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Unless I am misunderstanding a lot of works, it is my understanding that a finite and non negative measure $\mu=g\mathcal{H}^\alpha$, where $\mathcal{H}^\alpha$ is the $\alpha$-Haudorff measure, admits $\lfloor{\alpha}\rfloor$ Alberti representations.

This is the united effort of a papers by Bate, De Philippis, Schioppa and others.

In particular this should mean that any continuous curve $f:\mathbb R\to \mathbb R$ will be so that its graph has dimension $\alpha\geq 1$ and will always admit at least one Alberti representation.

What I fail to see is how the Lipschitz curves of this representation will be chosen as it feels like an unreasonable idea to be able to split the graph of the Weierstrass function as a superposition of Lipschitz fragments.

Am I misunderstanding something and this is impossible? Or given any continuous map $g:t\mapsto (t,f(t))$ we can construct a bounded nowhere zero vector field $X:\mathbb R^2\to \mathbb R^2$ such that $X g_\#\mathcal{L}^1$ is a metric current?

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  • $\begingroup$ I don't think $\mathcal H^\alpha|_E$ necessarily admits an Alberti rep. if $\mathcal H^\alpha(E)<\infty$ and $\alpha>1$, e.g. $E=$ Van Koch snowflake. $\endgroup$
    – Teri
    Commented Nov 7 at 8:20
  • $\begingroup$ @Teri but that is literally my question. Do you have a proof that the Koch snowflake has no Alberti representation? It would be very helpful to see it. $\endgroup$
    – Lolman
    Commented Nov 7 at 9:18
  • $\begingroup$ It should follow from the fact that $E$ is purely 1-unrectifiable. If $\mathcal H^\alpha|_E$ had an alberti representation, you could find a non-trivial bi-Lipschitz piece of a curve (1-rectifiable set) inside $E$. $\endgroup$
    – Teri
    Commented Nov 7 at 13:14
  • $\begingroup$ @Teri yes, that sounds very circular: a set is purely 1-unrectifiable if and only if it has no Alberti representation. Is there a proof that $E$ doesn't have fragments in it? $\endgroup$
    – Lolman
    Commented Nov 7 at 17:44
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    $\begingroup$ It can indeed happen that your measure is singular wrt $\mathcal H^1|_{\gamma}$ for all curve fragments $\gamma$, but this doesn't mean the space is purely 1-unrectifiable. For example, the Sierpinski gasket with the correct dimensional Hausdorff measure (and Euclidean metric) is quasiconvex, in particular contains full line segments (and thus isn't purely 1-unrectifiable). However the measure does not admit any Alberti representation. $\endgroup$
    – Teri
    Commented Nov 10 at 10:43

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The implication does not work. What it is true is that if a measure admits $k$ Alberti representations, it is absolutely continuous with respect to $\mathcal H^k$. Actually, it is absolutely continuous with respect to the $k$-integral geometric measure. In particular, if the measure is concentrated on a $\mathcal H^k$ $\sigma$-finte set and admits $k$ Alberti representations, it is $k$-rectifiable.

A counterexample to your claim is $\mathcal H^1$ restricted to $E=K\times K$ where $K$ is the $1/4$ Cantor set. Indeed $E$ is purely unrectiable.

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  • $\begingroup$ Yeah I have finally understood the 1/4 cantor set just this afternoon. What I don't understand is the two implications you have mentioned. If I take the measure on the standard 1/3 cantor set $\nu$ and consider $\nu\times \mathcal{L}^1$ restricted to $[0,1]^2$, won't this be a measure with only 1 Alberti representation but not absolutely continuous w.r.t. $\mathcal{H}^1$? Moreover the 1/4 cantor set times itself is a $\sigma$ finite set for $\mathcal{H}^1$, wouldn't your claim make it 1-rectifiable? $\endgroup$
    – Lolman
    Commented Nov 9 at 0:39
  • $\begingroup$ In your first example the measure is indeed absolutely continuous with respect to $\mathcal H^1$, any $\mathcal H^1$ negligible set, it is also $nu \times \mathcal L^1$ negligible. i probably do not understand the second question, the cartesian product of $1/4$ Cantor set with itself is is purely not rectifiable, and this is coherent with the fact that it has no Alberti representations. $\endgroup$
    – Guido De Philippis
    Commented Nov 11 at 10:35
  • $\begingroup$ Ah yes, i forget the importance of sigma finiteness for radon nykodym for my first example. The second question comes from your assertion "In particular, if the measure is concentrated on a $\mathcal{H}^k$ σ-finte set, it is k-rectifiable." I am missing something to make it coherent with the cantor set example. $\endgroup$
    – Lolman
    Commented Nov 11 at 11:10
  • $\begingroup$ I wil edit the answer to make it more clear "In particular if a measure is concentrated on a $k$-rectifiable set and it admits $k$ independent Alberti representations, it is $k$-rectifiable" $\endgroup$
    – Guido De Philippis
    Commented Nov 11 at 11:37

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