\neq y, \{1\}\subsetneq \{x,y\}.\end{array}\right.$$

Here, any two non-$1$ points are distance $1$ from each other, and $1$ is distance $2$ from everything else.

There are lots of isometries from $X$ to $Y$ (any bijection $F:X\to Y$ which has $F(0)=1$ is an isometry). But there are no linear isometries, because a linear operator takes $0$ to $0$ and an isometry from $X$ to $Y$ has to take $0$ to $1$.

I have another question : if a metric space $E$ of $n$-dimension (as we can consider $E$ as a topological manifold) is isometric to Euclidean space $\mathbb R^n$ with Euclidean distance, then is $E$ a vector space ? — PermQi, Jan 01 '24 at 16:08
Any set with cardinality at least the continuum can be made into a vector space. Indeed, for any cardinality $\mathfrak{a}\geqslant \mathfrak{c}$, choose a set $\Gamma$ with cardinality $\mathfrak{a}$ and let $c_0(\Gamma)$ denote the space of functions $f:\Gamma\to \mathbb{R}$ such that ${\gamma\in \Gamma:f(\gamma)\neq 0}$ is finite. Then $c_0(\Gamma)$ has the same cardinality as $\Gamma$. Moreover, $c_0(\Gamma)$ is a vector space with usual pointwise additional and scalar multiplication. If $T$ also has cardinality $\mathfrak{a}$, let $M:T\to c_0(\Gamma)$ be any bijection. — user114263, Jan 05 '24 at 22:45
We can then endow $T$ with a vector space structure by $$s+t:=f^{-1}(f(s)+f(t))$$ and $as = f^{-1}(af(s))$ for $a\in \mathbb{R}$. This makes $T$ into a vector space, and under these structures, $f$ becomes a linear bijection between $T$ and $c_0(\Gamma)$. So asking whether or not something is a vector space is not as meaningful of a question, because lots of things can be turned into vector spaces. — user114263, Jan 05 '24 at 22:48

If 2 vector space are isometric then they are isomorphic

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