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I am interested in vector bundles over a nonsingular complete algebraic curve $C$ over $\mathbb C$. For a vector bundle $E$, its Harder-Narasimhan filtration is a filtration of subbundles $$0=E_0\subset E_1\subset\cdots\subset E_n=E$$ such that each $E_i/E_{i-1}$ is semitable and $\frac{\deg(E_1/E_0)}{\mathrm{rank}(E_1/E_0)}>\cdots>\frac{\deg(E_n/E_{n-1})}{\mathrm{rank}(E_n/E_{n-1})}$.

Now if we have two vector bundles $E,F$ and we know their HN filtation $\{E_i\},\{F_j\}$, can we get any information about the HN filtration of $E\otimes F$?

One information I am interested in is the upper bound of $\frac{\deg}{\mathrm{rank}}$ of subbundles of $E\otimes F$.

Thanks.

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1 Answer 1

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It is a result of Narasimhan and Seshadri that if $V$ is semistable and $W$ is semistable then $V \otimes W$ is semistable.

If $E$ has a filtration with associated graded $E_i/ E_{i-1}$, and $F$ has a filtration with associated graded $F_j/ F_{j-1}$, then $E \otimes F$ has a filtration with associated graded $(E_i/ E_{i-1}) \otimes (F_j/ F_{j-1})$. By the previous claim, the associated graded pieces of this filtration are semistable, and we can choose the filtration so that these pieces are in order of increasing slope. Hence it is the Harder-Narasimhan filtration.

Thus the slopes of $E \otimes F$ are the slopes of $E$ plus the slope of $F$, and in particular the maximal slope of $E \otimes F$ is the sum of the maximal slope of $E$ and the maximal slope of $F$.

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  • $\begingroup$ Your claim states that if one is stable and the other one is semistable the tensor is semistable. Now how do you deduce that tensor product of two semistable is semistable? $\endgroup$
    – user127776
    Commented Jan 27, 2022 at 4:35
  • $\begingroup$ You can infer this from the Kobayashi-Hitchin correspondence $\endgroup$
    – diverietti
    Commented Jan 27, 2022 at 11:03
  • $\begingroup$ @user127776 I found a review for this result ems.press/content/serial-article-files/10191. May be helpful. $\endgroup$ Commented Jan 27, 2022 at 11:18
  • $\begingroup$ @user127776 I meant to say that both were semistable. (But the result would follow, since the tensor product would be an iterated extension of semistable bundles of the same slope.) $\endgroup$
    – Will Sawin
    Commented Jan 27, 2022 at 13:11
  • $\begingroup$ @diverietti Yes, the Kobayashi-Hitchin correspondence generalizes the Narasimhan–Seshadri theorem (from which this follows) from curves to higher-dimensional varieties. $\endgroup$
    – Will Sawin
    Commented Jan 27, 2022 at 13:13

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