I am trying to understand a basic computation of convolution. Throughout, $R$ is the real line as a topological group and $k$ is some base field. I would like to understand the computation of the convolution
$k_{(a,b)} \star k_{[0,\infty)}$.
I believe this convolution should be $k_{[a,\infty)}[-1]$, but I am unable to prove this.
In general, the stalk at $t \in R$ is (by Beck-Chevalley) the compactly supported cohomology of $k_{(a,b)} \boxtimes k_{[0,\infty)}$ restricted to the line $\{t_1 + t_2 = t\} \subset R^2$. In what follows, $A<B$ are real numbers:
- When $t>b$, the stalk is $R\Gamma_c(k_{(A,B)}) \simeq k[-1]$, the reduced cohomology of the circle, as one is computing the compactly supported cohomology of an open interval.
- When $a<t\leq b$, the stalk is $R\Gamma_c(k_{[A,B)})$. By the short exact sequence $k_{(-\infty,A)} \to k_{(-\infty,B)} \to k_{[A,B)}$, the stalk is thus zero. (The map $k_{(-\infty,A)} \to k_{(-\infty,B)}$ induces an isomorphism on compactly supported cohomology.)
- When $t=a$, the stalk is the compactly supported cohomology of $R$ with respect to a skyscraper sheaf. The skyscraper sheaf is flabby, so we conclude that the stalk at $t=0$ is a copy of $k$ in degree 0.
- When $t<a$, the stalk is zero.
Is this computation of stalks correct? I would like very much to know what I am doing wrong.
