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Let $X$ be a complex projective variety, $E$ be a rang $n$ bundle with $n<dim X$ and $s$ be a (holomorphic) section of $E$.

There is a relatively straightforward criterium to check if the space $s=0$ is non-empty. Namely it is enough to know that $c_n(E)\ne 0$.

Question. I would like to know how one could check that $s=0$ is connected. In the case that is of interest to me $X$ is a homogenious variety (i.e. it admits a transitive group action) and $E$ is an equivariant bundle.

Maybe there is some kind of Lefshetz principle that says that $s=0$ is connected if $E$ is "sufficiently" positive?

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Fulton-Lazarsfeld – Jason Starr Feb 5 '14 at 16:33
up vote 2 down vote accepted

Let $Z = \{s = 0\}$. It is connected if and only if $H^0(Z)$ is 1-dimensional. You can compute $H^0(Z)$ by using Koszul resolution $$ 0 \to \Lambda^n E^* \to \Lambda^{n-1}E^* \to \dots \to E^* \to O_X \to O_Y \to 0 $$ (which is indeed a resolution if $X$ is Cohen--Macaulay and $Y$ has codimension $n$ in $X$). So, if you know that $H^i(X,\Lambda^iE^*) = 0$ for $i > 0$ the connectedness of $Y$ follows. Even if for some $i$ the cohomology is nontrivial, it can be killed in the spectral sequence, and you can check it.

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Sasha, many thanks! Maybe this is what will work for me. Do I understand correctly that one should check $H^i(X,\Lambda^iE^*)=0$ (i.e. you forgot to put "*")? – aglearner Feb 6 '14 at 15:02
Yes, thank you, I edited the answer. Note also that even if there is a nonzero cohomology, still the zero locus can be connected, if the cohomology is killed in the spectral sequence. – Sasha Feb 6 '14 at 15:50
Sasha, thanks again. I realised that in the case that I consider $E$ is just a sum of line bundles. But I suspect indeed that some higher cohomology will not vanish. Since I am not very familiar with the spectral sequence you are talking about, I would like to ask you if in the case when $E$ is a sum of line bundles, the calculation simplifies. Would you advise me some pedagogical reference where I could read about this? Or maybe there is an instructive example worked out somewhere? – aglearner Feb 6 '14 at 17:50
If $E = L_1 \oplus \dots \oplus L_n$ is a sum of line bundles and so $s = (s_1,\dots,s_n)$, you can try to argue inductively --- just consider the sequence of subschemes $X = X_0 \supset X_1 \supset \dots \supset X_n = Y$ where $X_i$ is the zero locus of $s_i$ on $X_{i-1}$. Then each time you will have only a long exact sequence of cohomology. This is a way to avoid considering the spectral sequence. – Sasha Feb 6 '14 at 18:04

Yes. A (very) particular case of the connectedness theorem of Fulton-Lazarsfeld tells you that the zero locus of $s$ is connected if $E$ is ample (and of rank $< \dim(X)$). See Lazarsfeld's book Positivity in Algebraic Geometry II, ch. 7, §1.

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Thank you for this answer! Unfortunately in my situation, I suspect that $E$ is semi-ample, in fact it is a direct sum of semi-ample bundles. I wonder if there is a way to get connectedness in this case. Also, I would like to ask you if you could suggest some reference that gives a classification of homogeneous ample budnles, say, over full flag varieties (or grassmanians) – aglearner Feb 5 '14 at 21:50
Homogeneous vector bundles on $G/P$ correspond to representations of $P$, so they are well understood. The standard reference is Bott's paper in Ann. of Math. (2) 66 (1957), 203–248. – abx Feb 6 '14 at 10:59

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