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Suppose you are given a smooth algebraic variety $X$ inside a projective space $\mathbb{P}$ and that there is a linear action of a finite cyclic group $G$ on $\mathbb{P}$ which restricts to an action on $X$. Does there exists an hyperplane $H$ on $\mathbb{P}$ such that $H \cap X$ is smooth and stable under $G$? I guess it should be possible to prove such a result by using a variant of Bertini's theorem, but I don't see how. Thanks for your help |
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Typically the answer will be no, as QiL notes in his answer. I have had to consider exactly this kind of question in my research, and the solution was to look for a (possibly high degree) hypersurface that is $G$-invariant and cuts $X$ smoothly. For many purposes hypersurfaces are just as good as hyperplanes (e.g a degree $d$ hypersurface section becomes a hyperplane section after applying the $d$-uple embedding, and so theorems such as weak Lefschetz apply just as well to a smooth hypersurface section as to a smooth hyperplane section). Added: QiL asks in a comment whether it is really possible to find $G$-invariant smooth hypersurface sections. I would guess the answer is yes, but I haven't thought carefully about it. However, there is something very close to this that one can do: this paper proves an equivariant form of Bertini's theorem; see Theorem 9.7. (They use the Luna slice theorem, and hence restrict themselves to char. zero; but I wouldn't be surprised if some kind of char. $p$ analogue can be proved.) They don't get an equivariant hypersurface section, but rather produce a $G$-invariant smooth closed hypersurface $W$ on $X$ which is ample and which (consequently) has affine complement. This is enough for most applications where you might traditionally pass to a hyperplane section via the usual Bertini theorem. (More-or-less, they produce a $G$-invariant hypersurface on $X$ which is a hypersuface section for some projective embedding, but not necessarily the projective embedding that you start out with.) The argument proceeds by a applying an appropriate non-equivariant version of Bertini to the quotient $X/G$, and this is generally a useful method when trying to make $G$-equivariant constructions on $X$. |
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Edit3 ! Here is a simple proof I learnt from Michel Brion which works in characteristic $0$ and which gives a hypersurface (not a hyperplane) as desired. Consider the quotient $\mathbb P^n/G$. It is a projective variety of dimension $n$, so it has a finite morphism to the projective space $\mathbb P^n$. Now consider the finite morphism obtained by composition $$f: X\to \mathbb P^n\to \mathbb P^n/G\to \mathbb P^n.$$ By Bertini in characteristic $0$, $f^{-1}(H)$ is smooth for some hyperplane in the last $\mathbb P^n$. This is a smooth $G$-invariant hypersurface section in $X\subseteq \mathbb P^n$. Unfortunately this method doesn't work in positive characteristic. No this is not always possible. Consider a $G$ acting on $\mathbb P$ with finitely many fixed hyperplanes (e.g. cyclic permutation of the coordinates). For $d>>1$, there exists a smooth hypersurface $X$ of degree $d$, tangent to each of these fixed hyperplanes (in the space of hypersurfaces of degree $d$, to be tangent to a given hyperplane imposes two conditions, so it is enough to take $d$ bigger than twice the number of fixed hyperplanes). Let $H$ be such that $H\cap X$ is smooth and stable by $G$. Then for any $g\in G$, $g(H)$ and $H$ both contain $H\cap X$ which is a smooth hypersurface of degree $d$ in $H$ and in $g(H)$. In particular $H\cap X$ is not linear. This implies that $g(H)=H$ because otherwise $g(H)\cap H=H\cap X$ by comparing the dimensions. But by constructio $H$ is then tangent to $X$, so $H\cap X$ is not smooth. Contradiction. Edit It is more reasonable to ask the existence of $H$ such that $g(H)\cap X$ is smooth for all $g\in G$. Edit 2 I have to withdraw my "proof" because my reasoning on the existence of smooth hypersurfaces $X$ tangent to given hyperplanes is not complete, moreover $X$ must also be stable by $G$. But my feeling is still that the answer should be negative in general because it can happen that there are only finitely many invariant hyperplanes. |
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