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I cannot understand the definition of polynomials $s_I$ in the Milnor and Stasheff, Characteristic classes, page $188.$

In Milnor and Stasheff, Characteristic classes page $188$, the polynomials $S_I$ are defined as follows.

Let $t_1,...,t_n$ be indeterminates.

Now for any partition $I=i_1,...,i_r$ of $k$, define a polynomial $S_I$ in $k$ variables as follows. Choose $n \geq k$ so that the elementary symmetric functions $\sigma_1,...,\sigma_k$ of $t_1,...t_n$ are algebraically independent and let $s_I(\sigma_1,....,\sigma_k)= \sum t_1^{i_1}...t_r^{i_r}$.

The summation $s_I(\sigma_1,....,\sigma_k)= \sum t_1^{i_1}...t_r^{i_r}$ is taken over all monomials transformed from $t_1^{i_1}...t_r^{i_r}$ by the permuation group acting on the set $\{t_1,....,t_n\}$.

$s() = 1$

$s_1(\sigma_1) = \sigma_1$

$s_{2}(\sigma_1,\sigma_2) = \sigma_1^2-2\sigma_2$

$s_{1,1}(\sigma_1,\sigma_2) = \sigma_2$

$s_{3}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1^3 - 3 \sigma_1 \sigma_2 + 3\sigma_3$

$s_{1,2}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1 \sigma_2 - 3\sigma_3$

$s_{1,1,1}(\sigma_1,\sigma_2,\sigma_3) = \sigma_3$

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Edit after the answer of Eric

I can understand the above equations. For example, consider the following equation.

$s_{3}(\sigma_1,\sigma_2,\sigma_3) = \sigma_1^3 - 3 \sigma_1 \sigma_2 + 3\sigma_3$

This equation means the following.

$$s_3(t_1,t_2,t_3) := (t_1)^3+(t_2)^3+(t_3)^3 \\ = (t_1+t_2+t_3)^3 -3(t_1+t_2+t_3)(t_1t_2+t_2t_3+t_3t_1) -3t_1t_2t_3 \\ = \sigma_1^3 - 3 \sigma_1 \sigma_2 - 3\sigma_3$$

The last sign is minus instead of plus ... I guess the book misprinted?

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    $\begingroup$ Can you elaborate on what exactly you don't understand? $\endgroup$ Commented Jan 1, 2020 at 22:21
  • $\begingroup$ In particular there are quite a few typos in your examples at the end which might be related to why you are confused. $\endgroup$ Commented Jan 1, 2020 at 22:38
  • $\begingroup$ In page 188 of Milnor and Stasheff, it says that "Choose $n \geq k$ so that the elementary symmetric functions $\sigma_1,...,\sigma_k$ of $t_1,...t_n$ are algebraically independent ". But I cannnot understand what this means? $\endgroup$ Commented Jan 2, 2020 at 8:30

1 Answer 1

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Let $S$ be the subring of $\mathbb{Z}[t_1,\dots,t_n]$ consisting of polynomials that are symmetric in the $n$ variables (i.e., invariant under any permutation of the variables). It is a theorem that $S$ itself is a polynomial ring in the elementary symmetric polynomials $\sigma_1,\dots,\sigma_n$, where $\sigma_k$ is the sum of all the distinct monomials that can be obtained from $t_1\dots t_k$ by permuting the variables. In other words, every element of $S$ can be uniquely written as a polynomial in $\sigma_1,\dots,\sigma_n$.

So, here is what the definition of $s_I$ means. Take the polynomial $f(t_1,\dots,t_n)$ in $n$ variables given by summing up all distinct monomials that can be obtained from $t_1^{i_1}\dots t_r^{i_r}$ by permuting the variables. Then $f\in S$, so it can be written as a polynomial in $\sigma_1,\dots,\sigma_n$. In fact, it turns out that $f$ is a polynomial in just $\sigma_1,\dots,\sigma_k$, and that moreover the coefficients of this polynomial in $\sigma_1,\dots,\sigma_k$ do not depend on the choice of $n\geq k$. We write $s_I$ for the unique polynomial in $k$ variables such that $s_I(\sigma_1,\dots,\sigma_k)=f(t_1,\dots,t_n)$.

Let's look at an example. For instance, take $I$ to be the partition $(2)$ and $n=2$, so $f(t_1,t_2)=t_1^2+t_2^2$. To find $s_I$, we want to write $t_1^2+t_2^2$ in terms of $\sigma_1=t_1+t_2$ and $\sigma_2=t_1t_2$. To do that, we can simply observe that $(t_1+t_2)^2-2t_1t_2=t_1^2+t_2^2$, so $f=\sigma_1^2-2\sigma_2$. Thus the polynomial $s_I$ is $s_I(x_1,x_2)=x_1^2-2x_2$, since if we evaluate this polynomial at $(x_1,x_2)=(\sigma_1,\sigma_2)$ we get $f$.

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  • $\begingroup$ Thank you. I could not understand with the book, but in your explanation, I can understand it. Thank you. $\endgroup$ Commented Jan 2, 2020 at 23:17
  • $\begingroup$ How do we know that the coefficients of the polynomial does not depend on the choice of $n$? $\endgroup$
    – blancket
    Commented Aug 20, 2020 at 11:39

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