# Is there a relationship between Fourier transformations and cotangent spaces?

Maybe a trivial question but I can't seem to find it treated anywhere.

Consider a smooth manifold $Q$ (configuration space) and its cotangent bundle $T^*Q$ (phase space). Any function $F:Q\to\mathbb{R}$ induces a map $q\mapsto dF_q \in T^*_qQ$ which induces the Legendre involution when $F$ is convex/concave. I think of $(q,dF_q)$ as determining a Lagrangian submanifold $L_F\subset T^*Q$, and the Legendre transformation as lifting functions on $Q$ to $L_F$ and subsequently projecting them to a fibre. The geometric construction shows it is a symplectomorphism. In coordinates, this changes the variables from $q$ to $p=\partial F/\partial q$.

My question is how can I understand the Fourier transform in this picture. By the stationary phase approximation (Wiki) or similarly Laplace's method (also on Wiki) we have something like $\mathcal{F}(\exp(i f)) \approx \exp(i\mathcal{L}f)$ where $\mathcal{L}$ is the Legendre transform.

Furthermore, how does this relate to (quantum) field theory? Path integrals always seem to have the form $\int \exp(\frac{i}{h}S[\phi]) F(\phi) \mathcal{D}\phi$ where $S$ is the action functional and integration is over all paths $\phi$ with given endpoints. This action functional is'' the tautological one-form (Wiki yet again), so is path integration just some kind of Fourier transform?

Any reference will help, I am not an expert.

• The first part of your question strongly reminds me of microlocal analysis. But unfortunately I am not able to make the connection precise. Hopefully someone who can will come along and explain. Commented Apr 30, 2015 at 12:28
• Your path integral, as written maps $F$ to a number rather than another function, so a naive analogy with the Fourier transform fails. However, if you write it with an external source $J$, $\int \exp(\frac{i}{h}(S[\phi]+J\cdot\phi)) F(\phi) \mathcal{D}\phi$, where $J\cdot\phi = \int J(x) \phi(x) dx$, then the path integral, as a function of $J$, is essentially the Fourier transform of $\exp(\frac{i}{h}S[\phi]+\log F(\phi))$. I think that should make the formal analogy clear. Commented Apr 30, 2015 at 12:39
• @IgorKhavkine Thanks for the reference (available [online](www.freeinfosociety.com/media/pdf/2287.pdf)) which is very promising. Commented May 1, 2015 at 13:15
• @IgorKhavkine But. Your answer to my path integral question doesn't address the point I was trying to make. In coordinates, writing $\theta = p_idq^î$ for the tautological one-form, then $S[\phi]=\int_\phi \theta = \int_\phi p_i dq^i$. I meant that the action-exponent itself reminds me of a Fourier term. Do you know if G&S mention this? Commented May 1, 2015 at 13:22
• No. Unless I'm misunderstanding it, your analogy is flawed for the reason I commented on earlier and also because in the formula that you give for $S[\phi]$ is only valid when $\phi$ solves the equations of motion, so the $p_i$ are not independent variables like in the Fourier transform. What is true is that the path integral is an oscillatory integral. But not all oscillatory integrals are Fourier transforms and neither is the path integral without the external source term. Commented May 1, 2015 at 13:59

There is indeed a deep relation between Lagrangian submanifolds, the Fourier transformation and microlocal analysis. This is extensively discussed in Bates and Weinstein: Lectures on the Geometry of Quantization.

Let me illustrate the basics idea in the simplest case. Assume we are given an Lagrangian immersion $\iota: L \to T^* R = R^2$. Projecting on the $p$-component yields a map $\pi_p: L \to R$. If $\pi_p$ is a diffeomorphism then, for every half-density $a$ on $L$, we can define a function $B$ on the momentum line $R$ by $$B |dp|^{1/2} = \pi_p^* \, e^{i \tau} a .$$ Here $\tau: L \to \ R$ is a "generalized phase function" satisfying $d \tau = \iota^* \theta$ with $\theta$ being the canonical 1-form on $T^* R$. Taking the inverse Fourier transformation of $B(p)$ yields a function $F(q)$ living on the configuration space. Hence, we see that indeed the Fourier transformation of $F$ is related to the Lagrangian immersion $\iota$. However, your formula $\mathcal{F}(\exp(i f)) \approx \exp(i\mathcal{L}f)$ is a little bit to naive and just works under certain non-degeneracy assumptions. (If $\pi_p$ fails to be a diffeomorphism, then you need to include further corrections - these include the Maslov index of the Lagrangian immersion).

So what is the relation to quantum mechanics? If your $F: Q \to R$ satisfies the Hamilton-Jacobi equation, then it may serve as the phase function of a first-order approximate solution of Schrödinger’s equation. In other words, the WKB ansatz $e^{i / \hbar F(q)}$ is then a first-order approximation. More generally, you could also consider approximate solutions of the form $a\,e^{i / \hbar F}$, where the amplitude $a$ is essentially the half-density $a$ living on $L$ as above.

• Great. Let me paraphrase for my understanding. Given $H:T^*Q\to\mathbb{R}$ find $\tau$ that solves the HJ equations and use this to define the Lagrangian $L_\tau$ on which $i^*\theta=d\tau$. Half-densities on $L_\tau$--multiplied by $e^{i\tau}$--project to base and fibre: these projections are Fourier transforms of each other, and solve (approximately) the Schr\"{o}dinger equation. This $\tau$ gives the action I referred to in my question. Simplifying terrifically, Fourier is 'quantum Legendre', or 'Legendre for wave functions'. The uncertainty principle is a consequence... Commented May 9, 2015 at 15:05
• Correction: the projections are NOT Fourier transforms of one another, the difference involving the Maslov correction mentioned by Tobias. The notes of Bates & Weinstein can be found here. Commented May 28, 2015 at 6:40

Since the question asked for a reference for this relation between the Fourier and Legendre transforms, here's one:

Guillemin & Sternberg, Geometric Asymptotics (AMS, 1990). See in particular the discussion on pp.403-404 in section VII.1.

Writing the path integral with an external source, $\int \exp(\frac{i}{h}(S[\phi]+J\cdot\phi)) F(\phi) \mathcal{D}\phi$, where $J\cdot\phi = \int J(x) \phi(x) dx$, like I suggested in the comments, makes it formally look precisely like a Fourier integral (mapping functions of $\phi$ to functions of $J$).