Let $f:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be such that $\left\| f(x)-f(y)\right\| =\left\| x-y\right\|$. Is $f$ necessarily surjective?

If this is so, you can prove (Mazur-Ulam Theorem) that $f$ is affine, and hence you could classify all isometries of $\mathbb{R}^n$. However, at the moment, I can't think of any good ideas to prove that $f$ is surjective. For that matter, is it even the case that $f$ must be surjective?

Any ideas would be most welcomed.

Thanks much!

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    $\begingroup$ In the Euclidean case it easily follows from polarization that $f$ is of the form $f(x) = f(0) + Ux$ where $U$ is orthogonal. Do you want more general norms? $\endgroup$
    – t.b.
    May 2, 2011 at 20:20
  • 1
    $\begingroup$ One can prove that $T(x):=f(x)-f(0)$ defines an orthogonal linear map (i.e. $T$ is an isomorphism and $\|Tx\|=\|x\|$). See e.g. page 148 of math.brown.edu/~treil/papers/LADW/LADW.pdf $\endgroup$ May 2, 2011 at 20:22
  • $\begingroup$ If $f$ is an isometry it is not so hard, let $T_v$ be the translation over $v$, then set $g:= T_{-v} \circ f$, then $g$ is an isometry with $g(0) = 0$ Hence, $g$ is an orthogonal map. Hence $f$ is surjective. $\endgroup$
    – JT_NL
    May 2, 2011 at 20:25
  • $\begingroup$ Just to make my point more clear (which was repeated by wildildildlife and Jonas). If you're dealing with Euclidean $\mathbb{R}^n$ polarization is sufficient and you don't need Mazur-Ulam at all. By the way: For a neat proof of Mazur-Ulam I'd recommend Väisälä's recent paper. $\endgroup$
    – t.b.
    May 2, 2011 at 20:32
  • $\begingroup$ @Theo,@Jonas It seems as if you are both using what I am trying to prove. How do you know $U$ is orthogonal (in this case, orthogonal=linear isometry). Of course $U$ is an isometry, but how is it linear? This requires an application of Mazur-Ulam, which requires surjectivity, hence, the reason I posed the question. A proof that $U$ is linear without using surjectivity would be an solution to my problem, however. $\endgroup$ May 2, 2011 at 20:34

4 Answers 4


Let's assume WLOG that $f(0) = 0$. For every $r$, it follows that $f$ defines an isometry from the sphere of radius $r$ to the sphere of radius $r$.

Proposition: Any isometry $f : X \to X$ of a compact metric space is bijective.

Proof. $f$ is clearly injective. Suppose $f$ is not bijective. Then $f(X)$ is compact, so given $x \in X \setminus f(X)$ the distance $\text{dist}(x, f(X))$ is positive. Pick $\epsilon < \text{dist}(x, f(X))$. Let $N$ be the smallest positive integer for which $X$ admits a cover by $N$ open sets of diameter less than $\epsilon$. No such set containing $x$ can intersect $f(X)$, but by pulling back along $f$ it follows that we can find a cover of $X$ by $N-1$ open sets of diameter less than $\epsilon$; contradiction.

(In fact any isometry of a compact metric space is a homeomorphism, since a continuous bijection from a compact space to a Hausdorff space is necessarily closed.)

Apparently there are counterexamples to the above when $X$ is not compact, but I don't know any nice ones off the top of my head.

  • 14
    $\begingroup$ Re: Last comment: simply take the unilateral shift $e_{n} \mapsto e_{n+1}$ on $\ell^2{(\mathbb{N})}$ (and restrict it to the unit sphere, if you want). $\endgroup$
    – t.b.
    May 2, 2011 at 21:04
  • $\begingroup$ @Theo: yes, I just thought of a similar example (shifting the right half of a function to the right on $L^2(\mathbb{R})$). Do you know of a counterexample which is locally compact? $\endgroup$ May 2, 2011 at 21:06
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    $\begingroup$ Once you have that $f$ is bijective, you don't need to work so hard to prove it is a homeomorphism. $f^{-1}$ is also clearly an isometry and isometries are continuous, therefore $f$ is a homeomorphism. In fact, without surjectivity, you still have that it is a homeomorphism onto its image. Really nice argument BTW. $\endgroup$ May 2, 2011 at 21:10
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    $\begingroup$ I don't think so, it seems that the linear structure (or some unique geodesic extension property from a base point arbitrarily far in all directions of a sphere around some base point) is essential. The silliest counterexample I can think of is to take the positive reals and to push them in the positive direction. $\endgroup$
    – t.b.
    May 2, 2011 at 21:12
  • $\begingroup$ @Theo: right. That's a pretty easy counterexample; thanks. $\endgroup$ May 2, 2011 at 21:14

