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$\DeclareMathOperator\rk{rk}$Suppose $(p_1, \dotsc, p_n)$ and $(q_1, \dotsc, q_n)$ are two (non-strictly) decreasing sequences of positive real numbers. We say that $(q_1, \dotsc, q_n)$ dominates $(p_1, \dotsc, p_n)$ if $q_n \geq p_n$, $q_n+q_{n-1} \geq p_n+p_{n-1}$, …, $q_n + \dotsb + q_2 \geq p_n + \dotsb + q_2$, $q_n + \dotsb+ q_1 = p_n + \dotsb + q_1$.

For a vector bundle $F$ consider its slope $\mu(F)=\deg(F)/\rk(F)$. For a vector bundle $E$ with Harder–Narasimhan filtration $0 \neq E_1 \subset E_2 \subset \dotsb \subset E_m = E$ we consider the sequence $\mu(E_1),\dotsc,\mu(E_1), \dotsc, \mu(E_m/E_{m-1}), \dotsc, \mu(E_m/E_{m-1})$, where $\mu(E_i/E_{i-1})$ is repeated $\rk(E_i/E_{i-1})$ times, and call this sequence the degree of the Harder–Narasimhan filtration of $E$.

Suppose that $A \subset B$ are two vector bundles of the same rank $n$ such that $\deg B = \deg A + 1$. Assume that the degree of the Harder–Narasimhan filtration of $B$ is dominated by a (non-strictly) decreasing integer sequence $(d_1, \dotsc, d_n)$ such that $d_1>d_2$. Is it true that then the degree of the Harder–Narasimhan filtration of $A$ is dominated by $(d_1-1, d_2, \dotsc, d_n)$?

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  • $\begingroup$ Did you mean to have $q_n + \dotsb + q_2 \ge p_n + \dotsb + q_2$, with $q_2$ on the right, and similarly for the next inequality? $\endgroup$
    – LSpice
    Commented Sep 7 at 0:37

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Sorry, I remembered how to answer my own question. Consider the degree of a HN stratum as a Young diagram with rational lengths of rows. It is enough to show that the Young diagram of A is a subset of the Young diagram of B because we can just omit the last inequality in the definition of domination. Suppose the Young diagram of A is not a subset of the Young diagram of B. Then there is a subbundle C of A consisting of several members of HN filtration of highest slope whose Young diagram is not inside the Young diagram of B. The bundle C maps by 0 to the quotient E of B consisting of several members of the HN filtration of lowest slope whose slope is smaller than the slope of the parts of C. So the bundle C is a subbundle of the kernel of the surjection from B to E. But since the Young diagram of C is not inside the Young diagram of B, the kernel of the map from B to E has smaller rank than the rank of C. Contradiction.

Please see a related question which I still need Slopes of the Harder-Narasimhan filtration of a limit of vector bundles

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    $\begingroup$ Don't forget to accept your own answer, if it resolves the question. (Otherwise this question just keeps getting bumped.) $\endgroup$
    – LSpice
    Commented Sep 7 at 0:37

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