# Understanding square-root sheaf and similar sheaves

Let $n$ be a positive integer, and $\mathbb C^* = \mathbb C -\{0\}$. Let $f: \mathbb C^* \rightarrow \mathbb C^*$, sending $z$ to $z^n$. Let $\mathcal F$ be the constant sheaf on $\mathbb C^*$ with fibers $\cong \mathbb C$. Then $f_*\mathcal F \cong \mathcal F_0 \oplus \mathcal F_1 \oplus \cdots \oplus \mathcal F_{n-1}$, where $\mathcal F_0$ is the constant sheaf. Others are locally constant but not constant. There is a special case: when $n=2$, $\mathcal F_1$ is called the square-root sheaf.

The facts mentioned above are what I have learned, but I don't really understand them.

1. Is it true that $\mathcal F(U) \cong \mathbb C$ whenever $U$ is a connected open subset of $\mathbb C^*$? If this is true, taking $U = \mathbb C^*$, we have $f_*\mathcal F(U) = \mathcal F(f^{-1}(U)) = \mathcal F(U) \cong \mathbb C \cong \mathcal F_0(U)$, so other $\mathcal F_i$'s have no global sections?

1. For an open $U \subseteq \mathbb C^*$ and $i \in \{1,\cdots, n-1\}$, is it true that $\mathcal F_i(U) \neq 0$ implies that for any $x \neq y$ in $U$, $x^n \neq y^n$?

2. Generally, for an open subset $U$ of $\mathbb C^*$, connected or not, what exactly is $\mathcal F_i(U)$ for $i=1,2,\cdots, n-1$?

Thanks to everyone.

No it is not true that $\mathcal{F}(U)=\mathbb{C}$ for every open subset of $\mathbb{C}^*$. In fact, $U\mapsto\mathbb{C}$ is called the constant presheaf, and $\mathcal{F}$ is its sheafification. In other words, $\mathcal{F}(U)$ is the set of locally constant function $U\rightarrow\mathbb{C}$. So we have $\mathcal{F}(U)=\mathbb{C}^{\pi_0(U)}$.

However, this is true that the sheaves $\mathcal{F}_i$ have no global sections for $i>0$.

As you said, the sheaf $f_*\mathcal{F}$ is locally constant and so correspond to a linear representation $(V,\rho)$ of $\pi_1(\mathbb{C}^*)\simeq\mathbb{Z}\ni 1$. Here it is simply the one given on the vector space freely generated by $n$-vectors $(e_1,\dots,e_n)$ and such that $\rho(1)$ acts by a cyclic permutation.

Here, this matrix is diagonalizable, which leads to a decomposition $V=V_0\oplus\dots\oplus V_{n-1}$ of subrepresentation of dimension 1, with $V_0$ being the trivial one. This corresponds to the decomposition of the sheaf $f_*\mathcal{F}=\mathcal{F}_0\oplus\dots\oplus\mathcal{F}_{n-1}$.

For an open subset $U$, the sheaf $(f_*\mathcal{F})_{|U}$ is also a locally constant sheaf, so it also correspond to representation of $\pi_1(U)$. In fact, we have a morphism $\pi_1(U)\rightarrow\pi_1(\mathbb{C}^*)$, so the representation $(V,\rho)$ induces a representation of $\pi_1(U)$, and this is the representation corresponding to $(f_*\mathcal{F})_{|U}$.

Now, to compute global sections using this correspondence, $f_*\mathcal{F}(U)=V^{\pi_1(U)}$ is the set of invariant of $V$ under the group $\pi_1(U)$.

For example, $f_*\mathcal{F}(\mathbb{C}^*)=\mathbb{C}$ generated by the eigenvector $e_1+\dots+e_n$. Also, if $U$ is simply connected, then $\pi_1(U)=0$ and so that all vectors are invariant under $\pi_1(U)$, therefore $f_*\mathcal{F}(U)\simeq V\simeq\mathbb{C}^n$.

Note that I didn't write base point. I didn't to give the picture, but I should have, because the correspondence between locally constant sheaves and representation of $\pi_1(U,x)$ depends on $x$...

• Thank you very much. Now things become clearer to me. – Sunkist Jan 16 '16 at 7:58
• Dear Roland, this is a nice answer as always ! (+1 of course) Did you mean "locally constant" instead of "locally free" in the last line ? – Nicolas Hemelsoet Mar 1 '18 at 13:03
• @NicolasHemelsoet Oh yes, I'll edit. Thanks. – Roland Mar 1 '18 at 13:04