Difference between distance and norm In special relativity, different observers measure lengths of objects differently, depending on their respective velocities. However, all observers agree on the invariant inner product of two vectors (4-vectors, that is.) But doesn't the norm of a vector correspond to the distance between two points in space (or spacetime)? Even in geometry and topology, as far as I can tell, the norm and the metric are tightly connected objects. If $(V,\left\|.\right\|)$ is a normed vector space, then there's a corresponding metric $d$ such that $d(a,b)=\left\|a-b\right\|$ for all $a$ and $b$ in $V$.
So why is one, the norm, invariant while the other, the distance, isn't?
Or is there a difference between the length of a rod and the distance between the end points of the rod as points in space?
 A: First, it's important to note that "displacement vectors" - which one might interpret as beginning at one spacetime point and ending at another - are a generally untenable concept if the spacetime in question possesses curvature. When one makes the jump from special relativity to general relativity then they need to be dispensed with, and there is some merit to the opinion that this should be done sooner rather than later. Nevertheless, flat Minkowski spacetime can be regarded as an affine space, so this question can be answered in that context.

So why is one, the norm, invariant while the other, the distance, isn't?

Your mistake is in mixing up space with spacetime.  The points in Minkowski spacetime are events, not just positions.  The norm of a displacement vector provides a notion of distance$^\dagger$ between events, not points in space - for example, one might measure the spacetime distance between the events "a firecracker goes off on the left side of my desk" and "my neighbor's office door slams shut." In general, this is not the same as the spatial distance between the left side of my desk and my neighbor's office door.
Noting that $\Delta s^2:= -c^2\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2$, this spacetime distance does correspond to the physical distance between the left side of my desk and my neighbor's office door if and only if $\Delta t$=0; that is, spacetime distance coincides with spatial distance if the two events in question are simultaneous.
This is where the relativity of simultaneity comes into play.  If Alice observes two events to be simultaneous, then Bob (who is moving with respect to Alice) generically will not.  As a result, to Alice the spacetime distance between the events and the spatial distance between the positions at which they occur will be the same, while the same will not be true for Bob.  But since both observers agree on the spacetime distance between the events, they must necessarily disagree on the spatial distance between the positions.

$^\dagger$Really more of a pseudodistance. Ordinarily we require distances to be positive semidefinite - meaning that they must always be greater than or equal to zero - and if the distance between any two points is zero, then they are the same point.  This is relaxed in relativity because the Minkowski metric is not positive semidefinite, but rather only non-degenerate; the set of events at zero spacetime "distance" away from a specified event $p$ is called the light cone of $p$.
A: 
In special relativity, different observers measure lengths of objects differently, depending on their respective velocities.

No: Considering one particular object whose (two) ends, say $A$ and $B$, remain separate and at rest wrt. each other, such these two may be attributed a particular value of (non-zero) distance wrt. other to begin with,
then two members of some particular inertial system, say $P$ and $Q$, which are identified by a simultaneity projection of $A$, and of $B$ resp., into this inertial system have distance
$$PQ = AB \, \sqrt{ 1 - (\beta_{PQ}[ \, A \, ])^2 },$$
where of course $$0 \lt (\beta_{PQ}[ \, A \, ])^2 = (\beta_{PQ}[ \, B \, ])^2 = (\beta_{AB}[ \, P \, ])^2 = (\beta_{AB}[ \, Q \, ])^2 \lt 1, $$
such that $AB \lt PQ$ and where $\beta_{PQ}[ \, A \, ]$ denotes the ratio of the (constant) speed of $A$ wrt. $P$ and $Q$ (and all members of their joint inertial system) to "speed of light in vacuum" (signal front speed), and so on.
The result value which $P$ and $Q$ would determine for their measurement of the length of the given object is therefore $PQ / \sqrt{ 1 - (\beta_{PQ}[ \, A \, ])^2 }$;  i.e. in any case (equal to) the distance value $AB$, regardless of the specific value of $(\beta_{PQ}[ \, A \, ])^2$.

However, all observers agree on the invariant inner product of two vectors (4-vectors, that is.)

Yes. Therefore also: all observers agree on the norm (a.k.a. "magnitude") of any one given 4-vector; based on their understanding and shared agreement on how values of this norm are to be determined.
For example: considering two (space-like separated) events, $\varepsilon_{AJ}$ (identified as coincidence event of $A$ and another suitable particiant $J$) and $\varepsilon_{BK}$ (identified as coincidence event of $B$ and another suitable particiant $K$) specified such that $A$'s indication of having met and passed $J$ was simultaneous to $B$'s indication of having met and passed $J$,
then the magnitude of the spacetime 4-vector $\overset{\Large \longrightarrow}{\varepsilon_{AJ} \, \varepsilon_{BK} }$ is assigned the distance value $AB$.
Or likewise: the magnitude of the spacetime 4-vector $\overset{\Large\longrightarrow}{\varepsilon_{AP} \, \varepsilon_{BQ} }$, which are specified such that $P$'s indication of having met and passed $A$ was simultaneous to $Q$'s indication of having met and passed $B$, is assigned the distance value $PQ$.

So why is one, the norm, invariant while the other, the distance, isn't?

The supposed distinction does not pertain to begin with: all observers agree on the distance between two particular ends (which remain at rest wrt. each other); just as all observers agree on the norm of a given 4-vector, or on the scalar product of any two given 4-vectors.
However: any one given 4-vector has different decompositions wrt. different systems of (orthogonal) basis 4-vectors. Consequently any decomposition and especially the magnitudes of the corresponding orthogonal 4-vectors (whose sum equals the one given 4-vector) generally differ for different choices of bases.
