Explaining Mukai-Fourier transforms physically - MathOverflow most recent 30 from http://mathoverflow.net2013-05-19T21:38:08Zhttp://mathoverflow.net/feeds/question/97361http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/97361/explaining-mukai-fourier-transforms-physicallyExplaining Mukai-Fourier transforms physically Tom Copeland2012-05-19T01:19:28Z2012-09-30T19:12:20Z
<p>A core concept in mathematics, engineering, and physics is the Fourier Transform (FT) and its many variants (<a href="http://en.wikipedia.org/wiki/Generalized_fourier_series" rel="nofollow">Generalized Fourier Series</a>, <a href="http://en.wikipedia.org/wiki/Green%27s_function" rel="nofollow">Green's Function</a>, <a href="http://en.wikipedia.org/wiki/Pontryagin_duality" rel="nofollow">Pontryagin duality</a>). </p>
<p>The basic algorithm is to find dual sets of eigenvectors/eigenfunctions parametrized by a continuous (e.g., $\omega$ below) or discrete index (e.g., $n$ below), that satisfy completeness and orthogonality relations encapsulated in <a href="http://en.wikipedia.org/wiki/Dirac_delta_function" rel="nofollow">Dirac delta function</a> resolutions such as that for the FT</p>
<p>$$\delta(x-y)= \int_{-\infty}^{\infty}\exp(i2\pi \omega x)\exp(-i2\pi \omega y)d\omega$$ </p>
<p>giving</p>
<p>$$\int_{-\infty}^{\infty}f(y)\delta(x-y)dy=f(x)=\int_{-\infty}^{\infty}\exp(i2\pi \omega x)\int_{-\infty}^{\infty}f(y)\exp(-i2\pi \omega y) dy d\omega$$</p>
<p>$$=\int_{-\infty}^{\infty}\exp(i2\pi \omega x)\hat{f}(\omega) d\omega,$$</p>
<p>or that for the eigenvectors of <a href="http://en.wikipedia.org/wiki/Sturm-Liouville" rel="nofollow">Sturm-Liouville</a> differential operators over finite domains</p>
<p>$$\delta(x-y)=\sum_{n=0}^{\infty }\Psi_n(x)\Psi_n^*(y)$$</p>
<p>giving</p>
<p>$$f(x)=\sum_{n=0}^{\infty }\Psi_n(x)\int_{a}^{b}f(y)\Psi_n^*(y) dy,$$ </p>
<p>or <a href="http://en.wikipedia.org/wiki/Kronecker_delta" rel="nofollow">Kronecker delta</a> resolutions such as that for the <a href="http://en.wikipedia.org/wiki/Laguerre_polynomials" rel="nofollow">associated Laguerre functions</a> </p>
<p>$$\frac{(n+\alpha)!}{n!}\delta_{mn}=\int_{0}^{\infty}x^{\alpha}e^{-x}L_{n}^{\alpha}(x)L_{m}^{\alpha}(x)dx$$ </p>
<p>giving</p>
<p>$$f(x)=\sum_{n=0}^{\infty }\frac{n!L_{n}^{\alpha}(x)}{(n+\alpha)!}\hat{f}_n$$</p>
<p>with</p>
<p><code>$$\hat{f}_n=\int_{0}^{\infty}x^{\alpha}e^{-x}L_{n}^{\alpha}(x)f(x)\,dx.$$</code></p>
<p>The basic "physical" operation (BPO) at work here can be regarded as destructive/constructive interference; the product at a point of the value of the function (to be resolved) with the corresponding value of an eigenfunction has a negative or positive value (or phase factor) that may sum constructively or destructively with products at other points (seen as a <a href="http://en.wikipedia.org/wiki/Matched_filter" rel="nofollow">matched filtering</a> or correlation by replacing $y$ with $x-z$ above). Alternatively, the BPO may be viewed as projection of vectors onto a set of orthonormal axes. In addition, if the function and operations are discretized and/or the domains restricted (in one space or its dual or both, as for the DFT) <a href="http://en.wikipedia.org/wiki/Aliasing" rel="nofollow">aliasing</a> (which seems analogous to the introduction of equivalence classes) is introduced and periodicity imposed.</p>
<p><strong>Can you explain the machinery behind the Mukai-Fourier transform in terms of these BPOs or close analogies?</strong> </p>
http://mathoverflow.net/questions/97361/explaining-mukai-fourier-transforms-physically/108207#108207Answer by Dox for Explaining Mukai-Fourier transforms physically Dox2012-09-27T03:41:13Z2012-09-27T03:41:13Z<p>I'd suggest you to check the post <a href="http://mathoverflow.net/questions/9834/heuristic-behind-the-fourier-mukai-transform" rel="nofollow">HERE</a>.</p>
<p>As you will see, the analogy enters by thinking the pullback of the sheaf $\mathcal{F}$ on the $X$ variety, i.e. $p_1^*\mathcal{F}$, as Fourier coefficients, while the sheaf $\mathcal{P}$ on the product variety $X\times Y$ plays the role of integral kernel.</p>
<p>Cheers.</p>
http://mathoverflow.net/questions/97361/explaining-mukai-fourier-transforms-physically/108214#108214Answer by S. Carnahan for Explaining Mukai-Fourier transforms physically S. Carnahan2012-09-27T06:04:37Z2012-09-27T06:04:37Z<p>You can think of line bundles and skyscraper sheaves as sheaf-theoretic analogues to exponentials and delta functions, respectively. The Fourier-Mukai transform on an elliptic curve takes one type of sheaf to the other (with a homological shift that I will ignore). In higher dimension, you get some mixtures of these types, from complexes of sheaves with cohomology supported on subvarieties of positive dimension and positive codimension - you can think of these as an analogue of more general distributions. Vector bundles on an abelian variety get "orthogonally decomposed" into skyscrapers on the "frequency space", i.e., the dual abelian variety.</p>