Why is the geometric median called the $L_1$ estimator? My question is simply, why is the geometric median called the $L_1$ estimator? This always reminds of $L_p$ spaces but the distance being minimized in the geometric median's definition isn't $L_1$ but rather the $L_2$ (Euclidean) norm. What does the $L_1$ refer to?
Is it just a misnomer/an accident/a freak of nature/something historical?
 A: Given a collection $\{\mathbf x^{(i)}\}$ of $N$ points in $\mathbb R^m$, their geometric median is a point $\mathbf y$ minimizing the sum of Euclidean distances to each point: $$L = \sum_{i=1}^N \|\mathbf x^{(i)} - \mathbf y\|.$$ Each Euclidean distance in this sum is indeed a $L_2$ norm of a vector in $\mathbb R^m$. But  consider a vector $\mathbf{d} \in \mathbb R^N$, whose coordinates are given by these distances: $d_i = \|\mathbf x^{(i)} - \mathbf y\|$. Then the same cost function $L$ can be equivalently written as the $L_1$ norm of this vector: $$L=\sum_{i=1}^N d_i = \sum_{i=1}^N |d_i| = \|\mathbf d\|_1.$$
That is, I believe, why the geometric median is called a $L_1$ estimator.
A: [Response rewritten]
I think I was too confusing, I apologize for that. Now I am trying to give a proper answer.
We know that median minimize the $L_1$ norm. The formula is
$$ L_1 = \underset{y \in \mathbb{R}}{\operatorname{arg\,min}}\sum_{i=1}^{n}|x_i-y|$$
Also the mean minimize the $L_2$ norm. Again, the formula is
$$ L_2 = \underset{y \in \mathbb{R}}{\operatorname{arg\,min}}\sum_{i=1}^{n}(x_i-y)^2$$
In plain English we say that the median minimize the sum of distances and the mean minimize the sum of squares of those distances. We note also that we are in $\mathbb{R}$. 
My idea is that because we are in $\mathbb{R}$, the distance function can be any particular case of p-norm, the result would be the same. So I generalize by saying that the distance is p-norm (it might be any type of distance in fact) and to finish quicker we move to $\mathbb{R}^m$ at the same time
$$L_1 = \underset{y \in \mathbb{R}^m}{\operatorname{arg\,min}}\sum_{i=1}^{n}|(\sum_{j=1}^{m}(x_{i,j}^p-y_j^p))^\frac{1}{p}| = \underset{y \in \mathbb{R}^m}{\operatorname{arg\,min}}\sum_{i=1}^{n}\|x_i-y\|_p $$
What is important here is that it does not matter what is the value for $p$, it will be an $L_1$. [Note, as suggested by @amoeba, there are two norms, one inside another; the first one is $L_1$ applied on distances, and a nested one applied on the elements of the vectors in $\mathbb{R}^m$].
Going back to your original question, the geometric median is defined as the point in Euclidean space which minimize the sum of distances. I believe the reason for the word geometric comes from Euclidean space and Euclidean distance (which is $\|.\|_2$) and minimize the sum of distances (not the squares as in the case of an $L_2$ estimator), so
$$GM = L_1 = \underset{y \in \mathbb{R}^m}{\operatorname{arg\,min}}\sum_{i=1}^{n}\|x_i-y\|_2 $$
As a final note we might choose to minimize:


*

*the sum of Manhattan distances ($L_1$ and $\|.\|_1$ as distance)

*the sum of squares of Manhattan distances ($L_2$ and $\|.\|_1$ as distance)

*the sum of Euclidean distances (geometric median)

*the sum of squares of Euclidean distances ($L_2$ and $\|.\|_2$ as distance)
and so on.

