Why are some q-analogues more canonical than others? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:26:17Z http://mathoverflow.net/feeds/question/89996 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others Why are some q-analogues more canonical than others? spanferkel 2012-03-01T23:25:27Z 2012-03-02T17:37:27Z <p>It is striking that some q-analogs of functions, operators, identities and especially whole theorems seem quite "canonical", e.g.</p> <ul> <li>the factorial and the q-Gamma function</li> <li>the basic hypergeometric series (at least $_{r+1}\Phi_r$ and $_{r}\Psi_r$)</li> <li><a href="http://mathworld.wolfram.com/q-Pi.html" rel="nofollow">q-Pi and the q-Wallis formula</a></li> </ul> <p>In a strict sense, q-analogs can of course not be canonical, as we might throw in almost everywhere powers of $q$ without changing the limit if $q\to1$. </p> <blockquote> <p>I mean <strong><em>canonical</em></strong> in the sense that these are the forms that require the least extra powers of $q$, or, more importantly, that other q-identities/theorems using them also tend to avoid such extra powers at best, making the formulae shorter.</p> </blockquote> <ul> <li><em><strong>Does it really make sense to call these (and certain other) q-analogs "canonical"? And if so, is there an explanation why some are much more canonical than others?</em></strong></li> </ul> <p>(Or is there a better definition of canonical?)</p> <p>The canonicity of the <em>q-binomial coefficients</em> is obviously accounted for by their relationship with linear subspaces. (So this is not an analytical criteria using the limit $q\to1$.) </p> <p>What about the <em>q-Gamma function</em>? We may consider it canonical because of the (?!) q-analogue of the Bohr-Mollerup theorem proved by R. Askey, which states that for $0\lt q\lt 1$, the only logarithmically convex function satisfying $f(1)=1$ and $f(x+1)=\frac{q^x –1}{q–1}f(x)$ is the q-gamma function $\Gamma_q(z)=(1-q)^{1-x} \frac{(q\;;\; q)_{\infty}}{{(q^x;\; q) _\infty }}.$</p> <p>Also note that this formula looks at least as elegant as Euler's definition of $\Gamma(z)= \dfrac{1}{z} \prod\limits_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}.$ </p> <p>On the other hand, one of the most basic identities, the recursion of binomial coefficients, has only an "asymmetric" q-analog and thus two of them: ${\genfrac[]00{n}{k}}_q=q^k{\genfrac[]00{n-1}{k}}_q+{\genfrac[]00{n-1}{k-1}}_q={\genfrac[]00{n-1}{k}}_q+q^{n-k}{\genfrac[]00{n-1}{k-1}}_q$.</p> <p>For the <em>classical orthogonal polynomials</em>, it looks like there exist systematically "nice" q-analogs (see e.g. <a href="http://homepage.tudelft.nl/11r49/documents/as98.pdf" rel="nofollow">this survey</a>), but it is not clear if there is a certain sense in which those can be considered canonical. Maybe for the Chebyshev polynomials, <a href="http://arxiv.org/ftp/arxiv/papers/1201/1201.4703.pdf" rel="nofollow">there is one "best" q-analog.</a></p> <ul> <li><p><strong>Is there a reasonable way of considering certain q-analogs of orthogonal polynomials "canonical", e.g. their uniqueness w.r.t. to an appropriate criteria, as for most classical orthogonal polynomials?</strong></p></li> <li><p><strong>Has a q-analog of a polynomial identity (e.g. involving binomial coefficients) more chances of being canonical if it has a combinatorial interpretation?</strong></p></li> </ul> <p>For the q-derivative, there are at least two completely different approaches, both with their merits. So there is no use looking for canonicity there.<br> But nevertheless the next question:</p> <ul> <li><strong>Are q-analogs conceptually similar to an extension from $\mathbb R$ to $\mathbb C$?</strong></li> </ul> <p>By the latter I mean the following:<br> I wonder if generally speaking, the shift from an entity to its q-analog(s) can be likened, at least sometimes, to the shift of passing from $\mathbb R$ to $\mathbb C$, in the sense that some q-analogs provide a more complete picture than the entity itself (cp. for the $\mathbb R\to\mathbb C$ case the fundamental theorem of algebra or the meromorphic extension of the zeta function)? </p> <ul> <li><p>Many features in $\mathbb C$, e.g. the residue theorem, cannot be reduced to $\mathbb R$. Likewise for example, identities of "infinite q-polynomials" (i.e. of generating functions), cannot be taken to the limit $q\to1$.