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Let $f : X \to Y$ be a finite surjective morphism of smooth affine algebraic varieties over the complex numbers. Is it true that a function on $Y$ whose pullback via $f$ is an analytic function on $X$, is itself analytic?

I ask because I am interested in knowing that, for a reductive complex algebraic group $G$, an analytic $W$-invariant function on a Cartan subalgebra $\mathfrak{h}$ lifts to an analytic $G$-invariant function on the Lie algebra $\mathfrak{g}$. So, if I am not mistaken, it is enough for me to know that an analytic function on $\mathfrak{h}$, invariant under $W$, gives rise to an analytic function on $W \backslash \mathfrak{h}$.

Thanks, Sasha

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  • $\begingroup$ You should at least assume that your function is continuous, otherwise there are obvious counter-examples. $\endgroup$
    – abx
    Commented Nov 11, 2017 at 17:16
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    $\begingroup$ @abx: I had the same initial gut reaction, but it is unnecessary: since $f$ is a finite surjection, hence topologically a quotient map, a function on $Y$ whose composition with $f$ is continuous must itself be continuous. $\endgroup$
    – nfdc23
    Commented Nov 11, 2017 at 17:19
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    $\begingroup$ @Sándor Kovács: root $\circ$ power is not identity in general, that's your mistake if I am not mistaken? $\endgroup$
    – Sasha
    Commented Nov 12, 2017 at 2:49
  • $\begingroup$ @Sasha: Right, it would just mash everything into a wedge. I knew it was too simple... $\endgroup$
    – Sándor Kovács
    Commented Nov 12, 2017 at 18:42
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    $\begingroup$ @Sasha: the result is false for general $Y$ (I'll edit my answer later); it might be true if $Y$ is normal. $\endgroup$
    – Laurent Moret-Bailly
    Commented Nov 14, 2017 at 8:26

1 Answer 1

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Let $h:Y\to \mathbb{C}$ be such a function. If $U\subset Y$ is an open subset (for the complex topology) and $s:U\to X$ is an (analytic) local section of $f$ on $U$, then $h=h\circ f\circ s$ on $U$, hence $h_{\vert U}$ is analytic since $h\circ f$ is.

So, $h$ is analytic on $Y\smallsetminus B$ where $B\subset Y$ is the branch locus of $f$, which is a proper Zariski closed subset. Since $h$ is continuous on $Y$ by nfdc23's remark, it follows that it is analytic on $Y$.

[EDIT] It seems plausible that the argument works assuming $Y$ normal. Here is a counterexample where it isn't: take for $Y$ the cuspidal cubic in $\mathbb{C}^2$ (with equation $y^2=x^3$ and for $f:X=\mathbb{C}\to Y$ the normalization (mapping $t$ to $(t^2,t^3)$). Then $f$ is homeomorphism, so $t$ descends to a continuous function on $Y$, which is not analytic.

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  • $\begingroup$ Thank you. Probably very elementary, but where can I read about this fact - that a continuous function on an analytic manifold, being analytic outside a proper subvariety, is analytic? $\endgroup$
    – Sasha
    Commented Nov 12, 2017 at 19:05
  • $\begingroup$ OK, references can be found here: planetmath.org/… $\endgroup$
    – Sasha
    Commented Nov 12, 2017 at 19:07
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    $\begingroup$ @Laurent Moret-Bailly .The argument works for Y normal .See Grauert Remmert Coherent Analytic Sheaves page 144. In fact Y semi-normal is enough. $\endgroup$
    – Mohan Ramachandran
    Commented Nov 15, 2017 at 15:53

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