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Let $U_X(n,d)$ be the moduli space of semistable vector bundle of rank $n$ and degree $d$ over a smooth projective curves over the complex numbers.

How do I know if $U_X(n,d)$ contains direct sums of line bundles? I have found on a paper that this is true when $d=n(g-1)$ where $g$ is the genus of the curve, does any of you know why is it true? When does $U_X(n,d)$ not contain any direct sum of line bundles?

Thanks a lot!

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$\newcommand{\L}{\mathcal L} \DeclareMathOperator{\rk}{rk}$ Say you have some line bundles $\L_i$ of degree $d_i$. The degree is additive, so $E = \deg (\bigoplus \L_i) = \sum_i d_i$. So to get a vector bundle of rank $n$ and degree $d$, we whould have $n$ line bundles, such that $d_1+\cdots +d_n = d$.

Now, unless they all have the same degree the sum cannot be semistable. If they have different degree, suppose that $d_1$ is the maximum. Then $$ d_1 = \frac{\deg \L_1}{\rk \L_1} > \frac{\deg E}{\rk E} = \frac{d_1+\cdots + d_n}{n}. $$ In other words, $d_1$ is strictly bigger than the average of all the degrees. This shows that in order for $U_X(n,d)$ to contain a direct sum of line bundles, $d$ has to be a multiple of $n$. I don't understand why you would need $d=n(g-1)$ specifically (for example, if $n=1$ then all vector bundles are sums of line bundles, in that they are just line bundles). Which paper did you see this in?

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    $\begingroup$ Thank you! That is the answer I was waiting for :) I need $d=n(g-1)$ for specific reason, I am studing the generalized theta divisor. In the paper I was reading, the author claims that the result is trivial when $d=n(g-1)$ but I could not understand why, thanks again! :) $\endgroup$
    – Tribofilo
    Jan 29, 2020 at 9:35

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