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If $d(x,y)$ and $e(x,y)$ are metrics then $d(x,y)+e(x,y)$ and $\frac{d(x,y)}{1+d(x,y)}$ are metrics.

If $d_i(x,y)$ for $i=1,\dots,n$ are metrics then so is $\sqrt{\sum_{i=1}^n{d_i^2(x,y)}}$

Are there other interesting ways of constructing new metrics from old metrics?

Greg Kuperberg
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Kim Greene
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7 Answers7

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Personally, I prefer $\min(d(x,y),\epsilon)$ over the standard $d(x,y)/(1+d(x,y))$ trick if the goal is to turn a metric into an equivalent metric – it's a lot easier to prove the metric axioms for this one.

From that follows one way to turn a countably infinite number of metrics into a new metric: $d(x,y)=\sum\_{n=1}^\infty \min(d\_n(x,y),2^{-n})$. The point being that convergence in the new metric is the same as convergence in all of the original ones. So this can be used to prove stuff like $C(\mathbb{R})$ with the topology of uniform convergence on compact sets is a metrizable space, and similarly for the space of test functions used in distribution theory. (The individual $d\_n$ could even be pseudometrics rather than metrics; it doesn't matter, so long as $x\ne y$ implies $d\_n(x,y)\ne0$ for some $n$.)

Harald Hanche-Olsen
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Generalizing d(x,y)/(1+d(x,y)): Let d(x,y) be a metric, and let $f: R^+ \to R^+$ be a function with $f(0) = 0$, $f^\prime(x) > 0$ for all x, and $f^{\prime\prime}(x) < 0$ for all x. Then I believe h(x,y) = f(d(x,y)) is a metric. For example, the square root of a metric is a metric.

Michael Lugo
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  • Does it always induce the same topology as d? – Qiaochu Yuan Nov 16 '09 at 22:45
  • I think so. They have the same balls. – Michael Lugo Nov 16 '09 at 23:05
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    Generalizing slightly more: instead of differentiability assumptions on $f:\mathbb{R}^+\to\mathbb{R}^+$ you only need $f(0)=0$, $f$ is non-decreasing, and $f$ is concave. This then also subsumes the $\min(d,\varepsilon)$ example. – Konrad Swanepoel Nov 17 '09 at 15:39
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    I would like to add that the $d^\alpha$ metrics, with $0<\alpha<1$, have some interesting properties. For example, every path has infinite length for any of these metrics, $d$ being any metric.

    It is an open question to determine for which $n,\alpha, N$ one can find a bi-Lipschitz embedding of $(\mathbb{R}^n,d^\alpha_{eucl})$ into $(\mathbb{R}^N,d_{eucl})$.

     – Benoît Kloeckner Feb 18 '12 at 06:07
  • Concave defined on all of $\mathbb{R}^{+}$ (and valued in $\mathbb{R}^{+}$) implies non-decreasing. Moreover, one can apply the same process to a bounded d, in which case non-decreasing assumption is not necessary. – Max M Oct 23 '15 at 12:55
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If you have a smooth submanifold of a Riemannian manifold, of course you have the subspace metric, but you also have the induced path metric, which is usually the "right" metric to think about.

Autumn Kent
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I have to confess to a dislike of the word "interesting" for these constructions! So far, all that's been said is:

  1. Any metric is equivalent to a bounded metric, so "boundedness" is not a topological property.

  2. Countable products of metric spaces are metrisable, so the category of metric spaces has countable products. (Incidentally this construction fails for arbitrary small products.)

To that I can add arbitrary small coproducts: if $x,y$ are in the same component, take $d(x,y) = min(d_X(x,y),1)$ and if $x,y$ are in different components, take $d(x,y) = 1$.

What this shows is that from a topological point of view, it's usually best to have a bounded metric.

But I would say that an interesting construction is one that is not functorial, for example, something where you replace the given metric locally by another one. For example, on a manifold we can always replace a given metric by one which is flat near a point (indeed, flat near quite a large number of points) but we can't necessarily replace it by one which is flat everywhere.

Andrew Stacey
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This example can be found in Munkres (p. 133). If $X_i$ is a countable collection of metric spaces with metrics $d_i$, let $\bar{d_i} = \text{min}(d_i, 1)$ denote the standard bounded metric (which gives the same topology as $d_i$) and let $d = \sup \frac{ \bar{d_i} }{i}$ be defined on the product $\prod X_i$. Then this metric induces the product topology.

Qiaochu Yuan
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The second example in the original post generalizes a lot. Let $d_i$ be finitely or countably many pseudometrics (it's possible for $d_i(x,y)=0$ even if $x\neq y$) for $i\ge 1$, and assume $d_1$ is a metric. Let $X$ be a space of sequences with a 1-unconditional norm, i.e., $$ \lVert (x_1,x_2,\ldots) \rVert = \lVert (\pm x_1, \pm x_2, \ldots) \rVert$$ for every choice of signs. For example, $X=\ell^p$ works. Then $$d(x,y) = \lVert (d_1(x,y),d_2(x,y),\ldots ) \rVert$$ is a metric.

For example, the metrics on $C^k$ spaces are examples of this kind of construction.

Mark Meckes
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Associated to any metric is a length structure. The length of an absolutely continuous path is the integral of it's metric derivative. Associated to any length structure is a (pseudo)-metric. If this is an honest metric, it is also a path metric: for any $x, y\in X$ and any $\varepsilon>0$, there is some $z\in X$ such that $d(x, z)+d(z, y)\le d(x, y)+\varepsilon$.

Taken from Gromov, "Metric structures for Riemannian and non-Riemannian spaces".

Martijn
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