Nonconstant sheaf with microsupport contained in the zero section Consider the inclusion $i:B^n\to \mathbb{R}^n$ of the open ball into Euclidean space. I want to understand the proper pushforward $F=i_!\mathbb{C}_{B^n}$ of the constant sheaf. Now since $B^n$ is open, this is just extension by zero. So $F$ restricted to $B^n$ is the constant sheaf and on the complement $F$ is just the zero sheaf. So the support of the sheaf is $B^n$ itself. But if I want to compute the microsupport (Definition 5.1.2, Sheaves on Manifolds, Kashiwara and Schapira) of this sheaf then I'm stuck. For any point $x$ on the boundary, if I take the function $\varphi(y)= 1-\|y\|^2$ then the derivative of $\varphi$ corresponds to the normal vector pointing inwards toward the origin, based at $x$. But the sections over a neighbourhood cut by the set $\{\varphi<0\}$ are isomorphic to sections over the whole neighbourhood(Both are $0$). So it seems like no cotangent rays over points $x$ in the boundary belong in the microsupport. But that's not possible, since if the microsupport is contained in the zero section the sheaf must be locally constant. But that is not the case.
So I am not sure what I am doing wrong here. I would be very glad if someone could point out my mistake. Thank you.
 A: You have to work in the derived category to see the microsupport. Recall that for a closed set $Z \subset M$, $\Gamma_Z(\mathscr{F})$ is defined as the internal Hom-sheaf $\underline{Hom}(\Bbb C_Z, \mathscr{F})$ where $\mathbb{C}_Z = i_{Z, *} i_Z^* \Bbb C_{\Bbb R^n}$ is the extension-by-zero sheaf of the constant sheaf on $Z$. One may alternatively describe $\Gamma_Z \mathscr{F}$ as the sheaf of those sections of $\mathscr{F}$ which are supported on $Z$.
For a sheaf $\mathscr{F}$ on a manifold $M$, a point $(x, \xi) \in T^* M$ is contained in the microsupport of $\mathscr{F}$ if and only if there is a smooth function $\varphi$ near $x$ such that $\varphi(x) = 0, d\varphi_x = \xi$ and moreover $R\Gamma_{\varphi^{-1}[0, \infty)}(\mathscr{F})_x \neq 0$. Crucially, the nonvanishing of this stalk does not depend on $\varphi$ but only on the cotangent vector $\xi$. Note that even if it is true that $\Gamma_{\varphi^{-1}[0, \infty)}(\mathscr{F})_x = 0$, the derived sheaf $R\Gamma_{\varphi^{-1}[0, \infty)}(\mathscr{F})$ might have stalk at $0$. This is exactly what is occuring in your situation.
Let $U \subset M$ be an open subset and $Z = \Bbb C \setminus U$ its complement. I claim there is a short exact sequence $0 \to \Gamma_Z \mathscr{F} \to \mathscr{F} \to \Gamma_U \mathscr{F} \to 0$. To see this, observe that
$$\begin{align*}(\Gamma_U \mathscr{F})(V) = \underline{Hom}_{Sh/V}(\Bbb C_U, \mathscr{F})(V) &= Hom_{Sh/V}(\Bbb C_U|_V, \mathscr{F}|_V) \\ &= Hom_{Sh/V}(i_V^*\Bbb C_U, i_V^* \mathscr{F}) \\ &= Hom_{Sh/U}(\Bbb C_U, i_{V,*} i_V^* \mathscr{F}) \\ &= i_{V, *}i_V^*\mathscr{F}(U) \\ &= i_V^* \mathscr{F}(U \cap V) \\ &= \mathscr{F}(U \cap V)\end{align*}$$
Thus, we define a sheaf morphism $\mathscr{F} \to \Gamma_U \mathscr{F}$ by letting the map over $V$ be the restriction map $\mathscr{F}(V) \to \mathscr{F}(U \cap V)$. This is clearly stalkwise surjective, and the kernel consists of sections of $\mathscr{F}$ over $V$ which vanish on $U \cap V = V \setminus Z$, i.e., sections of $\mathscr{F}$ over $V$ supported on $Z$. This is exactly $\Gamma_Z \mathscr{F}$, as required.
