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If I collapse a (say, closed) set in a length space, I obtain a length space: is there some literature on this?

We consider length spaces as defined by Gromov and others. [However the case of a Riemannian distance already leads to interesting examples of what I am writing]. If $X$ is one such space, it has a distance $d$ which can be recovered by the length of curves. Suppose $X$ is path connected. Let $c:[a,b]\to X$ be a curve, define $|c'(t)|=\lim_{\epsilon\to0}\sup_{|u-t|,|v-t|\le\epsilon}\frac{d(c(u),c(v))}{|u-v|}$ and $$ L(c)=\int_a^b|c'(t)|dt, $$ and set $$ D(x,y)=\inf\{L(c) \text{ $c$ is a curve having endpoints $x$ and $y$}\}. $$ We are interested in cases where $d=D$ and where the distance $D=d$ is realized by lengths of geodesics (i.e. the hard work has been already done).

Now, let $E\subseteq X$ be closed and define a distance $D_X$ on $X\setminus E\cup\{E\}$ ($X$ with $E$ collapsed to a point): $$ L_E(c)=\int_a^b|c'(t)|\chi_{X\setminus E}(c(t))dt, $$ and set $$ D_E(x,y)=\inf\{L_E(c) \text{ $c$ is a curve having endpoints $x$ and $y$}\}, $$ if $x,y\in X\setminus E$, and $D_E(x,E)=D(x,E)=\inf_{y\in E}D(x,y)$.

The space $X\setminus E\cup\{E\}$ is clearly a length space w.r.t. the distance $D_E$.

The reason I find these objects interesting is that, if $E$ is the smooth boundary of an open subset of $X$, a nice Riemannian manifold, then the points of $E$ play the role of the unit vectors in the tangent space of a point; this meaning that they can parametrize those geodesics leaving $E$ which, at least locally, minimize the distance from $E$. Even in the case of the Euclidean plane one obtains interesting pictures.

I would be very surprised if no one had developed this viewpoint in the past.

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  • $\begingroup$ Post Scriptum: one may view the definition of $L_E$ and $D_E$ as a limiting case in optics, with the light having infinite speed inside $E$. $\endgroup$ – Nicola Arcozzi Jun 11 at 11:57
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    $\begingroup$ you mean $\chi_{X\setminus E}$ instead of $\chi_E$? $\endgroup$ – Emanuele Paolini Jun 11 at 13:36
  • $\begingroup$ Yes! Thanks Emanuele... I edit. $\endgroup$ – Nicola Arcozzi Jun 11 at 15:20
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    $\begingroup$ On a side note: for a continuous, non-constant curve of finite length $c:[a,b]\to X$ one may have $|c'(t)|=0$ a.e., thus $\int_a^b|c'(t)dt=0$ (e.g. for the Cantor function). In fact, I think this is always true up to reparametrizations, so you would have $D(x,y)=0$ for all $x,y$! However, the equality "integral of |c'(t)| = length" is true for Lipschitz curves, which is ok for many purposes since continuous curves, finite length curves are Lipschitz up to reparametrization. $\endgroup$ – Pietro Majer Jun 12 at 7:42
  • $\begingroup$ (more precisely: if $c:[0,1]\to X$ is any curve and $C:[0,1]\to[0,1]$ is the Cantor function and ${\bf C}\subset[0,1]$ is the Cantor set, then $|(c\circ C)'|=0$ a.e., because $c\circ C$ is locally constant in ${\bf C^c}$) $\endgroup$ – Pietro Majer Jun 12 at 8:14
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If you take $X=\mathbb R$ and $E=[1,2]\cup[3,4]$ you see that $D_E(0,5) = 3$ while the geodesic distance between $0$ and $5$ in $X\setminus E \cup\{E\}$ is $2$ because you can join $0$ to $E$ with a curve of length $1$ and $E$ to $5$ with a curve of length $1$. I think you need to require $E$ to be geodesic-convex if you want the resulting space to be a length space.

Maybe what you really mean is $$ d_E(x,y) = \min\{d(x,y), d(x,E) + d(y,E)\}. $$ This is the particular case of a quotient space. It is understood that the quotient space of a length space is a length space if the quotient has good properties (in your case $E$ being closed). See, for example, https://people.math.ethz.ch/~lang/LengthSpaces.pdf Proposition 3.2

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  • $\begingroup$ You are right Emanuele: with the definition I gave I am collapsing each connected component of $E$ to a distinct point (which is what I had in mind in this morning's mental picture), while you definition collapses the whole set to a point (which is my usual mental picture and which gives your definition). I have read the Proposition in Lang's notes. Both mental pictures correspond to his assumptions, with different equivalence relations. $\endgroup$ – Nicola Arcozzi Jun 11 at 15:33

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