2
$\begingroup$

Let $X$ be a complex projective normal variety and let $\mu:Y\to X$ be a desingularization. Note that each fiber $F$ of $\mu$ is connected. Assume that $\rho:\pi_1(Y)\to {\rm GL}_N(K)$ is a linear representation where $K$ is some field of characteristic zero such that for each fiber $F$ of $\mu$, $i^*\rho: \pi_1(F)\to {\rm GL}_N(K)$ is trivial where $i:F\to Y$ is the inclusive map. Question: can we conclude that there is a representation $\tau:\pi_1(X)\to {\rm GL}_N(K)$ such that $\rho=\mu^*\tau$?

$\endgroup$

1 Answer 1

3
$\begingroup$

Yes, this is true. To simplify the discussion, assume that $X$ has isolated singularities $p_1,\ldots, p_n$ (but this isn't essential). Choose sufficiently nice contractible neighbourhoods $B_i$ of $p_i$. Then $\tilde B_i = \mu^{-1}B_i$ is homotopy equivalent to $F_i= \mu^{-1}(p_i)$. Again, out of laziness, let me assume $n=1$. Then by Van Kampen $$\pi_1(Y) = \pi_1(X-p_1)*_{\pi_1(\partial B_1)}\pi_1(\tilde B_1)=\pi_1(X-p_1)*_{\pi_1(\partial B_1)}\pi_1(F_1)$$ and $$\pi_1(X) = \pi_1(X-p_1)*_{\pi_1(\partial B_1)}\pi_1(B_1)=\pi_1(X-p_1)*_{\pi_1(\partial B_1)}\{1\}$$ Then one can see immediately from these formulas that a representation of the first group $\pi_1(Y)$, trivial on $\pi_1(F_1)$, factors through the second group $\pi_1(X)$.

$\endgroup$
3
  • $\begingroup$ many thank for the answer! I have a question: if the singularties are isolated, then $\tilde{B}_i$ is homotopy equivarlent to $F_i$. But what is the case for general singularities? $\endgroup$ Commented Mar 4, 2023 at 6:32
  • 2
    $\begingroup$ In general, $B_i$ should be a tubular nbhd of a component of the singular set. (I noticed you asked about that previously.) The argument is bit more involved, but should follow the same strategy. $\endgroup$ Commented Mar 4, 2023 at 6:51
  • 2
    $\begingroup$ Right, except van Kampen's theorem gives a somewhat different answer (one needs to add more relators to the amalgams because $\pi_1(\partial B_i)$ does not necessarily embed in the factors). But this does not change the end result. $\endgroup$ Commented Mar 4, 2023 at 10:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .