Skip to main content
27 events
when toggle format what by license comment
Aug 2, 2021 at 9:53 vote accept Jules
Oct 17, 2020 at 22:29 history edited Jules CC BY-SA 4.0
added 22 characters in body
S Oct 2, 2020 at 8:01 history bounty ended CommunityBot
S Oct 2, 2020 at 8:01 history notice removed CommunityBot
Sep 24, 2020 at 14:57 answer added Carlo Beenakker timeline score: 11
S Sep 24, 2020 at 6:51 history bounty started Jules
S Sep 24, 2020 at 6:51 history notice added Jules Authoritative reference needed
Sep 23, 2020 at 1:38 history edited Jules CC BY-SA 4.0
added 23 characters in body
Sep 23, 2020 at 1:35 comment added Jules I put a concise version of my proof in the question.
Sep 23, 2020 at 1:33 history edited Jules CC BY-SA 4.0
added 388 characters in body
Sep 23, 2020 at 0:25 comment added Fedor Petrov @TimothyChow done, and I found an issue preparing this answer.
Sep 23, 2020 at 0:23 answer added Fedor Petrov timeline score: 11
Sep 22, 2020 at 16:48 comment added Timothy Chow @FedorPetrov : Perhaps you could collect your comments into an answer even though they don't directly answer the "reference request" part of the question.
Sep 22, 2020 at 14:31 comment added Jules Ah, now I understand. I misunderstood your initial comment and thought there was a particular known family of functions $H$ that you were talking about, but you are talking about the $p_i$ that are of that form for any $H$. Sorry for the confusion and thank you for the explanation.
Sep 22, 2020 at 14:20 comment added Fedor Petrov because the product of two such functions is again such a function.
Sep 22, 2020 at 13:15 comment added Jules Thanks! I think I understand all the steps except one. I could not understand why the linear span is the same as the algebra. The degree of these special $p_i$ seems to be $\leq n$ so I'd expect that the same holds true for their span, whereas in general the $p_i$ can have any degree. Could you help me understand where my reasoning went wrong? By the way, it is easy to prove the reverse implication, that any symmetric polynomial can be written in elementary symmetric polynomials, by taking $A$ to be the companion matrix of a polynomial in the identity :)
Sep 21, 2020 at 23:44 comment added Fedor Petrov At first, the linear span of such functions is the same as the algebra generated by them. Next, it suffices to consider $p$ with a finite support $A$. Then for $H(m)=m+\alpha$ for $m\in A$ (hereafter: and 0 otherwise) we get a function of the form $p_i=(i_1+\alpha)...(i_n+\alpha)$. Varying $\alpha$ and taking linear combinations we get any elementary symmetric polynomial in $i_1,..., i_n$. They generate an algebra of all symmetric polynomials, and any symmetric function on $A$ is represented by a symmetric polynomial.
Sep 21, 2020 at 23:36 history edited Jules CC BY-SA 4.0
edited body
Sep 21, 2020 at 23:25 history edited Jules CC BY-SA 4.0
added 128 characters in body
Sep 21, 2020 at 23:18 history edited Jules CC BY-SA 4.0
added 45 characters in body
Sep 21, 2020 at 23:13 comment added Jules That sounds cool! What is the name of these functions $H$ and where could I find a proof that they generate all symmetric functions of natural number tuples? (which I think means that any symmetric $p$ can be written as a linear combination of those special $p$'s?)
Sep 21, 2020 at 22:49 comment added Fedor Petrov That's nice. I do not know the reference. When $p_i=\prod_{k=1}^n H(i_k)$ for certain function $H$ defined on $\{0,1,\ldots\}$, both parts factorize and are equal to the determinant of $\sum_m H(m)A^m$. The functions of such type generate all symmetric functions $i=(i_1,\ldots,i_n)\mapsto p_i$, so the result follows.
Sep 21, 2020 at 22:33 history edited Jules CC BY-SA 4.0
added 114 characters in body
Sep 21, 2020 at 22:24 history edited Jules CC BY-SA 4.0
edited body
Sep 21, 2020 at 22:18 history edited Jules CC BY-SA 4.0
edited tags
Sep 21, 2020 at 21:45 history edited Jules CC BY-SA 4.0
added 1 character in body
Sep 21, 2020 at 21:30 history asked Jules CC BY-SA 4.0