</p></li> <li><p>In situations where there are several useful q-analogs, e.g. for the <a href="http://mathworld.wolfram.com/q-ExponentialFunction.html" rel="nofollow">q-exponential function</a> or the <a href="http://mathworld.wolfram.com/q-Cosine.html" rel="nofollow">q-cosine</a>, we might consider those corresponding to different panes of a Riemann surface like the one of $\sqrt{z}$ or $\ln z$.</p></li> <li><p>And we can take it further:<br> The next step after $\mathbb R$ to $\mathbb C$ are the quaternions.<br> The next step after q-analogs are <a href="http://www.emis.de/journals/INTEGERS/papers/i29/i29.pdf" rel="nofollow">p,q-analogs</a>. It looks like they haven't been studied a lot yet. </p></li> </ul> <p>Thank you for reading me so far.<br> Some of the questions may be rather subjective. As for the title, I had thought at first about "Why are most q-analogues canonical?" But then I figured that in the whole ocean of q-analogues, maybe there are just some "very" canonical islands, but for the vast majority the notion of canonicity is more or less fuzzy. Is that a feasible perception? Even though some of these thoughts are somewhat philosopical, anyway, here goes. Looking forward to your input!</p> http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others/90041#90041 Answer by Nicola Ciccoli for Why are some q-analogues more canonical than others? Nicola Ciccoli 2012-03-02T14:04:03Z 2012-03-02T14:04:03Z <p>As for orthogonal polynomials (and for some other functions like Bessel and $q$-Bessel) connected to the Askey-scheme you were referring to, there is a nice explanation of "canonical".</p> <p>In the classical case most of these orthogonal families of polynomials arise from representation theory of compact Lie groups, and, in fact, many of their properties (e.g. addition and multiplication formulas) can be explained from representation theory.</p> <p>Standard compact quantum groups are, in a sense, rather canonical. They quantize a very specific Poisson-Lie group structure (which can be characterized in between all Poisson-Lie group structures on a compact Lie group) and carry the property of leaving representation theory substantially undeformed. </p> <p>The $q$-orthogonal polynomials arising in the Askey scheme have the same connection with compact quantum groups as their classical analogues have with compact Lie groups. It is exactly for this reason that they share the same kind of properties like addition formulas,just to mention one.</p> http://mathoverflow.net/questions/89996/why-are-some-q-analogues-more-canonical-than-others/90060#90060 Answer by Johann Cigler for Why are some q-analogues more canonical than others? Johann Cigler 2012-03-02T17:37:27Z 2012-03-02T17:37:27Z <p>As I said above I have some difficulty to denote specific $q-$analogues as canonical. Consider as example the Catalan numbers $\frac{1}{{n + 1}}{2n\choose n}$ . They have a simple generating function $f(z)$ which satisfies $f(z) = 1 + zf(z)^2$ or equivalently $f(z) = \frac{{1 - \sqrt {1 - 4z} }}{{2z}}$ and they are characterized by the fact that all Hankel determinants are 1.</p> <p>Which of the following simple $q -$analogues should be called "canonical"? </p> <p>a) The polynomials $C_n (q)$ introduced by Carlitz with generating function $F(z) = 1 + zF(z)F(qz)$. They also have very simple Hankel determinants, but there is no known formula for the polynomials themselves.</p> <p>b) The polynomials $\frac{1}{{[n + 1]}}{2n\brack n}$ . They have a simple formula but no simple formula for their generating function and no simple Hankel determinants.</p> <p>c) The $q -$Catalan numbers $c_n (q)$ introduced by George Andrews. Their generating function $A(z)$ is a $q -$analogue of $\frac{{1 - \sqrt {1 - 4z} }}{{2z}}.$ Let $h(z)$ be the $q -$analogue of $\sqrt {1 + z}$ defined by $h(z)h(qz)=1+z$. Then $A(z)= \frac{1+q}{{4qz}}(1-h(-4qz))$. They have both simple formulas and a simple formula for the generating function and also simple Hankel determinants. But they are not polynomials in $q.$ </p>