Let $ f $ be an isometry of $ \mathbb{R}^n $ ( i.e. a map $ f : \mathbb{R}^n \rightarrow \mathbb{R}^n $ such that $ || f(x) - f(y) || = || x - y || $ for all $ x,y $ ). We can write $ f(x) = f(0) + ( f(x) - f(0) ) $ and now $ g(x) := f(x) - f(0) $ is an isometry fixing $ 0 $ ( especially $ g $ preserves norm, i.e. $ || g(x) || = || x || $ for all $ x $ ). We'll now focus on $ g $.

Notice $ g $ preserves inner product i.e. $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ for all $ x, y $ ( since expanding $ \langle g(x) - g(y), g(x) - g(y) \rangle = \langle x - y, x - y \rangle $ gives $ || g(x) ||^2 - 2 \langle g(x), g(y) \rangle + || g(y) ||^2 = || x ||^2 + || y ||^2 - 2 \langle x, y \rangle $ and so $ \langle g(x), g(y) \rangle = \langle x, y \rangle $ ).

Turns out $ g $ is linear too : Expanding $ || g(x+y) - g(x) - g(y) ||^2 $ as $ \langle g(x+y) - g(x) - g(y), g(x+y) - g(x) - g(y) \rangle $ gives \begin{align*} &|| g(x+y) ||^2 + || g(x) ||^2 + || g(y) ||^2 - 2 \langle g(x+y), g(x) \rangle - 2 \langle g(x+y), g(y) \rangle + 2 \langle g(x), g(y) \rangle \\ &= || x + y ||^2 + ||x||^2 + || y ||^2 - 2 \langle x + y, x \rangle - 2 \langle x + y, y \rangle + 2 \langle x, y \rangle\\ &= 0. \end{align*}
Therefore $ g(x+y) = g(x) + g(y) $ for all $ x,y $. Similarly we can show $ g(ax) = ag(x) $ for all $ a \in \mathbb{R} $ and $ x \in \mathbb{R}^n $.

So $ g(x) = Ax $ for some $ n \times n $ matrix $ A $. Notice $ g(e_j) = A_j $, where $ (e_1, \cdots, e_n) $ is the standard basis of $ \mathbb{R}^n $ and $ A_1, \cdots, A_n $ are the columns of $ A $. Since $ \langle g(e_i), g(e_j) \rangle = \langle e_i, e_j \rangle $, we see $ \langle A_i, A_j \rangle $ is $ 0 $ if $ i \neq j $ and $ 1 $ if $ i = j $. Therefore $ A^T A = I $, i.e. $ A $ is an orthogonal matrix.

To summarise : Let $ f $ be an isometry of $ \mathbb{R}^n $. Then $ f(x) = f(0) + Ax $ for some orthogonal matrix $ A $.
( So especially the $ f $ here is bijective, answering the original question )

Edit : Treil's "Linear Algebra done Wrong" mentioned in the comments seems to take the same approach ( which is also outlined in above answers ), making this quite redundant. Nevertheless I'll leave it undeleted.


First assume $f$ fixes the origin. Show that $f$ preserves the inner product. Then you can show that $f$ is linear. Since you can translate a general isometry $f$ to obtain a new isometry that fixes the origin, $f$ must be surjective, and in fact given by $Ax+b$, where $A$ is an orthogonal matrix and $b$ is a vector.

  • $\begingroup$ Sure, but the point was that, in showing that $f$ is linear, I needed to know a priori that $f$ was surjective. $\endgroup$ Nov 22, 2012 at 16:04


Yeah I thought that too at first, but you can show an isometry of $\mathbb{R}^n$ fixing the origin is linear without assuming that it's surjective. The key is the inner product, which, of course, you don't have in a general normed vector space. Preserving the inner product and fixing the origin implies that the map is linear (a great exercise). Then since it's injective and we're in finite and equal dimensions, it's also surjective. I found this thread because I had the exact same question as you. There are geometric proofs of surjectivity in $\mathbb{R}^2$ involving triangles or circles as well.


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