Take an injective resolution $0 \to \mathscr{F} \to I^\bullet$ and feed $I^\bullet$ through the above short exact sequence. This gives a short exact sequence of complexes of sheaves $0 \to \Gamma_Z I^\bullet \to I^\bullet \to \Gamma_U I^\bullet \to 0$. Next, feed this through the functor $\Gamma(V; -)$ for some open set $V \subset M$ around $x$ and take the homology. This will give rise to a long exact sequence (where I am deliberately writing the evaluation of a sheaf $\mathscr{A}(U)$ as $\Gamma(U, \mathscr{A})$ to not clutter up the notation with parentheses)
$$\cdots \to H^k\Gamma(V, \Gamma_Z I^\bullet) \to H^k\Gamma(V, I^\bullet) \to H^k\Gamma(V, \Gamma_U I^\bullet) \to \cdots$$
The middle term is the sheaf cohomology $H^k(V, \mathscr{F})$ by definition, and the third term is the sheaf cohomology $H^k(U \cap V, \mathscr{F})$ because as we computed above with $\mathscr{F}$ in place of $I^\bullet$,
$$\Gamma(V, \Gamma_U I^\bullet) = (\Gamma_U I^\bullet)(V) = I^\bullet(U \cap V) = \Gamma(U \cap V, I^\bullet)$$
Note that $H^k\Gamma(V, R\Gamma_ZI^\bullet) = H^k\Gamma(V, R\Gamma_Z\mathscr{F}) = R^k\Gamma_Z\mathscr{F}(V)$. Let us write the simplified long exact sequence:
$$\cdots \to R^k\Gamma_Z\mathscr{F}(V) \to H^k(V, \mathscr{F}) \to H^k(U \cap V, \mathscr{F}) \to \cdots$$
This implies that the cohomology of the stalk of $R\Gamma_Z \mathscr{F}$ at $x$ (which is a complex of abelian groups) is zero if and only if the map $\varinjlim H^k(V, \mathscr{F}) \to \varinjlim H^k(U \cap V, \mathscr{F})$ is an isomorphism, where limits are over open sets $V$ containing $x$. But also observe
$$\varinjlim H^k(V, \mathscr{F}) = \varinjlim H_k(\Gamma(V, I^\bullet)) = H_k(\varinjlim \Gamma(V, I^\bullet)) = H_k(I^\bullet_x) = \mathscr{F}_x[0]$$
Specialize all of the above to the case where $U = B \subset M = \Bbb R^n$ is the unit open ball, $\mathscr{F} = i_! \Bbb C_{U}$ where $i : U \to M$ is the inclusion map, and $Z = M \setminus U$. Let $\varphi : \Bbb R^n \to \Bbb R$, $\varphi(y) = \|y\|^2 - 1$ and let $x \in \partial U$. Note that $Z = \varphi^{-1}[0, \infty)$. Observe,
$$\varinjlim H^0(U \cap V, \mathscr{F}) = \varinjlim H^0(U \cap V, \Bbb C) = \Bbb C$$
whereas $\varinjlim H^k(V, \mathscr{F}) = 0$ as in degree $k = 0$ it is $\mathscr{F}_x = 0$. This means the microstalk is in fact nonzero, i.e., $R\Gamma_Z\mathscr{F}_x \neq 0$. Indeed, we get $R^1 \Gamma_Z \mathscr{F}_x = \Bbb C$.
Thus, $(x, d\varphi_x)$, which is the outward conormal covector to $\partial B$ at $x$, is contained in the microsupport $SS(i_! \Bbb C_B)$. This is opposite to the one you were suggesting with your test function.
Some remarks: (1) As you rightly pointed out, the underived sheaf has stalk $\Gamma_{\varphi^{-1}[0, \infty)}(\mathscr{F})_x = 0$ but after taking the derived functor $R\Gamma$ we might still be able to see stalks; essentially what we are seeing are cocycles (or forms, if you like) at $x$ which do not contribute to the cohomology on the half-space $\varphi < 0$, but contributes to the cohomology of the microstalk. Currently, my mental picture is that of a standing wavelet at the boundary $\varphi^{-1}(0)$. (2) Note that the cotangent vector corresponds to the outward normal vector to $\partial B$. Why is it not the outward normal vector? This is because any local section on $\mathscr{F}$ at $x$ is zero, which extends across an open neighborhood of $x$. This means they "micropropagate", which is exactly what the microsupport measures: the region on the phase space given by a ("point", "codirection") tuple where the local sections of the sheaves at the "point" do not micropropagate along the given "codirection". (3) The above proof goes through with $B$ replaced by any open submanifold with smooth boundary. In fact, it shows that
$$SS(\mathscr{F}) = B_0 \cup (T^*_{\partial B} M)_{+}$$
Here $B_0 \subset T^* M$ is $B \subset M$ in the zero section, $T^*_{\partial B} M$ denotes the conormal space to $\partial B \subset M$, and $+$ in the index indicating the outward-pointing conormal covectors.
