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I think another interpretation is in terms of Euclidean geometry instead of hyperbolic geometry. If $\omega$ is a holomorphic 1-form on a Riemann surface $\Sigma$, then $|\omega|$ defines a Euclidean metric on $\Sigma$ with singularities at the zeroes of $\omega$. Think of the 1-form as locally defining a map to $\mathbb{C}$, and pull back the Euclidean metric. The metric near the zero is a cone of angle $2\pi k$, where $k$ is the order of the zero, since this is modeled locally on the map $z\to z^k$. The number of zeroes, counted with multiplicity, must be $2g-2$. A point $P\in \Sigma$ is a Weierstrass point if and only if there is a holomorphic 1-form with a zero at $P$ of order at least $g$. Equivalently, there is a flat cone metric (conformally equivalent to the underlying Riemann surface and with all cone angles of the form $2\pi k$) with a cone angle of $2\pi g$ at $P$ and trivial holonomy. I found this in some course notes of McMullen, but they seem to be no longer available online.

Addendum: I've added a link to McMullen's course notes on Riemann surfaces. Theorem 10.19 and the preceding discussion connect the Weierstrass points with holomorphic 1-forms. See Section 3 of another of McMullen's papers for a discussion of holomorphic 1-forms and branched Euclidean metrics,

Correction: I forgot that a holomorphic 1-form is not quite equivalent to a flat $2\pi k$-cone metric. The problem is that if one develops the surface into the plane around a closed loop, there might be a rotational part - that is why I added that there needs to be trivial holonomy. So a flat $2\pi k$ cone metric corresponds to a holomorphic 1-form twisted by a $U(1)$ character. It also corresponds to a special sort of holomorphic quadratic differential. It's still possible that high order zeroes of such twisted 1-forms correspond to Weierstrass points, but it's not equivalent to McMullen's characterization. As pointed out below by Dmitri, Troyanov has shown that there is a Euclidean structure on a surface with a cone point at any chosen point.

Addendum: Schmutz-Schaller stated this problem, which maybe is akin to what you are searching for (and as far as I know has not been answered). He gave a characterization of hyperelliptic Riemann surfaces in terms of closed geodesics.

Problem. Find a geometric characterization (based on closed geodesics) of Weierstrass points.

On a closed hyperbolic surface, there are points which have an open neighborhood which misses every simple closed geodesic. So there is certain "focussing" of simple closed geodesics.

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I think another interpretation is in terms of Euclidean geometry instead of hyperbolic geometry. If $\omega$ is a holomorphic 1-form on a Riemann surface $\Sigma$, then $|\omega|$ defines a Euclidean metric on $\Sigma$ with singularities at the zeroes of $\omega$. Think of the 1-form as locally defining a map to $\mathbb{C}$, and pull back the Euclidean metric. The metric near the zero is a cone of angle $2\pi k$, where $k$ is the order of the zero, since this is modeled locally on the map $z\to z^k$. The number of zeroes, counted with multiplicity, must be $2g-2$. A point $P\in \Sigma$ is a Weierstrass point if and only if there is a holomorphic 1-form with a zero at $P$ of order at least $g$. Equivalently, there is a flat cone metric (conformally equivalent to the underlying Riemann surface and with all cone angles of the form $2\pi k$) with a cone angle of $2\pi g$ at $P$ and trivial holonomy. I found this in some course notes of McMullen, but they seem to be no longer available online.

Addendum: I've added a link to McMullen's course notes on Riemann surfaces. Theorem 10.19 and the preceding discussion connect the Weierstrass points with holomorphic 1-forms. See Section 3 of another of McMullen's papers for a discussion of holomorphic 1-forms and branched Euclidean metrics,

Correction: I forgot that a holomorphic 1-form is not quite equivalent to a flat $2\pi k$-cone metric. The problem is that if one develops the surface into the plane around a closed loop, there might be a rotational part - that is why I added that there needs to be trivial holonomy. So a flat $2\pi k$ cone metric corresponds to a holomorphic 1-form twisted by a $U(1)$ character. It also corresponds to a special sort of holomorphic quadratic differential. It's still possible that high order zeroes of such twisted 1-forms correspond to Weierstrass points, but it's not equivalent to McMullen's characterization.

Addendum: Schmutz-Schaller stated this problem, which maybe is akin to what you are searching for (and as far as I know has not been answered). He gave a characterization of hyperelliptic Riemann surfaces in terms of closed geodesics.

Problem. Find a geometric characterization (based on closed geodesics) of Weierstrass points.

On a closed hyperbolic surface, there are points which have an open neighborhood which misses every simple closed geodesic. So there is certain "focussing" of simple closed geodesics.

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I think another interpretation is in terms of Euclidean geometry instead of hyperbolic geometry. If $\omega$ is a holomorphic 1-form on a Riemann surface $\Sigma$, then $|\omega|$ defines a Euclidean metric on $\Sigma$ with singularities at the zeroes of $\omega$. Think of the 1-form as locally defining a map to $\mathbb{C}$, and pull back the Euclidean metric. The metric near the zero is a cone of angle $2\pi k$, where $k$ is the order of the zero, since this is modeled locally on the map $z\to z^k$. The number of zeroes, counted with multiplicity, must be $2g-2$. A point $P\in \Sigma$ is a Weierstrass point if and only if there is a holomorphic 1-form with a zero at $P$ of order at least $g$. Equivalently, there is a flat cone metric (conformally equivalent to the underlying Riemann surface and with all cone angles of the form $2\pi k$) with a cone angle of $2\pi g$ at $P$ and trivial holonomy. I found this in some course notes of McMullen, but they seem to be no longer available online.

Correction: I forgot that a holomorphic 1-form is not quite equivalent to a flat $2\pi k$-cone metric. The problem is that if one develops the surface into the plane around a closed loop, there might be a rotational part - that is why I added that there needs to be trivial holonomy. So a flat $2\pi k$ cone metric corresponds to a holomorphic 1-form twisted by a $U(1)$ character. It also corresponds to a special sort of holomorphic quadratic differential. It's still possible that high order zeroes of such twisted 1-forms correspond to Weierstrass points, but it's not equivalent to McMullen's characterization.

Addendum: Schmutz-Schaller stated this problem, which maybe is akin to what you are searching for (and as far as I know has not been answered). He gave a characterization of hyperelliptic Riemann surfaces in terms of closed geodesics.

Problem. Find a geometric characterization (based on closed geodesics) of Weierstrass points.

On a closed hyperbolic surface, there are points which have an open neighborhood which misses every simple closed geodesic. So there is certain "focussing" of simple closed geodesics.

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