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Let $X$ be a compact Riemann surface. Let $r(E), d(E)$ denote the rank and degree of a vector bundle $E$ on $X$.

From Narasimhan-Seshadri we have the following proposition

([Proposition 4.1) A vector bundle $W$ on $X$ is stable if for every proper subbundle $V$ of $W$, we have $d(W^*\otimes V)<0$.

The proof relies on this equality

$d(W^*\otimes V)=r(W)d(V) - r(V)d(W)$.

Where does the equality comes from?

Thank you.

Conjecture
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    I asked a very similar question not long ago: https://math.stackexchange.com/questions/4240156/functorial-properties-of-the-degree-in-vector-bundles The only thing the nice answer there does not cover is why the degree of the dual bundle flips signs: you can set up a specific representative of $c_1(W)$ through the curvature of some connection on $W$ and use that the curvature on the dual bundle is minus the transpose. – topolosaurus Feb 22 '22 at 19:52
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    @topolosaurus: That's one way to do it which is sufficient for this question, but it does not establish the relationship at the level of integral cohomology. In general, $c_i(W^) = (-1)^ic_i(W)$, see Lemma $14.9$ of Characteristic Classes* by Milnor & Stasheff. – Michael Albanese Feb 22 '22 at 21:53

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