# Semi-stability of Ulrich bundle

A vector bundle $E$ on a smooth projective variety $X$ is called Ulrich bundle if it is Arithmetically Cohen-Macaulay , i.e., $H^i(E(t)) = 0$ for all $t \in Z$ and $0 < i < k$ and with Hilbert polynomial $cr\binom{t+k}{k}$ for some linear projection $X\to \mathbb P^k$

The following statement appeard in the survey paper of Beauville https://link.springer.com/article/10.1007%2Fs40879-017-0154-4

Let $E$ be an Ulrich bundle of rank $r$ on a non-singular projective variety $X$ why $E$ is slope semi-stable?

• Use the fact that $\pi _*E$ is trivial, hence semi-stable -- where $\pi :X\rightarrow \mathbb{P}^k$ is a general linear projection.
– abx
Jun 8, 2017 at 16:49
• From the definition of Ulrich bundle, for some linear projection $\pi : X → \mathbb P^k$ we have $π_∗E \cong \mathcal O_{\mathbb P^k}^{cr}$ so its direct image is Mumford semi-stable(since it is trivial) and if a direct image of vector bundle be Mumford semi-stable then that vector bundle is ,Mumford semi-stable itself
– user21574
Jun 8, 2017 at 17:13
• Moreover you can define Ulrich bundle on a singular variety as Ulrich sheaf(see Hartshorne paper) and still Ulrich sheaf remain semi-stable in the sense of Mumford. The main question is related to existence of Ulrich sheaf which seems to be very hard
– user21574
Jun 8, 2017 at 17:21

From the definition of Ulrich bundle, for some linear projection $\pi : X → \mathbb P^k$ we have $π_∗E \cong \mathcal O_{\mathbb P^k}^{cr}$ so its direct image is Mumford semi-stable(since it is trivial) and if a direct image of vector bundle be Mumford semi-stable then that vector bundle is ,Mumford semi-stable itself.Moreover you can define Ulrich bundle on a singular variety as Ulrich sheaf(see Hartshorne paper) and still Ulrich sheaf remain semi-stable in the sense of Mumford. The main question is related to existence of Ulrich sheaf which seems to be very hard