Timeline for Symmetric polynoms are Hopf algebra ? What for one needs co-product ?
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22 events
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| Jun 9, 2017 at 10:58 | history | edited | darij grinberg | CC BY-SA 3.0 |
dead MIT link is dead
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| Mar 11, 2012 at 19:41 | history | edited | darij grinberg | CC BY-SA 3.0 |
URL updated
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| Feb 26, 2012 at 10:05 | vote | accept | Alexander Chervov | ||
| Jan 19, 2012 at 17:01 | comment | added | Alexander Chervov | What this coproduct means for finite n i.e. why it is natural ? By the way if we consider S(gln) with natiral Lie algebra co-product, will it induce the same co-product as you write on gln invariants = C[x1...xn]Sn ? For trace it is seems okay :) | |
| Jan 19, 2012 at 13:35 | comment | added | darij grinberg | Now that I am thinking about, actually you can define a bialgebra structure on $k\left[x_1,x_2,...,x_n\right]^{S_n}$ such that $\Delta\left(e_i\right)=\sum\limits_{j=0}^i e_j\otimes e_{i-j}$ for every $i\leq n$. This gives a Hopf subalgebra of $\mathbf{Symm}$, namely the one generated by $e_1$, $e_2$, ..., $e_n$. But one should be very careful to keep in mind that it is a Hopf subalgebra, not a quotient Hopf algebra, so one cannot blindly take an identity holding in $\mathbf{Symm}$ and set all $e_i$ to $0$ for $i > n$. | |
| Jan 19, 2012 at 13:29 | comment | added | darij grinberg | As for applications, take a look at this (suggested by Andy B): arxiv.org/PS_cache/arxiv/pdf/0908/0908.3714v3.pdf | |
| Jan 19, 2012 at 13:22 | comment | added | darij grinberg | For any natural $u$ and $v$, you can define a "modified coproduct" $k\left[x_1,...,x_{u+v}\right]^{S_{u+v}} \to k\left[x_1,...,x_u\right]^{S_u} \otimes k\left[x_1,...,x_v\right]^{S_v}$ in more or less the same way as I did for $\mathbf{Symm}$ above. But then, in order to formulate coassociativity, you will have to consider these "modified coproducts" for all $u$ and $v$ rather than just for one pair $\left(u,v\right)$, so there is not much gained in my opinion. | |
| Jan 19, 2012 at 13:04 | comment | added | Alexander Chervov | Is there the way to define and motivate co-product only in terms of $C[x_1...x_n]^S_n$, not appealing to anything-else? If co-product is defined only in the limit - then there should be "modified co-product" or something which exist for finite "n". Co-product means that Spec(Symm) is group. May be it is more easy to see that this group law and something on $Spec(C[x_1...x_n]^S_n)$ which gives this co-product ? | |
| Jan 19, 2012 at 12:58 | comment | added | Alexander Chervov | @Darij I appreciate Marc's answer. However sym. polynoms is something very simple and basic even at school one plays with them. So the fact that there is some structure which cannot be seen in elementary way seems strange... I want to understand if this co-product reflects some essenses about symmetric polynoms or what it reflects ? All the answers are great but they leaving feeling that it is artificial construction... | |
| Jan 19, 2012 at 11:57 | comment | added | darij grinberg | There are various "natural" coproducts on $\mathbb C\left[x_1,...,x_n\right]^{S_n}$, but none of them has anything to do with the natural coproduct on $\mathbf{Symm}$. The one on $\mathbf{Symm}$ is defined only for infinitely many variables. I think Marc van Leeuwen's answer gives a very vivid explanation of why we need them to be infinitely many. | |
| Jan 19, 2012 at 5:05 | comment | added | Alexander Chervov | Let me confirm my understanding - you mean that: there is NO co-product ("natural" co-product) on $C[x_1...x_n]^{S_n}$. Is it correct ? So the co-product on symmetric functions is related to this limit n->Inf ? | |
| Jan 18, 2012 at 15:50 | comment | added | darij grinberg | ... these equalities break when you apply the comultiplication. Don't let the infinite number of indeterminates make you think of $\mathbf{Symm}$ as a very big thing: it is not. The $n$-th graded component of $\mathbf{Symm}$ is generated, as a $k$-module, by the products $e_{i_1}e_{i_2}...e_{i_m}$ where $\left(i_1,i_2,...,i_m\right)$ is a partition of $n$. Every $n$ has only finitely many partitions, so the dimension $n$-th graded component of $\mathbf{Symm}$ is finite (and equals the number of partitions of $n$, which should not surprise you given Dan's posting). | |
| Jan 18, 2012 at 15:45 | comment | added | darij grinberg | By "avoiding the limit", do you maybe mean avoiding infinitely many indeterminates? Because when introducing $\mathbf{Symm}$ by means of power series in infinitely many indeterminates, the notion of a limit is never used. As I said, there is no good way to avoid the infinity in here, but there is no really good reason to do so either. Equalities that hold in $\mathbf{Symm}$ should hold for symmetric functions in arbitrarily many indeterminates; if you restrict yourself to $n$ indeterminates, you get some equalities like $e_{n+1}=0$ which don't hold if you add more indeterminates; and ... | |
| Jan 18, 2012 at 14:10 | comment | added | Alexander Chervov | I do not catch how you "do avoid", let me think some time... | |
| Jan 18, 2012 at 13:54 | comment | added | darij grinberg | It's just some blabber that was supposed to describe the equality (1). By the union of two multisets $\left(p_1,p_2,...,p_u\right)$ and $\left(q_1,q_2,...,q_v\right)$, I mean the multiset $\left(p_1,p_2,...,p_u,q_1,q_2,...,q_v\right)$. The intuition behind (1) is that to evaluate $p$ at the union of two multisets, we write the tensor $\Delta\left(p\right)$ in the form $\sum\limits_{i\in I} q_i\otimes r_i$, and sum (over all $i$) the product $q_i\left(\text{first multiset}\right)r_i\left(\text{second multiset}\right)$. | |
| Jan 18, 2012 at 13:48 | comment | added | Alexander Chervov | What means "correspond to the union of multisets" ? | |
| Jan 18, 2012 at 13:44 | history | edited | darij grinberg | CC BY-SA 3.0 |
edited body
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| Jan 18, 2012 at 13:43 | comment | added | darij grinberg | Here is the canonical source on the applications of $\mathbf{Symm}$ to symmetric groups: Andrey V. Zelevinsky, Representations of Finite Classical Groups - A Hopf algebra approach (LNM 869, Springer). | |
| Jan 18, 2012 at 13:41 | comment | added | darij grinberg | As I said, I do avoid this limit, so it's not impossible - it is just an alternative construction of $\mathbf{Symm}$. As for motivation, the underlying motivation in arxiv.org/abs/0804.3888 was to explain Witt vectors in a more algebraic (and less number-theoretic) context. But if you read Section 18 of arxiv.org/abs/0804.3888 you will learn of a second application probably more interesting to you: Equalities in $\mathbf{Symm}$ "encode" isomorphisms of representations of $S_n$. (See also Dan's reply for this.) | |
| Jan 18, 2012 at 13:36 | comment | added | Alexander Chervov | Thank You very much ! Great answer ! But are you sure that it is impossible to avoid this limit n->Inf ? (I do not know it and so do not like it:) It seems you introduce coproduct ad hoc just . What is motivation and use of it ? | |
| Jan 18, 2012 at 13:34 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 137 characters in body
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| Jan 18, 2012 at 13:25 | history | answered | darij grinberg | CC BY-SA 3.0 |