Q:

Degree of filtered vector bundle

Suppose I have the sheaf $\mathscr{M}$ defined by 
$$0\to \mathscr{M}\to \mathscr{O}_{\mathbf{P}^r}^{r+1}\to \mathscr{O}_{\mathbf{P}^r}(1)\to 0 $$
that is, $\mathscr{M}\simeq \Omega_{\mathbf{P}^r}^1(1)$ is the twisted cotangent sheaf. 
Now, I can prove that if $X\subseteq \mathbf{P}^r$ is a projective non singular curve then $\mathscr{M}|_X$ can be filtered as
$$\mathscr M = \mathscr M_0\supset \mathscr M_1\supset \ldots \supset \mathscr M_r\supset 0 $$
in a way such that $\mathscr M _i/\mathscr M _{i+1}=\mathscr L _i$ is an invertible sheaf of strictly negative degree.
In Eisenbud's book The Geometry of Syzygies at the end of chapter 5 it is assumed that 
$$\sum \deg \mathscr L _i = -\deg X$$
I can't figure out how to prove this. I acknowledge that it should be a really simple fact but I'm stuck. I don't think there's the need of more context from the book, it should be a self-contained question.

A:

First, there is a minor error in your sequence; the middle term should be $\mathcal{O}_{\mathbb{P}^r}^{r+1}$. Can you see that $\sum\deg\mathcal{L}_i=\deg \mathcal{M}_{|X}$? Then $\deg\mathcal{M}_{|X}=\deg \mathcal{O}_{\mathbb{P}^r}(-1)_{|X}$? The latter of course is just $-\deg X$.

