Let $E$ be a vector bundle on some smooth algebraic variety and $E^*$ its dual. Suppose $A$ is a sheaf (constructible or a $D$-module) on $E$. Given a linear function $f$ on $E$, we may compute the stalk at $f$ of the Fourier transform of $A$.

Now I’ve heard a slogan along the lines of “the stalk of the Fourier transform of $A$ is the vanishing cycles of $A$ along $f$.” Obviously this doesn’t really make sense; the stalk is a sheaf on the base and the vanishing cycles a sheaf on $f^{-1}(0)$. Nevertheless, my guess is that somehow this idea can be made precise.

So, is there a precise statement relating nearby and vanishing cycles to Fourier transforms of sheaves?


You probably want to say $A$ is a constructible sheaf / $D$-module with regular singularities. $D$-modules with irregular singularities (for example, those created by the Fourier transform) will behave very differently, even if $E$ is a vector bundle of rank $1$ on a point.

I'm going to pretend you asked the question for $\ell$-adic sheaves and describe the relevant phenomena there. I think the situation for $D$-modules will be analogous.

Let's start with the Fourier transform on $\mathbb A^1$. For an $\ell$-adic sheaf $A$ with tame ramification (analogue of regular singularities), the correct statement is:

The nearby cycles of the Fourier transform of $A$ at $\infty$ is isomorphic to a sum over points of $\mathbb A^1$ of the vanishing cycles of $A$ at that point. Furthermore, the Fourier transform is locally constant away from $0$, so its stalk at any point is non-canonically isomorphic to the nearby cycles of the Fourier transform at $\infty$ and thus non-canonically isomorphic to this sum.

Now let's consider the Fourier transform on $\mathbb A^n$, keeping the base still a point. The stalk of the Fourier transform at $\lambda f$ for a scalar $\lambda$ is going to be the stalk of the Fourier transform of $R f_* A$ at $\lambda$. Thus it will be non-canonically isomorphic to the sum of the stalks of the vanishing cycles of $R f_* A$ at each point of $\mathbb A^1$.

If $f$ were proper, it would then by isomorphic to the sum over $a$ of the cohomology of $f^{-1}(a)$ with coefficients in the vanishing cycles complex of $A$ along $f$. Maybe we can say that this is morally true in general. So we need to take $f^{-1}(a)$ for all $a$, not just $f^{-1}(0)$, and take the cohomology, to get the desired isomorphism. If the vanishing cycles are supported at finitely many points, this would just be the sum of the vanishing cycles at those points. Maybe that's what people who told you this were thinking of?

However, while this may be morally true, it will not be literally true in every case. The simplest counterexample I can think of is on $\mathbb A^n$, take $A$ to be the constant sheaf on the curve $xy=1$ and $f$ to be $y$. Then $A$ has no vanishing cycles on any point of $\mathbb A^2$ (none on $f^{-1}(0)$ since $A$ vanishes in a neighborhood of $f^{-1}(0)$, and none elsewhere because $A$ in each other fiber looks identical to the generic fiber) but its Fourier transform has a stalk of rank one at $y$.

Over a nontrivial base, I guess you want to take the vanishing cycles sheaf of $A$ along $f$ on $f^{-1}(a)$, push forward to the base, and then sum over $a$. This will be even more morally and less factually correct, but maybe there is some genericity condition under which it is literally true.

  • $\begingroup$ Thank you for your answer. Could you explain to me what “nearby cycles at infinity” means? $\endgroup$
    – leibnewtz
    Aug 2 at 14:37
  • $\begingroup$ @leibnewtz Maybe a better way to say it is "the stalk at a point near $\infty$". $\endgroup$
    – Will Sawin
    Aug 2 at 14:49

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