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The tittle of this question is the main question.

Why it is necessary to not consider these kind of functions in $\Lambda$. My teacher told me that if one consider these functions to be in the ring of symmetric functions, then the inner product would fail since this kind of functions would have infinitely non-zero homogeneous parts. But I do not see it clearly.

Can anyone explain me why it is necessary to exclude the functions which do not have bounded degree?

Thank you very much.

$$\prod_i (1+x_i) \notin \Lambda.$$

idriskameni
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  • His explanation is baloney. But you haven't defined a function. – logarithm May 14 '19 at 10:22
  • Your question makes no sense. What is a symmetric function? The obvious definition would be that if $X$ is a $G$-set, $G$ a group, then a symmetric function on $X$ with values in $Y$ would be a $G$-invariant map from $X$ to $Y$. If $Y$ is a ring, these functions have a natural ring structure. Not sure what the grading would be though. Also you appear to have written a polynomial as your symmetric function. However, what is the range of the indices in the product? Based on your question, there would appear to be infinitely many indices, otherwise the degree would appear to be finite. – jgon May 14 '19 at 13:07
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    It make sense since it is the example MacDonald gives in his book. Before writing such a comment please read. The index is obviously from i=0 to infinite as we always write when they are not specified. MacDonald says that this is not a symmetric function because it does not match his construction. But I would like to know why when you construct $\Lambda$ the ring of symmetric functions, you are not interested in including such functions. – idriskameni May 14 '19 at 16:44
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    @jgon Symmetric functions are a well-known concept, they are not literal functions. The range of indices if infinite. See https://en.wikipedia.org/wiki/Symmetric_function. – Mike Earnest May 14 '19 at 22:31
  • @MikeEarnest I did Google symmetric functions before commenting, and according to wiki they are literal functions. More specifically wiki gives the special case of the definition I gave where $X=A^n$ for some set $A$ and $G=S_n$, with $G$ acting on $X$ in the obvious way. Wiki then clarifies that this is most commonly adopted for polynomial functions. No idea what you mean when you say that the range of indices is infinite. – jgon May 14 '19 at 22:48
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    @jgon Oops, I gave the wrong link, here is what I meant. A symmmetric function can have infinitely many variables, like $x_1+x_2+x_3+\dots$. This is not a function, because $1+(-1)+1+(-1)+\dots$ makes no sense. When I say the range of indices is infinite, I mean that the product in OP refers to $\prod_{i=1}^\infty (1+x_i)$. – Mike Earnest May 14 '19 at 22:52
  • @MikeEarnest Thanks, that link cleared up my confusion as to what the question was referring to. – jgon May 14 '19 at 23:32
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    @idriskameni I did read the question and google symmetric function before commenting. I didn't come across the link Mike Earnest provided to the ring of symmetric functions, and was therefore confused, since the wiki article for symmetric functions doesn't mention this ring at all. As a suggestion, it might not hurt to define your terms in your questions. There are several benefits to doing so. Firstly, stating the definition might have helped you answer your own question. Secondly, those unfamiliar with a term, but capable of answering given the definition will be able to help. – jgon May 14 '19 at 23:46
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    One of the tags on the question is [tag:symmetric-functions], and clicking on that takes you to a short description of symmetric functions and a list of questions likewise tagged. Clicking on Learn more... gives a more detailed description. – robjohn May 15 '19 at 14:34
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    Let me say I strongly disagree that the linked question is a duplicate. There are three categories being discussed: [1] symmetric polynomials (finite variables, bounded degree) [2] symmetric functions (infinite variables, bounded degree) [3] symmetric power series (infinite variables, infinite degree). The linked question concerns the difference between [1] and [2], while this question asks about the difference between [3] and [2] (or rather, why the term "symmetric functions" excludes [3] by definition). – Mike Earnest May 15 '19 at 15:49
  • TOTALLY AGREE. I do not understand why are they doing that. They are creating such a toxic community doing that. They are only interested in closing questions before even reading them. MSE let them do it because they have high reputations. But it is unfair. – idriskameni May 15 '19 at 15:51

1 Answers1

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I agree that it is difficult to see how $f:=\prod_i(1+x_i)$ expands into homogenous poylnomials. But it is still easy to see why a contradiction arises.

If we were to expand $f$ into elementary symmetric functions, the result would be the infinite sum $$ f=1+e_1+e_2+\dots $$ This is the extension of the finite equality $$(1+x_1)(1+x_2)(1+x_3)=1+(x_1+x_2+x_3)+(x_1x_2+x_1x_3+x_2x_3)+x_1x_2x_3.$$ Now, applying the $\omega$ involution to both sides, $$ \omega(f)=1+h_1+h_2+\dots $$ Next, expand $\omega(f)$ into the $m$ basis. The result is $$ \omega(f)=\sum_{\lambda }m_\lambda, $$ since each $m_\lambda$ appears in exactly one $h_n$. Finally, we can see that (using the convention $h_{(0)}=m_{(0)}=1$ for convenience) $$ \langle \omega(f),\omega(f)\rangle=\left\langle \sum_{i=0}^\infty h_{(i)},\sum_\lambda m_\lambda\right\rangle=\sum_{i=0}^\infty \langle h_{(i)},m_{(i)}\rangle = \sum_{i=0}^\infty 1=\infty $$ We do not want infinite values of the inner product. It may not seem like problem, but if $\infty$ can arise, then so can $\infty-\infty$, which is a problem. For example, computing the inner product of $\sum_{i=1}^\infty (-1)^i h_{(i)}$ and $\sum_{\lambda}m_\lambda$ would be $1-1+1-1+\dots$, which is undefinable.

Mike Earnest
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  • That is a nice explanation. Thank you very much. Super clear. – idriskameni May 15 '19 at 15:26
  • One question. What do you mean when you write $m_{(i)}$. I do understand the $h_{(i)}$. But do not understand why can you do the steps in the middle. Thanks. – idriskameni May 15 '19 at 15:55
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    When you expand out the inner product, you would get the sum of $\langle h_{(i)},m_\lambda\rangle$ over all $i\ge 0$, and all partitions $\lambda$. In particular, $(i)$ is an example of a partition (with just one part, equal to $i$), so $\langle h_{(i)},m_{(i)}\rangle$ appears in this sum. These are the only partitions which contribute to the sum, since $\langle h_\mu,m_\lambda\rangle=0$ when $\mu\neq \lambda$, and the only $\lambda$ for which $h_\lambda$ appears are $\lambda$ of the form $(i)$. – Mike Earnest May 15 '19 at 15:59