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Let $ f: X \to Y$ be a continuous map between connected manifolds s.t. for all $y \in Y$ the fiber $f^{-1}(y)$ is homeomorphic to some fixed connected manifold $Z$.

Let $k$ be a ring and for every $j \ge 0$ let $\mathcal{H}^j:=R^{j}f_!(k_X)$, i.e. the shefification of the presheaf on $Y$ given by:

$$U \mapsto H_{c}^{j}(f^{-1}(U),k)$$

Question: Must $\mathcal{H}^j$ be a local system on $Y$? If not what's a counterexample?

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First of all, in case $f$ is not proper, the sheaf $\mathcal{H}^j = R^j f_!(k_X)$ is not defined as a sheaf associated to a presheaf $$U\mapsto H^j_c(f^{-1}(U),k),$$ since that rule is not a presheaf. Compactly supported cohomology is covariant for open inclusions, it is not contravariant (presheaves are contravariant).

Using the corrected definition, typically $\mathcal{H}^j$ is not a local system if $f$ is not proper. In case $f$ is a proper submersion of manifolds, then $f_!$ equals $f_*$. The rule that you wrote down for usual cohomology (not compactly supported cohomology) is a presheaf whose associated sheaf equals $R^jf_*(k_X).$ The fact that $R^jf_*(k_X)$ is a local system in the proper case follows from Ehresmann's theorem, for instance.

In the non-proper case, one family of counterexamples arises as in my answer to the following MathOverflow question.

Is an affine fibration over an affine space necessarily trivial?

Here is the construction. Let $Y$ equal the complex projective line as a complex manifold. Let $y$ and $y'$ be distinct points of $Y.$ Let $\overline{X}$ equal the compact complex manifold $Y\times Y.$ Let $\overline{f}:\overline{X}\to Y$ equal projection onto the first factor. Let $Z\subset Y\times Y$ equal the union of the following three irreducible closed, complex submanifolds: the diagonal $Z_1=\Delta,$ the constant section $Z_2=Y\times\{y\},$ and the singleton $Z_3=\{(y,y')\}.$ Denote by $Z'$ the union $Z_1\cup Z_2.$ Let $X,$ resp. $X',$ denote the open complement in $\overline{X}$ of $Z,$ resp. of $Z'.$ Let $f:X\to Y$ denote the restriction to $X$ of $\overline{f}.$ This is a holomorphic submersion.

Every fiber of $f$ is a complex manifold that is biholomorphic to $\mathbb{C}^\times.$ The compactly supported cohomology equals the reduced cohomology of the one-point compactification (a "nodal plane cubic"), $$H^0_c(\mathbb{C}^\times,\mathbb{Z}) = 0,\ \ H^1_c(\mathbb{C}^\times,\mathbb{Z}) = \mathbb{Z}, \ \ H^2_c(\mathbb{C}^\times,\mathbb{Z}) = \mathbb{Z}. $$ Restricted over $Y\setminus\{y\},$ the sheaves $\mathcal{H}^j$ on $Y$ are local systems.

Now let $U\subset Y$ be an open disk centered at $y.$ The inclusion $Z_3\subset X'$ with open complement $X$ induces a long exact sequence of cohomology with compact supports. In particular, $H^0_c(Z_3,\mathbb{Z})=\mathbb{Z}$ maps to a nonzero element in $\mathcal{H}^1(U)$ for every $U.$ In particular, the germ of this element in $\mathcal{H}^1_y=H^1_c(X_y,\mathbb{Z})$ is nonzero. However, for every $z\in U\setminus\{y\},$ the image of this element in the stalk $\mathcal{H}^1_z = H^1_c(X_z,\mathbb{Z})$ is zero. Thus, $\mathcal{H}^1$ is not a local system.

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Here is a proper example. Define two functions $f^+,f^-\colon \mathbb R\to R$ by $x^2$ and $-x^2$. Then let $f=f^+\cup f^-\colon \mathbb R\cup\mathbb R\to \mathbb R$. Every fiber of $f$ is 2 points, so the pushforward $f_*\mathbb k$ has stalk $k^2$ at every point, but it is not a local system because it is the sum of the two components: $f_*\mathbb k=f^+_*k\oplus f^-_*k$, which are supported on $[0,\infty)$ and $(-\infty,0]$.

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  • $\begingroup$ I forgot to make it connected, but just glue the ends. In other words $S^1\to [0,1]\to S^1$ where the first map is cosine and the second map is gluing the endpoints of the interval. $\endgroup$ – Ben Wieland Jun 6 '18 at 19:34
  • $\begingroup$ If you allow disconnected examples, and you allow the dimension to jump, then there is the holomorphic example of $\mathbb C\cup pt\to \mathbb C$, where the map on $\mathbb C$ is the squaring map and the point goes to $0$ to fix that fiber. $\endgroup$ – Ben Wieland Jun 6 '18 at 19:36

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