Suppose, $X$ is a normed linear space and $M$ be a closed subspace.Then we can define a norm on $X/M$ by $\|\tilde x\|=\|x-M\|=\inf\limits_{m\in M}\|x-m\|$.I want to know why this should be the natural norm on this space.I have some intuition but I am not satisfied by that.I have observed that for $\mathbb R^n,n\geq 3$ it coincides with the natural one.If we visualize that quotient operation as squishifying the cosets of $M$ to a single point ,then for any point $x\in X$,$x\in x_0+M$ for some $x_0\in X$,then image under the quotient operation is the orthogonal projection on the orthogonal complement $M^\perp$.and the norm of $\tilde x$ which is the distance of $\tilde x$ from $\tilde 0$,will be the distance of $x_1$ from the projection of $x$ onto $M^\perp$ which is nothing but the distance of the subspace $M$ from $x$ $($actually the distance between $x+M$ and $0+M$ as cosets(think of them as planes or lines)$)$.

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Is my intuition alright?I am not sure because in case of infinite dimensional normed spaces,our intuition of orthogonal complement may not hold good.

  • 1
    $\begingroup$ It’s the unique norm satisfying the universal property of the quotient, but for that you need to know what the correct morphisms of normed spaces are; they are linear maps of norm $\le 1$. For more on this you can look up Lawvere metric spaces. $\endgroup$ Oct 6, 2022 at 6:35
  • $\begingroup$ @QiaochuYuan Lawvere metric spaces! It is a surprise to see category theory here. $\endgroup$
    – FShrike
    Oct 6, 2022 at 6:41
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    $\begingroup$ In the quotient, the subspace $M$ is collapsed into the origin. So the quotient norm is just measuring distance from the (new) origin, in my intuition, which is a reasonable and natural consideration $\endgroup$
    – FShrike
    Oct 6, 2022 at 6:42
  • $\begingroup$ The quotient space has aseleents translates of a common linear space. In Calculus 101 we learn how to estimates distances between parallel planes, that geometric picture is what one should have in mind in this situation (similar to what FDhrike had in mind). Meditations about category theory are uncesserary here in my opinion. $\endgroup$
    – Mittens
    Oct 6, 2022 at 23:27

1 Answer 1


Here is a somewhat indirect answer which I believe is "philosophically correct." We'll answer the following questions in turn:

  1. What is a metric space, really?
  2. What is a morphism between metric spaces?
  3. What is the natural metric on an arbitrary quotient of a metric space by an equivalence relation?

The first question is somewhat philosophical but the second question has a precise technical answer (which is motivated by the first question) and given the answer to the second question the third question has a unique answer. All of what follows is thoroughly motivated by the notion of a Lawvere metric.

Q1: I claim that the correct way to interpret the "meaning" of the metric space axioms is that they are describing the abstract properties that the "length of the shortest path" should have, given a fairly arbitrary notion of "path." The key axiom is the triangle inequality, which means: if I join a path from $x$ to $y$ and a path from $y$ to $z$, then I get a path from $x$ to $z$. So the shortest path from $x$ to $z$ is at most as long as the shortest path from $x$ to $y$ joined with the shortest path from $y$ to $z$. Hence $d(x, y) + d(y, z) \le d(x, z)$.

Then the symmetry axiom $d(x, y) = d(y, x)$ "means" that reversing a path gives us another path, and the nondegeneracy axiom means that the only paths of length $0$ are the "identity" paths from a point to itself. This is reminiscent of the notion of a category, and we can in fact produce a large class of examples of metric spaces via the following construction: select a collection of points $X$ in $\mathbb{R}^n$ and a category of paths between them, meaning a collection of (rectifiable) paths $p_{x, y} : [0, 1] \to \mathbb{R}^n$ between points $x$ and $y$ such that

  • the identity path $p_{x, x}$ (the function with constant value $x$) is a path,
  • the reversal $p(1 - t)$ of a path is a path, and
  • the join of two paths is a path.

Then if we define $d(x, y)$ to be the infimum of the length of all paths between two points $x$ and $y$ we get a metric on $X$. (Strictly speaking for a "category of paths" we only need the first and third axioms and the second one can be ignored. The second one corresponds to a dagger category of paths. We can in fact drop the symmetry axiom from the definition of a metric space and we get a generalization called a quasimetric space. Intuitively this corresponds to, for example, the metric incorporating an "energy penalty," e.g. it is easier to walk downhill than uphill. So we might say that in this case the metric measures "difficulty" or "cost" of traveling along a path, in a way that generalizes "length.")

Q2: Given Q1, what should a morphism between two metric spaces $f : M \to N$ be? The intuitive idea here is that $f$ sends paths in $M$ to paths in $N$, so the length of a path between two points $n_1, n_2 \in N$ is at most the length of the paths given by images of paths in $M$.

This motivates the following definition. A weak contraction or short map is a map $f : M \to N$ between two metric spaces which is weakly distance-decreasing in the sense that

$$d_M(m_1, m_2) \ge d_N(f(m_1), f(m_2)).$$

Note that unlike continuous functions, taking these to be the morphisms in the category of metric spaces gives that an isomorphism of metric spaces is an isometry. So with this definition we can now hope that categorical operations on metric spaces exist, and in particular have unique metrics by the Yoneda lemma.

Q3: Now we consider a metric space $M$ and an arbitrary equivalence relation $\sim$ on it. What metric does the quotient $M/\sim$ inherit, if any? If $[m]$ denotes the equivalence class of $m \in M$ then we see that, in order for the quotient map $M \to M/\sim$ to be a weak contraction, we must have

$$d_{M/\sim}([m_1], [m_2]) \le d_M(m_1, m_2)$$

and since this must be true for any $m_1', m_2'$ equivalent to $m_1, m_2$ we must in fact have

$$d_{M/\sim}([m_1], [m_2]) \le \inf_{m_1' \sim m_1, m_2' \sim m_2} d_M(m_1', m_2').$$

Conversely, if we simply define the metric to be this infimum then the quotient map is a weak contraction and this metric satisfies the universal property of the quotient... except that it might be zero for distinct points. We can fix this by either dropping the nondegeneracy axiom (which is my personal preference) or taking the further quotient of $M/\sim$ by the equivalence relation given by two equivalence classes having distance zero from each other.

What this metric means, intuitively, is that the shortest path between two equivalence classes $[m_1], [m_2]$ is the shortest path between any two members of those equivalence classes. And this specializes to the special case of quotients of normed vector spaces by subspaces as desired.


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