Describing homs and tensors of the square-root sheaf

Question

Let $\mathbb{C}^\times$ have the classical topology, and $f: \mathbb{C}^\times \to \mathbb{C}^\times$, $z \to z^2$ the covering map (where the first $\mathbb{C}^\times$ is thought of as the double-cover of the second). The definition of the square-root sheaf $Q$ is that the pushforward sheaf $f_* \underline{\mathbb{C}}$ splits as a direct sum $f_* \underline{\mathbb{C}} \cong \underline{\mathbb{C}} \oplus Q$.

So far, I have described this splitting as follows: Let $U \subseteq \mathbb{C}^\times$ be open, and denote by $\widetilde{U}$ its preimage in the double cover. Then $$(f_*\underline{\mathbb{C}})(U) = \{\text{locally constant } f: \widetilde{U} \to \mathbb{C} \} = V_1 \oplus V_{-1}$$

where $V_1$ are the even locally constant functions, and $V_{-1}$ are the odd locally constant functions (this makes sense since $z \in \widetilde{U} \iff -z \in \widetilde{U}$). Then restriction respects this splitting, and I get the constant sheaf $\underline{\mathbb{C}}$ as the $V_1$ part (even functions), and $Q$ as the odd part.

I learn a lot about $Q$ from this description: it has no nonzero global sections, and I can calculate the representation of the fundamental group $\pi_1(\mathbb{C}^\times)$ associated to the local system $Q$. However, due probably to my inexperience with sheaves, I am having trouble calculating the hom-set $\mathrm{Hom}(Q, \underline{\mathbb{C}})$, the hom-sheaf $\mathcal{Hom}(Q, \underline{\mathbb{C}})$, and the sheaf tensor $Q \otimes Q$. (I am fairly sure the answers to these three questions should be $0$, $Q$, and $\underline{\mathbb{C}}$ respectively).

What method of attack should I use to answer these questions? I keep getting bogged down in lots of restriction diagrams and making complicated arguments to do with pulling back small sections to almost-global ones and so on. Is it possible to answer these questions by only ever considering open sets of the form open balls around a point?

Answer 1

A local system $L$ of rank $r$ is equivalent to the data of a representation $\rho : \pi_1(X) \to GL_r$. Here $r = 1$ and we have $\rho_{\underline{\Bbb C}} : \Bbb Z \to \Bbb C^*, 1 \mapsto 1$ and $\rho_Q : 1 \mapsto -1$.

$\mathcal{Hom}(Q, \underline{\Bbb{C}}) := L$ is a local system of rank $1$, so it is equivalent to understand the induced representation $\rho : \Bbb Z \to Aut(L)$. Let us take a small disk $U \subset \Bbb C^*$, and take a section $\phi \in Hom(Q(U), \underline{\Bbb C}(U))$. For example we will take $\phi : 1_U \mapsto 1$. Since $\rho_Q(1) = -1$ we see that we get $\rho_L(1) \cdot \phi = - \phi$. But this is exactly $\rho_Q$ and so we get an isomorphism $Q \cong \mathcal{Hom}(Q, \underline{\Bbb C})$.

We can apply the same method for $Q \otimes Q$ : this is a local system of rank $1$, and $\rho_Q$ was multiplication by $-1$. Of course $\rho_{Q \otimes Q} = \rho_Q \otimes \rho_Q = \text{id}$, and we obtain as you predicted that $Q \otimes Q \cong \underline{\Bbb{C}}$.

Answer 2

This should be a comment, but it is too long, so I posted it as an answer.

Nicolas Hemelsoet gives the right reason for your question, and I agree with him : you should get confortable with the equivalence locally constant sheaves = representations of $\pi_1$ instead of going back to the definition.

However, here is also another way to think about it without speaking of $\pi_1$ : see $\underline{\mathbb{C}}$ and $Q$ as subsheaves of $\mathcal{C}^\infty$ in the following way : $\underline{\mathbb{C}}(U)$ is the $\mathbb{C}$-vector subspace spanned by the identity function $z\mapsto z$ on $U$ (or more accurately the subspace of functions $f$ such that $f/z$ is locally constant), and $Q(U)$ is the $\mathbb{C}$-vector space spanned by the the square roots of $z\mapsto z$.

If $U$ contains a punctured disk $D(0,r)^*$, then there are no way to define continuously a square root of $z$, so $Q(U)=0$. Otherwise there are two of them, but since they are opposite, $Q(U)=\mathbb{C}.\sqrt{z}$ (if $U$ is connected).

Now consider the map $Q\otimes Q\rightarrow\mathcal{C}^\infty$ given by multiplication. Clearly this lands in $\underline{\mathbb{C}}$ and this is locally an isomorphism, hence you have an isomorphism $Q\otimes Q\overset{\sim}\longrightarrow\underline{\mathbb{C}}$.

(This also shows that $Q=Q^\vee=\mathcal{H}om(Q,\mathbb{C})$ and since $Q$ has no global section $\operatorname{Hom}(Q,\mathbb{C})=\Gamma(\mathbb{C}^*,\mathcal{H}om(Q,\mathbb{C}))=\Gamma(\mathbb{C}^*,Q)=0$)