(Real) algebraic geometry for (real) trigonometric polynomials? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:40:48Z http://mathoverflow.net/feeds/question/66842 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66842/real-algebraic-geometry-for-real-trigonometric-polynomials (Real) algebraic geometry for (real) trigonometric polynomials? Helge 2011-06-03T18:10:07Z 2011-06-03T20:54:33Z <p>Has somebody developed a comprehensive theory of the algebraic structure of trigonometric polynomials in several variables? If yes, where?</p> <hr> <p><strong>Background:</strong></p> <p>By a (real) trigonometric polynomial in $d$-variables, I mean a map $\mathbb{T}^d \to \mathbb{R}$ that is given by an expression of the form $$f(x) = \sum_{|k| \leq K} \hat{f}(k) \exp(2\pi\mathrm{i} k\cdot x)$$ where $k \in \mathbb{Z}^d$ and $|k| = \sup_{j=1,\dots,d} |k_j|$. Also $\mathbb{T} = \mathbb{R}/\mathbb{Z}$.</p> <p>These trigonometric polynomials have many of the properties of usual polynomials, but are NOT polynomials. So as far as I know it, one cannot apply the usual algebraic-geometry constructions.</p> <p>An example of a result, I would be interested in is: Given polynomials $f_1, \dots, f_{\ell}$ how does the dimension of their zero locus $$\{x \in \mathbb{T}^d:\quad f_j(x) = 0,\quad j=1,\dots,\ell\}$$ relate to the ideal generated by these polynomials?</p> <hr> <p><strong>One approach</strong></p> <p>In the Annals paper by Bourgain and Goldstein, a hint of how to do this is given. Write $$\exp(2\pi\mathrm{i} k \cdot x) = \prod_{j=1}^{d} \exp(2\pi\mathrm{i} x_j)^{k_j}.$$ Using that $\exp(2\pi\mathrm{i} x_j) = \cos(2\pi x_j) + \mathrm{i} \sin(2\pi x_j)$, one can write a trigonometric polynomial as a honest polynomial in the $2 d$ variables $C_j = \cos(2\pi x_j)$ and $S_j = \sin(2\pi x_j)$. A computation shows that this is a honest polynomial with real coefficients. Call this polynomial $\tilde{f}$.</p> <p>These set from the previous example can then be described as the zero locus of the polynomials $\tilde{f}_j$ and the polynomials $$(C_j)^2 + (S_j)^2 = 1.$$</p> <p>It seems to me that using this approach one can more or less carry over most results, but I am not very good at algebra, so I might miss subtleties. It would be nice if there was some work out of these things by somebody in the field.</p> http://mathoverflow.net/questions/66842/real-algebraic-geometry-for-real-trigonometric-polynomials/66851#66851 Answer by Thierry Zell for (Real) algebraic geometry for (real) trigonometric polynomials? Thierry Zell 2011-06-03T20:54:33Z 2011-06-03T20:54:33Z <p>The Bézout theorem as you describe should work in $\mathbb{T}^2$ for the reasons you outlined (do the change of variables, add in the Pythagorean conditions and apply the regular Bézout). I am more familiar with the situation over $\mathbb{R}^2$, where, since the functions are periodic, you cannot expect finitely many solutions. </p> <p>However, the natural setup in that case is given by Khovanskii's theory of <a href="http://en.wikipedia.org/wiki/Pfaffian_function" rel="nofollow">fewnomials</a>: if you restrict the arguments of your sines and cosines to some bounded interval, you can represent your trigonometric polynomials as an instance of the more general <strong>Pfaffian functions</strong>. These functions have a natural complexity associated to them, which is degree-like, though it's actually a vector of positive integers. The larger the restriction intervals, the bigger the complexity. (Any elementary function would work here, you don't need them to be trigonometric polynomials).</p> <p>Khovanskii's theorem is an explicit upper bound on the number of isolated solutions of such a system (with as many equations as variables). Unfortunately, this is only an upper-bound (the problem is conjectured to be decidable, but this is not known), and the bounds are very big, probably much too big.</p> <p>Khovanksii's theorem deals with more general functions than only trigonometric polynomials, so it may be possible to improve somewhat. </p>