Less discriminating discriminants - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:34:01Z http://mathoverflow.net/feeds/question/72268 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72268/less-discriminating-discriminants Less discriminating discriminants Igor Rivin 2011-08-07T01:34:19Z 2011-08-08T18:31:46Z <p>Suppose that I want to know whether a polynomial $P(z)$ has a root with multiplicity at least three. This is obviously an algebraic condition, but is there some reasonably concise set of conditions defining the variety (in the space of coefficients)? This must have been studied by the ancients. It is clearly necessary that the discriminant vanish, and also that the resultant of the polynomial $P$ and the second derivative $P^{\prime\prime}$ vanish, but, just as obviously, not sufficient...</p> <p><strong>EDIT</strong> Abhinav certainly gives a nice answer to the question, but the question I would REALLY like to know the answer to is: what is the degree of the variety as a function of the partition (as in @Gjergji's answer). Maybe I should read the reference...</p> http://mathoverflow.net/questions/72268/less-discriminating-discriminants/72269#72269 Answer by Abhinav Kumar for Less discriminating discriminants Abhinav Kumar 2011-08-07T02:14:20Z 2011-08-07T02:14:20Z <p>Here's one way to think of this. Let $\lambda$ be a new variable and let $Q(z) = \lambda P'(z) + P''(z)$. Let $R(\lambda)$ be the resultant of $P$ and $Q$. Then we want $R$ to identically vanish (i.e. want to see if there's value of $z$ for which $P$ and $Q$ have a common zero, for every value of $\lambda$). So the coefficients of $R$ will describe an ideal cutting out the variety you want. This argument should work at least in characteristic $0$.</p> http://mathoverflow.net/questions/72268/less-discriminating-discriminants/72270#72270 Answer by Gjergji Zaimi for Less discriminating discriminants Gjergji Zaimi 2011-08-07T02:24:48Z 2011-08-07T02:30:04Z <p>The situation becomes considerable harder beyond testing for a root of multiplicity $2$. The problem itself is very old. A recent article about this is <a href="http://arxiv.org/abs/math/0110224" rel="nofollow">"On equations defining coincident root loci"</a> by J. Chipalkatti. The result is that the algebraic conditions can be expressed through a cohomology group of a certain complex of $SL_2$ representations</p> <p>Note: This is addressing the general question of when a polynomial has roots of multiplicities indexed by a partition $\lambda$. The question itself is about $\lambda=(3,1^{n-3})$.</p> http://mathoverflow.net/questions/72268/less-discriminating-discriminants/72382#72382 Answer by SPG for Less discriminating discriminants SPG 2011-08-08T18:31:46Z 2011-08-08T18:31:46Z <p>For partitions of the special form $\lambda=(k,1,1,\dots,1)$ (i.e., hook shapes) there is a very explicit description of the ideal of definition of the corresponding set of polynomials (i.e., those polynomials with a root of order $\geq k$). Namely, it is the span of the set of (symmetric) Jack polynomials $f_\mu$, with parameter specialized to $-1/k$, and where $\mu$ ranges over all partitions satisfying</p> <p>$$\mu_i-\mu_{i+k-1} \geq 2$$ for all $i$. This is the main result of the paper <a href="http://arxiv.org/pdf/math/0112127" rel="nofollow">http://arxiv.org/pdf/math/0112127</a> by Feigin, Jimbo, Miwa, and Mukhin.</p> <p>For the application to degree, one might note that this computes the Hilbert function of this ideal quite explicitly in combinatorial terms, though for your problem you are probably interested in the grading where each elementary symmetric function has degree one, and I don't know, off the top of my head, how to extract that information. Maybe I'll post a separate question so we can see if anywhere else here knows.</p>